Chapter 16 from Textbook Rs Aggarwal 2021 2022 for Class 10 MATHS

Question 1:

Answer:

Circumference = 39.6 cm
We know:
Circumference of a circle = $2 π r$
$\Rightarrow 39.6 = 2 \times \frac{22}{7} \times r \Rightarrow \frac{39.6 \times 7}{2 \times 22} = r \Rightarrow r = 6.3 cm$

Also,
Area of the circle = $π r^{2}$
$= \frac{22}{7} \times 6.3 \times 6.3 = 124.74 {cm}^{2}$

Question 2:

Circumference = 39.6 cm
We know:
Circumference of a circle = $2 π r$
$\Rightarrow 39.6 = 2 \times \frac{22}{7} \times r \Rightarrow \frac{39.6 \times 7}{2 \times 22} = r \Rightarrow r = 6.3 cm$

Also,
Area of the circle = $π r^{2}$
$= \frac{22}{7} \times 6.3 \times 6.3 = 124.74 {cm}^{2}$

Answer:

Let the radius of the circle be r.
â€‹Now,
$Area = 98.56 \Rightarrow π r^{2} = 98.56 \Rightarrow \frac{22}{7} \times r^{2} = 98.56 \Rightarrow r = 5.6$
Now,
Circumference = $2 π r = 2 \times \frac{22}{7} \times 5.6 = 35.2 cm$
Hence, the circumference of the circle is 35.2 cm.

Question 3:

Let the radius of the circle be r.
â€‹Now,
$Area = 98.56 \Rightarrow π r^{2} = 98.56 \Rightarrow \frac{22}{7} \times r^{2} = 98.56 \Rightarrow r = 5.6$
Now,
Circumference = $2 π r = 2 \times \frac{22}{7} \times 5.6 = 35.2 cm$
Hence, the circumference of the circle is 35.2 cm.

Answer:

Let the radius of the circle be r.
Now,
Circumference = Diameter + 45
$\Rightarrow 2 π r = 2 r + 45 \Rightarrow 2 π r - 2 r = 45 \Rightarrow 2 r (\frac{22}{7} - 1) = 45 \Rightarrow 2 r \times \frac{15}{7} = 45 \Rightarrow r = 10.5 cm$
∴ Circumference = Diameter + 45 = 2(10.5) + 45 = 66 cm
Hence, the circumference of the circle is 66 cm.

Question 4:

Let the radius of the circle be r.
Now,
Circumference = Diameter + 45
$\Rightarrow 2 π r = 2 r + 45 \Rightarrow 2 π r - 2 r = 45 \Rightarrow 2 r (\frac{22}{7} - 1) = 45 \Rightarrow 2 r \times \frac{15}{7} = 45 \Rightarrow r = 10.5 cm$
∴ Circumference = Diameter + 45 = 2(10.5) + 45 = 66 cm
Hence, the circumference of the circle is 66 cm.

Answer:

Area of the circle = 484 cm²
Area of the square = ${Side}^{2}$
$\Rightarrow 484 = {Side}^{2} \Rightarrow 22^{2} = {Side}^{2} \Rightarrow Side = 22 cm$
Perimeter of the square = $4 \times Side$
Perimeter of the square = $4 \times 22$
= 88 cm
Length of the wire = 88 cm
Circumference of the circle = Length of the wire = 88 cm
Now, let the radius of the circle be r cm.
â€‹Thus, we have:
â€‹ $2 πr = 88 \Rightarrow 2 \times \frac{22}{7} \times r = 88 \Rightarrow r = 14$

Area of the circle = ${πr}^{2}$
$= \frac{22}{7} \times 14 \times 14 = 616 {cm}^{2}$

Thus, the area enclosed by the circle is 616 cm².

Question 5:

Area of the circle = 484 cm²
Area of the square = ${Side}^{2}$
$\Rightarrow 484 = {Side}^{2} \Rightarrow 22^{2} = {Side}^{2} \Rightarrow Side = 22 cm$
Perimeter of the square = $4 \times Side$
Perimeter of the square = $4 \times 22$
= 88 cm
Length of the wire = 88 cm
Circumference of the circle = Length of the wire = 88 cm
Now, let the radius of the circle be r cm.
â€‹Thus, we have:
â€‹ $2 πr = 88 \Rightarrow 2 \times \frac{22}{7} \times r = 88 \Rightarrow r = 14$

Area of the circle = ${πr}^{2}$
$= \frac{22}{7} \times 14 \times 14 = 616 {cm}^{2}$

Thus, the area enclosed by the circle is 616 cm².

Answer:

$Area of a n equilateral t riangle = \frac{\sqrt{3}}{4} \times {(Side)}^{2} \Rightarrow 121 \sqrt{3} = \frac{\sqrt{3}}{4} \times {(Side)}^{2} \Rightarrow 121 \times 4 = {(Side)}^{2} \Rightarrow Side = 22 cm$

$Perimeter of an equilateral triangle = 3 \times Side$
$= 3 \times 22 = 66 cm$

Length of the wire $= 66 cm$
Now, let the radius of the circle be r cm.
We know:
Circumference of the circle = Length of the wire
$2 π r = 66 \Rightarrow 2 \times \frac{22}{7} \times r = 66 \Rightarrow r = \frac{66 \times 7}{2 \times 22} \Rightarrow r = \frac{21}{2} \Rightarrow r = 10.5$
Thus, we have:
Area of the circle = $π r^{2}$
$= \frac{22}{7} \times 10.5 \times 10.5 = 346.5 sq . cm$
Area enclosed by the circle = 346.5 cm²

Question 6:

$Area of a n equilateral t riangle = \frac{\sqrt{3}}{4} \times {(Side)}^{2} \Rightarrow 121 \sqrt{3} = \frac{\sqrt{3}}{4} \times {(Side)}^{2} \Rightarrow 121 \times 4 = {(Side)}^{2} \Rightarrow Side = 22 cm$

$Perimeter of an equilateral triangle = 3 \times Side$
$= 3 \times 22 = 66 cm$

Length of the wire $= 66 cm$
Now, let the radius of the circle be r cm.
We know:
Circumference of the circle = Length of the wire
$2 π r = 66 \Rightarrow 2 \times \frac{22}{7} \times r = 66 \Rightarrow r = \frac{66 \times 7}{2 \times 22} \Rightarrow r = \frac{21}{2} \Rightarrow r = 10.5$
Thus, we have:
Area of the circle = $π r^{2}$
$= \frac{22}{7} \times 10.5 \times 10.5 = 346.5 sq . cm$
Area enclosed by the circle = 346.5 cm²

Answer:

Let the radius of the park be r.
Length of chain = Perimeter of the semicircular park
⇒ 108 = Length of the arc + Diameter
$\Rightarrow 108 = \frac{1}{2} \times 2 π r + 2 r \Rightarrow 108 = r (\frac{22}{7} + 2) \Rightarrow 108 = \frac{36}{7} r \Rightarrow r = 21 m$
Now, Area of park $= \frac{1}{2} π r^{2} = \frac{1}{2} \times \frac{22}{7} \times {(21)}^{2} = 693 m^{2}$
Hence, the area of the park is 693 m² .

Question 7:

Let the radius of the park be r.
Length of chain = Perimeter of the semicircular park
⇒ 108 = Length of the arc + Diameter
$\Rightarrow 108 = \frac{1}{2} \times 2 π r + 2 r \Rightarrow 108 = r (\frac{22}{7} + 2) \Rightarrow 108 = \frac{36}{7} r \Rightarrow r = 21 m$
Now, Area of park $= \frac{1}{2} π r^{2} = \frac{1}{2} \times \frac{22}{7} \times {(21)}^{2} = 693 m^{2}$
Hence, the area of the park is 693 m² .

Answer:

Let the radii of the two circles be r₁ cm and r₂cm.
Now,
Sum of the radii of the two circles = 7 cm
$r_{1 +} r_{2} = 7 . . . (i)$

Difference of the circumferences of the two circles = 88 cm
$\Rightarrow 2 {πr}_{1} - 2 {πr}_{2} = 8 \Rightarrow 2 π (r_{1} - r_{2}) = 8 \Rightarrow (r_{1} - r_{2}) = \frac{8}{2 π} \Rightarrow r_{1} - r_{2} = \frac{8}{2 \times \frac{22}{7}} \Rightarrow r_{1 -} r_{2} = \frac{8 \times 7}{44}$

$r_{1} - r_{2} = \frac{56}{44} r_{1} - r_{2} = \frac{14}{11} . . . (i i)$

Adding (i) and (ii), we get:
$2 r_{1} = \frac{91}{11} r_{1} = \frac{91}{22}$

∴ Circumference of the first circle = $2 {πr}_{1}$
$= 2 \times \frac{22}{7} \times \frac{91}{22} = 26 cm$
Also,
$r_{1} - r_{2} = \frac{14}{11} \frac{91}{22} - r_{2} = \frac{14}{11} \frac{91}{22} - \frac{14}{11} = r_{2} r_{2} = \frac{63}{22}$

∴ Circumference of the second circle = $2 {πr}_{2}$

$= 2 \times \frac{22}{7} \times \frac{63}{22} = 18 cm$

Therefore, circumferences of the first and second circles are 18 cm and 26 cm, respectively.

Question 8:

Let the radii of the two circles be r₁ cm and r₂cm.
Now,
Sum of the radii of the two circles = 7 cm
$r_{1 +} r_{2} = 7 . . . (i)$

Difference of the circumferences of the two circles = 88 cm
$\Rightarrow 2 {πr}_{1} - 2 {πr}_{2} = 8 \Rightarrow 2 π (r_{1} - r_{2}) = 8 \Rightarrow (r_{1} - r_{2}) = \frac{8}{2 π} \Rightarrow r_{1} - r_{2} = \frac{8}{2 \times \frac{22}{7}} \Rightarrow r_{1 -} r_{2} = \frac{8 \times 7}{44}$

$r_{1} - r_{2} = \frac{56}{44} r_{1} - r_{2} = \frac{14}{11} . . . (i i)$

Adding (i) and (ii), we get:
$2 r_{1} = \frac{91}{11} r_{1} = \frac{91}{22}$

∴ Circumference of the first circle = $2 {πr}_{1}$
$= 2 \times \frac{22}{7} \times \frac{91}{22} = 26 cm$
Also,
$r_{1} - r_{2} = \frac{14}{11} \frac{91}{22} - r_{2} = \frac{14}{11} \frac{91}{22} - \frac{14}{11} = r_{2} r_{2} = \frac{63}{22}$

∴ Circumference of the second circle = $2 {πr}_{2}$

$= 2 \times \frac{22}{7} \times \frac{63}{22} = 18 cm$

Therefore, circumferences of the first and second circles are 18 cm and 26 cm, respectively.

Answer:

Let r₁ cm and r₂ cm be the radii of the outer and inner boundaries of the ring, respectively.
We have:

$r_{1} = 23 cm r_{2} = 12 cm$

Now,
Area of the outer ring = $π {r_{1}}^{2}$

$= \frac{22}{7} \times 23 \times 23 = 1662.57 {cm}^{2}$

Area of the inner ring = $π {r_{2}}^{2}$

      $= \frac{22}{7} \times 12 \times 12 = 452.57 {cm}^{2}$

Area of the ring = Area of the outer ring $-$ Area of the inner ring
   = 1662.57 $-$ 452.57
       = 1210 ${cm}^{2}$

Question 9:

Let r₁ cm and r₂ cm be the radii of the outer and inner boundaries of the ring, respectively.
We have:

$r_{1} = 23 cm r_{2} = 12 cm$

Now,
Area of the outer ring = $π {r_{1}}^{2}$

$= \frac{22}{7} \times 23 \times 23 = 1662.57 {cm}^{2}$

Area of the inner ring = $π {r_{2}}^{2}$

      $= \frac{22}{7} \times 12 \times 12 = 452.57 {cm}^{2}$

Area of the ring = Area of the outer ring $-$ Area of the inner ring
   = 1662.57 $-$ 452.57
       = 1210 ${cm}^{2}$

Answer:

(i) The radius (r) of the inner circle is 17 m.
The radius (R) of the outer circle is 25 m.          [Includes path, i.e., (17 + 8)]

Area of the path = $π R^{2} - π r^{2}$
           $= π (R^{2} - r^{2}) = \frac{22}{7} (25^{2} - 17^{2}) = \frac{22}{7} \times (25 - 17) (25 + 17) = \frac{22}{7} \times 8 \times 42 = 1056 m^{2}$

∴ Area of the path = 1056 m²

(ii)

Diameter of the circular park = 7 m
∴ Radius of the circular park, r = $\frac{7}{2}$ = 3.5 m
Width of the path = 0.7 m
∴ Radius of the park including the path, R = 3.5 + 0.7 = 4.2 m
Area of the path
$= π R^{2} - π r^{2} = π (R^{2} - r^{2})$
$= \frac{22}{7} (4 . 2^{2} - 3 . 5^{2}) = \frac{22}{7} \times (4.2 - 3.5) \times (4.2 + 3.5) = \frac{22}{7} \times 0.7 \times 7.7 = 16.94 m^{2}$
Rate of cementing the path = Rs 110/m²      (Given)
∴ Total cost of cementing the path
= 16.94 × 110
= Rs 1863.40
Thus, the expenditure of cementing the path is Rs 1863.40.

Question 10:

(i) The radius (r) of the inner circle is 17 m.
The radius (R) of the outer circle is 25 m.          [Includes path, i.e., (17 + 8)]

Area of the path = $π R^{2} - π r^{2}$
           $= π (R^{2} - r^{2}) = \frac{22}{7} (25^{2} - 17^{2}) = \frac{22}{7} \times (25 - 17) (25 + 17) = \frac{22}{7} \times 8 \times 42 = 1056 m^{2}$

∴ Area of the path = 1056 m²

(ii)

Diameter of the circular park = 7 m
∴ Radius of the circular park, r = $\frac{7}{2}$ = 3.5 m
Width of the path = 0.7 m
∴ Radius of the park including the path, R = 3.5 + 0.7 = 4.2 m
Area of the path
$= π R^{2} - π r^{2} = π (R^{2} - r^{2})$
$= \frac{22}{7} (4 . 2^{2} - 3 . 5^{2}) = \frac{22}{7} \times (4.2 - 3.5) \times (4.2 + 3.5) = \frac{22}{7} \times 0.7 \times 7.7 = 16.94 m^{2}$
Rate of cementing the path = Rs 110/m²      (Given)
∴ Total cost of cementing the path
= 16.94 × 110
= Rs 1863.40
Thus, the expenditure of cementing the path is Rs 1863.40.

Answer:

Let r m and R m be the radii of the inner and outer tracks.
Now,
Circumference of the outer track = $2 π R$
$\Rightarrow 396 = 2 \times \frac{22}{7} \times R \Rightarrow R = \frac{396 \times 7}{44} \Rightarrow R = 63$

Circumference of the inner track = $2 π r$
$\Rightarrow 352 = 2 \times \frac{22}{7} \times r \Rightarrow r = \frac{352 \times 7}{44} \Rightarrow r = 56$

Width of the track = Radius of the outer track $-$ Radius of the inner track
$= 63 - 56 = 7 m$

Area of the outer circle = $π R^{2}$
    $= \frac{22}{7} \times 63 \times 63 = 12474 m^{2}$

Area of the inner circle = $π R^{2}$
        $= \frac{22}{7} \times 56 \times 56 = 9856 m^{2}$

Area of the track = 12474 $-$ 9856
   = 2618 $m^{2}$

Question 11:

Let r m and R m be the radii of the inner and outer tracks.
Now,
Circumference of the outer track = $2 π R$
$\Rightarrow 396 = 2 \times \frac{22}{7} \times R \Rightarrow R = \frac{396 \times 7}{44} \Rightarrow R = 63$

Circumference of the inner track = $2 π r$
$\Rightarrow 352 = 2 \times \frac{22}{7} \times r \Rightarrow r = \frac{352 \times 7}{44} \Rightarrow r = 56$

Width of the track = Radius of the outer track $-$ Radius of the inner track
$= 63 - 56 = 7 m$

Area of the outer circle = $π R^{2}$
    $= \frac{22}{7} \times 63 \times 63 = 12474 m^{2}$

Area of the inner circle = $π R^{2}$
        $= \frac{22}{7} \times 56 \times 56 = 9856 m^{2}$

Area of the track = 12474 $-$ 9856
   = 2618 $m^{2}$

Answer:

Given:
Radius = 2 cm
Angle of sector = $150^{\circ}$
Now,
$Length of the arc = \frac{2 πr θ}{360} = \frac{2 \times \frac{22}{7} \times 21 \times 150}{360} = 55 cm$

Area of the sector = $\frac{{πr}^{2} θ}{360}$

$= \frac{22}{7} \times 21 \times 21 \times \frac{150}{360} = 577.5 cm$

Question 12:

Given:
Radius = 2 cm
Angle of sector = $150^{\circ}$
Now,
$Length of the arc = \frac{2 πr θ}{360} = \frac{2 \times \frac{22}{7} \times 21 \times 150}{360} = 55 cm$

Area of the sector = $\frac{{πr}^{2} θ}{360}$

$= \frac{22}{7} \times 21 \times 21 \times \frac{150}{360} = 577.5 cm$

Answer:

Radius of the circle, r = 10 cm

Area of sector OPRQ
$= \frac{60 °}{360 °} \times π r^{2} = \frac{1}{6} \times 3.14 \times {(10)}^{2} = 52.33 {cm}^{2}$

In ΔOPQ,
∠OPQ = ∠OQP (As OP = OQ)
∠OPQ + ∠OQP + ∠POQ = 180°
2∠OPQ = 120°
∠OPQ = 60°
ΔOPQ is an equilateral triangle.
So, area of ΔOPQ
$= \frac{\sqrt{3}}{4} \times {(Side)}^{2} = \frac{\sqrt{3}}{4} \times {(10)}^{2} = \frac{100 \sqrt{3}}{4} {cm}^{2} = 43.30 {cm}^{2}$

Area of minor segment PRQ
= Area of sector OPRQ − Area of ΔOPQ
= 52.33 − 43.30
= 9.03 cm²

Area of major segment PSQ
= Area of circle − Area of minor segment PRQ
$= π {(10)}^{2} - 9.03 = 314 - 9.03 = 304.97 {cm}^{2}$

Question 13:

Radius of the circle, r = 10 cm

Area of sector OPRQ
$= \frac{60 °}{360 °} \times π r^{2} = \frac{1}{6} \times 3.14 \times {(10)}^{2} = 52.33 {cm}^{2}$

In ΔOPQ,
∠OPQ = ∠OQP (As OP = OQ)
∠OPQ + ∠OQP + ∠POQ = 180°
2∠OPQ = 120°
∠OPQ = 60°
ΔOPQ is an equilateral triangle.
So, area of ΔOPQ
$= \frac{\sqrt{3}}{4} \times {(Side)}^{2} = \frac{\sqrt{3}}{4} \times {(10)}^{2} = \frac{100 \sqrt{3}}{4} {cm}^{2} = 43.30 {cm}^{2}$

Area of minor segment PRQ
= Area of sector OPRQ − Area of ΔOPQ
= 52.33 − 43.30
= 9.03 cm²

Area of major segment PSQ
= Area of circle − Area of minor segment PRQ
$= π {(10)}^{2} - 9.03 = 314 - 9.03 = 304.97 {cm}^{2}$

Answer:

Length of the arc = 16.5 cm
$θ = 54^{\circ}$
Radius = ?
Circumference=?
We know:
Length of the arc $= \frac{2 πrθ}{360}$
$\Rightarrow 16.5 = \frac{2 \times \frac{22}{7} \times r \times 54}{360} \Rightarrow r = \frac{16.5 \times 360 \times 7}{44 \times 54} \Rightarrow r = 17.5 cm$

$c = 2 πr = 2 \times \frac{22}{7} \times 17.5 = 110 cm$

Circumference = 110 cm

Now,
Area of the circle = ${πr}^{2}$
$= \frac{22}{7} \times 17.5 \times 17.5 = 962.5 {cm}^{2}$

Question 14:

Length of the arc = 16.5 cm
$θ = 54^{\circ}$
Radius = ?
Circumference=?
We know:
Length of the arc $= \frac{2 πrθ}{360}$
$\Rightarrow 16.5 = \frac{2 \times \frac{22}{7} \times r \times 54}{360} \Rightarrow r = \frac{16.5 \times 360 \times 7}{44 \times 54} \Rightarrow r = 17.5 cm$

$c = 2 πr = 2 \times \frac{22}{7} \times 17.5 = 110 cm$

Circumference = 110 cm

Now,
Area of the circle = ${πr}^{2}$
$= \frac{22}{7} \times 17.5 \times 17.5 = 962.5 {cm}^{2}$

Answer:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB
$= \frac{90 °}{360 °} π {(OA)}^{2} - \frac{1}{2} \times OA \times OB = \frac{1}{4} \times \frac{22}{7} \times {(7)}^{2} - \frac{1}{2} \times 7 \times 7 = \frac{1}{4} \times \frac{22}{7} \times {(7)}^{2} - \frac{1}{2} \times 7 \times 7 = 38.5 - 24.5 = 14 {cm}^{2}$
Area of major segment APB = Area of circle − Area of minor segment
$= π {(OA)}^{2} - 14 = \frac{22}{7} \times {(7)}^{2} - 14 = 154 - 14 = 140 {cm}^{2}$
Hence, the area of major segment is 140 cm² .

Question 15:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB
$= \frac{90 °}{360 °} π {(OA)}^{2} - \frac{1}{2} \times OA \times OB = \frac{1}{4} \times \frac{22}{7} \times {(7)}^{2} - \frac{1}{2} \times 7 \times 7 = \frac{1}{4} \times \frac{22}{7} \times {(7)}^{2} - \frac{1}{2} \times 7 \times 7 = 38.5 - 24.5 = 14 {cm}^{2}$
Area of major segment APB = Area of circle − Area of minor segment
$= π {(OA)}^{2} - 14 = \frac{22}{7} \times {(7)}^{2} - 14 = 154 - 14 = 140 {cm}^{2}$
Hence, the area of major segment is 140 cm² .

Answer:

Let AB be the chord. Joining A and B to O, we get an equilateral triangle OAB.
Thus, we have:
$∠ O = ∠ A = ∠ B = 60 °$

Length of the arc ACB:
$2 π \times 12 \times \frac{60}{360} = 4 π = 12.56 cm$

Length of the arc ADB:

$Circumference of the circle - Length of the arc ACB = 2 π \times 12 - 4 π = 20 π cm = 62.80 cm$

Now,
Area of the minor segment:

$Area of the sector - Area of the triangle = [π \times {(12)}^{2} \times \frac{60}{360} - \frac{\sqrt{3}}{4} \times {(12)}^{2}] = 13.08 {cm}^{2}$

Question 16:

Let AB be the chord. Joining A and B to O, we get an equilateral triangle OAB.
Thus, we have:
$∠ O = ∠ A = ∠ B = 60 °$

Length of the arc ACB:
$2 π \times 12 \times \frac{60}{360} = 4 π = 12.56 cm$

Length of the arc ADB:

$Circumference of the circle - Length of the arc ACB = 2 π \times 12 - 4 π = 20 π cm = 62.80 cm$

Now,
Area of the minor segment:

$Area of the sector - Area of the triangle = [π \times {(12)}^{2} \times \frac{60}{360} - \frac{\sqrt{3}}{4} \times {(12)}^{2}] = 13.08 {cm}^{2}$

Answer:

Let O be the centre of the circle and AB be the chord.

Consider $∆$ OAB.

$OA = OB = 5 \sqrt{2} cm$

${OA}^{2} + {OB}^{2} = 50 + 50 = 100$

Now,
$\sqrt{100} = 10 cm = A B$

Thus, $∆$ OAB is a right isosceles triangle.

Thus, we have:
Area of $∆$ OAB = $\frac{1}{2} \times 5 \sqrt{2} \times 5 \sqrt{2} = 25 {cm}^{2}$

Area of the minor segment = Area of the sector $-$ Area of the triangle
$= (\frac{90}{360} \times π \times {(5 \sqrt{2})}^{2}) - 25 = 14.25 {cm}^{2}$

Area of the major segment = Area of the circle $-$ Area of the minor segment
$= π \times {(5 \sqrt{2})}^{2} - 14.25 = 142.75 {cm}^{2}$

Question 17:

Let O be the centre of the circle and AB be the chord.

Consider $∆$ OAB.

$OA = OB = 5 \sqrt{2} cm$

${OA}^{2} + {OB}^{2} = 50 + 50 = 100$

Now,
$\sqrt{100} = 10 cm = A B$

Thus, $∆$ OAB is a right isosceles triangle.

Thus, we have:
Area of $∆$ OAB = $\frac{1}{2} \times 5 \sqrt{2} \times 5 \sqrt{2} = 25 {cm}^{2}$

Area of the minor segment = Area of the sector $-$ Area of the triangle
$= (\frac{90}{360} \times π \times {(5 \sqrt{2})}^{2}) - 25 = 14.25 {cm}^{2}$

Area of the major segment = Area of the circle $-$ Area of the minor segment
$= π \times {(5 \sqrt{2})}^{2} - 14.25 = 142.75 {cm}^{2}$

Answer:

$Area of the minor sector = \frac{120}{360} \times π \times 42 \times 42 = \frac{1}{3} \times π \times 42 \times 42 = π \times 14 \times 42 = 1848 {cm}^{2}$

Area of the triangle = $\frac{1}{2} R^{2} \sin θ$
Here, R is the measure of the equal sides of the isosceles triangle and θ is the angle enclosed by the equal sides.
Thus, we have:
$\frac{1}{2} \times 42 \times 42 \times \sin (120 °) = 762.93 {cm}^{2}$

Area of the minor segment = Area of the sector $-$ Area of the triangle
= $1848 - 762.93 = 1085.07 {cm}^{2}$

Area of the major segment = Area of the circle $-$ Area of the minor segment
$= (π \times 42 \times 42) - 1085.07 = 5544 - 1085.07 = 4458.93 {cm}^{2}$

Question 18:

$Area of the minor sector = \frac{120}{360} \times π \times 42 \times 42 = \frac{1}{3} \times π \times 42 \times 42 = π \times 14 \times 42 = 1848 {cm}^{2}$

Area of the triangle = $\frac{1}{2} R^{2} \sin θ$
Here, R is the measure of the equal sides of the isosceles triangle and θ is the angle enclosed by the equal sides.
Thus, we have:
$\frac{1}{2} \times 42 \times 42 \times \sin (120 °) = 762.93 {cm}^{2}$

Area of the minor segment = Area of the sector $-$ Area of the triangle
= $1848 - 762.93 = 1085.07 {cm}^{2}$

Area of the major segment = Area of the circle $-$ Area of the minor segment
$= (π \times 42 \times 42) - 1085.07 = 5544 - 1085.07 = 4458.93 {cm}^{2}$

Answer:

Let the chord be AB. The ends of the chord are connected to the centre of the circle O to give the triangle OAB.

OAB is an isosceles triangle. The angle at the centre is 60 $°$

Area of the triangle = $\frac{1}{2} {(30)}^{2} \sin 60 ° = 450 \times \frac{\sqrt{3}}{2} = 389.25 {cm}^{2}$

Area of the sector OACBO = $\frac{60}{360} \times π \times 30 \times 30 = 150 π = 471 {cm}^{2}$

Area of the minor segment = Area of the sector $-$ Area of the triangle
= $471 - 389.25 = 81.75 {cm}^{2}$

Area of the major segment = Area of the circle $-$ Area of the minor segment
$= (π \times 30 \times 30) - 81.29 = 2744.71 {cm}^{2}$

Question 19:

Let the chord be AB. The ends of the chord are connected to the centre of the circle O to give the triangle OAB.

OAB is an isosceles triangle. The angle at the centre is 60 $°$

Area of the triangle = $\frac{1}{2} {(30)}^{2} \sin 60 ° = 450 \times \frac{\sqrt{3}}{2} = 389.25 {cm}^{2}$

Area of the sector OACBO = $\frac{60}{360} \times π \times 30 \times 30 = 150 π = 471 {cm}^{2}$

Area of the minor segment = Area of the sector $-$ Area of the triangle
= $471 - 389.25 = 81.75 {cm}^{2}$

Area of the major segment = Area of the circle $-$ Area of the minor segment
$= (π \times 30 \times 30) - 81.29 = 2744.71 {cm}^{2}$

Answer:

Let the length of the major arc be $x$ cm
Radius of the circle = 10.5 cm
∴ Length of the minor arc = $\frac{x}{5} cm$

Circumference = $(x + \frac{x}{5}) = \frac{6 x}{5} cm$

Using the given data, we get:

$\frac{6 x}{5} = 2 \times \frac{22}{7} \times \frac{21}{2} \Rightarrow \frac{6 x}{5} = 66 Or, x = 55$

∴ Area of the sector corresponding to the major arc = $(\frac{1}{2} \times 55 \times \frac{21}{2}) = 288.75 {cm}^{2}$

Question 20:

Let the length of the major arc be $x$ cm
Radius of the circle = 10.5 cm
∴ Length of the minor arc = $\frac{x}{5} cm$

Circumference = $(x + \frac{x}{5}) = \frac{6 x}{5} cm$

Using the given data, we get:

$\frac{6 x}{5} = 2 \times \frac{22}{7} \times \frac{21}{2} \Rightarrow \frac{6 x}{5} = 66 Or, x = 55$

∴ Area of the sector corresponding to the major arc = $(\frac{1}{2} \times 55 \times \frac{21}{2}) = 288.75 {cm}^{2}$

Answer:

In 2 days, the short hand will complete 4 rounds.
Length of the short hand = 4 cm

Distance covered by the short hand = $4 \times 2 π \times 4 = 32 π cm$

In the same 2 days, the long hand will complete 48 rounds.
Length of the long hand = 6 cm
Distance covered by the long hand = $48 \times 2 π \times 6 = 576 π cm$

∴ Total distance covered by both the hands = Distance covered by the short hand + Distance covered by the long hand

$= 32 π + 576 π = 608 π = 608 \times 3.14 = 1909.12 cm$

Question 21:

In 2 days, the short hand will complete 4 rounds.
Length of the short hand = 4 cm

Distance covered by the short hand = $4 \times 2 π \times 4 = 32 π cm$

In the same 2 days, the long hand will complete 48 rounds.
Length of the long hand = 6 cm
Distance covered by the long hand = $48 \times 2 π \times 6 = 576 π cm$

∴ Total distance covered by both the hands = Distance covered by the short hand + Distance covered by the long hand

$= 32 π + 576 π = 608 π = 608 \times 3.14 = 1909.12 cm$

Answer:

Let the radius of the circle be r.
â€‹Now,
$Circumference = 88 \Rightarrow 2 π r = 88 \Rightarrow r = 14 cm$
Now,
Area of quadrant = $\frac{1}{4} π r^{2} = \frac{1}{4} \times \frac{22}{7} \times {(14)}^{2} = 154 {cm}^{2}$
Hence, the area of the quadrant of the circle is 154 cm².

Question 22:

Let the radius of the circle be r.
â€‹Now,
$Circumference = 88 \Rightarrow 2 π r = 88 \Rightarrow r = 14 cm$
Now,
Area of quadrant = $\frac{1}{4} π r^{2} = \frac{1}{4} \times \frac{22}{7} \times {(14)}^{2} = 154 {cm}^{2}$
Hence, the area of the quadrant of the circle is 154 cm².

Answer:

r₁ = 16 m
r₂ = 23 m

Amount of additional ground available = Area of the bigger circle $-$ Area of the smaller circle
$= π ({r_{1}}^{2} - {r_{2}}^{2}) = π (23^{2} - 16^{2}) = π (23 + 16) (23 - 16) = 858 m^{2}$

Question 23:

r₁ = 16 m
r₂ = 23 m

Amount of additional ground available = Area of the bigger circle $-$ Area of the smaller circle
$= π ({r_{1}}^{2} - {r_{2}}^{2}) = π (23^{2} - 16^{2}) = π (23 + 16) (23 - 16) = 858 m^{2}$

Answer:

Radius of the quadrant of the circle = 21 m
The shaded portion shows the part of the field the horse can graze.

Area of the grazed field = Area of the quadrant OPQ
                                 $= \frac{1}{4} of the circle having radius OP$
   $= \frac{1}{4} {πr}^{2} = \frac{1}{4} \times \frac{22}{7} \times 21 \times 21 = 346.5 m^{2}$

Total area of the field = $70 \times 52 = 3640 m^{2}$

Area left ungrazed = Area of the field $-$ Area of the grazed field
                               = $3640 - 346.5 = 3293.5 m^{2}$

Question 24:

Radius of the quadrant of the circle = 21 m
The shaded portion shows the part of the field the horse can graze.

Area of the grazed field = Area of the quadrant OPQ
                                 $= \frac{1}{4} of the circle having radius OP$
   $= \frac{1}{4} {πr}^{2} = \frac{1}{4} \times \frac{22}{7} \times 21 \times 21 = 346.5 m^{2}$

Total area of the field = $70 \times 52 = 3640 m^{2}$

Area left ungrazed = Area of the field $-$ Area of the grazed field
                               = $3640 - 346.5 = 3293.5 m^{2}$

Answer:

Side of the equilateral triangle = 12 m
Area of the equilateral triangle = $\frac{\sqrt{3}}{4} \times (Side)^{2}$
$= \frac{\sqrt{3}}{4} \times 12 \times 12 = 62.28 m^{2}$
Length of the rope = 7 m
Area of the field the horse can graze is the area of the sector of radius 7 m .Also, the angle subtended at the centre is 60 $°$

= $\frac{θ}{360} \times π r^{2}$

$= \frac{60}{360} \times \frac{22}{7} \times (7)^{2} = 25.67 m^{2}$

Area of the field the horse cannot graze = Area of the equilateral triangle $-$ Area of the field the horse can graze
$= 62.28 - 25.67 = 36.61 m^{2}$

Question 25:

Side of the equilateral triangle = 12 m
Area of the equilateral triangle = $\frac{\sqrt{3}}{4} \times (Side)^{2}$
$= \frac{\sqrt{3}}{4} \times 12 \times 12 = 62.28 m^{2}$
Length of the rope = 7 m
Area of the field the horse can graze is the area of the sector of radius 7 m .Also, the angle subtended at the centre is 60 $°$

= $\frac{θ}{360} \times π r^{2}$

$= \frac{60}{360} \times \frac{22}{7} \times (7)^{2} = 25.67 m^{2}$

Area of the field the horse cannot graze = Area of the equilateral triangle $-$ Area of the field the horse can graze
$= 62.28 - 25.67 = 36.61 m^{2}$

Answer:

Each cow can graze a region that cannot be accessed by other cows.
∴ Radius of the region grazed by each cow = $\frac{50}{2} = 25 m$

Area that each cow grazes = $\frac{1}{4} \times π \times r^{2}$
$= \frac{1}{4} \times 3.14 \times 25 \times 25 = 490.625 {cm}^{2}$

Total area grazed = $4 \times 490.625 = 1963.49 m^{2}$
$Area of the square = (Side)^{2} = 50^{2} = 2500 {cm}^{2}$

Now,
Area left ungrazed = Area of the square $-$ Grazed area
= $2500 - 1963.49 = 536.51 m^{2}$

Question 26:

Each cow can graze a region that cannot be accessed by other cows.
∴ Radius of the region grazed by each cow = $\frac{50}{2} = 25 m$

Area that each cow grazes = $\frac{1}{4} \times π \times r^{2}$
$= \frac{1}{4} \times 3.14 \times 25 \times 25 = 490.625 {cm}^{2}$

Total area grazed = $4 \times 490.625 = 1963.49 m^{2}$
$Area of the square = (Side)^{2} = 50^{2} = 2500 {cm}^{2}$

Now,
Area left ungrazed = Area of the square $-$ Grazed area
= $2500 - 1963.49 = 536.51 m^{2}$

Answer:

In a rhombus, all sides are congruent to each other.

Thus, we have:
$O P = P Q = Q R = R O$

Now, consider $∆ Q O P$ .

$O Q = O P (Both are radii .)$

Therefore, $∆ Q O P$ is equilateral.

Similarly, $∆ Q O R$ is also equilateral and $∆ Q O P ≅ ∆ Q O R$ .

$Ar . (Q R O P) = Ar . (∆ Q O P) + A (∆ Q O R) = 2 Ar . [∆ Q O P]$

$Ar . (∆ QOP) = \frac{1}{2} \times 32 \sqrt{3} = 16 \sqrt{3} O r, 16 \sqrt{3} = \frac{\sqrt{3}}{4} s^{2} (where s is the side of the rhombus) O r, s^{2} = 16 \times 4 = 64 \Rightarrow s = 8 cm$

∴ OQ = 8 cm

Hence, the radius of the circle is 8 cm.

Question 27:

In a rhombus, all sides are congruent to each other.

Thus, we have:
$O P = P Q = Q R = R O$

Now, consider $∆ Q O P$ .

$O Q = O P (Both are radii .)$

Therefore, $∆ Q O P$ is equilateral.

Similarly, $∆ Q O R$ is also equilateral and $∆ Q O P ≅ ∆ Q O R$ .

$Ar . (Q R O P) = Ar . (∆ Q O P) + A (∆ Q O R) = 2 Ar . [∆ Q O P]$

$Ar . (∆ QOP) = \frac{1}{2} \times 32 \sqrt{3} = 16 \sqrt{3} O r, 16 \sqrt{3} = \frac{\sqrt{3}}{4} s^{2} (where s is the side of the rhombus) O r, s^{2} = 16 \times 4 = 64 \Rightarrow s = 8 cm$

∴ OQ = 8 cm

Hence, the radius of the circle is 8 cm.

Answer:

(â€‹i)â€‹ If a circle is inscribed in a square, then the side of the square is equal to the diameter of the circle.
Side of the square = 10 cm
Side = Diameter = 10
∴ Radius = 5 cm
Area of the inscribed circle = ${πr}^{2}$
      $= 3.14 \times 5 \times 5 = 78.5 {cm}^{2}$

(ii) If a circle is circumscribed in a square, then the diagonal of the square is equal to the diameter of the circle.
$Diagonal of the square = \sqrt{2} \times Side of the square = \sqrt{2} \times 10 = 10 \sqrt{2} cm$

Diagonal = Diameter = $10 \sqrt{2} cm$
∴ $r = 5 \sqrt{2}$ cm

Now,
Area of the circumscribed circle = ${πr}^{2}$
                             $= 3.14 \times {(5 \sqrt{2})}^{2} = 3.14 \times 50 = 157 {cm}^{2}$

Question 28:

(â€‹i)â€‹ If a circle is inscribed in a square, then the side of the square is equal to the diameter of the circle.
Side of the square = 10 cm
Side = Diameter = 10
∴ Radius = 5 cm
Area of the inscribed circle = ${πr}^{2}$
      $= 3.14 \times 5 \times 5 = 78.5 {cm}^{2}$

(ii) If a circle is circumscribed in a square, then the diagonal of the square is equal to the diameter of the circle.
$Diagonal of the square = \sqrt{2} \times Side of the square = \sqrt{2} \times 10 = 10 \sqrt{2} cm$

Diagonal = Diameter = $10 \sqrt{2} cm$
∴ $r = 5 \sqrt{2}$ cm

Now,
Area of the circumscribed circle = ${πr}^{2}$
                             $= 3.14 \times {(5 \sqrt{2})}^{2} = 3.14 \times 50 = 157 {cm}^{2}$

Answer:

If a square is inscribed in a circle, then the diagonals of the square are diameters of the circle.
Let the diagonal of the square be d cm.
Thus, we have:
Radius, r = $\frac{d}{2} cm$

$Area of the circle = π r^{2}$
= $π \frac{d^{2}}{4} {cm}^{2}$

$We know : d = \sqrt{2} \times Side \Rightarrow Side = \frac{d}{\sqrt{2}} cm$

$Area of the square = (Side)^{2} = {(\frac{d}{\sqrt{2}})}^{2} = \frac{d^{2}}{2} {cm}^{2}$

Ratio of the area of the circle to that of the square:
$= \frac{π \frac{d^{2}}{4}}{\frac{d^{2}}{2}} = \frac{π}{2}$
Thus, the ratio of the area of the circle to that of the square is $π : 2$ .

Question 29:

If a square is inscribed in a circle, then the diagonals of the square are diameters of the circle.
Let the diagonal of the square be d cm.
Thus, we have:
Radius, r = $\frac{d}{2} cm$

$Area of the circle = π r^{2}$
= $π \frac{d^{2}}{4} {cm}^{2}$

$We know : d = \sqrt{2} \times Side \Rightarrow Side = \frac{d}{\sqrt{2}} cm$

$Area of the square = (Side)^{2} = {(\frac{d}{\sqrt{2}})}^{2} = \frac{d^{2}}{2} {cm}^{2}$

Ratio of the area of the circle to that of the square:
$= \frac{π \frac{d^{2}}{4}}{\frac{d^{2}}{2}} = \frac{π}{2}$
Thus, the ratio of the area of the circle to that of the square is $π : 2$ .

Answer:

Let the radius of the inscribed circle be r cm.
Given:
Area of the circle = 154 ${cm}^{2}$
We know:
Area of the circle = $π r^{2}$
$\Rightarrow 154 = \frac{22}{7} r^{2} \Rightarrow \frac{154 \times 7}{22} = r^{2} \Rightarrow r^{2} = 49 \Rightarrow r = 7$
In a triangle, the centre of the inscribed circle is the point of intersection of the medians and altitudes of the triangle. The centroid divides the median of a triangle in the ratio 2:1.
Here,
AO:OD = 2:1

Now,
Let the altitude be h cm.
We have:
$∠ A D B = 90^{\circ} O D = \frac{1}{3} A D O D = \frac{h}{3}$
$\Rightarrow h = 3 r \Rightarrow h = 21$

Let each side of the triangle be a cm.
$In the right - angled ∆ ADB, we have : A B^{2} = A D^{2} + D B^{2} a^{2} = h^{2} + {(\frac{a}{2})}^{2} 4 a^{2} = 4 h^{2} + a^{2} 3 a^{2} = 4 h^{2} a^{2} = \frac{4 h^{2}}{3} a = \frac{2 h}{\sqrt{3}} a = \frac{42}{\sqrt{3}}$
∴ Perimeter of the triangle = 3a
â€‹ $= 3 \times \frac{42}{\sqrt{3}} = \sqrt{3} \times 42 = 72.66 cm$

Question 30:

Let the radius of the inscribed circle be r cm.
Given:
Area of the circle = 154 ${cm}^{2}$
We know:
Area of the circle = $π r^{2}$
$\Rightarrow 154 = \frac{22}{7} r^{2} \Rightarrow \frac{154 \times 7}{22} = r^{2} \Rightarrow r^{2} = 49 \Rightarrow r = 7$
In a triangle, the centre of the inscribed circle is the point of intersection of the medians and altitudes of the triangle. The centroid divides the median of a triangle in the ratio 2:1.
Here,
AO:OD = 2:1

Now,
Let the altitude be h cm.
We have:
$∠ A D B = 90^{\circ} O D = \frac{1}{3} A D O D = \frac{h}{3}$
$\Rightarrow h = 3 r \Rightarrow h = 21$

Let each side of the triangle be a cm.
$In the right - angled ∆ ADB, we have : A B^{2} = A D^{2} + D B^{2} a^{2} = h^{2} + {(\frac{a}{2})}^{2} 4 a^{2} = 4 h^{2} + a^{2} 3 a^{2} = 4 h^{2} a^{2} = \frac{4 h^{2}}{3} a = \frac{2 h}{\sqrt{3}} a = \frac{42}{\sqrt{3}}$
∴ Perimeter of the triangle = 3a
â€‹ $= 3 \times \frac{42}{\sqrt{3}} = \sqrt{3} \times 42 = 72.66 cm$

Answer:

Radius of the wheel = 42 cm
â€‹Circumference of the wheel = $2 πr$
$= 2 \times \frac{22}{7} \times 42 = 264 cm = \frac{264}{100} m = 2.64 m$

Distance covered by the wheel in 1 revolution = 2.64 m

Total distance = 19.8 km or 19800 m

∴ Number of revolutions taken by the wheel = $\frac{19800}{2.64} = 7500$

Question 31:

Radius of the wheel = 42 cm
â€‹Circumference of the wheel = $2 πr$
$= 2 \times \frac{22}{7} \times 42 = 264 cm = \frac{264}{100} m = 2.64 m$

Distance covered by the wheel in 1 revolution = 2.64 m

Total distance = 19.8 km or 19800 m

∴ Number of revolutions taken by the wheel = $\frac{19800}{2.64} = 7500$

Answer:

Radius of the wheel = 2.1 m
Circumference of the wheel = $2 πr$
   $= 2 \times \frac{22}{7} \times 2.1 = 13.2 m$
Distance covered by the wheel in 1 revolution = 13.2 m
Distance covered by the wheel in 75 revolutions = $(13.2 \times 75) = (990 \times \frac{1}{100}) m$
    $= (990 \times \frac{1}{1000}) km$
Distance covered by the wheel in 1 minute = Distance covered by the wheel in 75 revolutions = $\frac{990}{1000}$ km

∴ Distance covered by the wheel in 1 hour = $\frac{990}{1000} \times 60$
                = $59.4 km / h$

Question 32:

Radius of the wheel = 2.1 m
Circumference of the wheel = $2 πr$
   $= 2 \times \frac{22}{7} \times 2.1 = 13.2 m$
Distance covered by the wheel in 1 revolution = 13.2 m
Distance covered by the wheel in 75 revolutions = $(13.2 \times 75) = (990 \times \frac{1}{100}) m$
    $= (990 \times \frac{1}{1000}) km$
Distance covered by the wheel in 1 minute = Distance covered by the wheel in 75 revolutions = $\frac{990}{1000}$ km

∴ Distance covered by the wheel in 1 hour = $\frac{990}{1000} \times 60$
                = $59.4 km / h$

Answer:

Distance = 4.95 km = $4.95 \times 1000 \times 100 cm$
∴ Distance covered by the wheel in 1 revolution $= \frac{Total distance covered}{Number of revolutions}$

   $= \frac{4.95 \times 1000 \times 100}{2500} = 198 cm$

Now,
Circumference of the wheel = 198 cm
$\Rightarrow 2 π r = 198 \Rightarrow 2 \times \frac{22}{7} \times r = 198 \Rightarrow r = \frac{198 \times 7}{44} \Rightarrow r = 31.5 cm$

∴ Diameter of the wheel = 2r
   = 2(31.5)
   = 63 cm

Question 33:

Distance = 4.95 km = $4.95 \times 1000 \times 100 cm$
∴ Distance covered by the wheel in 1 revolution $= \frac{Total distance covered}{Number of revolutions}$

   $= \frac{4.95 \times 1000 \times 100}{2500} = 198 cm$

Now,
Circumference of the wheel = 198 cm
$\Rightarrow 2 π r = 198 \Rightarrow 2 \times \frac{22}{7} \times r = 198 \Rightarrow r = \frac{198 \times 7}{44} \Rightarrow r = 31.5 cm$

∴ Diameter of the wheel = 2r
   = 2(31.5)
   = 63 cm

Answer:

Diameter of the wheel = 60 cm
â€‹∴ Radius of the wheel = 30 cm
â€‹Circumference of the wheel = $2 πr$
$= 2 \times \frac{22}{7} \times 30 = \frac{1320}{7} cm$
Distance covered by the wheel in 1 revolution = $\frac{1320}{7} cm$
∴ Distance covered by the wheel in 140 revolutions = $(\frac{1320}{7} \times 140 \times \frac{1}{100}) m$
$= (\frac{1320 \times 140}{7 \times 100} \times \frac{1}{1000}) km = \frac{264}{1000} km$

Now,
Distance covered by the wheel in 1 minute = Distance covered by the wheel in 140 revolutions = $\frac{264}{1000} km$

∴ Distance covered by the wheel in 1 hour = $\frac{264}{1000} \times 60 = 15.84 km / h$

Hence, the speed at which the boy is cycling is 15.84 km/h.

Question 34:

Diameter of the wheel = 60 cm
â€‹∴ Radius of the wheel = 30 cm
â€‹Circumference of the wheel = $2 πr$
$= 2 \times \frac{22}{7} \times 30 = \frac{1320}{7} cm$
Distance covered by the wheel in 1 revolution = $\frac{1320}{7} cm$
∴ Distance covered by the wheel in 140 revolutions = $(\frac{1320}{7} \times 140 \times \frac{1}{100}) m$
$= (\frac{1320 \times 140}{7 \times 100} \times \frac{1}{1000}) km = \frac{264}{1000} km$

Now,
Distance covered by the wheel in 1 minute = Distance covered by the wheel in 140 revolutions = $\frac{264}{1000} km$

∴ Distance covered by the wheel in 1 hour = $\frac{264}{1000} \times 60 = 15.84 km / h$

Hence, the speed at which the boy is cycling is 15.84 km/h.

Answer:

The radius of wheel of a motorcycle = 35 cm = 0.35 m.
So, the distance covered by this wheel in 1 revolution will be equal to perimeter of wheel i.e. $2 π r = 2 \times \frac{22}{7} \times 0.35 = 2.2 m .$
Since speed is given to be $66 km / hr = 66 \times \frac{1000 m}{60 \min} = 1100 m / \min .$
As we know $speed = \frac{distance}{time}$
$\Rightarrow 1100 = \frac{number of revolutions \times perimeter of wheel}{time} \Rightarrow 1100 = \frac{number of revolutions \times 2.2}{1} \Rightarrow number of revolutions = \frac{1100}{2.2} = 500 revolutions .$

Question 35:

The radius of wheel of a motorcycle = 35 cm = 0.35 m.
So, the distance covered by this wheel in 1 revolution will be equal to perimeter of wheel i.e. $2 π r = 2 \times \frac{22}{7} \times 0.35 = 2.2 m .$
Since speed is given to be $66 km / hr = 66 \times \frac{1000 m}{60 \min} = 1100 m / \min .$
As we know $speed = \frac{distance}{time}$
$\Rightarrow 1100 = \frac{number of revolutions \times perimeter of wheel}{time} \Rightarrow 1100 = \frac{number of revolutions \times 2.2}{1} \Rightarrow number of revolutions = \frac{1100}{2.2} = 500 revolutions .$

Answer:

Radius of the front wheel = $40 cm = \frac{2}{5} m$

Circumference of the front wheel = $(2 π \times \frac{2}{5}) m = \frac{4 π}{5} m$

Distance covered by the front wheel in 800 revolutions = $(\frac{4 π}{5} \times 800) m = (640 π) m$
Radius of the rear wheel = 1 m
Circumference of the rear wheel = $(2 π \times 1) = 2 π m$

∴ Required number of revolutions = $\frac{Distance covered by the front wheel in 800 revolutions}{Circumference of the rear wheel}$
$= \frac{640 π}{2 π} = 320$

Question 36:

Radius of the front wheel = $40 cm = \frac{2}{5} m$

Circumference of the front wheel = $(2 π \times \frac{2}{5}) m = \frac{4 π}{5} m$

Distance covered by the front wheel in 800 revolutions = $(\frac{4 π}{5} \times 800) m = (640 π) m$
Radius of the rear wheel = 1 m
Circumference of the rear wheel = $(2 π \times 1) = 2 π m$

∴ Required number of revolutions = $\frac{Distance covered by the front wheel in 800 revolutions}{Circumference of the rear wheel}$
$= \frac{640 π}{2 π} = 320$

Answer:

Side of the square = 14 cm
Radius of the circle $= \frac{14}{2}$ = 7 cm
Area of the quadrant of one circle = $\frac{1}{4} {πr}^{2}$
            $= \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 = 38.5 {cm}^{2}$
Area of the quadrants of four circles = $38.5 \times 4 = 154 {cm}^{2}$

Now,
Area of the square = ${(Side)}^{2}$
   $= 14^{2} = 196 {cm}^{2}$

Area of the shaded region = Area of the square $-$ Area of the quadrants of four circles
           = 196 $-$ 154
             = 42 cm²

Question 37:

Side of the square = 14 cm
Radius of the circle $= \frac{14}{2}$ = 7 cm
Area of the quadrant of one circle = $\frac{1}{4} {πr}^{2}$
            $= \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 = 38.5 {cm}^{2}$
Area of the quadrants of four circles = $38.5 \times 4 = 154 {cm}^{2}$

Now,
Area of the square = ${(Side)}^{2}$
   $= 14^{2} = 196 {cm}^{2}$

Area of the shaded region = Area of the square $-$ Area of the quadrants of four circles
           = 196 $-$ 154
             = 42 cm²

Answer:

Radius = 5 cm
AB = BC = CD = AD = 10 cm
All sides are equal, so it is a square.
Area of a square = ${Side}^{2}$
Area of the square = $10^{2} = 100 {cm}^{2}$
Area of the quadrant of one circle = $\frac{1}{4} {πr}^{2}$

$= \frac{1}{4} \times \frac{22}{7} \times 5 \times 5 = 19.64 {cm}^{2}$
Area of the quadrants of four circles = $19.64 \times 4 = 78.57 {cm}^{2}$
Area of the shaded portion = Area of the square $-$ Area of the quadrants of four circles
$= 100 - 78.57 = 21.43 {cm}^{2}$

Question 38:

Radius = 5 cm
AB = BC = CD = AD = 10 cm
All sides are equal, so it is a square.
Area of a square = ${Side}^{2}$
Area of the square = $10^{2} = 100 {cm}^{2}$
Area of the quadrant of one circle = $\frac{1}{4} {πr}^{2}$

$= \frac{1}{4} \times \frac{22}{7} \times 5 \times 5 = 19.64 {cm}^{2}$
Area of the quadrants of four circles = $19.64 \times 4 = 78.57 {cm}^{2}$
Area of the shaded portion = Area of the square $-$ Area of the quadrants of four circles
$= 100 - 78.57 = 21.43 {cm}^{2}$

Answer:

When four circles touch each other, their centres form the vertices of a square. The sides of the square are 2a units.

Area of the square = ${(2 a)}^{2} = 4 a^{2} sq . units$

Area occupied by the four sectors
$= 4 \times \frac{90}{360} \times π \times a^{2} = π a^{2} sq . units$

Area between the circles = Area of the square $-$ Area of the four sectors
$= (4 - \frac{22}{7}) a^{2} = \frac{6}{7} a^{2} sq . units$

Question 39:

When four circles touch each other, their centres form the vertices of a square. The sides of the square are 2a units.

Area of the square = ${(2 a)}^{2} = 4 a^{2} sq . units$

Area occupied by the four sectors
$= 4 \times \frac{90}{360} \times π \times a^{2} = π a^{2} sq . units$

Area between the circles = Area of the square $-$ Area of the four sectors
$= (4 - \frac{22}{7}) a^{2} = \frac{6}{7} a^{2} sq . units$

Answer:

Join ABC. All sides are equal, so it is an equilateral triangle.
Now,
Area of the equilateral triangle = $\frac{\sqrt{3}}{4} \times {Side}^{2}$

   $= \frac{1.73}{4} \times 12 \times 12 = 62.28 {cm}^{2}$

$Area of the arc of the circle = \frac{60}{360} {πr}^{2} = \frac{1}{6} {πr}^{2} = \frac{1}{6} \times \frac{22}{7} \times 6 \times 6 = 18.86 {cm}^{2} Area of the three sectors = 3 \times 18.86 = 56.57 {cm}^{2}$


Area of the shaded portion = Area of the triangle $-$ Area of the three quadrants

$= 62.28 - 56.57 = 5.71 {cm}^{2}$

Question 40:

Join ABC. All sides are equal, so it is an equilateral triangle.
Now,
Area of the equilateral triangle = $\frac{\sqrt{3}}{4} \times {Side}^{2}$

   $= \frac{1.73}{4} \times 12 \times 12 = 62.28 {cm}^{2}$

$Area of the arc of the circle = \frac{60}{360} {πr}^{2} = \frac{1}{6} {πr}^{2} = \frac{1}{6} \times \frac{22}{7} \times 6 \times 6 = 18.86 {cm}^{2} Area of the three sectors = 3 \times 18.86 = 56.57 {cm}^{2}$


Area of the shaded portion = Area of the triangle $-$ Area of the three quadrants

$= 62.28 - 56.57 = 5.71 {cm}^{2}$

Answer:

When three circles touch each other, their centres form an equilateral triangle, with each side being 2a.

Area of the triangle = $\frac{\sqrt{3}}{4} \times 2 a \times 2 a = \sqrt{3} a^{2}$

Total area of the three sectors of circles = $3 \times \frac{60}{360} \times \frac{22}{7} \times a^{2} = \frac{1}{2} \times \frac{22}{7} \times a^{2} = \frac{11}{7} a^{2}$

Area of the region between the circles = $Area of the triangle - Area of the three sectors$
$= (\sqrt{3} - \frac{11}{7}) a^{2} = (1.73 - 1.57) a^{2} = 0.16 a^{2} = \frac{4}{25} a^{2}$

Question 41:

When three circles touch each other, their centres form an equilateral triangle, with each side being 2a.

Area of the triangle = $\frac{\sqrt{3}}{4} \times 2 a \times 2 a = \sqrt{3} a^{2}$

Total area of the three sectors of circles = $3 \times \frac{60}{360} \times \frac{22}{7} \times a^{2} = \frac{1}{2} \times \frac{22}{7} \times a^{2} = \frac{11}{7} a^{2}$

Area of the region between the circles = $Area of the triangle - Area of the three sectors$
$= (\sqrt{3} - \frac{11}{7}) a^{2} = (1.73 - 1.57) a^{2} = 0.16 a^{2} = \frac{4}{25} a^{2}$

Answer:

Area of trapezium = $\frac{1}{2} (AD + BC) \times AB$
$\Rightarrow 24.5 = \frac{1}{2} (10 + 4) \times AB \Rightarrow AB = 3.5 cm$
Area of shaded region = Area of trapezium ABCD − Area of quadrant ABE
$= 24.5 - \frac{1}{4} π {(AB)}^{2} = 24.5 - \frac{1}{4} \times \frac{22}{7} {(3.5)}^{2} = 24.5 - 9.625 = 14.875 {cm}^{2}$
Hence, the area of shaded region is 14.875 cm²

Question 42:

Area of trapezium = $\frac{1}{2} (AD + BC) \times AB$
$\Rightarrow 24.5 = \frac{1}{2} (10 + 4) \times AB \Rightarrow AB = 3.5 cm$
Area of shaded region = Area of trapezium ABCD − Area of quadrant ABE
$= 24.5 - \frac{1}{4} π {(AB)}^{2} = 24.5 - \frac{1}{4} \times \frac{22}{7} {(3.5)}^{2} = 24.5 - 9.625 = 14.875 {cm}^{2}$
Hence, the area of shaded region is 14.875 cm²

Answer:

(i) Area of fours sector = Area of sector having central angle 60° + Area of sector having central angle 90° + Area of sector having central angle 90° + Area of sector having central angle 120°
$= \frac{60 °}{360 °} π {(14)}^{2} + \frac{90 °}{360 °} π {(14)}^{2} + \frac{90 °}{360 °} π {(14)}^{2} + \frac{120 °}{360 °} π {(14)}^{2} = (\frac{60 °}{360 °} + \frac{90 °}{360 °} + \frac{90 °}{360 °} + \frac{120 °}{360 °}) π {(14)}^{2} = (\frac{360 °}{360 °}) π {(14)}^{2} = 616 m^{2}$

(ii) Area of the remaining portion = Area of trapezium ABCD − Area of four quadrants
$= \frac{1}{2} (AD + BC) \times AB - 616 = \frac{1}{2} (55 + 45) \times 30 - 616 = 1500 - 616 = 884 m^{2}$

Question 43:

(i) Area of fours sector = Area of sector having central angle 60° + Area of sector having central angle 90° + Area of sector having central angle 90° + Area of sector having central angle 120°
$= \frac{60 °}{360 °} π {(14)}^{2} + \frac{90 °}{360 °} π {(14)}^{2} + \frac{90 °}{360 °} π {(14)}^{2} + \frac{120 °}{360 °} π {(14)}^{2} = (\frac{60 °}{360 °} + \frac{90 °}{360 °} + \frac{90 °}{360 °} + \frac{120 °}{360 °}) π {(14)}^{2} = (\frac{360 °}{360 °}) π {(14)}^{2} = 616 m^{2}$

(ii) Area of the remaining portion = Area of trapezium ABCD − Area of four quadrants
$= \frac{1}{2} (AD + BC) \times AB - 616 = \frac{1}{2} (55 + 45) \times 30 - 616 = 1500 - 616 = 884 m^{2}$

Answer:

In equilateral traingle all the angles are of 60°
∴ ∠ABO = ∠AOB = 60°
Area of the shaded region = (Area of triangle AOB − Area of sector having central angle 60°) + Area of sector having central angle (360° − 60°)
$= \frac{\sqrt{3}}{4} {(AB)}^{2} - \frac{60 °}{360 °} π {(6)}^{2} + \frac{300 °}{360 °} π {(6)}^{2} = \frac{1.73}{4} {(12)}^{2} - \frac{1}{6} \times 3.14 {(6)}^{2} + \frac{5}{6} \times 3.14 {(6)}^{2} = 62.28 - 18.84 + 94.2 = 137.64 {cm}^{2}$
Hence, the area of shaded region is 137.64 cm²

Question 44:

In equilateral traingle all the angles are of 60°
∴ ∠ABO = ∠AOB = 60°
Area of the shaded region = (Area of triangle AOB − Area of sector having central angle 60°) + Area of sector having central angle (360° − 60°)
$= \frac{\sqrt{3}}{4} {(AB)}^{2} - \frac{60 °}{360 °} π {(6)}^{2} + \frac{300 °}{360 °} π {(6)}^{2} = \frac{1.73}{4} {(12)}^{2} - \frac{1}{6} \times 3.14 {(6)}^{2} + \frac{5}{6} \times 3.14 {(6)}^{2} = 62.28 - 18.84 + 94.2 = 137.64 {cm}^{2}$
Hence, the area of shaded region is 137.64 cm²

Answer:

We know that the opposite sides of a rectangle are equal
AD = BC = 70 cm
In right triangle AED
AE² = AD² − DE²
= (70)² − (42)²
= 4900 − 1764
= 3136
∴ AE² = 3136
⇒ AE = 56
= Area of the shaded region = Area of rectangle − (Area of triangle AED + Area of semicircle)
$= AB \times BC - [\frac{1}{2} \times AE \times DE + \frac{1}{2} π {(\frac{BC}{2})}^{2}] = 80 \times 70 - [\frac{1}{2} \times 56 \times 42 + \frac{1}{2} \times \frac{22}{7} {(\frac{70}{2})}^{2}] = 5600 - 3101 = 2499 {cm}^{2}$
Hence, the area of shaded region is 2499 cm²

Question 45:

We know that the opposite sides of a rectangle are equal
AD = BC = 70 cm
In right triangle AED
AE² = AD² − DE²
= (70)² − (42)²
= 4900 − 1764
= 3136
∴ AE² = 3136
⇒ AE = 56
= Area of the shaded region = Area of rectangle − (Area of triangle AED + Area of semicircle)
$= AB \times BC - [\frac{1}{2} \times AE \times DE + \frac{1}{2} π {(\frac{BC}{2})}^{2}] = 80 \times 70 - [\frac{1}{2} \times 56 \times 42 + \frac{1}{2} \times \frac{22}{7} {(\frac{70}{2})}^{2}] = 5600 - 3101 = 2499 {cm}^{2}$
Hence, the area of shaded region is 2499 cm²

Answer:

In right triangle AED
AD² = AE² + DE²
= (9)² + (12)²
= 81 + 144
= 225
∴ AD² = 225
⇒ AD = 15 cm
We know that the opposite sides of a rectangle are equal
AD = BC = 15 cm
= Area of the shaded region = Area of rectangle − Area of triangle AED + Area of semicircle
$= AB \times BC - \frac{1}{2} \times AE \times DE + \frac{1}{2} π {(\frac{BC}{2})}^{2} = 20 \times 15 - \frac{1}{2} \times 9 \times 12 + \frac{1}{2} \times 3.14 {(\frac{15}{2})}^{2} = 300 - 54 + 88.31 = 334.31 {cm}^{2}$
Hence, the area of shaded region is 334.31 cm²

Question 46:

In right triangle AED
AD² = AE² + DE²
= (9)² + (12)²
= 81 + 144
= 225
∴ AD² = 225
⇒ AD = 15 cm
We know that the opposite sides of a rectangle are equal
AD = BC = 15 cm
= Area of the shaded region = Area of rectangle − Area of triangle AED + Area of semicircle
$= AB \times BC - \frac{1}{2} \times AE \times DE + \frac{1}{2} π {(\frac{BC}{2})}^{2} = 20 \times 15 - \frac{1}{2} \times 9 \times 12 + \frac{1}{2} \times 3.14 {(\frac{15}{2})}^{2} = 300 - 54 + 88.31 = 334.31 {cm}^{2}$
Hence, the area of shaded region is 334.31 cm²

Answer:

In right triangle ABC
BC² = AB² + AC²
= (7)² + (24)²
= 49 + 576
= 625
∴ BC² = 625
⇒ BC = 25
Now, ∠COD + ∠BOD = 180° (Linear pair angles)
⇒∠COD = 180° − 90° = 90°
Now, Area of the shaded region = Area of sector having central angle (360° − 90°) − Area of triangle ABC
$= \frac{270 °}{360 °} π {(\frac{BC}{2})}^{2} - \frac{1}{2} AB \times AC = \frac{3}{4} \times 3.14 {(\frac{25}{2})}^{2} - \frac{1}{2} \times 7 \times 24 = 367.97 - 84 = 283.97 {cm}^{2}$
Hence, the area of shaded region is 283.97 cm²

Question 47:

In right triangle ABC
BC² = AB² + AC²
= (7)² + (24)²
= 49 + 576
= 625
∴ BC² = 625
⇒ BC = 25
Now, ∠COD + ∠BOD = 180° (Linear pair angles)
⇒∠COD = 180° − 90° = 90°
Now, Area of the shaded region = Area of sector having central angle (360° − 90°) − Area of triangle ABC
$= \frac{270 °}{360 °} π {(\frac{BC}{2})}^{2} - \frac{1}{2} AB \times AC = \frac{3}{4} \times 3.14 {(\frac{25}{2})}^{2} - \frac{1}{2} \times 7 \times 24 = 367.97 - 84 = 283.97 {cm}^{2}$
Hence, the area of shaded region is 283.97 cm²

Answer:

We can find the radius of the incircle by using the formula
$r = \frac{2 \times Area of triangle}{Perimeter of triangle} = \frac{2 \times \frac{\sqrt{3}}{4} \times {(12)}^{2}}{3 \times 12} = 2 \sqrt{3} cm$
Now, area of shaded region = Area of triangle − Area of circle
$= \frac{\sqrt{3}}{4} \times {(12)}^{2} - 3.14 \times {(2 \sqrt{3})}^{2} = 62.28 - 37.68 = 24.6 {cm}^{2}$
Hence, the area of shaded region is 24.6 cm²

Question 48:

We can find the radius of the incircle by using the formula
$r = \frac{2 \times Area of triangle}{Perimeter of triangle} = \frac{2 \times \frac{\sqrt{3}}{4} \times {(12)}^{2}}{3 \times 12} = 2 \sqrt{3} cm$
Now, area of shaded region = Area of triangle − Area of circle
$= \frac{\sqrt{3}}{4} \times {(12)}^{2} - 3.14 \times {(2 \sqrt{3})}^{2} = 62.28 - 37.68 = 24.6 {cm}^{2}$
Hence, the area of shaded region is 24.6 cm²

Answer:

Construction: Join AO and extend it to D on BC.

Radius of the circle, r = 42 cm
∠OCD= 30°

$\cos 30 ° = \frac{DC}{OC} \Rightarrow \frac{\sqrt{3}}{2} = \frac{DC}{42} \Rightarrow DC = 21 \sqrt{3} \Rightarrow BC = 2 \times DC = 42 \sqrt{3} = 72.66 cm \sin 30 ° = \frac{OD}{OC} \Rightarrow \frac{1}{2} = \frac{OD}{42} \Rightarrow OD = 21 cm Now, AD = AO + OD = 42 + 21 = 63 cm$

Area of shaded region = Area of circle − Area of triangle ABC
$= π {(OA)}^{2} - \frac{1}{2} \times AD \times AB = \frac{22}{7} {(42)}^{2} - \frac{1}{2} \times 63 \times 72.66 = 5544 - 2288.79 = 3255.21 {cm}^{2}$

Question 49:

Construction: Join AO and extend it to D on BC.

Radius of the circle, r = 42 cm
∠OCD= 30°

$\cos 30 ° = \frac{DC}{OC} \Rightarrow \frac{\sqrt{3}}{2} = \frac{DC}{42} \Rightarrow DC = 21 \sqrt{3} \Rightarrow BC = 2 \times DC = 42 \sqrt{3} = 72.66 cm \sin 30 ° = \frac{OD}{OC} \Rightarrow \frac{1}{2} = \frac{OD}{42} \Rightarrow OD = 21 cm Now, AD = AO + OD = 42 + 21 = 63 cm$

Area of shaded region = Area of circle − Area of triangle ABC
$= π {(OA)}^{2} - \frac{1}{2} \times AD \times AB = \frac{22}{7} {(42)}^{2} - \frac{1}{2} \times 63 \times 72.66 = 5544 - 2288.79 = 3255.21 {cm}^{2}$

Answer:

Let the radius of the circle be r
Now, Perimeter of quadrant = $\frac{1}{4} (2 π r) + 2 r$
$\Rightarrow 25 = \frac{1}{2} \times \frac{22}{7} \times r + 2 r \Rightarrow 25 = \frac{25 r}{7} \Rightarrow r = 7 cm$
Area of quadrant = $\frac{1}{4} π r^{2} = \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 = 38.5 {cm}^{2}$
Hence, the area of quadrant is 38.5 cm²

Question 50:

Let the radius of the circle be r
Now, Perimeter of quadrant = $\frac{1}{4} (2 π r) + 2 r$
$\Rightarrow 25 = \frac{1}{2} \times \frac{22}{7} \times r + 2 r \Rightarrow 25 = \frac{25 r}{7} \Rightarrow r = 7 cm$
Area of quadrant = $\frac{1}{4} π r^{2} = \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 = 38.5 {cm}^{2}$
Hence, the area of quadrant is 38.5 cm²

Answer:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB
$= \frac{θ}{360 °} π {(OA)}^{2} - \frac{1}{2} \times OA \times OB = \frac{90 °}{360 °} \times 3.14 {(10)}^{2} - \frac{1}{2} \times 10 \times 10 = 78.5 - 50 = 28.5 {cm}^{2}$
Hence, the area of minor segment is 28.5 cm²

Question 51:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB
$= \frac{θ}{360 °} π {(OA)}^{2} - \frac{1}{2} \times OA \times OB = \frac{90 °}{360 °} \times 3.14 {(10)}^{2} - \frac{1}{2} \times 10 \times 10 = 78.5 - 50 = 28.5 {cm}^{2}$
Hence, the area of minor segment is 28.5 cm²

Answer:

Area of the road = Area of outer circle − Area of inner circle
$= π R^{2} - π r^{2} = π [{(110)}^{2} - {(100)}^{2}] = 3.14 \times 2100 = 6594 m^{2}$
Cost of levelling the road = Area of the road â¨¯ Rate
= 6594 â¨¯ 20
= Rs 131880

Question 52:

Area of the road = Area of outer circle − Area of inner circle
$= π R^{2} - π r^{2} = π [{(110)}^{2} - {(100)}^{2}] = 3.14 \times 2100 = 6594 m^{2}$
Cost of levelling the road = Area of the road â¨¯ Rate
= 6594 â¨¯ 20
= Rs 131880

Answer:

$Area of equilateral triangle = 49 \sqrt{3} \Rightarrow \frac{\sqrt{3}}{4} \times {(Side)}^{2} = 49 \sqrt{3} \Rightarrow {(Side)}^{2} = 7^{2} \times 2^{2} \Rightarrow Side = 14 cm$

Radius of the circle = Half of the side of the triangle = 7 cm
Area of triangle not included in the circle = Area of triangle − Area of 3 sectors having central angle 60^âˆ˜
$= 49 \sqrt{3} - 3 \times \frac{60 °}{360 °} \times \frac{22}{7} \times 7 \times 7 = 84.77 - 77 = 7.77 {cm}^{2}$
Hence, the required area is 7.77 cm²

Question 53:

$Area of equilateral triangle = 49 \sqrt{3} \Rightarrow \frac{\sqrt{3}}{4} \times {(Side)}^{2} = 49 \sqrt{3} \Rightarrow {(Side)}^{2} = 7^{2} \times 2^{2} \Rightarrow Side = 14 cm$

Radius of the circle = Half of the side of the triangle = 7 cm
Area of triangle not included in the circle = Area of triangle − Area of 3 sectors having central angle 60^âˆ˜
$= 49 \sqrt{3} - 3 \times \frac{60 °}{360 °} \times \frac{22}{7} \times 7 \times 7 = 84.77 - 77 = 7.77 {cm}^{2}$
Hence, the required area is 7.77 cm²

Answer:

CD = 8 cm
BP = HQ = 4 cm
DE = EF = 5 cm
Area of the parallelogram ABCD = $B \times H$
           $= 8 \times 4 = 32 sq . cm$

Area of parallelogram FGHI = $B \times H$
   $= 8 \times 4 = 32 sq . cm$
Area of the square = ${Side}^{2}$
   = $8^{2} = 64 sq . cm$

In $∆$ ELF, we have:
${EL}^{2} = 5^{2} - 4^{2} {EL}^{2} = 9 EL = 3 cm$

Area of $△$ DEF = $\frac{1}{2} \times B \times H$
$= \frac{1}{2} \times 8 \times 3 = 12 sq . cm$

Area of the semicircle = $\frac{1}{2} {πr}^{2}$

          $= \frac{1}{2} \times \frac{22}{7} \times 16 = 25.12 sq . cm$
∴ Total Area = Area of the parallelogram ABCD + Area of the parallelogram FGHI + Area of the triangle DEF + Area of the semicircle CKI + Area of the square
Total Area = 165.12 cm²

Question 54:

CD = 8 cm
BP = HQ = 4 cm
DE = EF = 5 cm
Area of the parallelogram ABCD = $B \times H$
           $= 8 \times 4 = 32 sq . cm$

Area of parallelogram FGHI = $B \times H$
   $= 8 \times 4 = 32 sq . cm$
Area of the square = ${Side}^{2}$
   = $8^{2} = 64 sq . cm$

In $∆$ ELF, we have:
${EL}^{2} = 5^{2} - 4^{2} {EL}^{2} = 9 EL = 3 cm$

Area of $△$ DEF = $\frac{1}{2} \times B \times H$
$= \frac{1}{2} \times 8 \times 3 = 12 sq . cm$

Area of the semicircle = $\frac{1}{2} {πr}^{2}$

          $= \frac{1}{2} \times \frac{22}{7} \times 16 = 25.12 sq . cm$
∴ Total Area = Area of the parallelogram ABCD + Area of the parallelogram FGHI + Area of the triangle DEF + Area of the semicircle CKI + Area of the square
Total Area = 165.12 cm²

Answer:

Area of sector having central angle 150° = $= \frac{150 °}{360 °} π {(6)}^{2} = \frac{5}{12} \times Area of circular disc$

Now, Area of sector having central angle 90° : Area of sector having central angle 120° : Area of sector having central angle 150°
$= \frac{90 °}{360 °} π {(6)}^{2} : \frac{120 °}{360 °} π {(6)}^{2} : \frac{150 °}{360 °} π {(6)}^{2} = \frac{1}{4} : \frac{1}{3} : \frac{5}{12} = 3 : 4 : 5$

Question 55:

Area of sector having central angle 150° = $= \frac{150 °}{360 °} π {(6)}^{2} = \frac{5}{12} \times Area of circular disc$

Now, Area of sector having central angle 90° : Area of sector having central angle 120° : Area of sector having central angle 150°
$= \frac{90 °}{360 °} π {(6)}^{2} : \frac{120 °}{360 °} π {(6)}^{2} : \frac{150 °}{360 °} π {(6)}^{2} = \frac{1}{4} : \frac{1}{3} : \frac{5}{12} = 3 : 4 : 5$

Answer:

Join each vertex of the hexagon to the centre of the circle.

The hexagon is made up of six triangles.

$Total area of the design = Area of the circle - Area of six triangles = [\frac{22}{7} \times 35 \times 35] - [6 \times (\frac{\sqrt{3}}{4} \times 35 \times 35)] = (35 \times 35) [\frac{22}{7} - (\frac{6 \times 1.732}{4})] = 1225 (\frac{22}{7} - 2.598) = 1225 \times 0.542 = 663.95 {cm}^{2}$

Question 56:

Join each vertex of the hexagon to the centre of the circle.

The hexagon is made up of six triangles.

$Total area of the design = Area of the circle - Area of six triangles = [\frac{22}{7} \times 35 \times 35] - [6 \times (\frac{\sqrt{3}}{4} \times 35 \times 35)] = (35 \times 35) [\frac{22}{7} - (\frac{6 \times 1.732}{4})] = 1225 (\frac{22}{7} - 2.598) = 1225 \times 0.542 = 663.95 {cm}^{2}$

Answer:

In the right $∆$ RPQ, we have:
$R Q = \sqrt{R P^{2} + P Q^{2}} = \sqrt{7^{2} + 24^{2}} = \sqrt{49 + 576} = 25 cm$
â€‹
OR = OQ = 12.5 cm
Now,
Area of the circle = ${πr}^{2}$
$= 3.14 \times 12.5 \times 12.5 = 490.625 sq . cm$

Area of the semicircle = $\frac{490.625}{2} = 245.31 sq . cm$
Area of the triangle $= \frac{1}{2} \times b \times h = \frac{1}{2} \times 7 \times 24 = 84 sq . cm$
Thus, we have:

$Area of the shaded part = Area of the semicircle - Area of the triangle = 245.31 - 84 = 161.31 {cm}^{2}$

Question 57:

In the right $∆$ RPQ, we have:
$R Q = \sqrt{R P^{2} + P Q^{2}} = \sqrt{7^{2} + 24^{2}} = \sqrt{49 + 576} = 25 cm$
â€‹
OR = OQ = 12.5 cm
Now,
Area of the circle = ${πr}^{2}$
$= 3.14 \times 12.5 \times 12.5 = 490.625 sq . cm$

Area of the semicircle = $\frac{490.625}{2} = 245.31 sq . cm$
Area of the triangle $= \frac{1}{2} \times b \times h = \frac{1}{2} \times 7 \times 24 = 84 sq . cm$
Thus, we have:

$Area of the shaded part = Area of the semicircle - Area of the triangle = 245.31 - 84 = 161.31 {cm}^{2}$

Answer:

Using Pythagoras' theorem for triangle ABC, we have:

$C A^{2} + A B^{2} = B C^{2}$

$C A = \sqrt{B C^{2} - A B^{2}} = \sqrt{100 - 36} = \sqrt{64} = 8 cm$

Now, we must find the radius of the incircle. Draw OE, OD and OF perpendicular to AC, AB and BC, respectively.

Consider quadrilateral AEOD.
Here,
$E O = O D (Both are radii .)$

Because the circle is an incircle, AE and AD are tangents to the circle.
$∠ A E O = ∠ A D O = 90 °$

Also,
$∠ A = 90 °$
Therefore, AEOD is a square.
Thus, we can say that $A E = E O = O D = A D = r$ .

$C E = C F = 8 - r B F = B D = 6 - r C F + B F = 10 \Rightarrow (8 - r) + (6 - r) = 10 \Rightarrow 14 - 2 r = 10 \Rightarrow r = 2 cm$

Area of the shaded part = Area of the triangle $-$ Area of the circle
$= \{\frac{1}{2} \times 6 \times 8\} - \{π \times 2 \times 2\} = 24 - 12.56 = 11.44 {cm}^{2}$

Question 58:

Using Pythagoras' theorem for triangle ABC, we have:

$C A^{2} + A B^{2} = B C^{2}$

$C A = \sqrt{B C^{2} - A B^{2}} = \sqrt{100 - 36} = \sqrt{64} = 8 cm$

Now, we must find the radius of the incircle. Draw OE, OD and OF perpendicular to AC, AB and BC, respectively.

Consider quadrilateral AEOD.
Here,
$E O = O D (Both are radii .)$

Because the circle is an incircle, AE and AD are tangents to the circle.
$∠ A E O = ∠ A D O = 90 °$

Also,
$∠ A = 90 °$
Therefore, AEOD is a square.
Thus, we can say that $A E = E O = O D = A D = r$ .

$C E = C F = 8 - r B F = B D = 6 - r C F + B F = 10 \Rightarrow (8 - r) + (6 - r) = 10 \Rightarrow 14 - 2 r = 10 \Rightarrow r = 2 cm$

Area of the shaded part = Area of the triangle $-$ Area of the circle
$= \{\frac{1}{2} \times 6 \times 8\} - \{π \times 2 \times 2\} = 24 - 12.56 = 11.44 {cm}^{2}$

Answer:

Perimeter (circumference of the circle) = $2 πr$
We know:
Perimeter of a semicircular arc = $πr$
Now,
For the arc PTS, radius is 6 cm.
∴ Circumference of the semicircle PTS = $πr = 6 π cm$

For the arc QES, radius is 4 cm.
â€‹∴ Circumference of the semicircle QES = $πr = 4 π cm$

For the arc PBQ, radius is 2 cm.
∴ Circumference of the semicircle PBQ = $πr = 2 π cm$

Now,
Perimeter of the shaded region = $6 π + 4 π + 2 π$
   $= 12 πcm$
   $= 12 \times 3.14 = 37.68 cm$

Area of the semicircle PBQ = $\frac{1}{2} {πr}^{2}$

          $= \frac{1}{2} \times 3.14 \times 2 \times 2 = 6.28 {cm}^{2}$

Area of the semicircle PTS = $\frac{1}{2} {πr}^{2}$

$= \frac{1}{2} \times 3.14 \times 6 \times 6 = 56.52 {cm}^{2}$

Area of the semicircle QES = $\frac{1}{2} {πr}^{2}$

      $= \frac{1}{2} \times 3.14 \times 4 \times 4 = 25.12 {cm}^{2}$

Area of the shaded region = Area of the semicircle PBQ + Area of the semicircle PTS $-$ Area of the semicircle QES
   $= 6.28 + 56.52 - 25.12 = 37.68 {cm}^{2}$

Question 59:

Perimeter (circumference of the circle) = $2 πr$
We know:
Perimeter of a semicircular arc = $πr$
Now,
For the arc PTS, radius is 6 cm.
∴ Circumference of the semicircle PTS = $πr = 6 π cm$

For the arc QES, radius is 4 cm.
â€‹∴ Circumference of the semicircle QES = $πr = 4 π cm$

For the arc PBQ, radius is 2 cm.
∴ Circumference of the semicircle PBQ = $πr = 2 π cm$

Now,
Perimeter of the shaded region = $6 π + 4 π + 2 π$
   $= 12 πcm$
   $= 12 \times 3.14 = 37.68 cm$

Area of the semicircle PBQ = $\frac{1}{2} {πr}^{2}$

          $= \frac{1}{2} \times 3.14 \times 2 \times 2 = 6.28 {cm}^{2}$

Area of the semicircle PTS = $\frac{1}{2} {πr}^{2}$

$= \frac{1}{2} \times 3.14 \times 6 \times 6 = 56.52 {cm}^{2}$

Area of the semicircle QES = $\frac{1}{2} {πr}^{2}$

      $= \frac{1}{2} \times 3.14 \times 4 \times 4 = 25.12 {cm}^{2}$

Area of the shaded region = Area of the semicircle PBQ + Area of the semicircle PTS $-$ Area of the semicircle QES
   $= 6.28 + 56.52 - 25.12 = 37.68 {cm}^{2}$

Answer:

Length of the inner curved portion $(400 - 2 \times 90) = 220 m$
∴ Length of each inner curved path = $\frac{220}{2}$ = 110 m
â€‹Thus, we have:
$\Rightarrow πr = 110 \Rightarrow \frac{22}{7} r = 110 \Rightarrow r = \frac{110 \times 7}{22} \Rightarrow r = 35 m$

Inner radius = 35 m
Outer radius = (35 + 14) = 49 m
Area of track = {Area of the two rectangles [each $(90 \times 14) m$ ] + Area of the circular ring with R = 49 m and r = 35 m)}
$= (2 \times 90 \times 14) + \frac{22}{7} \times [{(49)}^{2} - {(35)}^{2}] = 2520 + \frac{22}{7} \times (2401 - 1225) = 2520 + \frac{22}{7} \times 1176 = 2520 + 3696 = 6216 m^{2}$

â€‹Length of the outer boundary of the track
$= (2 \times 90 + 2 \times \frac{22}{7} \times 49) = 488 m$

Therefore, the length of the outer boundary of the track is 488 m and the area of the track is 6216 sq. m.

Question 60:

Length of the inner curved portion $(400 - 2 \times 90) = 220 m$
∴ Length of each inner curved path = $\frac{220}{2}$ = 110 m
â€‹Thus, we have:
$\Rightarrow πr = 110 \Rightarrow \frac{22}{7} r = 110 \Rightarrow r = \frac{110 \times 7}{22} \Rightarrow r = 35 m$

Inner radius = 35 m
Outer radius = (35 + 14) = 49 m
Area of track = {Area of the two rectangles [each $(90 \times 14) m$ ] + Area of the circular ring with R = 49 m and r = 35 m)}
$= (2 \times 90 \times 14) + \frac{22}{7} \times [{(49)}^{2} - {(35)}^{2}] = 2520 + \frac{22}{7} \times (2401 - 1225) = 2520 + \frac{22}{7} \times 1176 = 2520 + 3696 = 6216 m^{2}$

â€‹Length of the outer boundary of the track
$= (2 \times 90 + 2 \times \frac{22}{7} \times 49) = 488 m$

Therefore, the length of the outer boundary of the track is 488 m and the area of the track is 6216 sq. m.

Answer:

Area of the shaded region

= Area of the rectangle − Area of the semicircle

$= 21 \times 14 - \{\frac{1}{2} \times π \times {(\frac{14}{2})}^{2}\} = 294 - \{\frac{1}{2} \times \frac{22}{7} \times 7 \times 7\} = 294 - 77 = 217 {cm}^{2}$

Therefore, area of shaded region is 217 cm².

Length of the boundary (or perimeter) of the shaded region

$= AB + AD + DC + arc BC = AB + AD + DC + \frac{1}{2} \times 2 π (\frac{14}{2}) = 21 + 14 + 21 + \frac{22}{7} \times 7 = 56 + 22 = 78 cm$

Therefore, the perimeter of the shaded region is 78 cm.

Question 61:

Area of the shaded region

= Area of the rectangle − Area of the semicircle

$= 21 \times 14 - \{\frac{1}{2} \times π \times {(\frac{14}{2})}^{2}\} = 294 - \{\frac{1}{2} \times \frac{22}{7} \times 7 \times 7\} = 294 - 77 = 217 {cm}^{2}$

Therefore, area of shaded region is 217 cm².

Length of the boundary (or perimeter) of the shaded region

$= AB + AD + DC + arc BC = AB + AD + DC + \frac{1}{2} \times 2 π (\frac{14}{2}) = 21 + 14 + 21 + \frac{22}{7} \times 7 = 56 + 22 = 78 cm$

Therefore, the perimeter of the shaded region is 78 cm.

Answer:

Given: Radius of the inner circle with radius OC, r = 21 cm
Radius of the inner circle with radius OA, R = 42 cm
∠AOB = 60°

Area of the circular ring

$= π R^{2} - π r^{2} = π [R^{2} - r^{2}] = π [42^{2} - 21^{2}] {cm}^{2}$

Area of ACDB = area of sector AOB − area of COD
$= \frac{60}{360} \times π \times R^{2} - \frac{60}{360} \times π \times r^{2} = \frac{60}{360} \times π [R^{2} - r^{2}] = \frac{60}{360} \times π [42^{2} - 21^{2}]$

Area of shaded region = area of circular ring − area of ACDB
$= π [42^{2} - 21^{2}] - \frac{60}{360} π [42^{2} - 21^{2}] = π [42^{2} - 21^{2}] [1 - \frac{60}{360}] = \frac{22}{7} (42 - 21) (42 + 21) \times \frac{300}{360} = 3465 {cm}^{2}$

Question 62:

Given: Radius of the inner circle with radius OC, r = 21 cm
Radius of the inner circle with radius OA, R = 42 cm
∠AOB = 60°

Area of the circular ring

$= π R^{2} - π r^{2} = π [R^{2} - r^{2}] = π [42^{2} - 21^{2}] {cm}^{2}$

Area of ACDB = area of sector AOB − area of COD
$= \frac{60}{360} \times π \times R^{2} - \frac{60}{360} \times π \times r^{2} = \frac{60}{360} \times π [R^{2} - r^{2}] = \frac{60}{360} \times π [42^{2} - 21^{2}]$

Area of shaded region = area of circular ring − area of ACDB
$= π [42^{2} - 21^{2}] - \frac{60}{360} π [42^{2} - 21^{2}] = π [42^{2} - 21^{2}] [1 - \frac{60}{360}] = \frac{22}{7} (42 - 21) (42 + 21) \times \frac{300}{360} = 3465 {cm}^{2}$

Answer:

Area of the shaded region

= Area of the semi-circle with diameter of 9 cm − Areas of two semi-circles with diameter 3 cm − Area of the circle with diameter 4.5 cm + Area of semi-circle with diameter 3 cm

= Area of the semi-circle with radius of 4.5 cm − 2 × Area of semi-circle with radius 1.5 cm − Area of the circle with radius 2.25 cm + Area of semi-circle with radius 1.5 cm

= $3.9375 \times \frac{22}{7}$

= 12.375 cm²

Thus, the area of the shaded region is 12.375 cm².

Question 63:

Area of the shaded region

= Area of the semi-circle with diameter of 9 cm − Areas of two semi-circles with diameter 3 cm − Area of the circle with diameter 4.5 cm + Area of semi-circle with diameter 3 cm

= Area of the semi-circle with radius of 4.5 cm − 2 × Area of semi-circle with radius 1.5 cm − Area of the circle with radius 2.25 cm + Area of semi-circle with radius 1.5 cm

= $3.9375 \times \frac{22}{7}$

= 12.375 cm²

Thus, the area of the shaded region is 12.375 cm².

Answer:

We have,
Side of square = 28 cm and radius of each circle = $\frac{28}{2}$ cm

Area of the shaded region

= Area of the square + Area of the two circles − Area of the two quadrants

$= {(28)}^{2} + 2 \times π \times {(\frac{28}{2})}^{2} - 2 \times \frac{1}{4} \times π \times {(\frac{28}{2})}^{2} = {(28)}^{2} + \frac{3}{2} \times π \times {(\frac{28}{2})}^{2} = {(28)}^{2} (1 + \frac{3}{2} \times \frac{22}{7} \times \frac{1}{2} \times \frac{1}{2}) = {(28)}^{2} (1 + \frac{33}{28}) = {(28)}^{2} \times \frac{61}{28} = 28 \times 61 = 1708 {cm}^{2}$

Therefore, the area of the shaded region is 1708 cm².

Question 64:

We have,
Side of square = 28 cm and radius of each circle = $\frac{28}{2}$ cm

Area of the shaded region

= Area of the square + Area of the two circles − Area of the two quadrants

$= {(28)}^{2} + 2 \times π \times {(\frac{28}{2})}^{2} - 2 \times \frac{1}{4} \times π \times {(\frac{28}{2})}^{2} = {(28)}^{2} + \frac{3}{2} \times π \times {(\frac{28}{2})}^{2} = {(28)}^{2} (1 + \frac{3}{2} \times \frac{22}{7} \times \frac{1}{2} \times \frac{1}{2}) = {(28)}^{2} (1 + \frac{33}{28}) = {(28)}^{2} \times \frac{61}{28} = 28 \times 61 = 1708 {cm}^{2}$

Therefore, the area of the shaded region is 1708 cm².

Answer:

Given that ABC is a triangle with sides AB = 14 cm, BC = 48 cm, CA = 50 cm.
Clearly it is a right angled triangle.
Area of $△ ABC = \frac{1}{2} \times base \times height = \frac{1}{2} \times 48 \times 14 = 336 {cm}^{2} .$

Now we need to remove the area of 3 arcs each of radius 5 cm from this area of triangle.
Here, $∠ B = 90 °, ∠ A + ∠ C = 90 ° .$
The area of these 3 arcs would exactly be equal to a sector of circle with radius 5 cm and angle = $∠ A + ∠ B + ∠ C = 180 °$ i.e. a semi circle.
Since we know that area of semi circle is $\frac{1}{2} π r^{2} = \frac{1}{2} \times 3.14 \times 5 \times 5 = \frac{1}{2} \times \frac{314}{100} \times 25 = \frac{314}{8} = \frac{157}{4} {cm}^{2} .$
Hence, the area of shaded region is = $area of △ ABC - combined area of 3 arcs = 336 - \frac{157}{4} = 336 - 39.25 = 296.75 {cm}^{2} .$

Question 65:

Given that ABC is a triangle with sides AB = 14 cm, BC = 48 cm, CA = 50 cm.
Clearly it is a right angled triangle.
Area of $△ ABC = \frac{1}{2} \times base \times height = \frac{1}{2} \times 48 \times 14 = 336 {cm}^{2} .$

Now we need to remove the area of 3 arcs each of radius 5 cm from this area of triangle.
Here, $∠ B = 90 °, ∠ A + ∠ C = 90 ° .$
The area of these 3 arcs would exactly be equal to a sector of circle with radius 5 cm and angle = $∠ A + ∠ B + ∠ C = 180 °$ i.e. a semi circle.
Since we know that area of semi circle is $\frac{1}{2} π r^{2} = \frac{1}{2} \times 3.14 \times 5 \times 5 = \frac{1}{2} \times \frac{314}{100} \times 25 = \frac{314}{8} = \frac{157}{4} {cm}^{2} .$
Hence, the area of shaded region is = $area of △ ABC - combined area of 3 arcs = 336 - \frac{157}{4} = 336 - 39.25 = 296.75 {cm}^{2} .$

Answer:

The diameters of concentric circles are in ratio 1 : 2 : 3.
let the diameters be d₁ = 1x, d₂= 2x, d₃= 3x.
So, the radii are $r_{1} = \frac{x}{2}, r_{2} = x, r_{3} = \frac{3 x}{2} .$
The areas of three regions are given be
$A_{1} = π {r_{1}}^{2} = π {(\frac{x}{2})}^{2} = \frac{π x^{2}}{4}, A_{2} = π {r_{2}}^{2} - π {r_{1}}^{2} = π x^{2} - \frac{π x^{2}}{4} = \frac{3 π x^{2}}{4}, A_{3} = π {r_{3}}^{2} - π {r_{2}}^{2} = π {(\frac{3 x}{2})}^{2} - π x^{2} = \frac{5 π x^{2}}{4}$
So, their ratio is $A_{1} : A_{2} : A_{3} = \frac{π x^{2}}{4} : \frac{3 π x^{2}}{4} : \frac{5 π x^{2}}{4} = 1 : 3 : 5 .$

Question 1:

The diameters of concentric circles are in ratio 1 : 2 : 3.
let the diameters be d₁ = 1x, d₂= 2x, d₃= 3x.
So, the radii are $r_{1} = \frac{x}{2}, r_{2} = x, r_{3} = \frac{3 x}{2} .$
The areas of three regions are given be
$A_{1} = π {r_{1}}^{2} = π {(\frac{x}{2})}^{2} = \frac{π x^{2}}{4}, A_{2} = π {r_{2}}^{2} - π {r_{1}}^{2} = π x^{2} - \frac{π x^{2}}{4} = \frac{3 π x^{2}}{4}, A_{3} = π {r_{3}}^{2} - π {r_{2}}^{2} = π {(\frac{3 x}{2})}^{2} - π x^{2} = \frac{5 π x^{2}}{4}$
So, their ratio is $A_{1} : A_{2} : A_{3} = \frac{π x^{2}}{4} : \frac{3 π x^{2}}{4} : \frac{5 π x^{2}}{4} = 1 : 3 : 5 .$

Answer:

Let the radius of the circle be r and circumference C.
â€‹
Now,
$C - r = 37 \Rightarrow 2 π r - r = 37 \Rightarrow r (2 \times \frac{22}{7} - 1) = 37 \Rightarrow r = \frac{37 \times 7}{37} = 7 cm$
Now, $C = 2 π r = 2 \times \frac{22}{7} \times 7 = 44 cm$
Hence, the circumference of the circle is 44 cm.

Question 2:

Let the radius of the circle be r and circumference C.
â€‹
Now,
$C - r = 37 \Rightarrow 2 π r - r = 37 \Rightarrow r (2 \times \frac{22}{7} - 1) = 37 \Rightarrow r = \frac{37 \times 7}{37} = 7 cm$
Now, $C = 2 π r = 2 \times \frac{22}{7} \times 7 = 44 cm$
Hence, the circumference of the circle is 44 cm.

Answer:

Let the radius of the circle be r.
â€‹Now,
$Circumference = 22 \Rightarrow 2 π r = 22 \Rightarrow r = \frac{22 \times 7}{44} = \frac{7}{2} cm$
Now, Area of quadrant = $\frac{1}{4} π r^{2} = \frac{1}{4} \times \frac{22}{7} \times {(\frac{7}{2})}^{2} = \frac{77}{8} {cm}^{2}$
Hence, the area of the quadrant of the circle is $\frac{77}{8}$ cm².

Question 3:

Let the radius of the circle be r.
â€‹Now,
$Circumference = 22 \Rightarrow 2 π r = 22 \Rightarrow r = \frac{22 \times 7}{44} = \frac{7}{2} cm$
Now, Area of quadrant = $\frac{1}{4} π r^{2} = \frac{1}{4} \times \frac{22}{7} \times {(\frac{7}{2})}^{2} = \frac{77}{8} {cm}^{2}$
Hence, the area of the quadrant of the circle is $\frac{77}{8}$ cm².

Answer:

Let the diameter of the required circle be d.
Now, Area of required circle = Area of circle having diameter 10 cm + Area of circle having diameter 24 cm

$\Rightarrow π {(\frac{d}{2})}^{2} = π {(\frac{10}{2})}^{2} + π {(\frac{24}{2})}^{2} \Rightarrow {(\frac{d}{2})}^{2} = 25 + 144 \Rightarrow {(\frac{d}{2})}^{2} = 13^{2} \Rightarrow \frac{d}{2} = 13 \Rightarrow d = 26 cm$
Hence, the diameter of the of the circle is 26 cm.

Question 4:

Let the diameter of the required circle be d.
Now, Area of required circle = Area of circle having diameter 10 cm + Area of circle having diameter 24 cm

$\Rightarrow π {(\frac{d}{2})}^{2} = π {(\frac{10}{2})}^{2} + π {(\frac{24}{2})}^{2} \Rightarrow {(\frac{d}{2})}^{2} = 25 + 144 \Rightarrow {(\frac{d}{2})}^{2} = 13^{2} \Rightarrow \frac{d}{2} = 13 \Rightarrow d = 26 cm$
Hence, the diameter of the of the circle is 26 cm.

Answer:

Let the diameter of the required circle be d.
Now, Area of circle = 2 â¨¯ Circumference of the circle
$\Rightarrow π {(\frac{d}{2})}^{2} = 2 (2 π \times \frac{d}{2}) \Rightarrow {(\frac{d}{2})}^{2} = 2 d \Rightarrow d^{2} = 8 d \Rightarrow d^{2} - 8 d = 0 \Rightarrow d (d - 8) = 0 \Rightarrow d = 8 cm [∵ d \neq 0]$
Hence, the diameter of the of the circle is 8 cm.

Question 5:

Let the diameter of the required circle be d.
Now, Area of circle = 2 â¨¯ Circumference of the circle
$\Rightarrow π {(\frac{d}{2})}^{2} = 2 (2 π \times \frac{d}{2}) \Rightarrow {(\frac{d}{2})}^{2} = 2 d \Rightarrow d^{2} = 8 d \Rightarrow d^{2} - 8 d = 0 \Rightarrow d (d - 8) = 0 \Rightarrow d = 8 cm [∵ d \neq 0]$
Hence, the diameter of the of the circle is 8 cm.

Answer:

We know that if a square circumscribes a circle, then the side of the square is equal to the diameter of the circle.
∴ Side of Square = 2a
Now, Perimeter of the square = 4 â¨¯ Side of square = 4 â¨¯ 2a = 8a cm
Hence, the perimeter of the square is 8a cm.

Question 6:

We know that if a square circumscribes a circle, then the side of the square is equal to the diameter of the circle.
∴ Side of Square = 2a
Now, Perimeter of the square = 4 â¨¯ Side of square = 4 â¨¯ 2a = 8a cm
Hence, the perimeter of the square is 8a cm.

Answer:

We have $r = \frac{diameter}{2} = \frac{42}{2} = 21 cm and θ = 60 °$
Length of arc = $\frac{θ}{360^{°}} \times 2 π r = \frac{60 °}{360 °} \times 2 \times \frac{22}{7} \times 21 = 22 cm$
Hence, the length of the arc of the circle is 22 cm.

Question 7:

We have $r = \frac{diameter}{2} = \frac{42}{2} = 21 cm and θ = 60 °$
Length of arc = $\frac{θ}{360^{°}} \times 2 π r = \frac{60 °}{360 °} \times 2 \times \frac{22}{7} \times 21 = 22 cm$
Hence, the length of the arc of the circle is 22 cm.

Answer:

Let the diameter of the required circle be d.
Now, Area of the required circle = Area of circle having radius 4 cm + Area of circle having radius 3 cm

$\Rightarrow π {(\frac{d}{2})}^{2} = π {(4)}^{2} + π {(3)}^{2} \Rightarrow {(\frac{d}{2})}^{2} = 16 + 9 \Rightarrow {(\frac{d}{2})}^{2} = 25 = {(5)}^{2} \Rightarrow \frac{d}{2} = 5 \Rightarrow d = 10 cm$

Hence, the diameter of the circle is 10 cm.

Question 8:

Let the diameter of the required circle be d.
Now, Area of the required circle = Area of circle having radius 4 cm + Area of circle having radius 3 cm

$\Rightarrow π {(\frac{d}{2})}^{2} = π {(4)}^{2} + π {(3)}^{2} \Rightarrow {(\frac{d}{2})}^{2} = 16 + 9 \Rightarrow {(\frac{d}{2})}^{2} = 25 = {(5)}^{2} \Rightarrow \frac{d}{2} = 5 \Rightarrow d = 10 cm$

Hence, the diameter of the circle is 10 cm.

Answer:

Let the radius of the circle be r.
â€‹Now,
$Circumference = 8 π \Rightarrow 2 π r = 8 π \Rightarrow r = 4 cm$
Now, Area of circle= $π r^{2} = π \times 4^{2} = 16 π {cm}^{2}$
Hence, the area of the circle is 16π cm².

Question 9:

Let the radius of the circle be r.
â€‹Now,
$Circumference = 8 π \Rightarrow 2 π r = 8 π \Rightarrow r = 4 cm$
Now, Area of circle= $π r^{2} = π \times 4^{2} = 16 π {cm}^{2}$
Hence, the area of the circle is 16π cm².

Answer:

Perimeter of a semicircular protractor = Circumference of semicircular protractor + diameter of semicircular protractor
$= \frac{1}{2} (2 π r) + d = π r + d = π \frac{d}{2} + d = \frac{22}{7} \times 7 + 14 = 22 + 14 = 36 cm$
Hence, the perimeter of a semicircular protractor is 36 cm.

Question 10:

Perimeter of a semicircular protractor = Circumference of semicircular protractor + diameter of semicircular protractor
$= \frac{1}{2} (2 π r) + d = π r + d = π \frac{d}{2} + d = \frac{22}{7} \times 7 + 14 = 22 + 14 = 36 cm$
Hence, the perimeter of a semicircular protractor is 36 cm.

Answer:

Let the radius of the required circle be r.
Now, Area of circle = Perimeter of the circle
$\Rightarrow π r^{2} = 2 π \times r \Rightarrow r^{2} = 2 r \Rightarrow r^{2} - 2 r = 0 \Rightarrow r (r - 2) = 0 \Rightarrow r - 2 = 0 [∵ r \neq 0] \Rightarrow r = 2 units$
Hence, the radius of the circle is 2 units.

Question 11:

Let the radius of the required circle be r.
Now, Area of circle = Perimeter of the circle
$\Rightarrow π r^{2} = 2 π \times r \Rightarrow r^{2} = 2 r \Rightarrow r^{2} - 2 r = 0 \Rightarrow r (r - 2) = 0 \Rightarrow r - 2 = 0 [∵ r \neq 0] \Rightarrow r = 2 units$
Hence, the radius of the circle is 2 units.

Answer:

Let the radius of the required circle be r.
Now, Circumference of the required circle = Circumference of circle having radius 19 cm + Circumference of circle having radius 9 cm

$\Rightarrow 2 π r = 2 π \times 19 + 2 π \times 9 \Rightarrow r = 19 + 9 \Rightarrow r = 28 cm$
Hence, the radius of the required circle is 28 cm.

Question 12:

Let the radius of the required circle be r.
Now, Circumference of the required circle = Circumference of circle having radius 19 cm + Circumference of circle having radius 9 cm

$\Rightarrow 2 π r = 2 π \times 19 + 2 π \times 9 \Rightarrow r = 19 + 9 \Rightarrow r = 28 cm$
Hence, the radius of the required circle is 28 cm.

Answer:

Let the radius of the required circle be r.
Now, Area of the required circle = Area of circle having radius 8 cm + Area of circle having radius 6 cm

$\Rightarrow π r^{2} = π {(8)}^{2} + π {(6)}^{2} \Rightarrow r^{2} = 64 + 36 \Rightarrow r^{2} = 10^{2} \Rightarrow r = 10 cm$
Hence, the radius of the circle is 10 cm.

Question 13:

Let the radius of the required circle be r.
Now, Area of the required circle = Area of circle having radius 8 cm + Area of circle having radius 6 cm

$\Rightarrow π r^{2} = π {(8)}^{2} + π {(6)}^{2} \Rightarrow r^{2} = 64 + 36 \Rightarrow r^{2} = 10^{2} \Rightarrow r = 10 cm$
Hence, the radius of the circle is 10 cm.

Answer:

We have $r = 6 cm and θ = 30 °$
Now, Area of sector = $\frac{θ}{360^{°}} \times π r^{2} = \frac{30 °}{360 °} \times 3.14 \times 36 = 9.42 {cm}^{2}$
Hence, the area of the sector of the circle is 9.42 cm².

Question 14:

We have $r = 6 cm and θ = 30 °$
Now, Area of sector = $\frac{θ}{360^{°}} \times π r^{2} = \frac{30 °}{360 °} \times 3.14 \times 36 = 9.42 {cm}^{2}$
Hence, the area of the sector of the circle is 9.42 cm².

Answer:

We have $r = 21 cm and θ = 60 °$
Length of arc = $\frac{θ}{360^{°}} \times 2 π r = \frac{60 °}{360 °} \times 2 \times \frac{22}{7} \times 21 = 22 cm$
Hence, the length of the arc of the circle is 22 cm.

Question 15:

We have $r = 21 cm and θ = 60 °$
Length of arc = $\frac{θ}{360^{°}} \times 2 π r = \frac{60 °}{360 °} \times 2 \times \frac{22}{7} \times 21 = 22 cm$
Hence, the length of the arc of the circle is 22 cm.

Answer:

Let the the radii of the two circles be r and R, the circumferences of the circles be c and C and the areas of the two circles be a and A.
Now,
$\frac{c}{C} = \frac{2}{3} \Rightarrow \frac{2 π r}{2 π R} = \frac{2}{3} \Rightarrow \frac{r}{R} = \frac{2}{3}$
Now, the ratio between their areas is given by
$\frac{a}{A} = \frac{π r^{2}}{π R^{2}} = {(\frac{r}{R})}^{2} = {(\frac{2}{3})}^{2} = \frac{4}{9}$
Hence, the ratio between their areas is 4 : 9.

Question 16:

Let the the radii of the two circles be r and R, the circumferences of the circles be c and C and the areas of the two circles be a and A.
Now,
$\frac{c}{C} = \frac{2}{3} \Rightarrow \frac{2 π r}{2 π R} = \frac{2}{3} \Rightarrow \frac{r}{R} = \frac{2}{3}$
Now, the ratio between their areas is given by
$\frac{a}{A} = \frac{π r^{2}}{π R^{2}} = {(\frac{r}{R})}^{2} = {(\frac{2}{3})}^{2} = \frac{4}{9}$
Hence, the ratio between their areas is 4 : 9.

Answer:

Let the radii of the two circles be r and R, the circumferences of the circles be c and C and the areas of the two circles be a and A.
Now,
$\frac{a}{A} = \frac{4}{9} \Rightarrow \frac{π r^{2}}{π R^{2}} = {(\frac{2}{3})}^{2} \Rightarrow \frac{r}{R} = \frac{2}{3}$
Now, the ratio between their circumferences is given by
$\frac{c}{C} = \frac{2 π r}{2 π R} = \frac{r}{R} = \frac{2}{3}$
Hence, the ratio between their circumferences is 2 : 3.

Question 17:

Let the radii of the two circles be r and R, the circumferences of the circles be c and C and the areas of the two circles be a and A.
Now,
$\frac{a}{A} = \frac{4}{9} \Rightarrow \frac{π r^{2}}{π R^{2}} = {(\frac{2}{3})}^{2} \Rightarrow \frac{r}{R} = \frac{2}{3}$
Now, the ratio between their circumferences is given by
$\frac{c}{C} = \frac{2 π r}{2 π R} = \frac{r}{R} = \frac{2}{3}$
Hence, the ratio between their circumferences is 2 : 3.

Answer:

Let the side of the square be a and radius of the circle be r
We know that if a circle circumscribes a square, then the diameter of the circle is equal to the diagonal of the square.
$∴ \sqrt{2} a = 2 r \Rightarrow a = \sqrt{2} r$
Now,
$\frac{Area of circle}{Area of square} = \frac{π r^{2}}{a^{2}} = \frac{{πr}^{2}}{{(\sqrt{2} r)}^{2}} = \frac{π r^{2}}{2 r^{2}} = \frac{π}{2}$
Hence, the ratio of the areas of the circle and the square is π : 2

Question 18:

Let the side of the square be a and radius of the circle be r
We know that if a circle circumscribes a square, then the diameter of the circle is equal to the diagonal of the square.
$∴ \sqrt{2} a = 2 r \Rightarrow a = \sqrt{2} r$
Now,
$\frac{Area of circle}{Area of square} = \frac{π r^{2}}{a^{2}} = \frac{{πr}^{2}}{{(\sqrt{2} r)}^{2}} = \frac{π r^{2}}{2 r^{2}} = \frac{π}{2}$
Hence, the ratio of the areas of the circle and the square is π : 2

Answer:

Let the radius of the circle be r.
â€‹Now,
$Circumference = 8 \Rightarrow 2 π r = 8 \Rightarrow r = \frac{14}{11} cm$
We have $r = \frac{14}{11} cm and θ = 72 °$
Area of sector = $\frac{θ}{360^{°}} \times π r^{2} = \frac{72 °}{360 °} \times \frac{22}{7} \times {(\frac{14}{11})}^{2} = 1.02 {cm}^{2}$
Hence, the area of the sector of the circle is 1.02 cm².

Disclaimer : If we take the circumference of the circle is 8 cm then the area of the sector will be 1.02 cm². But if we take the circumference of the circle is 88 cm then the area of the sector will be 123.2 cm²

Question 19:

Let the radius of the circle be r.
â€‹Now,
$Circumference = 8 \Rightarrow 2 π r = 8 \Rightarrow r = \frac{14}{11} cm$
We have $r = \frac{14}{11} cm and θ = 72 °$
Area of sector = $\frac{θ}{360^{°}} \times π r^{2} = \frac{72 °}{360 °} \times \frac{22}{7} \times {(\frac{14}{11})}^{2} = 1.02 {cm}^{2}$
Hence, the area of the sector of the circle is 1.02 cm².

Disclaimer : If we take the circumference of the circle is 8 cm then the area of the sector will be 1.02 cm². But if we take the circumference of the circle is 88 cm then the area of the sector will be 123.2 cm²

Answer:

Given:
Length of the arc = 8.8 cm
And,
$θ = 30^{\circ}$

Now,
Length of the arc = $\frac{2 πrθ}{360}$
$\Rightarrow 8.8 = \frac{2 \times \frac{22}{7} \times r \times 30}{360} \Rightarrow r = \frac{8.8 \times 360 \times 7}{44 \times 30} \Rightarrow r = 16.8 cm$

∴ Length of the pendulum = 16.8 cm

Question 20:

Given:
Length of the arc = 8.8 cm
And,
$θ = 30^{\circ}$

Now,
Length of the arc = $\frac{2 πrθ}{360}$
$\Rightarrow 8.8 = \frac{2 \times \frac{22}{7} \times r \times 30}{360} \Rightarrow r = \frac{8.8 \times 360 \times 7}{44 \times 30} \Rightarrow r = 16.8 cm$

∴ Length of the pendulum = 16.8 cm

Answer:

Angle inscribed by the minute hand in 60 minutes = $360^{\circ}$
Angle inscribed by the minute hand in 20 minutes = $\frac{360}{60} \times 20 = 120^{\circ}$

We have:
$θ = 120^{\circ} and r = 15 cm$

∴ Required area swept by the minute hand in 20 minutes = Area of the sector with r = 15 cm and $θ = 120^{\circ}$
$= \frac{{πr}^{2} θ}{360}$

$= 3.14 \times 15 \times 15 \times \frac{120}{360} = 235.5 {cm}^{2}$

Question 21:

Angle inscribed by the minute hand in 60 minutes = $360^{\circ}$
Angle inscribed by the minute hand in 20 minutes = $\frac{360}{60} \times 20 = 120^{\circ}$

We have:
$θ = 120^{\circ} and r = 15 cm$

∴ Required area swept by the minute hand in 20 minutes = Area of the sector with r = 15 cm and $θ = 120^{\circ}$
$= \frac{{πr}^{2} θ}{360}$

$= 3.14 \times 15 \times 15 \times \frac{120}{360} = 235.5 {cm}^{2}$

Answer:

Area of the sector =17.6 cm²
Area of the sector $= \frac{{πr}^{2} θ}{360}$

$\Rightarrow 17.6 = \frac{22}{7} \times r^{2} \times \frac{56}{360} \Rightarrow r^{2} = \frac{17.6 \times 7 \times 360}{22 \times 56} \Rightarrow r^{2} = 36 \Rightarrow r = 6 cm$

∴ Radius of the circle = 6 cm

Question 22:

Area of the sector =17.6 cm²
Area of the sector $= \frac{{πr}^{2} θ}{360}$

$\Rightarrow 17.6 = \frac{22}{7} \times r^{2} \times \frac{56}{360} \Rightarrow r^{2} = \frac{17.6 \times 7 \times 360}{22 \times 56} \Rightarrow r^{2} = 36 \Rightarrow r = 6 cm$

∴ Radius of the circle = 6 cm

Answer:

Given:
Area of the sector = 63 cm²
Radius = 10.5 cm

Now,
Area of the sector $= \frac{{πr}^{2} θ}{360}$
$\Rightarrow 69.3 = \frac{22}{7} \times 10.5 \times 10.5 \times \frac{θ}{360} \Rightarrow θ = \frac{69.3 \times 7 \times 360}{22 \times 10.5 \times 10.5} \Rightarrow θ = 72^{\circ}$

∴ Central angle of the sector = $72^{\circ}$

Question 23:

Given:
Area of the sector = 63 cm²
Radius = 10.5 cm

Now,
Area of the sector $= \frac{{πr}^{2} θ}{360}$
$\Rightarrow 69.3 = \frac{22}{7} \times 10.5 \times 10.5 \times \frac{θ}{360} \Rightarrow θ = \frac{69.3 \times 7 \times 360}{22 \times 10.5 \times 10.5} \Rightarrow θ = 72^{\circ}$

∴ Central angle of the sector = $72^{\circ}$

Answer:

Given:
Radius = 6.5 cm

Let O be the centre of the circle with radius 6.5 cm and OACBO be its sector with perimeter 31 cm.
Thus, we have:
OA + OB + arc AB = 31 cm

$\Rightarrow 6.5 + 6.5 + a r c A B = 31 \Rightarrow a r c A B = 31 - 13 \Rightarrow a r c A B = 18 cm$

Now,
Area of the sector OACBO = $\frac{1}{2} \times Radius \times Arc$
$= \frac{1}{2} \times 6.5 \times 18 = 58.5 {cm}^{2}$
â€‹

Question 24:

Given:
Radius = 6.5 cm

Let O be the centre of the circle with radius 6.5 cm and OACBO be its sector with perimeter 31 cm.
Thus, we have:
OA + OB + arc AB = 31 cm

$\Rightarrow 6.5 + 6.5 + a r c A B = 31 \Rightarrow a r c A B = 31 - 13 \Rightarrow a r c A B = 18 cm$

Now,
Area of the sector OACBO = $\frac{1}{2} \times Radius \times Arc$
$= \frac{1}{2} \times 6.5 \times 18 = 58.5 {cm}^{2}$
â€‹

Answer:

Given:
Radius = 17.5 cm
Length of the arc = 44 cm

Now,
Length of the arc $= \frac{2 πrθ}{360}$

$\Rightarrow 44 = 2 \times \frac{22}{7} \times 17.5 \times \frac{θ}{360} \Rightarrow θ = \frac{44 \times 7 \times 360}{44 \times 17.5} \Rightarrow θ = 144^{\circ}$
Also,
Area of the sector = $\frac{{πr}^{2} θ}{360}$

$= \frac{22}{7} \times 17.5 \times 17.5 \times 144 = 385 {cm}^{2}$

Question 25:

Given:
Radius = 17.5 cm
Length of the arc = 44 cm

Now,
Length of the arc $= \frac{2 πrθ}{360}$

$\Rightarrow 44 = 2 \times \frac{22}{7} \times 17.5 \times \frac{θ}{360} \Rightarrow θ = \frac{44 \times 7 \times 360}{44 \times 17.5} \Rightarrow θ = 144^{\circ}$
Also,
Area of the sector = $\frac{{πr}^{2} θ}{360}$

$= \frac{22}{7} \times 17.5 \times 17.5 \times 144 = 385 {cm}^{2}$

Answer:

We know that we can cut two circular pieces of equal radii and maximum area from the rectangular cardboard whose diameter is equal to the width of the rectangular cardboard.
∴ Radii of two circuar pieces = Half of the width of the rectangular cardboard = 3.5 cm
Now,
Area of remaining cardboard = Area of rectangular cardboard − 2 â¨¯ Area of circular piece having radius 3.5 cm
$= 14 \times 7 - 2 (\frac{22}{7} \times 3.5 \times 3.5) = 98 - 77 = 21 {cm}^{2}$
Hence, the area of the remaining cardboard is 21 cm²

Question 26:

We know that we can cut two circular pieces of equal radii and maximum area from the rectangular cardboard whose diameter is equal to the width of the rectangular cardboard.
∴ Radii of two circuar pieces = Half of the width of the rectangular cardboard = 3.5 cm
Now,
Area of remaining cardboard = Area of rectangular cardboard − 2 â¨¯ Area of circular piece having radius 3.5 cm
$= 14 \times 7 - 2 (\frac{22}{7} \times 3.5 \times 3.5) = 98 - 77 = 21 {cm}^{2}$
Hence, the area of the remaining cardboard is 21 cm²

Answer:

Area of the square ABCD = $(Side)^{2}$
            $= 4^{2} = 16 {cm}^{2}$

Area of the circle = ${πr}^{2}$
Radius = 1 cm
$Area = 3.14 \times (1)^{2} = 3.14 {cm}^{2}$

Area of the quadrant of one circle = $\frac{1}{4} {πr}^{2}$
         $= \frac{1}{4} \times 3.14 \times 1^{2} = 0.785 {cm}^{2}$

Area of the quadrants of four circles = $0.785 \times 4 = 3.14 {cm}^{2}$
Area of the shaded region = Area of the square $-$ Area of the circle $-$ Area of the quadrants of four circles
   = $16 - 3.14 - 3.14 = 9.72 {cm}^{2}$

Question 27:

Area of the square ABCD = $(Side)^{2}$
            $= 4^{2} = 16 {cm}^{2}$

Area of the circle = ${πr}^{2}$
Radius = 1 cm
$Area = 3.14 \times (1)^{2} = 3.14 {cm}^{2}$

Area of the quadrant of one circle = $\frac{1}{4} {πr}^{2}$
         $= \frac{1}{4} \times 3.14 \times 1^{2} = 0.785 {cm}^{2}$

Area of the quadrants of four circles = $0.785 \times 4 = 3.14 {cm}^{2}$
Area of the shaded region = Area of the square $-$ Area of the circle $-$ Area of the quadrants of four circles
   = $16 - 3.14 - 3.14 = 9.72 {cm}^{2}$

Answer:

We know that the opposite sides of rectangle are equal
∴ AD = BC = 28 cm
Now, Radius of semicircular portion = $\frac{1}{2} BC = 14 cm$
∴ Area of remaining paper = Area of rectangular sheet −Area of semicircular portion
$= 40 \times 28 - \frac{1}{2} (\frac{22}{7} \times 14 \times 14) = 1120 - 308 = 812 {cm}^{2}$
Hence, the area of the remaining paper is 812 cm²

Question 28:

We know that the opposite sides of rectangle are equal
∴ AD = BC = 28 cm
Now, Radius of semicircular portion = $\frac{1}{2} BC = 14 cm$
∴ Area of remaining paper = Area of rectangular sheet −Area of semicircular portion
$= 40 \times 28 - \frac{1}{2} (\frac{22}{7} \times 14 \times 14) = 1120 - 308 = 812 {cm}^{2}$
Hence, the area of the remaining paper is 812 cm²

Answer:

Area of shaded region = Area of square OABC − Area of quadrant COPB having radius OC
$= {(Side)}^{2} - \frac{1}{4} (π \times r^{2}) = {(7)}^{2} - \frac{1}{4} [\frac{22}{7} \times 7^{2}] = 49 - 38.5 = 10.5 {cm}^{2}$
Hence, the area of the shaded region is 10.5 cm²

Question 29:

Area of shaded region = Area of square OABC − Area of quadrant COPB having radius OC
$= {(Side)}^{2} - \frac{1}{4} (π \times r^{2}) = {(7)}^{2} - \frac{1}{4} [\frac{22}{7} \times 7^{2}] = 49 - 38.5 = 10.5 {cm}^{2}$
Hence, the area of the shaded region is 10.5 cm²

Answer:

Area of the shaded region = Area of sector having central angle 60^âˆ˜ + Area of sector having central angle 80^âˆ˜ + Area of sector having central angle 40^âˆ˜
$= [\frac{60 °}{360^{°}} \times π {(7)}^{2}] + [\frac{80 °}{360^{°}} \times π {(7)}^{2}] + [\frac{40 °}{360^{°}} \times π {(7)}^{2}] = π {(7)}^{2} (\frac{60 °}{360^{°}} + \frac{80 °}{360^{°}} + \frac{40 °}{360^{°}}) = π {(7)}^{2} (\frac{180 °}{360^{°}}) = \frac{22}{7} {(7)}^{2} (\frac{1}{2}) = 77 {cm}^{2}$
Hence, the area of the shaded region is 77 cm².

Question 30:

Area of the shaded region = Area of sector having central angle 60^âˆ˜ + Area of sector having central angle 80^âˆ˜ + Area of sector having central angle 40^âˆ˜
$= [\frac{60 °}{360^{°}} \times π {(7)}^{2}] + [\frac{80 °}{360^{°}} \times π {(7)}^{2}] + [\frac{40 °}{360^{°}} \times π {(7)}^{2}] = π {(7)}^{2} (\frac{60 °}{360^{°}} + \frac{80 °}{360^{°}} + \frac{40 °}{360^{°}}) = π {(7)}^{2} (\frac{180 °}{360^{°}}) = \frac{22}{7} {(7)}^{2} (\frac{1}{2}) = 77 {cm}^{2}$
Hence, the area of the shaded region is 77 cm².

Answer:

Area of the shaded portion = Area of sector OPQ − Area of sector OAB
$= \frac{30 °}{360^{°}} \times π {(7)}^{2} - \frac{30 °}{360^{°}} \times π {(3.5)}^{2} = \frac{22}{7} \times \frac{1}{12} [{(7)}^{2} - {(3.5)}^{2}] = \frac{22}{7} \times \frac{1}{12} [{(7)}^{2} - {(\frac{7}{2})}^{2}] = \frac{22}{7} \times \frac{1}{12} \times \frac{147}{4} = \frac{77}{8} {cm}^{2}$
Hence, the area of the shaded portion is $\frac{77}{8}$ cm².

Question 31:

Area of the shaded portion = Area of sector OPQ − Area of sector OAB
$= \frac{30 °}{360^{°}} \times π {(7)}^{2} - \frac{30 °}{360^{°}} \times π {(3.5)}^{2} = \frac{22}{7} \times \frac{1}{12} [{(7)}^{2} - {(3.5)}^{2}] = \frac{22}{7} \times \frac{1}{12} [{(7)}^{2} - {(\frac{7}{2})}^{2}] = \frac{22}{7} \times \frac{1}{12} \times \frac{147}{4} = \frac{77}{8} {cm}^{2}$
Hence, the area of the shaded portion is $\frac{77}{8}$ cm².

Answer:

Area of the shaded region = Area of Square ABCD − (Area of semicircle APD + Area of semicircle BPC)
$= {(14)}^{2} - [\frac{1}{2} π {(\frac{14}{2})}^{2} + \frac{1}{2} π {(\frac{14}{2})}^{2}] = {(14)}^{2} - \frac{22}{7} {(7)}^{2} = 42 {cm}^{2}$
Hence, the area of the shaded region is 42 cm².

Question 32:

Area of the shaded region = Area of Square ABCD − (Area of semicircle APD + Area of semicircle BPC)
$= {(14)}^{2} - [\frac{1}{2} π {(\frac{14}{2})}^{2} + \frac{1}{2} π {(\frac{14}{2})}^{2}] = {(14)}^{2} - \frac{22}{7} {(7)}^{2} = 42 {cm}^{2}$
Hence, the area of the shaded region is 42 cm².

Answer:

We have $r = 42 cm and θ = 360 ° - 90 ° = 270 °$
Perimeter of the top of the table = Length of the major arc AB + Length of OA + Length of OB
$= \frac{θ}{360^{°}} \times 2 π r + 42 + 42 = \frac{270 °}{360 °} \times 2 \times \frac{22}{7} \times 42 + 84 = 198 + 84 = 282 cm$

Hence, the perimeter of the top of the table is 282 cm.

Question 33:

We have $r = 42 cm and θ = 360 ° - 90 ° = 270 °$
Perimeter of the top of the table = Length of the major arc AB + Length of OA + Length of OB
$= \frac{θ}{360^{°}} \times 2 π r + 42 + 42 = \frac{270 °}{360 °} \times 2 \times \frac{22}{7} \times 42 + 84 = 198 + 84 = 282 cm$

Hence, the perimeter of the top of the table is 282 cm.

Answer:

Area of the shaded portion = (Area of quadrant DPBA + Area of quadrant DQBC) − Area of Square ABCD
$= [\frac{1}{4} π {(7)}^{2} + \frac{1}{4} π {(7)}^{2}] - {(7)}^{2} = \frac{1}{2} \times \frac{22}{7} {(7)}^{2} - 49 = 28 {cm}^{2}$
Hence, the area of the shaded portion is 28 cm².

Question 34:

Area of the shaded portion = (Area of quadrant DPBA + Area of quadrant DQBC) − Area of Square ABCD
$= [\frac{1}{4} π {(7)}^{2} + \frac{1}{4} π {(7)}^{2}] - {(7)}^{2} = \frac{1}{2} \times \frac{22}{7} {(7)}^{2} - 49 = 28 {cm}^{2}$
Hence, the area of the shaded portion is 28 cm².

Answer:

Area of the right-angled $∆$ COD = $\frac{1}{2} \times b \times h$
   = $\frac{1}{2} \times 3.5 \times 2 = 3.5 {cm}^{2}$

Area of the sector AOC = $\frac{θ}{360} \times π \times r^{2}$
                                       $= \frac{90}{360} \times \frac{22}{7} \times {(3.5)}^{2} = 9.625 {cm}^{2}$

Area of the shaded region = Area of the $∆$ COD $-$ Area of the sector AOC

                                           $= 9.625 - 3.50 = 6.125 {cm}^{2}$

Question 35:

Area of the right-angled $∆$ COD = $\frac{1}{2} \times b \times h$
   = $\frac{1}{2} \times 3.5 \times 2 = 3.5 {cm}^{2}$

Area of the sector AOC = $\frac{θ}{360} \times π \times r^{2}$
                                       $= \frac{90}{360} \times \frac{22}{7} \times {(3.5)}^{2} = 9.625 {cm}^{2}$

Area of the shaded region = Area of the $∆$ COD $-$ Area of the sector AOC

                                           $= 9.625 - 3.50 = 6.125 {cm}^{2}$

Answer:

Permieter of shaded region = Length of the arc APB + Length of the arc CPD + Length of AD + Length of BC
$= \frac{1}{2} \times 2 π r + \frac{1}{2} \times 2 π r + 14 + 14 = 2 π r + 28 = 2 \times \frac{22}{7} \times \frac{14}{2} + 28 = 72 cm$

Hence, the perimeter of the shaded region is 72 cm.

Question 36:

Permieter of shaded region = Length of the arc APB + Length of the arc CPD + Length of AD + Length of BC
$= \frac{1}{2} \times 2 π r + \frac{1}{2} \times 2 π r + 14 + 14 = 2 π r + 28 = 2 \times \frac{22}{7} \times \frac{14}{2} + 28 = 72 cm$

Hence, the perimeter of the shaded region is 72 cm.

Answer:

Let the diagonal of the square be d.
We know that if a circle circumscribes a square, then the diameter of the circle is equal to the diagonal of the square.
∴ d = 2 â¨¯ 7 = 14 cm
Now,
Area of required region = Area of circle − Area of square
$= π r^{2} - \frac{1}{2} d^{2} = \frac{22}{7} \times {(7)}^{2} - \frac{1}{2} \times {(14)}^{2} = 56 {cm}^{2}$

Hence, the required area is 56 cm² .

Question 37:

Let the diagonal of the square be d.
We know that if a circle circumscribes a square, then the diameter of the circle is equal to the diagonal of the square.
∴ d = 2 â¨¯ 7 = 14 cm
Now,
Area of required region = Area of circle − Area of square
$= π r^{2} - \frac{1}{2} d^{2} = \frac{22}{7} \times {(7)}^{2} - \frac{1}{2} \times {(14)}^{2} = 56 {cm}^{2}$

Hence, the required area is 56 cm² .

Answer:

(i) Perimeter of shaded region = Length of the arc APB + Length of the arc ARC + Length of the arc BSD + Length of the arc CQD
$= \frac{1}{2} \times 2 π r_{1} + \frac{1}{2} \times 2 π r_{2} + \frac{1}{2} \times 2 π r_{3} + \frac{1}{2} \times 2 π r_{4} = \frac{1}{2} \times 2 π (\frac{7}{2}) + \frac{1}{2} \times 2 π (7) + \frac{1}{2} \times 2 π (7) + \frac{1}{2} \times 2 π (\frac{7}{2}) = 2 π (\frac{7}{2}) + 2 π (7) = 2 π (\frac{7}{2} + 7) = 2 \times \frac{22}{7} \times \frac{21}{2} = 66 cm$

(ii) Area of shaded region = Area of the arc ARC + Area of the arc BSD − (Area of the arc APB + Area of the arc CQD)
$= \frac{1}{2} \times π {(r_{1})}^{2} + \frac{1}{2} \times π {(r_{2})}^{2} - [\frac{1}{2} \times π {(r_{3})}^{2} + \frac{1}{2} \times π {(r_{4})}^{2}] = \frac{1}{2} \times π {(7)}^{2} + \frac{1}{2} \times π {(7)}^{2} - [\frac{1}{2} \times π {(\frac{7}{2})}^{2} + \frac{1}{2} \times π {(\frac{7}{2})}^{2}] = π {(7)}^{2} - π {(\frac{7}{2})}^{2} = π (49 - \frac{49}{4}) = \frac{22}{7} \times \frac{147}{4} = 115.5 {cm}^{2}$

Question 38:

(i) Perimeter of shaded region = Length of the arc APB + Length of the arc ARC + Length of the arc BSD + Length of the arc CQD
$= \frac{1}{2} \times 2 π r_{1} + \frac{1}{2} \times 2 π r_{2} + \frac{1}{2} \times 2 π r_{3} + \frac{1}{2} \times 2 π r_{4} = \frac{1}{2} \times 2 π (\frac{7}{2}) + \frac{1}{2} \times 2 π (7) + \frac{1}{2} \times 2 π (7) + \frac{1}{2} \times 2 π (\frac{7}{2}) = 2 π (\frac{7}{2}) + 2 π (7) = 2 π (\frac{7}{2} + 7) = 2 \times \frac{22}{7} \times \frac{21}{2} = 66 cm$

(ii) Area of shaded region = Area of the arc ARC + Area of the arc BSD − (Area of the arc APB + Area of the arc CQD)
$= \frac{1}{2} \times π {(r_{1})}^{2} + \frac{1}{2} \times π {(r_{2})}^{2} - [\frac{1}{2} \times π {(r_{3})}^{2} + \frac{1}{2} \times π {(r_{4})}^{2}] = \frac{1}{2} \times π {(7)}^{2} + \frac{1}{2} \times π {(7)}^{2} - [\frac{1}{2} \times π {(\frac{7}{2})}^{2} + \frac{1}{2} \times π {(\frac{7}{2})}^{2}] = π {(7)}^{2} - π {(\frac{7}{2})}^{2} = π (49 - \frac{49}{4}) = \frac{22}{7} \times \frac{147}{4} = 115.5 {cm}^{2}$

Answer:

Perimeter of shaded region = Length of the arc PAQ + Length of the arc PSR + Length of the arc RTQ
$= \frac{1}{2} \times 2 π r_{1} + \frac{1}{2} \times 2 π r_{2} + \frac{1}{2} \times 2 π r_{3} = \frac{1}{2} \times 2 π (\frac{7}{2}) + \frac{1}{2} \times 2 π (\frac{10}{2}) + \frac{1}{2} \times 2 π (\frac{3}{2}) = \frac{7}{2} π + 5 π + \frac{3}{2} π = \frac{7}{2} π + \frac{3}{2} π + 5 π = 5 π + 5 π = 10 π = 31.4 cm$
Hence, the perimeter of shaded region is 31.4 cm.

Question 39:

Perimeter of shaded region = Length of the arc PAQ + Length of the arc PSR + Length of the arc RTQ
$= \frac{1}{2} \times 2 π r_{1} + \frac{1}{2} \times 2 π r_{2} + \frac{1}{2} \times 2 π r_{3} = \frac{1}{2} \times 2 π (\frac{7}{2}) + \frac{1}{2} \times 2 π (\frac{10}{2}) + \frac{1}{2} \times 2 π (\frac{3}{2}) = \frac{7}{2} π + 5 π + \frac{3}{2} π = \frac{7}{2} π + \frac{3}{2} π + 5 π = 5 π + 5 π = 10 π = 31.4 cm$
Hence, the perimeter of shaded region is 31.4 cm.

Answer:

Construction: Join OB

In right triangle AOB
OB² = OA² + AB²
= 20² + 20²
= 400 + 400
= 800
∴ OB² = 800
Area of the shaded region = Area of quadrant OPBQ − Area of Square OABC
$= \frac{1}{4} π {(OB)}^{2} - {(OA)}^{2} = \frac{1}{4} \times 3.14 \times 800 - 400 = 628 - 400 = 228 {cm}^{2}$
Hence, the area of the shaded region is 228 cm².

Question 40:

Construction: Join OB

In right triangle AOB
OB² = OA² + AB²
= 20² + 20²
= 400 + 400
= 800
∴ OB² = 800
Area of the shaded region = Area of quadrant OPBQ − Area of Square OABC
$= \frac{1}{4} π {(OB)}^{2} - {(OA)}^{2} = \frac{1}{4} \times 3.14 \times 800 - 400 = 628 - 400 = 228 {cm}^{2}$
Hence, the area of the shaded region is 228 cm².

Answer:

Permieter of shaded region = Length of the arc AQO + Length of the arc APB + Length of OB
$\Rightarrow 40 = \frac{1}{2} \times 2 π (\frac{AO}{2}) + \frac{1}{2} \times 2 π (OB) + OB \Rightarrow 40 = \frac{11}{7} AO + \frac{22}{7} OB + OB \Rightarrow 40 = \frac{11}{7} OB + \frac{22}{7} OB + OB [∵ AO = OB] \Rightarrow 40 = \frac{40}{7} OB \Rightarrow OB = 7 cm$
Area of the shaded portion = Area of semicircle AQO + Area of semicircle APB
$= \frac{1}{2} π {(\frac{7}{2})}^{2} + \frac{1}{2} π {(7)}^{2} = \frac{1}{2} \times \frac{22}{7} \times {(\frac{7}{2})}^{2} + \frac{1}{2} \times \frac{22}{7} \times {(7)}^{2} = 96.25 {cm}^{2}$
Hence, the area of the shaded portion is 96.25 cm².

Question 41:

Permieter of shaded region = Length of the arc AQO + Length of the arc APB + Length of OB
$\Rightarrow 40 = \frac{1}{2} \times 2 π (\frac{AO}{2}) + \frac{1}{2} \times 2 π (OB) + OB \Rightarrow 40 = \frac{11}{7} AO + \frac{22}{7} OB + OB \Rightarrow 40 = \frac{11}{7} OB + \frac{22}{7} OB + OB [∵ AO = OB] \Rightarrow 40 = \frac{40}{7} OB \Rightarrow OB = 7 cm$
Area of the shaded portion = Area of semicircle AQO + Area of semicircle APB
$= \frac{1}{2} π {(\frac{7}{2})}^{2} + \frac{1}{2} π {(7)}^{2} = \frac{1}{2} \times \frac{22}{7} \times {(\frac{7}{2})}^{2} + \frac{1}{2} \times \frac{22}{7} \times {(7)}^{2} = 96.25 {cm}^{2}$
Hence, the area of the shaded portion is 96.25 cm².

Answer:

Let the radius of the circle be r.
â€‹Now,
$Circumference = 44 \Rightarrow 2 π r = 44 \Rightarrow r = 7 cm$
Now,
Area of quadrant = $\frac{1}{4} π r^{2} = \frac{1}{4} \times \frac{22}{7} \times {(7)}^{2} = 38.5 {cm}^{2}$
Hence, the area of the quadrant of the circle is 38.5 cm².

Question 42:

Let the radius of the circle be r.
â€‹Now,
$Circumference = 44 \Rightarrow 2 π r = 44 \Rightarrow r = 7 cm$
Now,
Area of quadrant = $\frac{1}{4} π r^{2} = \frac{1}{4} \times \frac{22}{7} \times {(7)}^{2} = 38.5 {cm}^{2}$
Hence, the area of the quadrant of the circle is 38.5 cm².

Answer:

Area of the square = ${Side}^{2} = 14^{2} = 196 sq . cm$

Area of the circles = $4 \times π \times 3.5 \times 3.5 = 154 sq . cm$

Area of the shaded region = Area of the square $-$ Area of four circles
$= 196 - 154 = 42 {cm}^{2}$

Question 43:

Area of the square = ${Side}^{2} = 14^{2} = 196 sq . cm$

Area of the circles = $4 \times π \times 3.5 \times 3.5 = 154 sq . cm$

Area of the shaded region = Area of the square $-$ Area of four circles
$= 196 - 154 = 42 {cm}^{2}$

Answer:

In right triangle ABC
AC² = AB² + BC²
= 8² +6²
= 64 + 36
= 100
∴ AC² = 100
⇒ AC = 10 cm
Now, Radius of circle(OA) $= \frac{1}{2} AC = 5 cm$
Area of the shaded region = Area of circle − Area of rectangle OABC
$= π {(OA)}^{2} - AB \times BC = \frac{22}{7} \times {(5)}^{2} - 8 \times 6 = 78.57 - 48 = 30.57 {cm}^{2}$
Hence, the area of the shaded region is 30.57 cm².

Question 44:

In right triangle ABC
AC² = AB² + BC²
= 8² +6²
= 64 + 36
= 100
∴ AC² = 100
⇒ AC = 10 cm
Now, Radius of circle(OA) $= \frac{1}{2} AC = 5 cm$
Area of the shaded region = Area of circle − Area of rectangle OABC
$= π {(OA)}^{2} - AB \times BC = \frac{22}{7} \times {(5)}^{2} - 8 \times 6 = 78.57 - 48 = 30.57 {cm}^{2}$
Hence, the area of the shaded region is 30.57 cm².

Answer:

Area of the circle = 484 cm²
Area of the square = ${Side}^{2}$
$\Rightarrow 484 = {Side}^{2} \Rightarrow 22^{2} = {Side}^{2} \Rightarrow Side = 22 cm$
Perimeter of the square = $4 \times Side$
Perimeter of the square = $4 \times 22$
= 88 cm
Length of the wire = 88 cm
Circumference of the circle = Length of the wire = 88 cm
Now, let the radius of the circle be r cm.
â€‹Thus, we have:
â€‹ $2 πr = 88 \Rightarrow 2 \times \frac{22}{7} \times r = 88 \Rightarrow r = 14$

Area of the circle = ${πr}^{2}$
$= \frac{22}{7} \times 14 \times 14 = 616 {cm}^{2}$

Thus, the area enclosed by the circle is 616 cm².

Question 45:

Area of the circle = 484 cm²
Area of the square = ${Side}^{2}$
$\Rightarrow 484 = {Side}^{2} \Rightarrow 22^{2} = {Side}^{2} \Rightarrow Side = 22 cm$
Perimeter of the square = $4 \times Side$
Perimeter of the square = $4 \times 22$
= 88 cm
Length of the wire = 88 cm
Circumference of the circle = Length of the wire = 88 cm
Now, let the radius of the circle be r cm.
â€‹Thus, we have:
â€‹ $2 πr = 88 \Rightarrow 2 \times \frac{22}{7} \times r = 88 \Rightarrow r = 14$

Area of the circle = ${πr}^{2}$
$= \frac{22}{7} \times 14 \times 14 = 616 {cm}^{2}$

Thus, the area enclosed by the circle is 616 cm².

Answer:

Let the diameter of the square be d and having circumscribed circle of radius r.
We know that if a circle circumscribes a square, then the diameter of the circle is equal to the diagonal of the square.
∴ d = 2r
Now,
$Area of square = \frac{1}{2} d^{2} = \frac{1}{2} {(2 r)}^{2} = 2 r^{2} sq units$
Hence, the area of the square ABCD is 2r² sq units.

Question 46:

Let the diameter of the square be d and having circumscribed circle of radius r.
We know that if a circle circumscribes a square, then the diameter of the circle is equal to the diagonal of the square.
∴ d = 2r
Now,
$Area of square = \frac{1}{2} d^{2} = \frac{1}{2} {(2 r)}^{2} = 2 r^{2} sq units$
Hence, the area of the square ABCD is 2r² sq units.

Answer:

$Circumference = \frac{Total cost of fencing}{Rate of fencing} = \frac{5500}{25} = 220$
Let the radius of the circle be r.
â€‹Now,
$Circumference = 220 \Rightarrow 2 π r = 220 \Rightarrow r = 35 cm$
Now,
Area of field = $π r^{2} = \frac{22}{7} \times {(35)}^{2} = 3850 {cm}^{2}$
Cost of ploughing = Rate â¨¯ Area of field = 0.5 â¨¯ 3850 = Rs 1925
Hence, the cost of ploughing the field is Rs 1925.

Question 47:

$Circumference = \frac{Total cost of fencing}{Rate of fencing} = \frac{5500}{25} = 220$
Let the radius of the circle be r.
â€‹Now,
$Circumference = 220 \Rightarrow 2 π r = 220 \Rightarrow r = 35 cm$
Now,
Area of field = $π r^{2} = \frac{22}{7} \times {(35)}^{2} = 3850 {cm}^{2}$
Cost of ploughing = Rate â¨¯ Area of field = 0.5 â¨¯ 3850 = Rs 1925
Hence, the cost of ploughing the field is Rs 1925.

Answer:

Area of the rectangle = $l \times b$
           $= 120 \times 90 = 10800 sq . m$

Area of the park excluding the lawn = 2950 m²

Area of the circular lawn = Area of the park $-$ Area of the park excluding the lawn
                                          = 10800 $-$ 2950
                                          = 7850 m²
Area of the circular lawn = $π r^{2}$
$\Rightarrow 7850 = \frac{22}{7} \times r^{2} \Rightarrow \frac{7850 \times 7}{22} = r^{2} \Rightarrow r^{2} = 2497.72 \Rightarrow r = 49.97 Or, r \approx 50 m$

Thus, the radius of the circular lawn is 50 m.

Question 48:

Area of the rectangle = $l \times b$
           $= 120 \times 90 = 10800 sq . m$

Area of the park excluding the lawn = 2950 m²

Area of the circular lawn = Area of the park $-$ Area of the park excluding the lawn
                                          = 10800 $-$ 2950
                                          = 7850 m²
Area of the circular lawn = $π r^{2}$
$\Rightarrow 7850 = \frac{22}{7} \times r^{2} \Rightarrow \frac{7850 \times 7}{22} = r^{2} \Rightarrow r^{2} = 2497.72 \Rightarrow r = 49.97 Or, r \approx 50 m$

Thus, the radius of the circular lawn is 50 m.

Answer:

Area of the flower bed is the difference between the areas of sectors OPQ and ORS.

$Area of the flower bed = \frac{θ}{360} \times π ({PO}^{2} - {OR}^{2}) = \frac{90}{360} \times \frac{22}{7} (21^{2} - 14^{2}) = \frac{1}{4} \times \frac{22}{7} \times 35 \times 7 = 192.5 m^{2}$

Question 49:

Area of the flower bed is the difference between the areas of sectors OPQ and ORS.

$Area of the flower bed = \frac{θ}{360} \times π ({PO}^{2} - {OR}^{2}) = \frac{90}{360} \times \frac{22}{7} (21^{2} - 14^{2}) = \frac{1}{4} \times \frac{22}{7} \times 35 \times 7 = 192.5 m^{2}$

Answer:

We have:
OA = OC = 27 cm
AB = AC $-$ BC
= 54 $-$ 10
= 44
AB is the diameter of the smaller circle.
Thus, we have:
Radius of the smaller circle = $\frac{AB}{2} = \frac{44}{2} = 22 cm$
Area of the smaller circle = ${πr}^{2}$
            $= \frac{22}{7} \times 22 \times 22 = 1521.14 {cm}^{2}$

Radius of the larger circle = $\frac{AC}{2} = \frac{54}{2} = 27 cm$
Area of the larger circle = ${πr}^{2}$
$= \frac{22}{7} \times 27 \times 27 = 2291.14 {cm}^{2}$
∴ Area of the shaded region = Area of the larger circle $-$ Area of the smaller circle
   = 2291.14 $-$ 1521.14
                 = 770 cm²

Question 50:

We have:
OA = OC = 27 cm
AB = AC $-$ BC
= 54 $-$ 10
= 44
AB is the diameter of the smaller circle.
Thus, we have:
Radius of the smaller circle = $\frac{AB}{2} = \frac{44}{2} = 22 cm$
Area of the smaller circle = ${πr}^{2}$
            $= \frac{22}{7} \times 22 \times 22 = 1521.14 {cm}^{2}$

Radius of the larger circle = $\frac{AC}{2} = \frac{54}{2} = 27 cm$
Area of the larger circle = ${πr}^{2}$
$= \frac{22}{7} \times 27 \times 27 = 2291.14 {cm}^{2}$
∴ Area of the shaded region = Area of the larger circle $-$ Area of the smaller circle
   = 2291.14 $-$ 1521.14
                 = 770 cm²

Answer:

Since, BFEC is a quarter of a circle.
Hence, BC = EC = 3.5 cm
Now, DC = DE + EC = 2 + 3.5 = 5.5 cm
Area of shaded region = Area of the trapezium ABCD − Area of the quadrant BFEC
$= \frac{1}{2} \times (AB + DC) \times BC - \frac{1}{4} \times π {(EC)}^{2} = \frac{1}{2} \times (3.5 + 5.5) \times 3.5 - \frac{1}{4} \times \frac{22}{7} \times {(3.5)}^{2} = 6.125 {cm}^{2}$
Hence, the area of the shaded region is 6.125 cm² .

Question 51:

Since, BFEC is a quarter of a circle.
Hence, BC = EC = 3.5 cm
Now, DC = DE + EC = 2 + 3.5 = 5.5 cm
Area of shaded region = Area of the trapezium ABCD − Area of the quadrant BFEC
$= \frac{1}{2} \times (AB + DC) \times BC - \frac{1}{4} \times π {(EC)}^{2} = \frac{1}{2} \times (3.5 + 5.5) \times 3.5 - \frac{1}{4} \times \frac{22}{7} \times {(3.5)}^{2} = 6.125 {cm}^{2}$
Hence, the area of the shaded region is 6.125 cm² .

Answer:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB
$= \frac{90 °}{360 °} \times π {(OA)}^{2} - \frac{1}{2} \times OA \times OB = \frac{1}{4} \times \frac{22}{7} \times {(35)}^{2} - \frac{1}{2} \times 35 \times 35 = \frac{1}{4} \times \frac{22}{7} \times {(35)}^{2} - \frac{1}{2} \times 35 \times 35 = 962.5 - 612.5 = 350 {cm}^{2}$
Area of major segment APB = Area of circle − Area of minor segment
$= π {(OA)}^{2} - 350 = \frac{22}{7} \times {(35)}^{2} - 350 = 3850 - 350 = 3500 {cm}^{2}$
Hence, the area of major segment is 3500 cm²

Question 1:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB
$= \frac{90 °}{360 °} \times π {(OA)}^{2} - \frac{1}{2} \times OA \times OB = \frac{1}{4} \times \frac{22}{7} \times {(35)}^{2} - \frac{1}{2} \times 35 \times 35 = \frac{1}{4} \times \frac{22}{7} \times {(35)}^{2} - \frac{1}{2} \times 35 \times 35 = 962.5 - 612.5 = 350 {cm}^{2}$
Area of major segment APB = Area of circle − Area of minor segment
$= π {(OA)}^{2} - 350 = \frac{22}{7} \times {(35)}^{2} - 350 = 3850 - 350 = 3500 {cm}^{2}$
Hence, the area of major segment is 3500 cm²

Answer:

(d) 22 cm
Let the radius be r cm.
We know:
Area of a circle $= π r^{2} {cm}^{2}$
Thus, we have:
$π r^{2} = 38.5$
$\Rightarrow \frac{22}{7} \times r^{2} = 38.5 \Rightarrow r^{2} = (38.5 \times \frac{7}{22}) \Rightarrow r^{2} = (\frac{385}{10} \times \frac{7}{22}) \Rightarrow r^{2} = \frac{49}{4} \Rightarrow r = \frac{7}{2}$
Now,
Circumference of the circle $= 2 π r$
$= 2 \times \frac{22}{7} \times r = (2 \times \frac{22}{7} \times \frac{7}{2}) = 22 cm$

Question 2:

(d) 22 cm
Let the radius be r cm.
We know:
Area of a circle $= π r^{2} {cm}^{2}$
Thus, we have:
$π r^{2} = 38.5$
$\Rightarrow \frac{22}{7} \times r^{2} = 38.5 \Rightarrow r^{2} = (38.5 \times \frac{7}{22}) \Rightarrow r^{2} = (\frac{385}{10} \times \frac{7}{22}) \Rightarrow r^{2} = \frac{49}{4} \Rightarrow r = \frac{7}{2}$
Now,
Circumference of the circle $= 2 π r$
$= 2 \times \frac{22}{7} \times r = (2 \times \frac{22}{7} \times \frac{7}{2}) = 22 cm$

Answer:

(b) 14π cm
Let the radius be r cm.
We know:
Area of a circle $= π r^{2}$
Thus, we have:
$π r^{2} = 49 π \Rightarrow r^{2} = 49 \Rightarrow r = \sqrt{49} \Rightarrow r = 7$

Now,
Circumference of the circle $= 2 πr$
$= (2 \times \frac{22}{7} \times 7) cm = 14 π cm$

Question 3:

(b) 14π cm
Let the radius be r cm.
We know:
Area of a circle $= π r^{2}$
Thus, we have:
$π r^{2} = 49 π \Rightarrow r^{2} = 49 \Rightarrow r = \sqrt{49} \Rightarrow r = 7$

Now,
Circumference of the circle $= 2 πr$
$= (2 \times \frac{22}{7} \times 7) cm = 14 π cm$

Answer:

(c) 154 cm²
Let the radius be r cm.
We know:
Circumference of the circle $= 2 πr$
Thus, we have:
$2 πr - r = 37 \Rightarrow r (2 π - 1) = 37 \Rightarrow r (2 \times \frac{22}{7} - 1) = 37 \Rightarrow r (\frac{37}{7}) = 37 \Rightarrow r = (37 \times \frac{7}{37}) \Rightarrow r = 7 cm$

Radius = 7 cm
Now,
Area of the circle $= π r^{2}$
$= (\frac{22}{7} \times 7 \times 7) {cm}^{2} = 154 {cm}^{2}$

Question 4:

(c) 154 cm²
Let the radius be r cm.
We know:
Circumference of the circle $= 2 πr$
Thus, we have:
$2 πr - r = 37 \Rightarrow r (2 π - 1) = 37 \Rightarrow r (2 \times \frac{22}{7} - 1) = 37 \Rightarrow r (\frac{37}{7}) = 37 \Rightarrow r = (37 \times \frac{7}{37}) \Rightarrow r = 7 cm$

Radius = 7 cm
Now,
Area of the circle $= π r^{2}$
$= (\frac{22}{7} \times 7 \times 7) {cm}^{2} = 154 {cm}^{2}$

Answer:

(c) 4658.5 m²
Let the radius be r m.
We know:
Perimeter of a circle $= 2 π r$
Thus, we have:
$2 π r = 242$
$\Rightarrow 2 \times \frac{22}{7} \times r = 242 \Rightarrow \frac{44}{7} \times r = 242 \Rightarrow r = (242 \times \frac{7}{44}) \Rightarrow r = \frac{77}{2}$

∴ Area of the circle $= π r^{2}$
$= (\frac{22}{7} \times \frac{77}{2} \times \frac{77}{2}) m^{2} = 4658.5 m^{2}$

Question 5:

(c) 4658.5 m²
Let the radius be r m.
We know:
Perimeter of a circle $= 2 π r$
Thus, we have:
$2 π r = 242$
$\Rightarrow 2 \times \frac{22}{7} \times r = 242 \Rightarrow \frac{44}{7} \times r = 242 \Rightarrow r = (242 \times \frac{7}{44}) \Rightarrow r = \frac{77}{2}$

∴ Area of the circle $= π r^{2}$
$= (\frac{22}{7} \times \frac{77}{2} \times \frac{77}{2}) m^{2} = 4658.5 m^{2}$

Answer:

(c) 96%
Let d be the original diameter.
Radius $= \frac{d}{2}$
Thus, we have:
Original area $= π \times {(\frac{d}{2})}^{2}$
$= \frac{π d^{2}}{4}$
New diameter $= 140 % of d$
     $= (\frac{140}{100} \times d) = \frac{7 d}{5}$
Now,
New radius $= \frac{7 d}{5 \times 2}$
   $= \frac{7 d}{10}$
New area $= π \times {(\frac{7 d}{10})}^{2}$
    $= \frac{49 π d^{2}}{10}$
Increase in the area $= (\frac{49 π d^{2}}{10} - \frac{π d^{2}}{4})$
$= \frac{24 π d^{2}}{100} = \frac{6 π d^{2}}{25}$
We have:
Increase in the area $= (\frac{6 π d^{2}}{25} \times \frac{4}{π d^{2}} \times 100) %$
= 96%

Question 6:

(c) 96%
Let d be the original diameter.
Radius $= \frac{d}{2}$
Thus, we have:
Original area $= π \times {(\frac{d}{2})}^{2}$
$= \frac{π d^{2}}{4}$
New diameter $= 140 % of d$
     $= (\frac{140}{100} \times d) = \frac{7 d}{5}$
Now,
New radius $= \frac{7 d}{5 \times 2}$
   $= \frac{7 d}{10}$
New area $= π \times {(\frac{7 d}{10})}^{2}$
    $= \frac{49 π d^{2}}{10}$
Increase in the area $= (\frac{49 π d^{2}}{10} - \frac{π d^{2}}{4})$
$= \frac{24 π d^{2}}{100} = \frac{6 π d^{2}}{25}$
We have:
Increase in the area $= (\frac{6 π d^{2}}{25} \times \frac{4}{π d^{2}} \times 100) %$
= 96%

Answer:

(d) None of these
Let r be the original radius.
Thus, we have:
Original area $= π r^{2}$
Also,
New radius $= 70 % of r$
$= (\frac{70}{100} \times r) = \frac{7 r}{10}$
New area $= π \times {(\frac{7 r}{10})}^{2}$
$= \frac{49 π r^{2}}{100}$
Decrease in the area $= (π r^{2} - \frac{49 π r^{2}}{100})$
$= \frac{59 π r^{2}}{100}$
Thus, we have:
Decrease in the area $= (\frac{59 π r^{2}}{100} \times \frac{1}{π r^{2}} \times 100) %$
=51%

Question 7:

(d) None of these
Let r be the original radius.
Thus, we have:
Original area $= π r^{2}$
Also,
New radius $= 70 % of r$
$= (\frac{70}{100} \times r) = \frac{7 r}{10}$
New area $= π \times {(\frac{7 r}{10})}^{2}$
$= \frac{49 π r^{2}}{100}$
Decrease in the area $= (π r^{2} - \frac{49 π r^{2}}{100})$
$= \frac{59 π r^{2}}{100}$
Thus, we have:
Decrease in the area $= (\frac{59 π r^{2}}{100} \times \frac{1}{π r^{2}} \times 100) %$
=51%

Answer:

(d) $\sqrt{π} : 2$
Let a be the side of the square.
We know:
Area of a square $= a^{2}$
Let r be the radius of the circle.
We know:
Area of a circle $= π r^{2}$
Because the area of the square is the same as the area of the circle, we have:
$a^{2} = π r^{2} \Rightarrow \frac{r^{2}}{a^{2}} = \frac{1}{π} \Rightarrow \frac{r}{a} = \frac{1}{\sqrt{π}}$
∴ Ratio of their perimeters $= \frac{2 π r}{4 a} [Because perimeter of the circle is 2 π r and perimeter of the square is 4 a]$

$= \frac{π r}{2 a} = \frac{π}{2} \times \frac{r}{a} = \frac{π}{2} \times \frac{1}{\sqrt{π}} [Since \frac{r}{a} = \frac{1}{\sqrt{π}}] = \frac{\sqrt{π}}{2} = \sqrt{π} : 2$

Question 8:

(d) $\sqrt{π} : 2$
Let a be the side of the square.
We know:
Area of a square $= a^{2}$
Let r be the radius of the circle.
We know:
Area of a circle $= π r^{2}$
Because the area of the square is the same as the area of the circle, we have:
$a^{2} = π r^{2} \Rightarrow \frac{r^{2}}{a^{2}} = \frac{1}{π} \Rightarrow \frac{r}{a} = \frac{1}{\sqrt{π}}$
∴ Ratio of their perimeters $= \frac{2 π r}{4 a} [Because perimeter of the circle is 2 π r and perimeter of the square is 4 a]$

$= \frac{π r}{2 a} = \frac{π}{2} \times \frac{r}{a} = \frac{π}{2} \times \frac{1}{\sqrt{π}} [Since \frac{r}{a} = \frac{1}{\sqrt{π}}] = \frac{\sqrt{π}}{2} = \sqrt{π} : 2$

Answer:

(b) 28 cm
Let r cm be the radius of the new circle.
We know:
Circumference of the new circle = Circumference of the circle with diameter 36 cm + Circumference of the circle with diameter 20 cm
Thus, we have:
$2 π r = 2 π r_{1} + 2 π r_{2}$
$\Rightarrow 2 π r = (2 π \times 18) + (2 π \times 10) \Rightarrow 2 π r = 2 π \times (18 + 10) \Rightarrow 2 π r = (2 π \times 28) \Rightarrow 2 π r = (2 \times \frac{22}{7} \times 28)$
$\Rightarrow 2 π r = 176 \Rightarrow 2 \times \frac{22}{7} \times r = 176 \Rightarrow r = (176 \times \frac{7}{44}) \Rightarrow r = 28 cm$

Question 9:

(b) 28 cm
Let r cm be the radius of the new circle.
We know:
Circumference of the new circle = Circumference of the circle with diameter 36 cm + Circumference of the circle with diameter 20 cm
Thus, we have:
$2 π r = 2 π r_{1} + 2 π r_{2}$
$\Rightarrow 2 π r = (2 π \times 18) + (2 π \times 10) \Rightarrow 2 π r = 2 π \times (18 + 10) \Rightarrow 2 π r = (2 π \times 28) \Rightarrow 2 π r = (2 \times \frac{22}{7} \times 28)$
$\Rightarrow 2 π r = 176 \Rightarrow 2 \times \frac{22}{7} \times r = 176 \Rightarrow r = (176 \times \frac{7}{44}) \Rightarrow r = 28 cm$

Answer:

(c) 50 cm
Let r cm be the radius of the new circle.
Now,
Area of the new circle = Area of the circle with radius 24 cm + Area of the circle with radius 7 cm
Thus, we have:
$π r^{2} = π {r_{1}}^{2} + π {r_{2}}^{2}$
$\Rightarrow π r^{2} = [π \times {(24)}^{2} + π \times {(7)}^{2}] {cm}^{2} \Rightarrow π r^{2} = [π \times 576 + π \times 49] {cm}^{2} \Rightarrow π r^{2} = π \times (576 + 49) {cm}^{2} \Rightarrow r^{2} = 625 π {cm}^{2} \Rightarrow r^{2} = 625 \Rightarrow r = 25$

∴ Diameter of the new circle $= (25 \times 2) cm$
= 50 cm

Question 10:

(c) 50 cm
Let r cm be the radius of the new circle.
Now,
Area of the new circle = Area of the circle with radius 24 cm + Area of the circle with radius 7 cm
Thus, we have:
$π r^{2} = π {r_{1}}^{2} + π {r_{2}}^{2}$
$\Rightarrow π r^{2} = [π \times {(24)}^{2} + π \times {(7)}^{2}] {cm}^{2} \Rightarrow π r^{2} = [π \times 576 + π \times 49] {cm}^{2} \Rightarrow π r^{2} = π \times (576 + 49) {cm}^{2} \Rightarrow r^{2} = 625 π {cm}^{2} \Rightarrow r^{2} = 625 \Rightarrow r = 25$

∴ Diameter of the new circle $= (25 \times 2) cm$
= 50 cm

Answer:

Let the side of the square be a and the radius of the circle be r.
Now, Perimeter of circle = Circumference of the circle
$\Rightarrow 4 a = 2 π r \Rightarrow \frac{a}{r} = \frac{π}{2} Now, \frac{Area of square}{Area of circle} = \frac{a^{2}}{π r^{2}} = \frac{1}{π} \times {(\frac{a}{r})}^{2} = \frac{1}{π} \times \frac{π^{2}}{4} = \frac{π}{4}$
Hence, the correct answer is option (b).

Question 11:

Let the side of the square be a and the radius of the circle be r.
Now, Perimeter of circle = Circumference of the circle
$\Rightarrow 4 a = 2 π r \Rightarrow \frac{a}{r} = \frac{π}{2} Now, \frac{Area of square}{Area of circle} = \frac{a^{2}}{π r^{2}} = \frac{1}{π} \times {(\frac{a}{r})}^{2} = \frac{1}{π} \times \frac{π^{2}}{4} = \frac{π}{4}$
Hence, the correct answer is option (b).

Answer:

(d) $R_{1}^{2} + R_{2}^{2} = R^{2}$
Because the sum of the areas of two circles with radii $R_{1} and R_{2}$ is equal to the area of a circle with radius R, we have:
$π {R_{1}}^{2} + π {R_{2}}^{2} = π R^{2} \Rightarrow π ({R_{1}}^{2} + {R_{2}}^{2}) = π R^{2} \Rightarrow {R_{1}}^{2} + {R_{2}}^{2} = R^{2}$

Question 12:

(d) $R_{1}^{2} + R_{2}^{2} = R^{2}$
Because the sum of the areas of two circles with radii $R_{1} and R_{2}$ is equal to the area of a circle with radius R, we have:
$π {R_{1}}^{2} + π {R_{2}}^{2} = π R^{2} \Rightarrow π ({R_{1}}^{2} + {R_{2}}^{2}) = π R^{2} \Rightarrow {R_{1}}^{2} + {R_{2}}^{2} = R^{2}$

Answer:

(a) $R_{1} + R_{2} = R$
Because the sum of the circumferences of two circles with radii $R_{1} and R_{2}$ is equal to the circumference of a circle with radius R, we have:

$2 π R_{1} + 2 π R_{2} = 2 π R \Rightarrow 2 π (R_{1} + R_{2}) = 2 π R \Rightarrow R_{1} + R_{2} = R$

Question 13:

(a) $R_{1} + R_{2} = R$
Because the sum of the circumferences of two circles with radii $R_{1} and R_{2}$ is equal to the circumference of a circle with radius R, we have:

$2 π R_{1} + 2 π R_{2} = 2 π R \Rightarrow 2 π (R_{1} + R_{2}) = 2 π R \Rightarrow R_{1} + R_{2} = R$

Answer:

(b) Area of the circle > Area of the square
Let r be the radius of the circle.
We know:
Circumference of the circle $= 2 π r$
Now,
Let a be the side of the square.
We know:
Perimeter of the square = 4a
Now,
$2 π r = 4 a \Rightarrow r = \frac{4 a}{2 π}$
∴ Area of the circle $= π r^{2}$
$= π \times {(\frac{4 a}{2 π})}^{2} = π \times \frac{16 a^{2}}{4 π^{2}} = \frac{4 a^{2}}{π} = \frac{4 \times 7 a^{2}}{22} = \frac{14 a^{2}}{11}$

Also,
Area of the square $= a^{2}$
Clearly, $\frac{14 a^{2}}{11} > a$ ².
∴ Area of the circle > Area of the square

Question 14:

(b) Area of the circle > Area of the square
Let r be the radius of the circle.
We know:
Circumference of the circle $= 2 π r$
Now,
Let a be the side of the square.
We know:
Perimeter of the square = 4a
Now,
$2 π r = 4 a \Rightarrow r = \frac{4 a}{2 π}$
∴ Area of the circle $= π r^{2}$
$= π \times {(\frac{4 a}{2 π})}^{2} = π \times \frac{16 a^{2}}{4 π^{2}} = \frac{4 a^{2}}{π} = \frac{4 \times 7 a^{2}}{22} = \frac{14 a^{2}}{11}$

Also,
Area of the square $= a^{2}$
Clearly, $\frac{14 a^{2}}{11} > a$ ².
∴ Area of the circle > Area of the square

Answer:

(b) 330 cm²
Let:
R = 19 cm and r = 16 cm
Thus, we have:
Area of the ring $= π (R^{2} - r^{2})$
$= π (R + r) (R - r) = |\frac{22}{7} \times (19 + 16) \times (19 - 16)| {cm}^{2} = (\frac{22}{7} \times 35 \times 3) {cm}^{2} = 330 {cm}^{2}$

Question 15:

(b) 330 cm²
Let:
R = 19 cm and r = 16 cm
Thus, we have:
Area of the ring $= π (R^{2} - r^{2})$
$= π (R + r) (R - r) = |\frac{22}{7} \times (19 + 16) \times (19 - 16)| {cm}^{2} = (\frac{22}{7} \times 35 \times 3) {cm}^{2} = 330 {cm}^{2}$

Answer:

(b) 3.5 cm
Let r cm and R cm be the radii of two concentric circles.
Thus, we have:
${πR}^{2} = 1386$
$\Rightarrow \frac{22}{7} \times R^{2} = 1386 \Rightarrow R^{2} = (1386 \times \frac{7}{22}) {cm}^{2} \Rightarrow R^{2} = 441 {cm}^{2} \Rightarrow R = 21 cm$

Also,
${πr}^{2} = 962.2 \Rightarrow \frac{22}{7} \times r^{2} = 962.2 \Rightarrow r^{2} = (962.2 \times \frac{7}{22}) {cm}^{2} \Rightarrow r^{2} = (\frac{962.2}{10} \times \frac{7}{22}) {cm}^{2} \Rightarrow r^{2} = \frac{1225}{4} {cm}^{2} \Rightarrow r = \frac{35}{2} cm$
∴ Width of the ring $= (R - r)$
$= (21 - \frac{35}{2}) cm = \frac{7}{2} cm = 3.5 cm$

Question 16:

(b) 3.5 cm
Let r cm and R cm be the radii of two concentric circles.
Thus, we have:
${πR}^{2} = 1386$
$\Rightarrow \frac{22}{7} \times R^{2} = 1386 \Rightarrow R^{2} = (1386 \times \frac{7}{22}) {cm}^{2} \Rightarrow R^{2} = 441 {cm}^{2} \Rightarrow R = 21 cm$

Also,
${πr}^{2} = 962.2 \Rightarrow \frac{22}{7} \times r^{2} = 962.2 \Rightarrow r^{2} = (962.2 \times \frac{7}{22}) {cm}^{2} \Rightarrow r^{2} = (\frac{962.2}{10} \times \frac{7}{22}) {cm}^{2} \Rightarrow r^{2} = \frac{1225}{4} {cm}^{2} \Rightarrow r = \frac{35}{2} cm$
∴ Width of the ring $= (R - r)$
$= (21 - \frac{35}{2}) cm = \frac{7}{2} cm = 3.5 cm$

Answer:

Let the the radii of the two circles be r and R, the circumferences of the circles be c and C and the areas of the two circles be a and A.
Now,
$\frac{c}{C} = \frac{3}{4} \Rightarrow \frac{2 π r}{2 π R} = \frac{3}{4} \Rightarrow \frac{r}{R} = \frac{3}{4}$
Now, the ratio between their areas is given by
$\frac{a}{A} = \frac{π r^{2}}{π R^{2}} = {(\frac{r}{R})}^{2} = {(\frac{3}{4})}^{2} = \frac{9}{16}$
Hence, the correct answer is option (c).

Question 17:

Let the the radii of the two circles be r and R, the circumferences of the circles be c and C and the areas of the two circles be a and A.
Now,
$\frac{c}{C} = \frac{3}{4} \Rightarrow \frac{2 π r}{2 π R} = \frac{3}{4} \Rightarrow \frac{r}{R} = \frac{3}{4}$
Now, the ratio between their areas is given by
$\frac{a}{A} = \frac{π r^{2}}{π R^{2}} = {(\frac{r}{R})}^{2} = {(\frac{3}{4})}^{2} = \frac{9}{16}$
Hence, the correct answer is option (c).

Answer:

Let the the radii of the two circles be r and R, the circumferences of the circles be c and C and the areas of the two circles be a and A.
Now,
$\frac{a}{A} = \frac{9}{4} \Rightarrow \frac{π r^{2}}{π R^{2}} = {(\frac{3}{2})}^{2} \Rightarrow \frac{r}{R} = \frac{3}{2}$
Now, the ratio between their circumferences is given by
$\frac{c}{C} = \frac{2 π r}{2 π R} = \frac{r}{R} = \frac{3}{2}$
Hence, the correct answer is option (a)

Question 18:

Let the the radii of the two circles be r and R, the circumferences of the circles be c and C and the areas of the two circles be a and A.
Now,
$\frac{a}{A} = \frac{9}{4} \Rightarrow \frac{π r^{2}}{π R^{2}} = {(\frac{3}{2})}^{2} \Rightarrow \frac{r}{R} = \frac{3}{2}$
Now, the ratio between their circumferences is given by
$\frac{c}{C} = \frac{2 π r}{2 π R} = \frac{r}{R} = \frac{3}{2}$
Hence, the correct answer is option (a)

Answer:

(d) 7000
Distance covered in 1 revolution $= 2 π r$
$= (2 \times \frac{22}{7} \times 0.25) m = (2 \times \frac{22}{7} \times \frac{25}{100}) m = \frac{11}{7} m$
Number of revolutions taken to cover 11 km $= (11 \times 1000 \times \frac{7}{11})$
= 7000

Question 19:

(d) 7000
Distance covered in 1 revolution $= 2 π r$
$= (2 \times \frac{22}{7} \times 0.25) m = (2 \times \frac{22}{7} \times \frac{25}{100}) m = \frac{11}{7} m$
Number of revolutions taken to cover 11 km $= (11 \times 1000 \times \frac{7}{11})$
= 7000

Answer:

(a) 140
Distance covered by the wheel in 1 revolution $= π d$
$= (\frac{22}{7} \times 40) cm = \frac{880}{7} cm = \frac{880}{7 \times 100} m$

Number of revolutions required to cover 176 m $= (\frac{176}{\frac{880}{7 \times 100}})$
$= (176 \times 100 \times \frac{7}{880})$
=140

Question 20:

(a) 140
Distance covered by the wheel in 1 revolution $= π d$
$= (\frac{22}{7} \times 40) cm = \frac{880}{7} cm = \frac{880}{7 \times 100} m$

Number of revolutions required to cover 176 m $= (\frac{176}{\frac{880}{7 \times 100}})$
$= (176 \times 100 \times \frac{7}{880})$
=140

Answer:

(c) 28 m
Distance covered by the wheel in 1 revolution $= (\frac{88 \times 1000}{1000}) m$
= 88 m
We have:
Circumference of the wheel = 88 m
Now, let the diameter of the wheel be d m.
Thus, we have:

$π d = 88 \Rightarrow \frac{22}{7} \times d = 88 \Rightarrow d = (88 \times \frac{7}{22}) \Rightarrow d = 28 m$

Question 21:

(c) 28 m
Distance covered by the wheel in 1 revolution $= (\frac{88 \times 1000}{1000}) m$
= 88 m
We have:
Circumference of the wheel = 88 m
Now, let the diameter of the wheel be d m.
Thus, we have:

$π d = 88 \Rightarrow \frac{22}{7} \times d = 88 \Rightarrow d = (88 \times \frac{7}{22}) \Rightarrow d = 28 m$

Answer:

(d) $\frac{π R^{2} θ}{360}$

Question 22:

(d) $\frac{π R^{2} θ}{360}$

Answer:

(b) $\frac{2 π R θ}{360}$

Question 23:

(b) $\frac{2 π R θ}{360}$

Answer:

Angle subtends by the minute hand in 1 minute = 6^âˆ˜
∴ Angle subtends by the minute hand in 10 minutes = 60^âˆ˜
Now,
Area of the sector $= \frac{θ}{360 °} π r^{2} = \frac{60 °}{360 °} \times \frac{22}{7} {(21)}^{2} = 231 {cm}^{2}$
Hence, the correct answer is option (a).

Question 24:

Angle subtends by the minute hand in 1 minute = 6^âˆ˜
∴ Angle subtends by the minute hand in 10 minutes = 60^âˆ˜
Now,
Area of the sector $= \frac{θ}{360 °} π r^{2} = \frac{60 °}{360 °} \times \frac{22}{7} {(21)}^{2} = 231 {cm}^{2}$
Hence, the correct answer is option (a).

Answer:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB
$= \frac{90 °}{360 °} π {(OA)}^{2} - \frac{1}{2} \times OA \times OB = \frac{1}{4} \times 3.14 \times {(10)}^{2} - \frac{1}{2} \times 10 \times 10 = 78.5 - 50 = 28.5 {cm}^{2}$
Hence, the correct answer is option (c).

Question 25:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB
$= \frac{90 °}{360 °} π {(OA)}^{2} - \frac{1}{2} \times OA \times OB = \frac{1}{4} \times 3.14 \times {(10)}^{2} - \frac{1}{2} \times 10 \times 10 = 78.5 - 50 = 28.5 {cm}^{2}$
Hence, the correct answer is option (c).

Answer:

We have $r = 21 cm and θ = 60 °$
Length of arc = $\frac{θ}{360^{°}} \times 2 π r = \frac{60 °}{360 °} \times 2 \times \frac{22}{7} \times 21 = 22 cm$
Hence, the correct answer is option (b)

Question 26:

We have $r = 21 cm and θ = 60 °$
Length of arc = $\frac{θ}{360^{°}} \times 2 π r = \frac{60 °}{360 °} \times 2 \times \frac{22}{7} \times 21 = 22 cm$
Hence, the correct answer is option (b)

Answer:

Radius of the circle, r = 14 cm
Draw a perpendicular OD to chord AB. It will bisect AB.
∠A = 180° − (90° + 60°) = 30°

$\cos 30 ° = \frac{AD}{OA} \Rightarrow \frac{\sqrt{3}}{2} = \frac{AD}{14} \Rightarrow AD = 7 \sqrt{3} \Rightarrow AB = 2 \times AD = 14 \sqrt{3} cm \sin 30 ° = \frac{OD}{OA} \Rightarrow \frac{1}{2} = \frac{OD}{14} \Rightarrow OD = 7 cm$

Area of minor segment = Area of sector OAPB − Area of triangle AOB
$= \frac{θ}{360 °} π {(OA)}^{2} - \frac{1}{2} \times OD \times AB = \frac{120 °}{360 °} \times \frac{22}{7} {(14)}^{2} - \frac{1}{2} \times 7 \times 14 \sqrt{3} = 205.33 - 84.77 = 120.56 {cm}^{2}$
Hence, the correct answer is option (a).

Question 1:

Radius of the circle, r = 14 cm
Draw a perpendicular OD to chord AB. It will bisect AB.
∠A = 180° − (90° + 60°) = 30°

$\cos 30 ° = \frac{AD}{OA} \Rightarrow \frac{\sqrt{3}}{2} = \frac{AD}{14} \Rightarrow AD = 7 \sqrt{3} \Rightarrow AB = 2 \times AD = 14 \sqrt{3} cm \sin 30 ° = \frac{OD}{OA} \Rightarrow \frac{1}{2} = \frac{OD}{14} \Rightarrow OD = 7 cm$

Area of minor segment = Area of sector OAPB − Area of triangle AOB
$= \frac{θ}{360 °} π {(OA)}^{2} - \frac{1}{2} \times OD \times AB = \frac{120 °}{360 °} \times \frac{22}{7} {(14)}^{2} - \frac{1}{2} \times 7 \times 14 \sqrt{3} = 205.33 - 84.77 = 120.56 {cm}^{2}$
Hence, the correct answer is option (a).

Answer:

(b) 228 cm²
Join OB.
Now, OB is the radius of the circle.
$We have : O B^{2} = O A^{2} + A B^{2} [By Pythagoras' theorem] \Rightarrow O B^{2} = \{{(20)}^{2} + {(20)}^{2}\} {cm}^{2}$

$\Rightarrow O B^{2} = (400 + 400) {cm}^{2} \Rightarrow O B^{2} = 800 {cm}^{2} \Rightarrow O B = 20 \sqrt{2} cm$

Hence, the radius of the circle is $20 \sqrt{2} cm$ .
Now,
Area of the shaded region = Area of the quadrant $-$ Area of the square OABC
$= |(\frac{1}{4} \times 3.14 \times 20 \sqrt{2} \times 20 \sqrt{2}) - (20 \times 20)| {cm}^{2} = |(\frac{1}{4} \times \frac{314}{100} \times 800) - 400| {cm}^{2} = (628 - 400) {cm}^{2} = 228 {cm}^{2}$

Question 2:

(b) 228 cm²
Join OB.
Now, OB is the radius of the circle.
$We have : O B^{2} = O A^{2} + A B^{2} [By Pythagoras' theorem] \Rightarrow O B^{2} = \{{(20)}^{2} + {(20)}^{2}\} {cm}^{2}$

$\Rightarrow O B^{2} = (400 + 400) {cm}^{2} \Rightarrow O B^{2} = 800 {cm}^{2} \Rightarrow O B = 20 \sqrt{2} cm$

Hence, the radius of the circle is $20 \sqrt{2} cm$ .
Now,
Area of the shaded region = Area of the quadrant $-$ Area of the square OABC
$= |(\frac{1}{4} \times 3.14 \times 20 \sqrt{2} \times 20 \sqrt{2}) - (20 \times 20)| {cm}^{2} = |(\frac{1}{4} \times \frac{314}{100} \times 800) - 400| {cm}^{2} = (628 - 400) {cm}^{2} = 228 {cm}^{2}$

Answer:

(c) 300

Let d cm be the diameter of the wheel.
We know:
Circumference of the wheel $= π \times d$
$= (\frac{22}{7} \times 84) cm = 264 cm$

Now,
Number of revolutions to cover 792 m $= (\frac{792 \times 1000}{264})$
=300

Question 3:

(c) 300

Let d cm be the diameter of the wheel.
We know:
Circumference of the wheel $= π \times d$
$= (\frac{22}{7} \times 84) cm = 264 cm$

Now,
Number of revolutions to cover 792 m $= (\frac{792 \times 1000}{264})$
=300

Answer:

(d) $\frac{x}{360} \times π r^{2}$

The area of a sector of a circle with radius r making an angle of $x °$ at the centre is $\frac{x}{360} \times π r^{2}$ .

Question 4:

(d) $\frac{x}{360} \times π r^{2}$

The area of a sector of a circle with radius r making an angle of $x °$ at the centre is $\frac{x}{360} \times π r^{2}$ .

Answer:

All options are incorrect; the correct answer is 30.5 cm.

Join AC.
Now, AC is the diameter of the circle.
$We have : A C^{2} = A B^{2} + B C^{2} [By Pythagoras' theorem] \Rightarrow A C^{2} = \{{(8)}^{2} + {(6)}^{2}\} {cm}^{2} \Rightarrow A C^{2} = (64 + 36) {cm}^{2} \Rightarrow A C^{2} = 100 {cm}^{2} \Rightarrow A C = 10 cm$
∴ Radius of the circle $= \frac{10}{2} cm$
=5 cm
Now,
Area of the shaded region = Area of the circle with radius 5 cm $-$ Area of the rectangle ABCD
$= |(3.14 \times 5 \times 5) - (8 \times 6)| {cm}^{2} = |(\frac{314}{100} \times 25) - 48| {cm}^{2} = (\frac{157}{2} - 48) {cm}^{2} = \frac{61}{2} {cm}^{2} = 30.5 {cm}^{2}$

Question 5:

All options are incorrect; the correct answer is 30.5 cm.

Join AC.
Now, AC is the diameter of the circle.
$We have : A C^{2} = A B^{2} + B C^{2} [By Pythagoras' theorem] \Rightarrow A C^{2} = \{{(8)}^{2} + {(6)}^{2}\} {cm}^{2} \Rightarrow A C^{2} = (64 + 36) {cm}^{2} \Rightarrow A C^{2} = 100 {cm}^{2} \Rightarrow A C = 10 cm$
∴ Radius of the circle $= \frac{10}{2} cm$
=5 cm
Now,
Area of the shaded region = Area of the circle with radius 5 cm $-$ Area of the rectangle ABCD
$= |(3.14 \times 5 \times 5) - (8 \times 6)| {cm}^{2} = |(\frac{314}{100} \times 25) - 48| {cm}^{2} = (\frac{157}{2} - 48) {cm}^{2} = \frac{61}{2} {cm}^{2} = 30.5 {cm}^{2}$

Answer:

Let r cm be the radius of the circle.
Now,
Circumference of the circle:
$2 π r = 22 \Rightarrow 2 \times \frac{22}{7} \times r = 22 \Rightarrow r = (22 \times \frac{7}{44}) cm \Rightarrow r = \frac{7}{2}$

Also,
Area of the circle $= π r^{2}$
$= (\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}) {cm}^{2} = \frac{77}{2} {cm}^{2} = 38.5 {cm}^{2}$

Question 6:

Let r cm be the radius of the circle.
Now,
Circumference of the circle:
$2 π r = 22 \Rightarrow 2 \times \frac{22}{7} \times r = 22 \Rightarrow r = (22 \times \frac{7}{44}) cm \Rightarrow r = \frac{7}{2}$

Also,
Area of the circle $= π r^{2}$
$= (\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}) {cm}^{2} = \frac{77}{2} {cm}^{2} = 38.5 {cm}^{2}$

Answer:

Let ACB be the given arc subtending at an angle of $60 °$ at the centre.
Now, we have:
$r = 21 cm and θ = 60 °$
∴ Length of the arc ACB $= \frac{2 π r}{360}$
$= (2 \times \frac{22}{7} \times 21 \times \frac{60}{360}) = 22 cm$

Question 7:

Let ACB be the given arc subtending at an angle of $60 °$ at the centre.
Now, we have:
$r = 21 cm and θ = 60 °$
∴ Length of the arc ACB $= \frac{2 π r}{360}$
$= (2 \times \frac{22}{7} \times 21 \times \frac{60}{360}) = 22 cm$

Answer:

Angle described by the minute hand in 60 minutes $= 360 °$
Angle described by the minute hand in 35 minutes $= (\frac{360}{60} \times 35) °$
$= 210 °$
Now,
$r = 12 cm and θ = 210 °$
∴ Required area swept by the minute hand in 35 minutes = Area of the sector with $r = 12 cm and θ = 210 °$
$= \frac{{πr}^{2} θ}{360} = (\frac{22}{7} \times 12 \times 12 \times \frac{210}{360}) {cm}^{2} = 264 {cm}^{2}$

Question 8:

Angle described by the minute hand in 60 minutes $= 360 °$
Angle described by the minute hand in 35 minutes $= (\frac{360}{60} \times 35) °$
$= 210 °$
Now,
$r = 12 cm and θ = 210 °$
∴ Required area swept by the minute hand in 35 minutes = Area of the sector with $r = 12 cm and θ = 210 °$
$= \frac{{πr}^{2} θ}{360} = (\frac{22}{7} \times 12 \times 12 \times \frac{210}{360}) {cm}^{2} = 264 {cm}^{2}$

Answer:

Let O be the centre of the circle with radius 5.6 cm and OACB be its sector with perimeter 27.2 cm.
Thus, we have:
$O A + O B + a r c A B = 27.2 \Rightarrow 5.6 + 5.6 + a r c A B = 27.2 \Rightarrow a r c A B = 16 cm$
Now,
Area of the sector OACBO $= (\frac{1}{2} \times Radius \times l) square units$
$= (\frac{1}{2} 5.6 \times 16) {cm}^{2} = 44.8 {cm}^{2}$

Question 9:

Let O be the centre of the circle with radius 5.6 cm and OACB be its sector with perimeter 27.2 cm.
Thus, we have:
$O A + O B + a r c A B = 27.2 \Rightarrow 5.6 + 5.6 + a r c A B = 27.2 \Rightarrow a r c A B = 16 cm$
Now,
Area of the sector OACBO $= (\frac{1}{2} \times Radius \times l) square units$
$= (\frac{1}{2} 5.6 \times 16) {cm}^{2} = 44.8 {cm}^{2}$

Answer:

Let r cm be the radius of the circle and $θ$ be the angle.
We have:
$r = 14 cm and θ = 90 °$
Area of the sector $= \frac{π r^{2} θ}{360}$
$= (\frac{22}{7} \times 14 \times 14 \times \frac{90}{360}) {cm}^{2} = 154 {cm}^{2}$

Question 10:

Let r cm be the radius of the circle and $θ$ be the angle.
We have:
$r = 14 cm and θ = 90 °$
Area of the sector $= \frac{π r^{2} θ}{360}$
$= (\frac{22}{7} \times 14 \times 14 \times \frac{90}{360}) {cm}^{2} = 154 {cm}^{2}$

Answer:

Area of the shaded region = (Area of the sector with $r = 7 cm and θ = 30 °$ ) $-$ (Area of the sector with $r = 3.5 cm and θ = 30 °$ )
$= |(\frac{22}{7} \times 7 \times 7 \times \frac{30}{360}) - (\frac{22}{7} \times 3.5 \times 3.5 \times \frac{30}{360})| {cm}^{2} = (\frac{77}{6} - \frac{77}{24}) {cm}^{2} = \frac{77}{8} {cm}^{2} = 9.625 {cm}^{2}$

Question 11:

Area of the shaded region = (Area of the sector with $r = 7 cm and θ = 30 °$ ) $-$ (Area of the sector with $r = 3.5 cm and θ = 30 °$ )
$= |(\frac{22}{7} \times 7 \times 7 \times \frac{30}{360}) - (\frac{22}{7} \times 3.5 \times 3.5 \times \frac{30}{360})| {cm}^{2} = (\frac{77}{6} - \frac{77}{24}) {cm}^{2} = \frac{77}{8} {cm}^{2} = 9.625 {cm}^{2}$

Answer:

Let a cm be the side of the equilateral triangle.
Now,
Area of the equilateral triangle $= \frac{\sqrt{3}}{4} a^{2}$
We have:
$\frac{\sqrt{3}}{4} a^{2} = 121 \sqrt{3} \Rightarrow \frac{a^{2}}{4} = 121 \Rightarrow a^{2} = 484 \Rightarrow a = 22$

Perimeter of the triangle = Circumference of the circle
Perimeter of the triangle = (22 + 22 + 22) cm
= 66 cm
Now, let r cm be the radius of the circle.
We know:
Circumference of the circle $= 2 π r$
$\Rightarrow 2 \times \frac{22}{7} \times r = 66 \Rightarrow r = (66 \times \frac{7}{44}) cm \Rightarrow r = \frac{21}{2} cm$
Also,
Area of the circle $= π r^{2}$
$= (\frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}) {cm}^{2} = \frac{693}{2} {cm}^{2} = 346.5 {cm}^{2}$

Question 12:

Let a cm be the side of the equilateral triangle.
Now,
Area of the equilateral triangle $= \frac{\sqrt{3}}{4} a^{2}$
We have:
$\frac{\sqrt{3}}{4} a^{2} = 121 \sqrt{3} \Rightarrow \frac{a^{2}}{4} = 121 \Rightarrow a^{2} = 484 \Rightarrow a = 22$

Perimeter of the triangle = Circumference of the circle
Perimeter of the triangle = (22 + 22 + 22) cm
= 66 cm
Now, let r cm be the radius of the circle.
We know:
Circumference of the circle $= 2 π r$
$\Rightarrow 2 \times \frac{22}{7} \times r = 66 \Rightarrow r = (66 \times \frac{7}{44}) cm \Rightarrow r = \frac{21}{2} cm$
Also,
Area of the circle $= π r^{2}$
$= (\frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}) {cm}^{2} = \frac{693}{2} {cm}^{2} = 346.5 {cm}^{2}$

Answer:

Distance covered in 1 revolution $= π \times d$
      $= (\frac{22}{7} \times 84) cm = 264 cm$
Distance covered in 1 second $= (5 \times 264) cm$
     = 1320 cm
Distance covered in 1 hour $= (60 \times 60 \times 1320) cm$

   $= 4752000 cm = (\frac{4752000}{1000 \times 100}) km = 47.52 km$

Question 13:

Distance covered in 1 revolution $= π \times d$
      $= (\frac{22}{7} \times 84) cm = 264 cm$
Distance covered in 1 second $= (5 \times 264) cm$
     = 1320 cm
Distance covered in 1 hour $= (60 \times 60 \times 1320) cm$

   $= 4752000 cm = (\frac{4752000}{1000 \times 100}) km = 47.52 km$

Answer:

(i) Area of the quadrant OACB $= (\frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5) {cm}^{2}$
$= (\frac{1}{4} \times \frac{22}{7} \times \frac{35}{10} \times \frac{35}{10}) {cm}^{2} = \frac{77}{8} {cm}^{2} = 9.625 {cm}^{2}$

(ii) Area of the shaded region = Area of the quadrant OACB $-$ Area of $∆ A O D$
$= |(\frac{77}{8}) - (\frac{1}{2} \times 3.5 \times 2)| {cm}^{2} = (\frac{77}{8} - \frac{35}{10}) {cm}^{2} = \frac{49}{8} {cm}^{2} = 6.125 {cm}^{2}$

Question 14:

(i) Area of the quadrant OACB $= (\frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5) {cm}^{2}$
$= (\frac{1}{4} \times \frac{22}{7} \times \frac{35}{10} \times \frac{35}{10}) {cm}^{2} = \frac{77}{8} {cm}^{2} = 9.625 {cm}^{2}$

(ii) Area of the shaded region = Area of the quadrant OACB $-$ Area of $∆ A O D$
$= |(\frac{77}{8}) - (\frac{1}{2} \times 3.5 \times 2)| {cm}^{2} = (\frac{77}{8} - \frac{35}{10}) {cm}^{2} = \frac{49}{8} {cm}^{2} = 6.125 {cm}^{2}$

Answer:

Let r be the radius of the circle.
Thus, we have:
$r = \frac{28}{2} cm$
=14 cm
Now,
Area of the shaded region = (Area of the square ABCD) $-$ 4(Area of the sector where $r = 14 cm and θ = 90 °$ )
$= |(28 \times 28) - 4 (\frac{22}{7} \times 14 \times 14 \times \frac{90}{360})| {cm}^{2} = |784 - 4 (154)| {cm}^{2} = (784 - 616) {cm}^{2} = 168 {cm}^{2}$

Question 15:

Let r be the radius of the circle.
Thus, we have:
$r = \frac{28}{2} cm$
=14 cm
Now,
Area of the shaded region = (Area of the square ABCD) $-$ 4(Area of the sector where $r = 14 cm and θ = 90 °$ )
$= |(28 \times 28) - 4 (\frac{22}{7} \times 14 \times 14 \times \frac{90}{360})| {cm}^{2} = |784 - 4 (154)| {cm}^{2} = (784 - 616) {cm}^{2} = 168 {cm}^{2}$

Answer:

Draw $O D ⊥ B C$ .
Because $∆ A B C$ is equilateral, $∠ A = ∠ B = ∠ C = 60 °$ .
Thus, we have:

$∠ O B D = 30 ° \Rightarrow \frac{O D}{O B} = \sin 30 ° \Rightarrow \frac{O D}{O B} = \frac{1}{2} \Rightarrow O D = (\frac{1}{2} \times 4) cm [∵ O B = radius] \Rightarrow O D = 2 cm$

${BD}^{2} = ({OB}^{2} - {OD}^{2}) [B y Pythagoras' theorem] \Rightarrow {BD}^{2} = (4^{2} - 2^{2}) {cm}^{2} \Rightarrow {BD}^{2} = (16 - 4) {cm}^{2} \Rightarrow {BD}^{2} = 12 {cm}^{2} \Rightarrow BD = 2 \sqrt{3} cm$

Also,
$BC = 2 \times BD = (2 \times 2 \sqrt{3}) cm = 4 \sqrt{3} cm$
∴ Area of the shaded region = (Area of the circle) $-$ (Area of $∆ A B C$ )
$= |(3.14 \times 4 \times 4) - (\frac{\sqrt{3}}{4} \times 4 \sqrt{3} \times 4 \sqrt{3})| {cm}^{2} = |50.24 - (12 \times 1.73)| {cm}^{2} = (50.24 - 20.76) {cm}^{2} = 29.48 {cm}^{2}$

Question 16:

Draw $O D ⊥ B C$ .
Because $∆ A B C$ is equilateral, $∠ A = ∠ B = ∠ C = 60 °$ .
Thus, we have:

$∠ O B D = 30 ° \Rightarrow \frac{O D}{O B} = \sin 30 ° \Rightarrow \frac{O D}{O B} = \frac{1}{2} \Rightarrow O D = (\frac{1}{2} \times 4) cm [∵ O B = radius] \Rightarrow O D = 2 cm$

${BD}^{2} = ({OB}^{2} - {OD}^{2}) [B y Pythagoras' theorem] \Rightarrow {BD}^{2} = (4^{2} - 2^{2}) {cm}^{2} \Rightarrow {BD}^{2} = (16 - 4) {cm}^{2} \Rightarrow {BD}^{2} = 12 {cm}^{2} \Rightarrow BD = 2 \sqrt{3} cm$

Also,
$BC = 2 \times BD = (2 \times 2 \sqrt{3}) cm = 4 \sqrt{3} cm$
∴ Area of the shaded region = (Area of the circle) $-$ (Area of $∆ A B C$ )
$= |(3.14 \times 4 \times 4) - (\frac{\sqrt{3}}{4} \times 4 \sqrt{3} \times 4 \sqrt{3})| {cm}^{2} = |50.24 - (12 \times 1.73)| {cm}^{2} = (50.24 - 20.76) {cm}^{2} = 29.48 {cm}^{2}$

Answer:

The length of minute hand of clock = 7.5 cm.
The angle made by minute hand in 60 minutes = $360 ° .$
The angle made by minute hand in 1 minute = $6 ° .$
The angle made by minute hand in 56 minutes = $56 \times 6 ° = 336 ° .$
So, the area of clock described by minute hand in 56 minutes = area of sector with angle $336 ° = π r^{2} \times \frac{θ}{360 °} = \frac{22}{7} \times 7.5 \times 7.5 \times \frac{336 °}{360 °} = 165 {cm}^{2} .$

Question 17:

The length of minute hand of clock = 7.5 cm.
The angle made by minute hand in 60 minutes = $360 ° .$
The angle made by minute hand in 1 minute = $6 ° .$
The angle made by minute hand in 56 minutes = $56 \times 6 ° = 336 ° .$
So, the area of clock described by minute hand in 56 minutes = area of sector with angle $336 ° = π r^{2} \times \frac{θ}{360 °} = \frac{22}{7} \times 7.5 \times 7.5 \times \frac{336 °}{360 °} = 165 {cm}^{2} .$

Answer:

Let r m and R m be the inner and outer boundaries, respectively.
Thus, we have:

$2 π r = 352 \Rightarrow r = \frac{352}{2 π} Also, 2 πR = 396 \Rightarrow R = \frac{396}{2 π}$

Width of the track $= (R - r)$
$= (\frac{396}{2 π} - \frac{352}{2 π}) m = \frac{1}{2 π} (396 - 352) m = (\frac{1}{2} \times \frac{7}{22} \times 44) m = 7 m$

Area of the track $= π (R^{2} - r^{2})$
$= π (R + r) (R - r) = [π (\frac{396}{2 π} + \frac{352}{2 π}) \times (\frac{396}{2 π} - \frac{352}{2 π})] m^{2} = (π \times \frac{748}{2 π} \times 7) m^{2} = \frac{748}{2} \times 7 m^{2} = 2618 m^{2}$

Question 18:

Let r m and R m be the inner and outer boundaries, respectively.
Thus, we have:

$2 π r = 352 \Rightarrow r = \frac{352}{2 π} Also, 2 πR = 396 \Rightarrow R = \frac{396}{2 π}$

Width of the track $= (R - r)$
$= (\frac{396}{2 π} - \frac{352}{2 π}) m = \frac{1}{2 π} (396 - 352) m = (\frac{1}{2} \times \frac{7}{22} \times 44) m = 7 m$

Area of the track $= π (R^{2} - r^{2})$
$= π (R + r) (R - r) = [π (\frac{396}{2 π} + \frac{352}{2 π}) \times (\frac{396}{2 π} - \frac{352}{2 π})] m^{2} = (π \times \frac{748}{2 π} \times 7) m^{2} = \frac{748}{2} \times 7 m^{2} = 2618 m^{2}$

Answer:

Let AB be the chord of a circle with centre O and radius 30 cm such that $∠ A O B = 60 °$ .
Area of the sector OACBO $= \frac{π r^{2} θ}{360}$
     $= (3.14 \times 30 \times 30 \times \frac{60}{360}) {cm}^{2} = 471 {cm}^{2}$
Area of $∆ O A B$ $= \frac{1}{2} r^{2} \sin θ$
   $= (\frac{1}{2} \times 30 \times 30 \times \sin 60 °) {cm}^{2} = (225 \times 1.732) {cm}^{2} = 389.7 {cm}^{2}$

Area of the minor segment = (Area of the sector OACBO) $-$ (Area of $∆ O A B$ )
            $= (471 - 389.7) {cm}^{2} = 81.3 {cm}^{2}$

Area of the major segment = (Area of the circle) $-$ (Area of the minor segment)
            $= |(3.14 \times 30 \times 30) - 81.3| {cm}^{2} = 2744.7 {cm}^{2}$

Question 19:

Let AB be the chord of a circle with centre O and radius 30 cm such that $∠ A O B = 60 °$ .
Area of the sector OACBO $= \frac{π r^{2} θ}{360}$
     $= (3.14 \times 30 \times 30 \times \frac{60}{360}) {cm}^{2} = 471 {cm}^{2}$
Area of $∆ O A B$ $= \frac{1}{2} r^{2} \sin θ$
   $= (\frac{1}{2} \times 30 \times 30 \times \sin 60 °) {cm}^{2} = (225 \times 1.732) {cm}^{2} = 389.7 {cm}^{2}$

Area of the minor segment = (Area of the sector OACBO) $-$ (Area of $∆ O A B$ )
            $= (471 - 389.7) {cm}^{2} = 81.3 {cm}^{2}$

Area of the major segment = (Area of the circle) $-$ (Area of the minor segment)
            $= |(3.14 \times 30 \times 30) - 81.3| {cm}^{2} = 2744.7 {cm}^{2}$

Answer:

Let r be the radius of the circle.
Thus, we have:
$r = \frac{50}{2} m$
= 25 m
Area left ungrazed = (Area of the square) $-$ 4(Area of the sector where $r = 25 m and θ = 90 °$ )

$= |(50 \times 50) - 4 (3.14 \times 25 \times 25 \times \frac{90}{360})| m^{2} = |2500 - (4 \times \frac{1962.5}{4})| m^{2} = (2500 - 1962.5) m^{2} = 537.5 m^{2}$

Question 20:

Let r be the radius of the circle.
Thus, we have:
$r = \frac{50}{2} m$
= 25 m
Area left ungrazed = (Area of the square) $-$ 4(Area of the sector where $r = 25 m and θ = 90 °$ )

$= |(50 \times 50) - 4 (3.14 \times 25 \times 25 \times \frac{90}{360})| m^{2} = |2500 - (4 \times \frac{1962.5}{4})| m^{2} = (2500 - 1962.5) m^{2} = 537.5 m^{2}$

Answer:

Let a m be the side of the square.
Area of the square $= a^{2}$
Thus, we have:
$a^{2} = 1600 \Rightarrow a = 40$
Area of the plots = 4(Area of the semicircle of radius 20 m)
$= |4 (\frac{1}{2} π r^{2})| m^{2} = |4 (\frac{1}{2} \times 3.14 \times 20 \times 20)| m^{2} = 2512 m^{2}$

∴ Cost of turfing the plots at $12.50 {per m}^{2} = Rs (2512 \times 12.50)$
= Rs 31400

RS Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 16 - Area Of Circle, Sector And Segment

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