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#### Question 1:

Circumference = 39.6 cm
We know:
Circumference of a circle = $2\mathrm{\pi }r$

Also,
Area of the circle = $\pi {r}^{2}$

#### Question 2:

Circumference = 39.6 cm
We know:
Circumference of a circle = $2\mathrm{\pi }r$

Also,
Area of the circle = $\pi {r}^{2}$

Let the radius of the circle be r.
â€‹Now,
$\mathrm{Area}=98.56\phantom{\rule{0ex}{0ex}}⇒\mathrm{\pi }{r}^{2}=98.56\phantom{\rule{0ex}{0ex}}⇒\frac{22}{7}×{r}^{2}=98.56\phantom{\rule{0ex}{0ex}}⇒r=5.6$
Now,
Circumference =
Hence, the circumference of the circle is 35.2 cm.

#### Question 3:

Let the radius of the circle be r.
â€‹Now,
$\mathrm{Area}=98.56\phantom{\rule{0ex}{0ex}}⇒\mathrm{\pi }{r}^{2}=98.56\phantom{\rule{0ex}{0ex}}⇒\frac{22}{7}×{r}^{2}=98.56\phantom{\rule{0ex}{0ex}}⇒r=5.6$
Now,
Circumference =
Hence, the circumference of the circle is 35.2 cm.

Let the radius of the circle be r.
Now,
Circumference = Diameter + 45

∴ 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

∴ Circumference = Diameter + 45 = 2(10.5) + 45 = 66 cm
Hence, the circumference of the circle is 66 cm.

Area of the circle = 484 cm2
Area of the square = ${\mathrm{Side}}^{2}$

Perimeter of the square = $4×\mathrm{Side}$
Perimeter of the square = $4×22$
= 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€‹

Area of the circle = ${\mathrm{\pi r}}^{2}$

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

#### Question 5:

Area of the circle = 484 cm2
Area of the square = ${\mathrm{Side}}^{2}$

Perimeter of the square = $4×\mathrm{Side}$
Perimeter of the square = $4×22$
= 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€‹

Area of the circle = ${\mathrm{\pi r}}^{2}$

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

Length of the wire
Now, let the radius of the circle be r cm.
We know:
Circumference of the circle = Length of the wire

Thus, we have:
Area of the circle =$\mathrm{\pi }{r}^{\mathit{2}}$

Area enclosed by the circle = 346.5 cm2

#### Question 6:

Length of the wire
Now, let the radius of the circle be r cm.
We know:
Circumference of the circle = Length of the wire

Thus, we have:
Area of the circle =$\mathrm{\pi }{r}^{\mathit{2}}$

Area enclosed by the circle = 346.5 cm2

Let the radius of the park be r.
Length of chain = Perimeter of the semicircular park
⇒ 108 = Length of the arc + Diameter

Now, Area of park
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

Now, Area of park
Hence, the area of the park is 693 m2 .

Let the radii of the two circles be r1 cm and r2cm.
Now,
Sum of the radii of the two circles = 7 cm

Difference of the circumferences of the two circles = 88 cm

Adding (i) and (ii), we get:
$2{r}_{1}=\frac{91}{11}\phantom{\rule{0ex}{0ex}}{r}_{1}=\frac{91}{22}$

∴ Circumference of the first circle = $2{\mathrm{\pi r}}_{1}$

Also,
${r}_{1}-{r}_{2}=\frac{14}{11}\phantom{\rule{0ex}{0ex}}\frac{91}{22}-{r}_{2}=\frac{14}{11}\phantom{\rule{0ex}{0ex}}\frac{91}{22}-\frac{14}{11}={r}_{2}\phantom{\rule{0ex}{0ex}}{r}_{2}=\frac{63}{22}$

∴ Circumference of the second circle = $2{\mathrm{\pi r}}_{2}$

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

Difference of the circumferences of the two circles = 88 cm

Adding (i) and (ii), we get:
$2{r}_{1}=\frac{91}{11}\phantom{\rule{0ex}{0ex}}{r}_{1}=\frac{91}{22}$

∴ Circumference of the first circle = $2{\mathrm{\pi r}}_{1}$

Also,
${r}_{1}-{r}_{2}=\frac{14}{11}\phantom{\rule{0ex}{0ex}}\frac{91}{22}-{r}_{2}=\frac{14}{11}\phantom{\rule{0ex}{0ex}}\frac{91}{22}-\frac{14}{11}={r}_{2}\phantom{\rule{0ex}{0ex}}{r}_{2}=\frac{63}{22}$

∴ Circumference of the second circle = $2{\mathrm{\pi r}}_{2}$

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

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

Now,
Area of the outer ring = $\mathrm{\pi }{{r}_{1}}^{2}$

Area of the inner ring = $\mathrm{\pi }{{r}_{2}}^{2}$

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

#### Question 9:

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

Now,
Area of the outer ring = $\mathrm{\pi }{{r}_{1}}^{2}$

Area of the inner ring = $\mathrm{\pi }{{r}_{2}}^{2}$

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

(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 = $\pi {R}^{2}-\pi {r}^{2}$

∴ Area of the path = 1056 m2

(ii)

Diameter of the circular park = 7 m
∴ Radius of the circular park, $\frac{7}{2}$ = 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
$=\mathrm{\pi }{R}^{2}-\mathrm{\pi }{r}^{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi }\left({R}^{2}-{r}^{2}\right)$

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 = $\pi {R}^{2}-\pi {r}^{2}$

∴ Area of the path = 1056 m2

(ii)

Diameter of the circular park = 7 m
∴ Radius of the circular park, $\frac{7}{2}$ = 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
$=\mathrm{\pi }{R}^{2}-\mathrm{\pi }{r}^{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi }\left({R}^{2}-{r}^{2}\right)$

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.

Let r m and R m be the radii of the inner and outer tracks.
Now,
Circumference of the outer track = $2\mathrm{\pi }R$
$⇒396=2×\frac{22}{7}×R\phantom{\rule{0ex}{0ex}}⇒R=\frac{396×7}{44}\phantom{\rule{0ex}{0ex}}⇒R=63$

Circumference of the inner track = $2\mathrm{\pi }r$
$⇒352=2×\frac{22}{7}×r\phantom{\rule{0ex}{0ex}}⇒r=\frac{352×7}{44}\phantom{\rule{0ex}{0ex}}⇒r=56$

Width of the track = Radius of the outer track $-$ Radius of the inner track

Area of the outer circle = $\mathrm{\pi }{R}^{2}$

Area of the inner circle = $\mathrm{\pi }{R}^{2}$

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

#### Question 11:

Let r m and R m be the radii of the inner and outer tracks.
Now,
Circumference of the outer track = $2\mathrm{\pi }R$
$⇒396=2×\frac{22}{7}×R\phantom{\rule{0ex}{0ex}}⇒R=\frac{396×7}{44}\phantom{\rule{0ex}{0ex}}⇒R=63$

Circumference of the inner track = $2\mathrm{\pi }r$
$⇒352=2×\frac{22}{7}×r\phantom{\rule{0ex}{0ex}}⇒r=\frac{352×7}{44}\phantom{\rule{0ex}{0ex}}⇒r=56$

Width of the track = Radius of the outer track $-$ Radius of the inner track

Area of the outer circle = $\mathrm{\pi }{R}^{2}$

Area of the inner circle = $\mathrm{\pi }{R}^{2}$

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

Given:
Angle of sector = ${150}^{\circ }$
Now,

Area of the sector =$\frac{{\mathrm{\pi r}}^{2}\theta }{360}$

#### Question 12:

Given:
Angle of sector = ${150}^{\circ }$
Now,

Area of the sector =$\frac{{\mathrm{\pi r}}^{2}\theta }{360}$

Radius of the circle, r = 10 cm

Area of sector OPRQ

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

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

#### Question 13:

Radius of the circle, r = 10 cm

Area of sector OPRQ

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

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

Length of the arc = 16.5 cm
$\theta ={54}^{\circ }$
Circumference=?
We know:
Length of the arc $=\frac{2\mathrm{\pi r\theta }}{360}$

Circumference = 110 cm

Now,
Area of the circle =${\mathrm{\pi r}}^{2}$

#### Question 14:

Length of the arc = 16.5 cm
$\theta ={54}^{\circ }$
Circumference=?
We know:
Length of the arc $=\frac{2\mathrm{\pi r\theta }}{360}$

Circumference = 110 cm

Now,
Area of the circle =${\mathrm{\pi r}}^{2}$

Area of minor segment = Area of sector AOBC − Area of right triangle AOB

Area of major segment APB = Area of circle − Area of minor segment

Hence, the area of major segment is 140 cm2 .

#### Question 15:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB

Area of major segment APB = Area of circle − Area of minor segment

Hence, the area of major segment is 140 cm2 .

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

Length of the arc ACB:

Now,
Area of the minor segment:

#### Question 16:

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

Length of the arc ACB:

Now,
Area of the minor segment:

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

Consider $∆$OAB.

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

Now,

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

Thus, we have:
Area of $∆$OAB =

Area of the minor segment = Area of the sector $-$ Area of the triangle

Area of the major segment = Area of the circle $-$ Area of the minor segment

#### Question 17:

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

Consider $∆$OAB.

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

Now,

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

Thus, we have:
Area of $∆$OAB =

Area of the minor segment = Area of the sector $-$ Area of the triangle

Area of the major segment = Area of the circle $-$ Area of the minor segment

Area of the triangle = $\frac{1}{2}{R}^{2}\mathrm{sin}\theta$
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:

Area of the minor segment = Area of the sector $-$ Area of the triangle
=

Area of the major segment = Area of the circle $-$ Area of the minor segment

#### Question 18:

Area of the triangle = $\frac{1}{2}{R}^{2}\mathrm{sin}\theta$
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:

Area of the minor segment = Area of the sector $-$ Area of the triangle
=

Area of the major segment = Area of the circle $-$ Area of the minor segment

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 =

Area of the sector OACBO =

Area of the minor segment = Area of the sector $-$ Area of the triangle
=

Area of the major segment = Area of the circle $-$ Area of the minor segment

#### 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 =

Area of the sector OACBO =

Area of the minor segment = Area of the sector $-$ Area of the triangle
=

Area of the major segment = Area of the circle $-$ Area of the minor segment

Let the length of the major arc be $x$ cm
Radius of the circle = 10.5 cm
∴ Length of the minor arc =

Circumference =

Using the given data, we get:

∴ Area of the sector corresponding to the major arc =

#### Question 20:

Let the length of the major arc be $x$ cm
Radius of the circle = 10.5 cm
∴ Length of the minor arc =

Circumference =

Using the given data, we get:

∴ Area of the sector corresponding to the major arc =

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

Distance covered by the short hand =

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 =

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

#### 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 =

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 =

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

Let the radius of the circle be r.
â€‹Now,

Now,
Hence, the area of the quadrant of the circle is 154 cm2.

#### Question 22:

Let the radius of the circle be r.
â€‹Now,

Now,
Hence, the area of the quadrant of the circle is 154 cm2.

r1 = 16 m
r2 = 23 m

Amount of additional ground available = Area of the bigger circle $-$ Area of the smaller circle

#### Question 23:

r1 = 16 m
r2 = 23 m

Amount of additional ground available = Area of the bigger circle $-$ Area of the smaller circle

The shaded portion shows the part of the field the horse can graze.

Area of the grazed field = Area of the quadrant OPQ

Total area of the field =

Area left ungrazed = Area of the field $-$ Area of the grazed field
=

#### Question 24:

The shaded portion shows the part of the field the horse can graze.

Area of the grazed field = Area of the quadrant OPQ

Total area of the field =

Area left ungrazed = Area of the field $-$ Area of the grazed field
=

Side of the equilateral triangle = 12 m
Area of the equilateral triangle =$\frac{\sqrt{3}}{4}×\left(\mathrm{Side}{\right)}^{2}$

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$°$

=$\frac{\theta }{360}×\mathrm{\pi }{r}^{\mathit{2}}$

Area of the field the horse cannot graze = Area of the equilateral triangle $-$ Area of the field the horse can graze

#### Question 25:

Side of the equilateral triangle = 12 m
Area of the equilateral triangle =$\frac{\sqrt{3}}{4}×\left(\mathrm{Side}{\right)}^{2}$

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$°$

=$\frac{\theta }{360}×\mathrm{\pi }{r}^{\mathit{2}}$

Area of the field the horse cannot graze = Area of the equilateral triangle $-$ Area of the field the horse can graze

Each cow can graze a region that cannot be accessed by other cows.
∴ Radius of the region grazed by each cow =

Area that each cow grazes =  $\frac{1}{4}×\mathrm{\pi }×{r}^{2}$

Total area grazed =

Now,
Area left ungrazed = Area of the square $-$ Grazed area
=

#### Question 26:

Each cow can graze a region that cannot be accessed by other cows.
∴ Radius of the region grazed by each cow =

Area that each cow grazes =  $\frac{1}{4}×\mathrm{\pi }×{r}^{2}$

Total area grazed =

Now,
Area left ungrazed = Area of the square $-$ Grazed area
=

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

Thus, we have:
$OP=PQ=QR=RO$

Now, consider $∆QOP$.

Therefore, $∆QOP$ is equilateral.

Similarly, $∆QOR$ is also equilateral and .

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:
$OP=PQ=QR=RO$

Now, consider $∆QOP$.

Therefore, $∆QOP$ is equilateral.

Similarly, $∆QOR$ is also equilateral and .

OQ = 8 cm

Hence, the radius of the circle is 8 cm.

(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
Area of the inscribed circle = ${\mathrm{\pi r}}^{2}$

(ii) If a circle is circumscribed in a square, then the diagonal of the square is equal to the diameter of the circle.

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

Now,
Area of the circumscribed circle = ${\mathrm{\pi r}}^{2}$

#### 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
Area of the inscribed circle = ${\mathrm{\pi r}}^{2}$

(ii) If a circle is circumscribed in a square, then the diagonal of the square is equal to the diameter of the circle.

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

Now,
Area of the circumscribed circle = ${\mathrm{\pi r}}^{2}$

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:

=

Ratio of the area of the circle to that of the square:
$=\frac{\pi \frac{{d}^{2}}{4}}{\frac{{d}^{2}}{2}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{2}$
Thus, the ratio of the area of the circle to that of the square is $\mathrm{\pi }:2$.

#### 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:

=

Ratio of the area of the circle to that of the square:
$=\frac{\pi \frac{{d}^{2}}{4}}{\frac{{d}^{2}}{2}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{2}$
Thus, the ratio of the area of the circle to that of the square is $\mathrm{\pi }:2$.

Let the radius of the inscribed circle be r cm.
Given:
Area of the circle = 154 ${\mathrm{cm}}^{2}$
We know:
Area of the circle =$\pi {r}^{2}$
$⇒154=\frac{22}{7}{r}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{154×7}{22}={r}^{2}\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=49\phantom{\rule{0ex}{0ex}}⇒r=7$
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:

$⇒h=3r\phantom{\rule{0ex}{0ex}}⇒h=21$

Let each side of the triangle be a cm.

∴ Perimeter of the triangle = 3a
â€‹

#### Question 30:

Let the radius of the inscribed circle be r cm.
Given:
Area of the circle = 154 ${\mathrm{cm}}^{2}$
We know:
Area of the circle =$\pi {r}^{2}$
$⇒154=\frac{22}{7}{r}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{154×7}{22}={r}^{2}\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=49\phantom{\rule{0ex}{0ex}}⇒r=7$
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:

$⇒h=3r\phantom{\rule{0ex}{0ex}}⇒h=21$

Let each side of the triangle be a cm.

∴ Perimeter of the triangle = 3a
â€‹

Radius of the wheel = 42 cm
â€‹Circumference of the wheel = $2\mathrm{\pi r}$

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 = $2\mathrm{\pi r}$

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 = $\frac{19800}{2.64}=7500$

Radius of the wheel = 2.1 m
Circumference of the wheel = $2\mathrm{\pi r}$

Distance covered by the wheel in 1 revolution = 13.2 m
Distance covered by the wheel in 75 revolutions =

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 = $\frac{990}{1000}×60$
=

#### Question 32:

Radius of the wheel = 2.1 m
Circumference of the wheel = $2\mathrm{\pi r}$

Distance covered by the wheel in 1 revolution = 13.2 m
Distance covered by the wheel in 75 revolutions =

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 = $\frac{990}{1000}×60$
=

Distance = 4.95 km =
∴ Distance covered by the wheel in 1 revolution

Now,
Circumference of the wheel = 198 cm
$⇒2\pi r=198\phantom{\rule{0ex}{0ex}}⇒2×\frac{22}{7}×r=198\phantom{\rule{0ex}{0ex}}⇒r=\frac{198×7}{44}\phantom{\rule{0ex}{0ex}}⇒r=31.5\mathrm{cm}$

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

#### Question 33:

Distance = 4.95 km =
∴ Distance covered by the wheel in 1 revolution

Now,
Circumference of the wheel = 198 cm
$⇒2\pi r=198\phantom{\rule{0ex}{0ex}}⇒2×\frac{22}{7}×r=198\phantom{\rule{0ex}{0ex}}⇒r=\frac{198×7}{44}\phantom{\rule{0ex}{0ex}}⇒r=31.5\mathrm{cm}$

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

Diameter of the wheel = 60 cm
â€‹∴ Radius of the wheel = 30 cm
â€‹Circumference of the wheel = $2\mathrm{\pi r}$

Distance covered by the wheel in 1 revolution =
∴ Distance covered by the wheel in 140 revolutions =

Now,
Distance covered by the wheel in 1 minute = Distance covered by the wheel in 140 revolutions =

∴ Distance covered by the wheel in 1 hour =

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 = $2\mathrm{\pi r}$

Distance covered by the wheel in 1 revolution =
∴ Distance covered by the wheel in 140 revolutions =

Now,
Distance covered by the wheel in 1 minute = Distance covered by the wheel in 140 revolutions =

∴ Distance covered by the wheel in 1 hour =

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

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.
Since speed is given to be
As we know $\mathrm{speed}=\frac{\mathrm{distance}}{\mathrm{time}}$

#### 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.
Since speed is given to be
As we know $\mathrm{speed}=\frac{\mathrm{distance}}{\mathrm{time}}$

Radius of the front wheel =

Circumference of the front wheel =

Distance covered by the front wheel in 800 revolutions =
Radius of the rear wheel = 1 m
Circumference of the rear wheel =

∴ Required number of revolutions =
$=\frac{640\mathrm{\pi }}{2\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}=320$

#### Question 36:

Radius of the front wheel =

Circumference of the front wheel =

Distance covered by the front wheel in 800 revolutions =
Radius of the rear wheel = 1 m
Circumference of the rear wheel =

∴ Required number of revolutions =
$=\frac{640\mathrm{\pi }}{2\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}=320$

Side of the square = 14 cm
Radius of the circle $=\frac{14}{2}$= 7 cm
Area of the quadrant of one circle = $\frac{1}{4}{\mathrm{\pi r}}^{2}$

Area of the quadrants of four circles =

Now,
Area of the square = ${\left(\mathrm{Side}\right)}^{2}$

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 = $\frac{1}{4}{\mathrm{\pi r}}^{2}$

Area of the quadrants of four circles =

Now,
Area of the square = ${\left(\mathrm{Side}\right)}^{2}$

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

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

Area of the quadrants of four circles =
Area of the shaded portion = Area of the square $-$ Area of the quadrants of four circles

#### Question 38:

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

Area of the quadrants of four circles =
Area of the shaded portion = Area of the square $-$ Area of the quadrants of four circles

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 =

Area occupied by the four sectors

Area between the circles = Area of the square $-$ Area of the four sectors

#### 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 =

Area occupied by the four sectors

Area between the circles = Area of the square $-$ Area of the four sectors

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

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

#### Question 40:

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

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

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

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

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

Area of the region between the circles =
$=\left(\sqrt{3}-\frac{11}{7}\right){a}^{2}\phantom{\rule{0ex}{0ex}}=\left(1.73-1.57\right){a}^{2}\phantom{\rule{0ex}{0ex}}=0.16{a}^{2}\phantom{\rule{0ex}{0ex}}=\frac{4}{25}{a}^{2}$

#### Question 41:

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

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

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

Area of the region between the circles =
$=\left(\sqrt{3}-\frac{11}{7}\right){a}^{2}\phantom{\rule{0ex}{0ex}}=\left(1.73-1.57\right){a}^{2}\phantom{\rule{0ex}{0ex}}=0.16{a}^{2}\phantom{\rule{0ex}{0ex}}=\frac{4}{25}{a}^{2}$

Area of trapezium = $\frac{1}{2}\left(\mathrm{AD}+\mathrm{BC}\right)×\mathrm{AB}$

Area of shaded region = Area of trapezium ABCD − Area of quadrant ABE

Hence, the area of shaded region is 14.875 cm2

#### Question 42:

Area of trapezium = $\frac{1}{2}\left(\mathrm{AD}+\mathrm{BC}\right)×\mathrm{AB}$

Area of shaded region = Area of trapezium ABCD − Area of quadrant ABE

Hence, the area of shaded region is 14.875 cm2

(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°

(ii) Area of the remaining portion = Area of trapezium ABCD − Area of four quadrants

#### 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°

(ii) Area of the remaining portion = Area of trapezium ABCD − Area of four quadrants

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°)

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°)

Hence, the area of shaded region is 137.64 cm2

We know that the opposite sides of a rectangle are equal
AD = BC =  70 cm
In right triangle AED
= (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)

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
= (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)

Hence, the area of shaded region is 2499 cm2

In right triangle AED
= (9)2 + (12)2
= 81 + 144
= 225
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

Hence, the area of shaded region is 334.31 cm2

#### Question 46:

In right triangle AED
= (9)2 + (12)2
= 81 + 144
= 225
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

Hence, the area of shaded region is 334.31 cm2

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

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

Hence, the area of shaded region is 283.97 cm2

We can find the radius of the incircle by using the formula

Now, area of shaded region = Area of triangle − Area of circle

Hence, the area of shaded region is 24.6 cm2

#### Question 48:

We can find the radius of the incircle by using the formula

Now, area of shaded region = Area of triangle − Area of circle

Hence, the area of shaded region is 24.6 cm2

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

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

Area of shaded region = Area of circle − Area of triangle ABC

#### Question 49:

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

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

Area of shaded region = Area of circle − Area of triangle ABC

Let the radius of the circle be r
Now, Perimeter of quadrant = $\frac{1}{4}\left(2\mathrm{\pi }r\right)+2r$

Hence, the area of quadrant is 38.5 cm2

#### Question 50:

Let the radius of the circle be r
Now, Perimeter of quadrant = $\frac{1}{4}\left(2\mathrm{\pi }r\right)+2r$

Hence, the area of quadrant is 38.5 cm2

Area of minor segment = Area of sector AOBC − Area of right triangle AOB

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

Hence, the area of minor segment is 28.5 cm2

Area of the road = Area of outer circle − Area of inner circle

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

Cost of levelling the road = Area of the road â¨¯ Rate
= 6594 â¨¯ 20
= Rs 131880

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 60âˆ˜

Hence, the required area is 7.77 cm2

#### Question 53:

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 60âˆ˜

Hence, the required area is 7.77 cm2

CD = 8 cm
BP = HQ = 4 cm
DE = EF = 5 cm
Area of the parallelogram ABCD = $B×H$

Area of parallelogram FGHI = $B\mathit{×}H$

Area of the square = ${\mathrm{Side}}^{2}$
=

In $∆$ELF, we have:

Area of $△$DEF = $\frac{1}{2}×B×H$

Area of the semicircle =$\frac{1}{2}{\mathrm{\pi r}}^{2}$

∴ 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 = $B×H$

Area of parallelogram FGHI = $B\mathit{×}H$

Area of the square = ${\mathrm{Side}}^{2}$
=

In $∆$ELF, we have:

Area of $△$DEF = $\frac{1}{2}×B×H$

Area of the semicircle =$\frac{1}{2}{\mathrm{\pi r}}^{2}$

∴ 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

Area of sector having central angle 150° =

Now,  Area of sector having central angle 90° : Area of sector having central angle 120° : Area of sector having central angle 150°
$=\frac{90°}{360°}\mathrm{\pi }{\left(6\right)}^{2}:\frac{120°}{360°}\mathrm{\pi }{\left(6\right)}^{2}:\frac{150°}{360°}\mathrm{\pi }{\left(6\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}:\frac{1}{3}:\frac{5}{12}\phantom{\rule{0ex}{0ex}}=3:4:5$

#### Question 55:

Area of sector having central angle 150° =

Now,  Area of sector having central angle 90° : Area of sector having central angle 120° : Area of sector having central angle 150°
$=\frac{90°}{360°}\mathrm{\pi }{\left(6\right)}^{2}:\frac{120°}{360°}\mathrm{\pi }{\left(6\right)}^{2}:\frac{150°}{360°}\mathrm{\pi }{\left(6\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}:\frac{1}{3}:\frac{5}{12}\phantom{\rule{0ex}{0ex}}=3:4:5$

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

The hexagon is made up of six triangles.

#### Question 56:

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

The hexagon is made up of six triangles.

In the right $∆$RPQ, we have:

â€‹
OR = OQ = 12.5 cm
Now,
Area of the circle = ${\mathrm{\pi r}}^{2}$

Area of the semicircle =
Area of the triangle
Thus, we have:

#### Question 57:

In the right $∆$RPQ, we have:

â€‹
OR = OQ = 12.5 cm
Now,
Area of the circle = ${\mathrm{\pi r}}^{2}$

Area of the semicircle =
Area of the triangle
Thus, we have:

Using Pythagoras' theorem for triangle ABC, we have:

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

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

Here,

Because the circle is an incircle, AE and AD are tangents to the circle.

Also,
$\angle A=90°$
Therefore, AEOD is a square.
Thus, we can say that $AE=EO=OD=AD=r$.

Area of the shaded part = Area of the triangle $-$ Area of the circle

#### Question 58:

Using Pythagoras' theorem for triangle ABC, we have:

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

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

Here,

Because the circle is an incircle, AE and AD are tangents to the circle.

Also,
$\angle A=90°$
Therefore, AEOD is a square.
Thus, we can say that $AE=EO=OD=AD=r$.

Area of the shaded part = Area of the triangle $-$ Area of the circle

Perimeter (circumference of the circle) = $2\mathrm{\pi r}$
We know:
Perimeter of a semicircular arc = $\mathrm{\pi r}$
Now,
For the arc PTS, radius is 6 cm.
∴ Circumference of the semicircle PTS =

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

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

Now,
Perimeter of the shaded region = $6\mathrm{\pi }+4\mathrm{\pi }+2\mathrm{\pi }$
$=12\mathrm{\pi cm}$

Area of the semicircle PBQ = $\frac{1}{2}{\mathrm{\pi r}}^{2}$

Area of the semicircle PTS = $\frac{1}{2}{\mathrm{\pi r}}^{2}$

Area of the semicircle QES = $\frac{1}{2}{\mathrm{\pi r}}^{2}$

Area of the shaded region = Area of the semicircle PBQ + Area of the semicircle PTS $-$ Area of the semicircle QES

#### Question 59:

Perimeter (circumference of the circle) = $2\mathrm{\pi r}$
We know:
Perimeter of a semicircular arc = $\mathrm{\pi r}$
Now,
For the arc PTS, radius is 6 cm.
∴ Circumference of the semicircle PTS =

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

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

Now,
Perimeter of the shaded region = $6\mathrm{\pi }+4\mathrm{\pi }+2\mathrm{\pi }$
$=12\mathrm{\pi cm}$

Area of the semicircle PBQ = $\frac{1}{2}{\mathrm{\pi r}}^{2}$

Area of the semicircle PTS = $\frac{1}{2}{\mathrm{\pi r}}^{2}$

Area of the semicircle QES = $\frac{1}{2}{\mathrm{\pi r}}^{2}$

Area of the shaded region = Area of the semicircle PBQ + Area of the semicircle PTS $-$ Area of the semicircle QES

Length of the inner curved portion
∴ Length of each inner curved path = $\frac{220}{2}$ = 110 m
â€‹Thus, we have:

Outer radius = (35 + 14) = 49 m
Area of track = {Area of the two rectangles [each ] + Area of the circular ring with R = 49 m and r = 35 m)}

â€‹Length of the outer boundary of the track

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
∴ Length of each inner curved path = $\frac{220}{2}$ = 110 m
â€‹Thus, we have:

Outer radius = (35 + 14) = 49 m
Area of track = {Area of the two rectangles [each ] + Area of the circular ring with R = 49 m and r = 35 m)}

â€‹Length of the outer boundary of the track

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

= Area of the rectangle − Area of the semicircle

Therefore, area of shaded region is 217  cm2.

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

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

#### Question 61:

= Area of the rectangle − Area of the semicircle

Therefore, area of shaded region is 217  cm2.

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

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

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

Area of ACDB = area of sector AOB − area of COD

Area of shaded region = area of circular ring − area of ACDB

#### 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

Area of ACDB = area of sector AOB − area of COD

Area of shaded region = area of circular ring − area of ACDB

= 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

$3.9375×\frac{22}{7}$

= 12.375 cm2

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

#### Question 63:

= 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

$3.9375×\frac{22}{7}$

= 12.375 cm2

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

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

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

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

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

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

Now we need to remove the area of 3 arcs each of radius 5 cm from this area of triangle.
Here,
The area of these 3 arcs would exactly be equal to a sector of circle with radius 5 cm and angle = $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°$ i.e. a semi circle.
Since we know that area of semi circle is
Hence, the area of shaded region is =

#### 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

Now we need to remove the area of 3 arcs each of radius 5 cm from this area of triangle.
Here,
The area of these 3 arcs would exactly be equal to a sector of circle with radius 5 cm and angle = $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°$ i.e. a semi circle.
Since we know that area of semi circle is
Hence, the area of shaded region is =

The diameters of concentric circles are in ratio 1 : 2 : 3.
let the diameters be d1 = 1xd2 = 2xd3 = 3x.
The areas of three regions are given be

So, their ratio is ${A}_{1}:{A}_{2}:{A}_{3}=\frac{\mathrm{\pi }{x}^{\mathit{2}}}{4}:\frac{3\mathrm{\pi }{x}^{2}}{4}:\frac{5\mathrm{\pi }{x}^{2}}{4}=1:3:5.$

#### Question 1:

The diameters of concentric circles are in ratio 1 : 2 : 3.
let the diameters be d1 = 1xd2 = 2xd3 = 3x.
The areas of three regions are given be

So, their ratio is ${A}_{1}:{A}_{2}:{A}_{3}=\frac{\mathrm{\pi }{x}^{\mathit{2}}}{4}:\frac{3\mathrm{\pi }{x}^{2}}{4}:\frac{5\mathrm{\pi }{x}^{2}}{4}=1:3:5.$

Let the radius of the circle be r and circumference C.
â€‹
Now,

Now,
Hence, the circumference of the circle is 44 cm.

#### Question 2:

Let the radius of the circle be r and circumference C.
â€‹
Now,

Now,
Hence, the circumference of the circle is 44 cm.

Let the radius of the circle be r.
â€‹Now,

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,

Hence, the area of the quadrant of the circle is $\frac{77}{8}$ cm2.

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

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

Hence, the diameter of the of the circle is 26 cm.

Let the diameter of the required circle be d.
Now, Area of circle = 2 â¨¯ Circumference of the circle

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

Hence, the diameter of the of the circle is 8 cm.

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.

We have
Length of arc =
Hence, the length of the arc of the circle is 22 cm.

#### Question 7:

We have
Length of arc =
Hence, the length of the arc of the circle is 22 cm.

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

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

Hence, the diameter of the circle is 10 cm.

Let the radius of the circle be r.
â€‹Now,

Now, Area of circle=
Hence, the area of the circle is 16π cm2.

#### Question 9:

Let the radius of the circle be r.
â€‹Now,

Now, Area of circle=
Hence, the area of the circle is 16π cm2.

Perimeter of a semicircular protractor = Circumference of semicircular protractor + diameter of semicircular protractor

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

Hence, the perimeter of a semicircular protractor is 36 cm.

Let the radius of the required circle be r.
Now, Area of circle = Perimeter of the circle

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

Hence, the radius of the circle is 2 units.

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

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

Hence, the radius of the required circle is 28 cm.

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

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

Hence, the radius of the circle is 10 cm.

We have
Now, Area of sector =
Hence, the area of the sector of the circle is 9.42 cm2.

#### Question 14:

We have
Now, Area of sector =
Hence, the area of the sector of the circle is 9.42 cm2.

We have
Length of arc =
Hence, the length of the arc of the circle is 22 cm.

#### Question 15:

We have
Length of arc =
Hence, the length of the arc of the circle is 22 cm.

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,
$\frac{c}{C}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{2\mathrm{\pi }r}{2\mathrm{\pi }R}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\frac{2}{3}$
Now, the ratio between their areas is given by
$\frac{a}{A}=\frac{\mathrm{\pi }{r}^{2}}{\mathrm{\pi }{R}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{r}{R}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{2}{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{4}{9}$
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,
$\frac{c}{C}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{2\mathrm{\pi }r}{2\mathrm{\pi }R}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\frac{2}{3}$
Now, the ratio between their areas is given by
$\frac{a}{A}=\frac{\mathrm{\pi }{r}^{2}}{\mathrm{\pi }{R}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{r}{R}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{2}{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{4}{9}$
Hence, the ratio between their areas is 4 : 9.

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,
$\frac{a}{A}=\frac{4}{9}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{\pi }{r}^{2}}{\mathrm{\pi }{R}^{2}}={\left(\frac{2}{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\frac{2}{3}$
Now, the ratio between their circumferences is given by
$\frac{c}{C}=\frac{2\mathrm{\pi }r}{2\mathrm{\pi }R}\phantom{\rule{0ex}{0ex}}=\frac{r}{R}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}$
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,
$\frac{a}{A}=\frac{4}{9}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{\pi }{r}^{2}}{\mathrm{\pi }{R}^{2}}={\left(\frac{2}{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\frac{2}{3}$
Now, the ratio between their circumferences is given by
$\frac{c}{C}=\frac{2\mathrm{\pi }r}{2\mathrm{\pi }R}\phantom{\rule{0ex}{0ex}}=\frac{r}{R}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}$
Hence, the ratio between their circumferences is 2 : 3.

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.
$\therefore \sqrt{2}a=2r\phantom{\rule{0ex}{0ex}}⇒a=\sqrt{2}r$
Now,

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.
$\therefore \sqrt{2}a=2r\phantom{\rule{0ex}{0ex}}⇒a=\sqrt{2}r$
Now,

Hence, the ratio of the areas of the circle and the square is π : 2

Let the radius of the circle be r.
â€‹Now,

We have
Area of sector =
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,

We have
Area of sector =
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

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

Now,
Length of the arc =$\frac{2\mathrm{\pi r\theta }}{360}$

∴ Length of the pendulum = 16.8 cm

#### Question 20:

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

Now,
Length of the arc =$\frac{2\mathrm{\pi r\theta }}{360}$

∴ Length of the pendulum = 16.8 cm

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

We have:

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

#### Question 21:

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

We have:

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

Area of the sector =17.6 cm2
Area of the sector$=\frac{{\mathrm{\pi r}}^{2}\mathrm{\theta }}{360}$

∴ Radius of the circle = 6 cm

#### Question 22:

Area of the sector =17.6 cm2
Area of the sector$=\frac{{\mathrm{\pi r}}^{2}\mathrm{\theta }}{360}$

∴ Radius of the circle = 6 cm

Given:
Area of the sector = 63 cm2

Now,
Area of the sector $=\frac{{\mathrm{\pi r}}^{2}\mathrm{\theta }}{360}$
$⇒69.3=\frac{22}{7}×10.5×10.5×\frac{\theta }{360}\phantom{\rule{0ex}{0ex}}⇒\theta =\frac{69.3×7×360}{22×10.5×10.5}\phantom{\rule{0ex}{0ex}}⇒\theta ={72}^{\circ }$

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

#### Question 23:

Given:
Area of the sector = 63 cm2

Now,
Area of the sector $=\frac{{\mathrm{\pi r}}^{2}\mathrm{\theta }}{360}$
$⇒69.3=\frac{22}{7}×10.5×10.5×\frac{\theta }{360}\phantom{\rule{0ex}{0ex}}⇒\theta =\frac{69.3×7×360}{22×10.5×10.5}\phantom{\rule{0ex}{0ex}}⇒\theta ={72}^{\circ }$

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

Given:

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

Now,
Area of the sector OACBO = $\frac{1}{2}×\mathrm{Radius}×\mathrm{Arc}$

â€‹

#### Question 24:

Given:

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

Now,
Area of the sector OACBO = $\frac{1}{2}×\mathrm{Radius}×\mathrm{Arc}$

â€‹

Given:
Length of the arc = 44 cm

Now,
Length of the arc $=\frac{2\mathrm{\pi r\theta }}{360}$

$⇒44=2×\frac{22}{7}×17.5×\frac{\theta }{360}\phantom{\rule{0ex}{0ex}}⇒\theta =\frac{44×7×360}{44×17.5}\phantom{\rule{0ex}{0ex}}⇒\theta ={144}^{\circ }$
Also,
Area of the sector =$\frac{{\mathrm{\pi r}}^{2}\mathrm{\theta }}{360}$

#### Question 25:

Given:
Length of the arc = 44 cm

Now,
Length of the arc $=\frac{2\mathrm{\pi r\theta }}{360}$

$⇒44=2×\frac{22}{7}×17.5×\frac{\theta }{360}\phantom{\rule{0ex}{0ex}}⇒\theta =\frac{44×7×360}{44×17.5}\phantom{\rule{0ex}{0ex}}⇒\theta ={144}^{\circ }$
Also,
Area of the sector =$\frac{{\mathrm{\pi r}}^{2}\mathrm{\theta }}{360}$

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

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

Hence, the area of the remaining cardboard is 21 cm2

Area of the square ABCD =

Area of the circle = ${\mathrm{\pi r}}^{2}$

Area of the quadrant of one circle = $\frac{1}{4}{\mathrm{\pi r}}^{2}$

Area of the quadrants of four circles =
Area of the shaded region = Area of the square $-$ Area of the circle $-$ Area of the quadrants of four circles
=

#### Question 27:

Area of the square ABCD =

Area of the circle = ${\mathrm{\pi r}}^{2}$

Area of the quadrant of one circle = $\frac{1}{4}{\mathrm{\pi r}}^{2}$

Area of the quadrants of four circles =
Area of the shaded region = Area of the square $-$ Area of the circle $-$ Area of the quadrants of four circles
=

We know that the opposite sides of rectangle are equal
∴ AD = BC = 28 cm
Now, Radius of semicircular portion =
∴ Area of remaining paper = Area of rectangular sheet −Area of semicircular portion

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 =
∴ Area of remaining paper = Area of rectangular sheet −Area of semicircular portion

Hence, the area of the remaining paper is 812 cm2

Area of shaded region = Area of square OABC − Area of quadrant COPB having radius OC

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

Hence, the area of the shaded region is 10.5 cm2

Area of the shaded region = Area of sector having central angle 60âˆ˜ + Area of sector having central angle 80âˆ˜ + Area of sector having central angle 40âˆ˜

Hence, the area of the shaded region is 77 cm2.

#### Question 30:

Area of the shaded region = Area of sector having central angle 60âˆ˜ + Area of sector having central angle 80âˆ˜ + Area of sector having central angle 40âˆ˜

Hence, the area of the shaded region is 77 cm2.

Area of the shaded portion = Area of sector OPQ − Area of sector OAB

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

Hence, the area of the shaded portion is $\frac{77}{8}$ cm2.

Area of the shaded region = Area of Square ABCD − (Area of semicircle APD + Area of semicircle BPC)

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)

Hence, the area of the shaded region is 42 cm2.

We have
Perimeter of the top of the table = Length of the major arc AB + Length of OA + Length of OB

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

#### Question 33:

We have
Perimeter of the top of the table = Length of the major arc AB + Length of OA + Length of OB

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

Area of the shaded portion = (Area of quadrant DPBA + Area of quadrant DQBC) − Area of Square ABCD

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

Hence, the area of the shaded portion is 28 cm2.

Area of the right-angled $∆$COD = $\frac{1}{2}×b×h$
=

Area of the sector AOC  =$\frac{\theta }{360}×\mathrm{\pi }×{r}^{2}$

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

#### Question 35:

Area of the right-angled $∆$COD = $\frac{1}{2}×b×h$
=

Area of the sector AOC  =$\frac{\theta }{360}×\mathrm{\pi }×{r}^{2}$

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

Permieter of shaded region = Length of the arc APB + Length of the arc CPD + Length of AD + Length of BC

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

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

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

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

Hence, the required area is 56 cm2 .

(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

(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)

#### 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

(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)

Perimeter of shaded region = Length of the arc PAQ + Length of the arc PSR + Length of the arc RTQ

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

Hence, the perimeter of shaded region is 31.4 cm.

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

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

Hence, the area of the shaded region is 228 cm2.

Permieter of shaded region = Length of the arc AQO + Length of the arc APB + Length of OB

Area of the shaded portion = Area of semicircle AQO + Area of semicircle APB

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

Area of the shaded portion = Area of semicircle AQO + Area of semicircle APB

Hence, the area of the shaded portion is 96.25 cm2.

Let the radius of the circle be r.
â€‹Now,

Now,
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,

Now,
Hence, the area of the quadrant of the circle is 38.5 cm2.

Area of the square =

Area of the circles =

Area of the shaded region = Area of the square $-$ Area of four circles

#### Question 43:

Area of the square =

Area of the circles =

Area of the shaded region = Area of the square $-$ Area of four circles

In right triangle ABC
AC2 = AB2 + BC2
= 82 +62
= 64 + 36
= 100
∴ AC2 = 100
⇒ AC = 10 cm
Area of the shaded region = Area of circle − Area of rectangle OABC

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
Area of the shaded region = Area of circle − Area of rectangle OABC

Hence, the area of the shaded region is 30.57 cm2.

Area of the circle = 484 cm2
Area of the square =

Perimeter of the square = $4×\mathrm{Side}$
Perimeter of the square = $4×22$
= 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€‹

Area of the circle = ${\mathrm{\pi r}}^{2}$

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

#### Question 45:

Area of the circle = 484 cm2
Area of the square =

Perimeter of the square = $4×\mathrm{Side}$
Perimeter of the square = $4×22$
= 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€‹

Area of the circle = ${\mathrm{\pi r}}^{2}$

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

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,

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,

Hence, the area of the square ABCD is 2r2 sq units.

Let the radius of the circle be r.
â€‹Now,

Now,
Area of field =
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:

Let the radius of the circle be r.
â€‹Now,

Now,
Area of field =
Cost of ploughing = Rate â¨¯ Area of field = 0.5 â¨¯ 3850 = Rs 1925
Hence, the cost of ploughing the field is Rs 1925.

Area of the rectangle = $l×b$

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 = $\mathrm{\pi }{r}^{2}$

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

#### Question 48:

Area of the rectangle = $l×b$

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 = $\mathrm{\pi }{r}^{2}$

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

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

#### Question 49:

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

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 =
Area of the smaller circle = ${\mathrm{\pi r}}^{2}$

Radius of the larger circle =
Area of the larger circle = ${\mathrm{\pi r}}^{2}$

∴ 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 =
Area of the smaller circle = ${\mathrm{\pi r}}^{2}$

Radius of the larger circle =
Area of the larger circle = ${\mathrm{\pi r}}^{2}$

∴ Area of the shaded region = Area of the larger circle $-$ Area of the smaller circle
= 2291.14 $-$ 1521.14
= 770 cm2

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

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

Hence, the area of the shaded region is 6.125 cm2 .

Area of minor segment = Area of sector AOBC − Area of right triangle AOB

Area of major segment APB = Area of circle − Area of minor segment

Hence, the area of major segment is 3500 cm2

#### Question 1:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB

Area of major segment APB = Area of circle − Area of minor segment

Hence, the area of major segment is 3500 cm2

(d) 22 cm
Let the radius be r cm.
We know:
Area of a circle
Thus, we have:
$\mathrm{\pi }{r}^{2}=38.5$
$⇒\frac{22}{7}×{r}^{2}=38.5\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=\left(38.5×\frac{7}{22}\right)\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=\left(\frac{385}{10}×\frac{7}{22}\right)\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=\frac{49}{4}\phantom{\rule{0ex}{0ex}}⇒r=\frac{7}{2}$
Now,
Circumference of the circle$=2\mathrm{\pi }r$

#### Question 2:

(d) 22 cm
Let the radius be r cm.
We know:
Area of a circle
Thus, we have:
$\mathrm{\pi }{r}^{2}=38.5$
$⇒\frac{22}{7}×{r}^{2}=38.5\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=\left(38.5×\frac{7}{22}\right)\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=\left(\frac{385}{10}×\frac{7}{22}\right)\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=\frac{49}{4}\phantom{\rule{0ex}{0ex}}⇒r=\frac{7}{2}$
Now,
Circumference of the circle$=2\mathrm{\pi }r$

(b) 14π cm
Let the radius be r cm.
We know:
Area of a circle$=\mathrm{\pi }{r}^{2}\phantom{\rule{0ex}{0ex}}$
Thus, we have:
$\mathrm{\pi }{r}^{2}=49\mathrm{\pi }\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=49\phantom{\rule{0ex}{0ex}}⇒r=\sqrt{49}\phantom{\rule{0ex}{0ex}}⇒r=7$

Now,
Circumference of the circle$=2\mathrm{\pi r}$

#### Question 3:

(b) 14π cm
Let the radius be r cm.
We know:
Area of a circle$=\mathrm{\pi }{r}^{2}\phantom{\rule{0ex}{0ex}}$
Thus, we have:
$\mathrm{\pi }{r}^{2}=49\mathrm{\pi }\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=49\phantom{\rule{0ex}{0ex}}⇒r=\sqrt{49}\phantom{\rule{0ex}{0ex}}⇒r=7$

Now,
Circumference of the circle$=2\mathrm{\pi r}$

(c) 154 cm2
Let the radius be r cm.
We know:
Circumference of the circle$=2\mathrm{\pi r}$
Thus, we have:

Now,
Area of the circle$=\mathrm{\pi }{r}^{2}$

#### Question 4:

(c) 154 cm2
Let the radius be r cm.
We know:
Circumference of the circle$=2\mathrm{\pi r}$
Thus, we have:

Now,
Area of the circle$=\mathrm{\pi }{r}^{2}$

(c) 4658.5 m2
Let the radius be r m.
We know:
Perimeter of a circle
Thus, we have:
$2\mathrm{\pi }r=242$
$⇒2×\frac{22}{7}×r=242\phantom{\rule{0ex}{0ex}}⇒\frac{44}{7}×r=242\phantom{\rule{0ex}{0ex}}⇒r=\left(242×\frac{7}{44}\right)\phantom{\rule{0ex}{0ex}}⇒r=\frac{77}{2}$

∴ Area of the circle$=\mathrm{\pi }{r}^{2}$

#### Question 5:

(c) 4658.5 m2
Let the radius be r m.
We know:
Perimeter of a circle
Thus, we have:
$2\mathrm{\pi }r=242$
$⇒2×\frac{22}{7}×r=242\phantom{\rule{0ex}{0ex}}⇒\frac{44}{7}×r=242\phantom{\rule{0ex}{0ex}}⇒r=\left(242×\frac{7}{44}\right)\phantom{\rule{0ex}{0ex}}⇒r=\frac{77}{2}$

∴ Area of the circle$=\mathrm{\pi }{r}^{2}$

(c) 96%
Let d be the original diameter.
Radius$=\frac{d}{2}$
Thus, we have:
Original area$=\mathrm{\pi }×{\left(\frac{d}{2}\right)}^{2}$
$=\frac{\mathrm{\pi }{d}^{2}}{4}$
New diameter
$=\left(\frac{140}{100}×d\right)\phantom{\rule{0ex}{0ex}}=\frac{7d}{5}$
Now,
New radius$=\frac{7d}{5×2}$
$=\frac{7d}{10}$
New area$=\mathrm{\pi }×{\left(\frac{7d}{10}\right)}^{2}$
$=\frac{49\mathrm{\pi }{d}^{2}}{10}$
Increase in the area$=\left(\frac{49\mathrm{\pi }{d}^{2}}{10}-\frac{\mathrm{\pi }{d}^{2}}{4}\right)$
$=\frac{24\mathrm{\pi }{d}^{2}}{100}\phantom{\rule{0ex}{0ex}}=\frac{6\mathrm{\pi }{d}^{2}}{25}$
We have:
Increase in the area$=\left(\frac{6\mathrm{\pi }{d}^{2}}{25}×\frac{4}{\mathrm{\pi }{d}^{2}}×100\right)%$
= 96%

#### Question 6:

(c) 96%
Let d be the original diameter.
Radius$=\frac{d}{2}$
Thus, we have:
Original area$=\mathrm{\pi }×{\left(\frac{d}{2}\right)}^{2}$
$=\frac{\mathrm{\pi }{d}^{2}}{4}$
New diameter
$=\left(\frac{140}{100}×d\right)\phantom{\rule{0ex}{0ex}}=\frac{7d}{5}$
Now,
New radius$=\frac{7d}{5×2}$
$=\frac{7d}{10}$
New area$=\mathrm{\pi }×{\left(\frac{7d}{10}\right)}^{2}$
$=\frac{49\mathrm{\pi }{d}^{2}}{10}$
Increase in the area$=\left(\frac{49\mathrm{\pi }{d}^{2}}{10}-\frac{\mathrm{\pi }{d}^{2}}{4}\right)$
$=\frac{24\mathrm{\pi }{d}^{2}}{100}\phantom{\rule{0ex}{0ex}}=\frac{6\mathrm{\pi }{d}^{2}}{25}$
We have:
Increase in the area$=\left(\frac{6\mathrm{\pi }{d}^{2}}{25}×\frac{4}{\mathrm{\pi }{d}^{2}}×100\right)%$
= 96%

(d) None of these
Let r be the original radius.
Thus, we have:
Original area$=\mathrm{\pi }{r}^{2}$
Also,
$=\left(\frac{70}{100}×r\right)\phantom{\rule{0ex}{0ex}}=\frac{7r}{10}$
New area$=\mathrm{\pi }×{\left(\frac{7r}{10}\right)}^{2}$
$=\frac{49\mathrm{\pi }{r}^{2}}{100}$
Decrease in the area$=\left(\mathrm{\pi }{r}^{2}-\frac{49\mathrm{\pi }{r}^{2}}{100}\right)$
$=\frac{59\mathrm{\pi }{r}^{2}}{100}$
Thus, we have:
Decrease in the area$=\left(\frac{59\mathrm{\pi }{r}^{2}}{100}×\frac{1}{\mathrm{\pi }{r}^{2}}×100\right)%$
=51%

#### Question 7:

(d) None of these
Let r be the original radius.
Thus, we have:
Original area$=\mathrm{\pi }{r}^{2}$
Also,
$=\left(\frac{70}{100}×r\right)\phantom{\rule{0ex}{0ex}}=\frac{7r}{10}$
New area$=\mathrm{\pi }×{\left(\frac{7r}{10}\right)}^{2}$
$=\frac{49\mathrm{\pi }{r}^{2}}{100}$
Decrease in the area$=\left(\mathrm{\pi }{r}^{2}-\frac{49\mathrm{\pi }{r}^{2}}{100}\right)$
$=\frac{59\mathrm{\pi }{r}^{2}}{100}$
Thus, we have:
Decrease in the area$=\left(\frac{59\mathrm{\pi }{r}^{2}}{100}×\frac{1}{\mathrm{\pi }{r}^{2}}×100\right)%$
=51%

(d) $\sqrt{\mathrm{\pi }}:2$
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$=\mathrm{\pi }{r}^{2}$
Because the area of the square is the same as the area of the circle, we have:
${a}^{2}=\mathrm{\pi }{r}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{{r}^{2}}{{a}^{2}}=\frac{1}{\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{a}=\frac{1}{\sqrt{\mathrm{\pi }}}$
∴ Ratio of their perimeters

#### Question 8:

(d) $\sqrt{\mathrm{\pi }}:2$
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$=\mathrm{\pi }{r}^{2}$
Because the area of the square is the same as the area of the circle, we have:
${a}^{2}=\mathrm{\pi }{r}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{{r}^{2}}{{a}^{2}}=\frac{1}{\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{a}=\frac{1}{\sqrt{\mathrm{\pi }}}$
∴ Ratio of their perimeters

(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:
$2\pi r=2\pi {r}_{1}+2\pi {r}_{2}$
$⇒2\pi r=\left(2\pi ×18\right)+\left(2\pi ×10\right)\phantom{\rule{0ex}{0ex}}⇒2\pi r=2\pi ×\left(18+10\right)\phantom{\rule{0ex}{0ex}}⇒2\pi r=\left(2\pi ×28\right)\phantom{\rule{0ex}{0ex}}⇒2\pi r=\left(2×\frac{22}{7}×28\right)$

#### 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:
$2\pi r=2\pi {r}_{1}+2\pi {r}_{2}$
$⇒2\pi r=\left(2\pi ×18\right)+\left(2\pi ×10\right)\phantom{\rule{0ex}{0ex}}⇒2\pi r=2\pi ×\left(18+10\right)\phantom{\rule{0ex}{0ex}}⇒2\pi r=\left(2\pi ×28\right)\phantom{\rule{0ex}{0ex}}⇒2\pi r=\left(2×\frac{22}{7}×28\right)$

(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:
$\pi {r}^{2}=\pi {{r}_{1}}^{2}+\pi {{r}_{2}}^{2}$

∴ Diameter of the new circle
= 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:
$\pi {r}^{2}=\pi {{r}_{1}}^{2}+\pi {{r}_{2}}^{2}$

∴ Diameter of the new circle
= 50 cm

Let the side of the square be a and the radius of the circle be r.
Now, Perimeter of circle = Circumference of the circle

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

Hence, the correct answer is option (b).

(d) ${R}_{1}^{2}+{R}_{2}^{2}={R}^{2}$
Because the sum of the areas of two circles with radii  is equal to the area of a circle with radius R, we have:
$\mathrm{\pi }{{R}_{1}}^{2}+\mathrm{\pi }{{R}_{2}}^{2}=\mathrm{\pi }{R}^{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{\pi }\left({{\mathrm{R}}_{1}}^{2}+{{\mathrm{R}}_{2}}^{2}\right)=\mathrm{\pi }{\mathrm{R}}^{2}\phantom{\rule{0ex}{0ex}}⇒{{\mathrm{R}}_{1}}^{2}+{{\mathrm{R}}_{2}}^{2}={\mathrm{R}}^{2}$

#### Question 12:

(d) ${R}_{1}^{2}+{R}_{2}^{2}={R}^{2}$
Because the sum of the areas of two circles with radii  is equal to the area of a circle with radius R, we have:
$\mathrm{\pi }{{R}_{1}}^{2}+\mathrm{\pi }{{R}_{2}}^{2}=\mathrm{\pi }{R}^{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{\pi }\left({{\mathrm{R}}_{1}}^{2}+{{\mathrm{R}}_{2}}^{2}\right)=\mathrm{\pi }{\mathrm{R}}^{2}\phantom{\rule{0ex}{0ex}}⇒{{\mathrm{R}}_{1}}^{2}+{{\mathrm{R}}_{2}}^{2}={\mathrm{R}}^{2}$

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

$2\mathrm{\pi }{R}_{1}+2\mathrm{\pi }{R}_{2}=2\mathrm{\pi }R\phantom{\rule{0ex}{0ex}}⇒2\mathrm{\pi }\left({R}_{1}+{R}_{2}\right)=2\mathrm{\pi }R\phantom{\rule{0ex}{0ex}}⇒{R}_{1}+{R}_{2}=R$

#### Question 13:

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

$2\mathrm{\pi }{R}_{1}+2\mathrm{\pi }{R}_{2}=2\mathrm{\pi }R\phantom{\rule{0ex}{0ex}}⇒2\mathrm{\pi }\left({R}_{1}+{R}_{2}\right)=2\mathrm{\pi }R\phantom{\rule{0ex}{0ex}}⇒{R}_{1}+{R}_{2}=R$

(b) Area of the circle > Area of the square
Let r be the radius of the circle.
We know:
Circumference of the circle$=2\mathrm{\pi }r\phantom{\rule{0ex}{0ex}}$
Now,
Let a be the side of the square.
We know:
Perimeter of the square = 4a
Now,
$2\mathrm{\pi }r=4a\phantom{\rule{0ex}{0ex}}⇒r=\frac{4a}{2\mathrm{\pi }}$
∴ Area of the circle$=\mathrm{\pi }{r}^{2}$
$=\mathrm{\pi }×{\left(\frac{4a}{2\mathrm{\pi }}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi }×\frac{16{a}^{\mathit{2}}}{4{\mathrm{\pi }}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4{a}^{2}}{\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}=\frac{4×7{a}^{2}}{22}\phantom{\rule{0ex}{0ex}}=\frac{14{a}^{2}}{11}$

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$=2\mathrm{\pi }r\phantom{\rule{0ex}{0ex}}$
Now,
Let a be the side of the square.
We know:
Perimeter of the square = 4a
Now,
$2\mathrm{\pi }r=4a\phantom{\rule{0ex}{0ex}}⇒r=\frac{4a}{2\mathrm{\pi }}$
∴ Area of the circle$=\mathrm{\pi }{r}^{2}$
$=\mathrm{\pi }×{\left(\frac{4a}{2\mathrm{\pi }}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi }×\frac{16{a}^{\mathit{2}}}{4{\mathrm{\pi }}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4{a}^{2}}{\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}=\frac{4×7{a}^{2}}{22}\phantom{\rule{0ex}{0ex}}=\frac{14{a}^{2}}{11}$

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

(b) 330 cm2
Let:
R = 19 cm and r = 16 cm
Thus, we have:
Area of the ring$=\mathrm{\pi }\left({R}^{2}-{r}^{2}\right)$

#### Question 15:

(b) 330 cm2
Let:
R = 19 cm and r = 16 cm
Thus, we have:
Area of the ring$=\mathrm{\pi }\left({R}^{2}-{r}^{2}\right)$

(b) 3.5 cm
Let r cm and R cm be the radii of two concentric circles.
Thus, we have:
${\mathrm{\pi R}}^{2}=1386$

Also,

∴ Width of the ring$=\left(R-r\right)$

#### Question 16:

(b) 3.5 cm
Let r cm and R cm be the radii of two concentric circles.
Thus, we have:
${\mathrm{\pi R}}^{2}=1386$

Also,

∴ Width of the ring$=\left(R-r\right)$

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,
$\frac{c}{C}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{2\mathrm{\pi }r}{2\mathrm{\pi }R}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\frac{3}{4}$
Now, the ratio between their areas is given by
$\frac{a}{A}=\frac{\mathrm{\pi }{r}^{2}}{\mathrm{\pi }{R}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{r}{R}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{3}{4}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{9}{16}$
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,
$\frac{c}{C}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{2\mathrm{\pi }r}{2\mathrm{\pi }R}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\frac{3}{4}$
Now, the ratio between their areas is given by
$\frac{a}{A}=\frac{\mathrm{\pi }{r}^{2}}{\mathrm{\pi }{R}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{r}{R}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{3}{4}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{9}{16}$
Hence, the correct answer is option (c).

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,
$\frac{a}{A}=\frac{9}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{\pi }{r}^{2}}{\mathrm{\pi }{R}^{2}}={\left(\frac{3}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\frac{3}{2}$
Now, the ratio between their circumferences is given by
$\frac{c}{C}=\frac{2\mathrm{\pi }r}{2\mathrm{\pi }R}\phantom{\rule{0ex}{0ex}}=\frac{r}{R}\phantom{\rule{0ex}{0ex}}=\frac{3}{2}$
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,
$\frac{a}{A}=\frac{9}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{\pi }{r}^{2}}{\mathrm{\pi }{R}^{2}}={\left(\frac{3}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\frac{3}{2}$
Now, the ratio between their circumferences is given by
$\frac{c}{C}=\frac{2\mathrm{\pi }r}{2\mathrm{\pi }R}\phantom{\rule{0ex}{0ex}}=\frac{r}{R}\phantom{\rule{0ex}{0ex}}=\frac{3}{2}$
Hence, the correct answer is option (a)

(d) 7000
Distance covered in 1 revolution$=2\mathrm{\pi }r$

Number of revolutions taken to cover 11 km$=\left(11×1000×\frac{7}{11}\right)$
= 7000

#### Question 19:

(d) 7000
Distance covered in 1 revolution$=2\mathrm{\pi }r$

Number of revolutions taken to cover 11 km$=\left(11×1000×\frac{7}{11}\right)$
= 7000

(a) 140
Distance covered by the wheel in 1 revolution$=\mathrm{\pi }d$

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

#### Question 20:

(a) 140
Distance covered by the wheel in 1 revolution$=\mathrm{\pi }d$

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

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

#### Question 21:

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

(d) $\frac{\mathrm{\pi }{R}^{2}\theta }{360}$

#### Question 22:

(d) $\frac{\mathrm{\pi }{R}^{2}\theta }{360}$

(b) $\frac{2\mathrm{\pi }R\theta }{360}$

#### Question 23:

(b) $\frac{2\mathrm{\pi }R\theta }{360}$

Angle subtends by the minute hand in 1 minute = 6âˆ˜
∴ Angle subtends by the minute hand in 10 minutes = 60âˆ˜
Now,
Area of the sector
Hence, the correct answer is option (a).

#### Question 24:

Angle subtends by the minute hand in 1 minute = 6âˆ˜
∴ Angle subtends by the minute hand in 10 minutes = 60âˆ˜
Now,
Area of the sector
Hence, the correct answer is option (a).

Area of minor segment = Area of sector AOBC − Area of right triangle AOB

Hence, the correct answer is option (c).

#### Question 25:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB

Hence, the correct answer is option (c).

We have
Length of arc =
Hence, the correct answer is option (b)

#### Question 26:

We have
Length of arc =
Hence, the correct answer is option (b)

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

Area of minor segment = Area of sector OAPB − Area of triangle AOB

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°

Area of minor segment = Area of sector OAPB − Area of triangle AOB

Hence, the correct answer is option (a).

(b) 228 cm2
Join OB
Now, OB is the radius of the circle.

Hence, the radius of the circle is .
Now,
Area of the shaded region = Area of the quadrant $-$ Area of the square OABC

#### Question 2:

(b) 228 cm2
Join OB
Now, OB is the radius of the circle.

Hence, the radius of the circle is .
Now,
Area of the shaded region = Area of the quadrant $-$ Area of the square OABC

(c) 300

Let d cm be the diameter of the wheel.
We know:
Circumference of the wheel$=\mathrm{\pi }×d$

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

#### Question 3:

(c) 300

Let d cm be the diameter of the wheel.
We know:
Circumference of the wheel$=\mathrm{\pi }×d$

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

(d) $\frac{x}{360}×\mathrm{\pi }{r}^{2}$

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

#### Question 4:

(d) $\frac{x}{360}×\mathrm{\pi }{r}^{2}$

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

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

Join AC.
Now, AC is the diameter of the circle.

=5 cm
Now,
Area of the shaded region = Area of the circle with radius 5 cm $-$ Area of the rectangle ABCD

#### Question 5:

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

Join AC.
Now, AC is the diameter of the circle.

=5 cm
Now,
Area of the shaded region = Area of the circle with radius 5 cm $-$ Area of the rectangle ABCD

Let r cm be the radius of the circle.
Now,
Circumference of the circle:

Also,
Area of the circle$=\mathrm{\pi }{r}^{2}$

#### Question 6:

Let r cm be the radius of the circle.
Now,
Circumference of the circle:

Also,
Area of the circle$=\mathrm{\pi }{r}^{2}$

Let ACB be the given arc subtending at an angle of $60°$ at the centre.
Now, we have:

∴ Length of the arc ACB$=\frac{2\mathrm{\pi }r}{360}$

#### Question 7:

Let ACB be the given arc subtending at an angle of $60°$ at the centre.
Now, we have:

∴ Length of the arc ACB$=\frac{2\mathrm{\pi }r}{360}$

Angle described by the minute hand in 60 minutes$=360°\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Angle described by the minute hand in 35 minutes$=\left(\frac{360}{60}×35\right)°$
$=210°$
Now,

∴ Required area swept by the minute hand in 35 minutes = Area of the sector with

#### Question 8:

Angle described by the minute hand in 60 minutes$=360°\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Angle described by the minute hand in 35 minutes$=\left(\frac{360}{60}×35\right)°$
$=210°$
Now,

∴ Required area swept by the minute hand in 35 minutes = Area of the sector with

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:

Now,
Area of the sector OACBO

#### 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:

Now,
Area of the sector OACBO

Let r cm be the radius of the circle and $\theta$ be the angle.
We have:

Area of the sector$=\frac{\mathrm{\pi }{r}^{2}\theta }{360}$

#### Question 10:

Let r cm be the radius of the circle and $\theta$ be the angle.
We have:

Area of the sector$=\frac{\mathrm{\pi }{r}^{2}\theta }{360}$

Area of the shaded region = (Area of the sector with $-$ (Area of the sector with )

#### Question 11:

Area of the shaded region = (Area of the sector with $-$ (Area of the sector with )

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

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$=2\mathrm{\pi }r$

Also,
Area of the circle$=\mathrm{\pi }{r}^{2}$

#### 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:

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$=2\mathrm{\pi }r$

Also,
Area of the circle$=\mathrm{\pi }{r}^{2}$

Distance covered in 1 revolution$=\mathrm{\pi }×d$

Distance covered in 1 second
= 1320 cm
Distance covered in 1 hour

#### Question 13:

Distance covered in 1 revolution$=\mathrm{\pi }×d$

Distance covered in 1 second
= 1320 cm
Distance covered in 1 hour

(i) Area of the quadrant OACB

(ii) Area of the shaded region = Area of the quadrant OACB $-$ Area of $∆AOD$

#### Question 14:

(i) Area of the quadrant OACB

(ii) Area of the shaded region = Area of the quadrant OACB $-$ Area of $∆AOD$

Let r be the radius of the circle.
Thus, we have:

=14 cm
Now,
Area of the shaded region = (Area of the square ABCD$-$ 4(Area of the sector where )

#### Question 15:

Let r be the radius of the circle.
Thus, we have:

=14 cm
Now,
Area of the shaded region = (Area of the square ABCD$-$ 4(Area of the sector where )

Draw $OD\perp BC\phantom{\rule{0ex}{0ex}}$.
Because $∆ABC$ is equilateral, $\angle A=\angle B=\angle C=60°$.
Thus, we have:

Also,

∴ Area of the shaded region = (Area of the circle) $-$ (Area of  $∆ABC$)

#### Question 16:

Draw $OD\perp BC\phantom{\rule{0ex}{0ex}}$.
Because $∆ABC$ is equilateral, $\angle A=\angle B=\angle C=60°$.
Thus, we have:

Also,

∴ Area of the shaded region = (Area of the circle) $-$ (Area of  $∆ABC$)

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 = $56×6°=336°.$
So, the area of clock described by minute hand in 56 minutes = area of sector with angle

#### 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 = $56×6°=336°.$
So, the area of clock described by minute hand in 56 minutes = area of sector with angle

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

Width of the track$=\left(R-r\right)$

Area of the track$=\mathrm{\pi }\left({\mathrm{R}}^{2}-{\mathrm{r}}^{2}\right)$

#### Question 18:

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

Width of the track$=\left(R-r\right)$

Area of the track$=\mathrm{\pi }\left({\mathrm{R}}^{2}-{\mathrm{r}}^{2}\right)$

Let AB be the chord of a circle with centre O and radius 30 cm such that $\angle AOB=60°$.
Area of the sector OACBO $=\frac{\mathrm{\pi }{r}^{2}\theta }{360}$

Area of $∆OAB$

Area of the minor segment = (Area of the sector OACBO$-$ (Area of $∆OAB$)

Area of the major segment = (Area of the circle) $-$ (Area of the minor segment)

#### Question 19:

Let AB be the chord of a circle with centre O and radius 30 cm such that $\angle AOB=60°$.
Area of the sector OACBO $=\frac{\mathrm{\pi }{r}^{2}\theta }{360}$

Area of $∆OAB$

Area of the minor segment = (Area of the sector OACBO$-$ (Area of $∆OAB$)

Area of the major segment = (Area of the circle) $-$ (Area of the minor segment)

Let r be the radius of the circle.
Thus, we have:

= 25 m
Area left ungrazed = (Area of the square) $-$ 4(Area of the sector where )

#### Question 20:

Let r be the radius of the circle.
Thus, we have:

= 25 m
Area left ungrazed = (Area of the square) $-$ 4(Area of the sector where )

Let a m be the side of the square.
Area of the square$={a}^{2}\phantom{\rule{0ex}{0ex}}$
Thus, we have:

Area of the plots = 4(Area of the semicircle of radius 20 m)

∴ Cost of turfing the plots at
= Rs 31400

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