Rs Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 16 Area Of Circle, Sector And Segment are provided here with simple step-by-step explanations. These solutions for Area Of Circle, Sector And Segment are extremely popular among Class 10 students for Maths Area Of Circle, Sector And Segment Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 10 Maths Chapter 16 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 730:

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

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

Page No 730:

Let the radius of the circle be r.
​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.

Page No 730:

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.

Page No 730:

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.
​Thus, we have:
​

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

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

Page No 730:

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

Page No 730:

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 .

Page No 730:

Let the radii of the two circles be r1 cm and r2 cm.
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.

Page No 730:

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 ${\mathrm{cm}}^{2}$

Page No 730:

(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.

Page No 730:

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 ${\mathrm{m}}^{2}$

Page No 731:

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

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

Page No 731:

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

Page No 731:

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 =${\mathrm{\pi r}}^{2}$

Page No 731:

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 .

Page No 731:

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:

Page No 731:

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

Page No 731:

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

Page No 731:

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

Page No 731:

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 =

Page No 731:

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

Page No 731:

Let the radius of the circle be r.
​Now,

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

Page No 731:

r1 = 16 m
r2 = 23 m

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

Page No 731:

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
=

Page No 731:

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

Page No 732:

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
=

Page No 732:

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.

Page No 732:

(​i)​ 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 = ${\mathrm{\pi r}}^{2}$

Page No 732:

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 $\mathrm{\pi }:2$.

Page No 732:

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
​

Page No 732:

Radius of the wheel = 42 cm
​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 = $\frac{19800}{2.64}=7500$

Page No 732:

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 = $\frac{990}{1000}×60$
=

Page No 732:

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

Page No 732:

Diameter of the wheel = 60 cm
​∴ Radius of the wheel = 30 cm
​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.

Page No 732:

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 $\mathrm{speed}=\frac{\mathrm{distance}}{\mathrm{time}}$

Page No 732:

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 =
$=\frac{640\mathrm{\pi }}{2\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}=320$

Page No 732:

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

Page No 732:

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

Page No 733:

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

Page No 733:

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

Page No 733:

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

Page No 733:

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

Page No 733:

(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

Page No 733:

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

Page No 734:

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

Page No 734:

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

Page No 734:

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

Page No 734:

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

Page No 734:

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

Page No 734:

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

Page No 734:

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

Hence, the area of minor segment is 28.5 cm2

Page No 734:

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

Cost of levelling the road = Area of the road ⨯ Rate
= 6594 ⨯ 20
= Rs 131880

Page No 735:

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

Hence, the required area is 7.77 cm2

Page No 735:

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

Page No 735:

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°
$=\frac{90°}{360°}\mathrm{\pi }{\left(6\right)}^{2}:\frac{120°}{360°}\mathrm{\pi }{\left(6\right)}^{2}:\frac{150°}{360°}\mathrm{\pi }{\left(6\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}:\frac{1}{3}:\frac{5}{12}\phantom{\rule{0ex}{0ex}}=3:4:5$

Page No 735:

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

The hexagon is made up of six triangles.

Page No 735:

In the right $∆$RPQ, we have:

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:

Page No 735:

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

Page No 736:

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.
​∴ 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

Page No 736:

Length of the inner curved portion
∴ Length of each inner curved path = $\frac{220}{2}$ = 110 m
​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)}

​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.

Page No 736:

= 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.

Page No 736:

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

Page No 736:

= 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.

Page No 736:

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.

Page No 737:

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 =

Page No 737:

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

Page No 747:

Let the radius of the circle be r and circumference C.

Now,

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

Page No 747:

Let the radius of the circle be r.
​Now,

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

Page No 747:

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.

Page No 747:

Let the diameter of the required circle be d.
Now, Area of circle = 2 ⨯ Circumference of the circle

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

Page No 747:

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 ⨯ Side of square = 4 ⨯ 2a = 8a cm
Hence, the perimeter of the square is 8a cm.

Page No 747:

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

Page No 747:

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.

Page No 747:

Let the radius of the circle be r.
​Now,

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

Page No 747:

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

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

Page No 747:

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.

Page No 747:

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.

Page No 747:

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.

Page No 747:

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

Page No 747:

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

Page No 747:

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.

Page No 747:

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.

Page No 748:

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

Page No 748:

Let the radius of the circle be r.
​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

Page No 748:

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

Page No 748:

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 $\theta ={120}^{\circ }$
$=\frac{{\mathrm{\pi r}}^{2}\mathrm{\theta }}{360}$

Page No 748:

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

Page No 748:

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 = ${72}^{\circ }$

Page No 748:

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 = $\frac{1}{2}×\mathrm{Radius}×\mathrm{Arc}$

​

Page No 748:

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

Page No 748:

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 ⨯ Area of circular piece having radius 3.5 cm

Hence, the area of the remaining cardboard is 21 cm2

Page No 748:

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
=

Page No 748:

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

Page No 748:

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

Page No 749:

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

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

Page No 749:

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.

Page No 749:

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.

Page No 749:

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.

Page No 749:

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.

Page No 749:

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

Page No 750:

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.

Page No 750:

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 ⨯ 7 = 14 cm
Now,
Area of required region = Area of circle − Area of square

Hence, the required area is 56 cm2 .

Page No 750:

(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)

Page No 750:

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.

Page No 750:

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.

Page No 751:

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.

Page No 751:

Let the radius of the circle be r.
​Now,

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

Page No 751:

Area of the square =

Area of the circles =

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

Page No 751:

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.

Page No 751:

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.
​Thus, we have:
​

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

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

Page No 751:

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.

Page No 751:

Let the radius of the circle be r.
​Now,

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

Page No 751:

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.

Page No 752:

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

Page No 752:

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

Page No 752:

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 .

Page No 752:

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

Page No 758:

(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$=2\mathrm{\pi }r$

Page No 758:

(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$=2\mathrm{\pi r}$

Page No 758:

(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$=\mathrm{\pi }{r}^{2}$

Page No 759:

(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$=\mathrm{\pi }{r}^{2}$

Page No 759:

(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%

Page No 759:

(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%

Page No 759:

(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

Page No 759:

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

Page No 759:

(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

Page No 759:

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).

Page No 759:

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

Page No 759:

(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:

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

Page No 759:

(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

Page No 759:

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

Page No 760:

(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$=\left(R-r\right)$

Page No 760:

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).

Page No 760:

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)

Page No 760:

(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

Page No 760:

(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

Page No 760:

(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:

Page No 760:

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

Page No 760:

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

Page No 760:

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

Page No 760:

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

Hence, the correct answer is option (c).

Page No 760:

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

Page No 760:

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).

Page No 764:

(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

Page No 764:

(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

Page No 764:

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

Page No 764:

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

Page No 764:

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

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

Page No 764:

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

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

Page No 764:

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

Page No 764:

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

Page No 764:

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

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

Page No 764:

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

Page No 765:

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$=\mathrm{\pi }{r}^{2}$

Page No 765:

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

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

Page No 765:

(i) Area of the quadrant OACB

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

Page No 765:

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 )

Page No 765:

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  $∆ABC$)

Page No 765:

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

Page No 765:

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

Page No 765:

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)

Page No 765:

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 )

Page No 766:

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

View NCERT Solutions for all chapters of Class 10