NCERT Exemplar Class 10 Maths Solutions Chapter 11 Areas Related to Circles

Question 2:

If the sum of the circumferences of two circles with radii R₁ and R₂ is equal to the circumference of a circle of radius R, then
(A) R₁ +R₂=R
(B) R₁ +R₂>R
(C) R₁ +R₂<R
(D) Nothing definite can be said about the relation among R₁, R₂ and R.

Answer:

[A] R₁ +R₂=R

R₁ + R₂=R
The radius of first circle = R₁
circumference of first circle =2πR₁
Radius of second circle =R₂
circumference of the second circle =2πR₂
The radius of the third circle = R
circumference of the third circle=2πR
According to the question
2πR₁ + 2πR₂ = 2πR
2π(R₁ + R₂) = 2πR
R₁ + R₂ = R
Hence, option A is correct.

Question 3

If the circumference of a circle and the perimeter of a square are equal, then
(A) Area of the circle = Area of the square
(B) Area of the circle > Area of the square
(C) Area of the circle < Area of the square
(D) Nothing definite can be said about the relation between the areas of the circle and the square.

Answer:

[A] Area of the circle = Area of the square

Area of the circle > Area of the square
circumference of a circle=$2\pi r$

Let the radius of the circle = r
perimeter of a square =$4 \times side$

let the side of a square = a
According to the question
circumference of a circle = perimeter of a square
$\begin{aligned} & 2 \pi r=4 a \\ & \pi r=2 a \\ & a=\frac{\pi r}{2} \\ & \therefore \frac{\text { Area of circle }}{\text { Area of square }}=\frac{\pi r^2}{\left(\frac{\pi r}{2}\right)^2}=\frac{\pi r^2}{\pi r^2} \times \frac{4}{\pi}=\frac{4 \times 7}{22}=\frac{14}{11}\end{aligned}$

And $\frac{14}{11}>1$

Hence, Area of the circle > Area of the square.

Question 4

Area of the largest triangle that can be inscribed in a semi-circle of radius r units is
(A) $r^{2}$ square unit

(B) $\frac{1}{2}r^{2}$ square unit

(C) 2$r^{2}$ square unit

(D) $\sqrt{2}r^{2}$ square unit

Answer:

[A] $r^{2}$ square unit

$r^{2}$ square unit
The base of the triangle = diameter of the triangle
= 2 x r
=2r {r is radius}
Height of triangle = r
$\\ Area=\frac{1}{2}\times base \times height\\ =\frac{1}{2} \times 2r \times r\\\\ =r^{2} \text{ Square unit}$

Question 5

If the perimeter of a circle is equal to that of a square, then the ratio of their areas is
(A) 22: 7

(B) 14: 11

(C) 7: 22

(D) 11: 14

Answer

[B] 14:11
Solution
According to the question
$2 \pi r=4a$ (Because perimeter of circle = 2πr Perimeter of square =4 $\times$ side)
$\pi r=2a$ (here side of square =a)
$a=\frac{\pi r}{2}\\\\ \therefore \frac{Area \; of \; circle}{Area \; of\; square}=\frac{\pi r^{2}}{\left (\frac{\pi r}{2} \right )^{2}}=\frac{\pi r^{2}}{\pi r^{2}}\times \frac{4}{\pi}=\frac{4 \times 7}{22}=\frac{14}{11}$ (Using area of square = a²)
Hence, the ratio of their areas is 14: 11.

Question 6

It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be
(A) 10 m

(B) 15 m

(C) 20 m

(D) 24 m

Answer:

[A] 10 m
Solution
Diameter of the first circle (D) = 16 m
Radius(R) =$\frac{16}{2}=8m$
Area =$\pi r^{2}=\pi \times 8 \times 8=64\pi$
Diameter of second circle (d) = 12 m
Radius(r) =$\frac{12}{2}=6m$
Area =$\pi r^{2}=\pi \times 6 \times 6=36\pi$
Let the radius of the new park = R₁
Area =$\pi R_{1}^{2}$
According to the question
$\begin{aligned} & \quad 64 \pi+36 \pi=\pi R_1^2 \\ & 100 \pi=\pi R_1^2 \\ & 100=R_1^2 \\ & \pi R_1= \pm 10\end{aligned}$

R = – 10 is not possible because the Radius must be positive.
Hence, the Radius is 10m

Question 7

The area of the circle that can be inscribed in a square of side 6 cm is
(A) 36 $\pi$ cm²

(B) 18 $\pi$ cm²

(C) 12 $\pi$ cm²

(D) 9 $\pi$ cm²

Answer:

(D) 9 $\pi$ cm²
Solution

Diameter of circle (d) = 6 cm
Radius (r)=$\frac{d}{2}=\frac{6}{2}=3cm$
Area =$\pi r^{2}$ (area of circle = πr²)
$=\pi \times 3 \times 3\\ =9 \pi cm^{2}$

Question 8

The area of the square that can be inscribed in a circle of radius 8 cm is
(A) 256 cm² (B) 128 cm² (C) 64$\sqrt{2}$ cm² (D) 64 cm²

Answer:

(B) 128 cm²
Area of square =a²

Diagonal of a square = Diameter of a circle
Diagonal of square =8 $\times$ 2 =16cm
Let the side of the square = a cm
Using Pythagoras' theorem in ABC
(16)²=a²+a²
2a²=256
a²=128
Area of square ABCD = a²
=128 cm²

Question 9

The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36cm and 20 cm is
(A) 56 cm (B) 42 cm (C) 28 cm (D) 16 cm

Answer:

(C) 28 cm
Circumference of circle = $2 \pi r$
Diameter of first circle (d₁) = 36
Radius (r₁) =$\frac{d_{1}}{2}=\frac{36}{2}=18$
Diameter of second circle (d₂) = 20 cm
Radius (r₂)=$\frac{d_2}{2}=\frac{20}{2}=10$

Let the Radius of 3^rd circle = R cm
According to the question
$\\2 \pi (18)+2 \pi (10)=2 \pi R\\ 2 \pi (18+10)=2 \pi R \\ R=28 cm$

Question 10

The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is
(A) 31 cm (B) 25 cm (C) 62 cm (D) 50 cm

Answer:

(D) 50 cm
area of circle = $\pi r^{2}$
Radius of first circle (r1) = 24
Area $=\pi r_1^2=\pi(24)^2$

The radius of the second circle (r₂) = 7 cm
Area $\pi r_{2}^{2}=\pi(7)^{2}$
Radius of third circle = R
Area of third circle = $\pi R^{2}$
According to the question
$\\ \pi (24)^{2}+\pi (7)^{2}=\pi R^{2}\\ \pi(576+49)=\pi R^{2}\\ 625=R^{2}\\ R=\pm 25\\ R=25$(Because radius is always positive)
Radius of circle = 25 cm
Diameter = 2$\times$R = 2$\times$25 = 50cm

Question 1

Is the area of the circle inscribed in a square of side a cm, $\pi$ a²cm²? Give reasons for your answer.

Answer:

[False]

Use area of circle =$\pi r^{2}$
Side of square = a
Diameter of a circle = a {$\because$ circle inscribed in square}
Radius =$\frac{a}{2}$
Area =$\pi r^{2}$
= $\frac{\pi a^{2}}{4} cm^{2}$ (Because Radius = $\frac{a}{2}$ )
Hence, the given statement is not true because the area of a circle inscribed in a square of side a cm is $\frac{\pi a^{2}}{4} cm^{2}$

Question 2

Will it be true to say that the perimeter of a square circumscribing a circle of radius a cm is 8a cm? Give reasons for your answer.

Answer:

[true]

Perimeter of a square = 4 $\times$ side
The radius of the circle = a
Diameter of circle =2 $\times$ radius= 2a
Side of square =diameter of circle= 2a
Perimeter of square =4 $\times$ side=4 $\times$ 2a=8a
The perimeter of square is 8a
Hence, given statement is true.

Question 3

In Figure, a square is inscribed in a circle of diameter ‘d’ and another square is circumscribing the circle. Is the area of the outer square four times the area of the inner square? Give reasons for your answer

Answer:

Diameter of circle = d
Side of the biggest square = d
The area of the biggest square is = side × side
=d $\times$ d=d²
The diagonal of the smallest square = d
Let side = a
d²=a²+a² {using Pythagoras' theorem}
d²=2a²
$\frac{d}{\sqrt{2}}=a$
Area of smallest square $\frac{d}{\sqrt{2}} \times \frac{d}{\sqrt{2}}=\frac{d^{2}}{2}$
Here we found that the area of the outer square is not 4 times the area of the inner square.

Question 4

Is it true to say that the area of a segment of a circle is less than the area of its corresponding sector? Why?

Answer:

[False]

From the above figure, it is clear that the given statement is true only in the case of the minor segment. But the area of the major segment is always the greatest
Hence, given statement is False.

Question 5

Is it true that the distance travelled by a circular wheel of diameter d cm in one revolution is 2$\pi$d cm? Why?

Answer:

[False]
Circumference of circle =2$\pi$r
Diameter = d
Radius =$\frac{d}{2}$
Circumference =2$\pi$r
= $2 \pi \frac{d}{2}=\pi d$
Here we found that the distance travelled by a circular wheel of diameter d cm in one revolution is πd, which is not equal to 2πd.
Hence, the given statement is False.

Question 6

In covering a distance s metres, a circular wheel of radius r metres makes $\frac{s}{2 \pi r}$ revolution. Is this statement true? Why?

Answer:

[True]
$\because$ Circumference of circle =$2 \pi r$
The radius of the circular wheel = r m
Circumference of wheel = $2 \pi r$
Distance covered in One revolution= circumference of wheel =$2 \pi r$
In covering a distance of s number of revolution required =$\frac{s}{2 \pi r}$
Hence, the given statement is True.

Question 7

The numerical value of the area of a circle is greater than the numerical value of its circumference. Is this statement true? Why?

Answer:

[False]
Area of circle =$\pi r^{2}$
Circumference of circle =$2\pi r$
Case 1:
Let r = 1
Area of circle =$\pi r^{2}$=$\pi(1)^{2}=\pi$
Circumference of circle = $2\pi r$= $2\pi (1)=2 \pi$
Case 2:
Let r = 3
Area of circle = πr2 = π(3)2 = 9π
Circumference of circle = 2πr = 2π(3) = 6π
Conclusion:- In case (1), we found that the area is less than the circumference, but in case (2,) we found that the area is greater than the circumference.
So, from the conclusion, we observe that it depends on the value of the radius of the circle.
Hence, the given statement is false.

Question 8

If the length of an arc of a circle of radius r is equal to that of an arc of a circle of radius 2r, then the angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle. Is this statement false? Why?

Answer:

[True]
$\because$ Formula of length of arc= $\frac{2 \pi r\theta }{360}$
Let the Radius of the first circle = r
Length of arc =$\frac{2 \pi r\theta_{1} }{360}$ ….. (1) {$\theta _{1}$ is the angle of first circle}
The radius of the second circle = 2r
Length of arc= $\frac{2 \pi (2r)\theta_{2} }{360}$
=$\frac{4\pi r\theta_{2} }{360}$ …..(2) {$\theta _{2}$ is the angle of second circle}
According to the question
$\\\frac{2 \pi r\theta_{1} }{360}=\frac{4\pi r\theta_{2} }{360}\\\\ \theta_{1} =2\theta_{2}$
No, this statement is True

Question 9

The areas of the two sectors of two different circles with equal corresponding arc lengths are equal. Is this statement true? Why?

Answer:

[False]
Let the radius of the first circle be r₁ and that of the other be r₂
The length of the arcs of both circles is the same.
Let the arc length = a.
length of arc (a)=$2\pi r \times \frac{\theta }{360^{\circ}}$
Area of sector of first circle = $a \times \frac{r_{1} }{2}$ (because area of sector = $\pi r^{2}\times \frac{\theta }{360^{\circ}}=\left [ 2\pi r \times \frac{\theta }{360^{\circ}} \right ] \times \frac{r}{2}$ )
Area of sector of second circle = $a \times \frac{r_{2} }{2}$
Here we found that the area of the sector depends on the radius of the circles.
When the circle is the same, then the radius is also the same, then the given statement is true.
But in the case of different circles then the radius is also different
Hence, the given statement is False.

Question 10

The areas of two sectors of two different circles are equal. Is it necessary that their corresponding arc lengths are equal? Why?

Answer:

[True]
Let the radius of the first circle is r₁ and of the other is r₂
Let the arc length of both circles be the same.
Let the arc length be a.
length of arc (a)= $2\pi r \times \frac{\theta }{360^{\circ}}$
Area of sector of first circle = $a \times \frac{r_{1} }{2}$
(because area of sector = $\pi r^{2}\times \frac{\theta }{360^{\circ}}=\left [ 2\pi r \times \frac{\theta }{360^{\circ}} \right ] \times \frac{r}{2}=\frac{r \times a}{2}$ )
Area of sector of other circle = $a \times \frac{r_{2} }{2}$
Here we found that both areas are equal in the case of when r₁ = r₂
Hence, the area of two sectors of two different circles would be equal only in case of both circles have equal radii and equal corresponding arc lengths.
Hence, it is necessary that their corresponding arc lengths are equal.

Question 11

Is the area of the largest circle that can be drawn inside a rectangle of length a cm and breadth b cm (a >b) $\pi$b² cm²?Why?

Answer:

[False]

Diameter of circle = b
Radius =$\frac{b}{2}$
Area =$\pi r^{2}=\pi \left (\frac{b}{2} \right )^{2}=\frac{1}{4}\pi b^{2}cm^{2}$
Here we found that the area of the largest circle is not equal πb² cm².
Hence, the given statement is False.

Question 12

Circumferences of two circles are equal. Is it necessary that their areas be equal? Why?

Answer:

[True]
Use circumference of circle =$2 \pi r$
Let two circles have radii r₁ and r₂
Here, it is given that their circumferences are equal
$2 \pi r_{1}=2 \pi r_{2}\\ \Rightarrow r_{1}= r_{2}$
We know that area of circle = $\pi r^{2}$
Area of circle with radius r₁ = πr₁²
Area of circle with radius r₂ = πr₂² ……..(1)
Put r₂ = r₁ in (1), we get
πr₂² = πr₁²
Hence, the area of the given circles are also equal because two circles with equal radii will also have equal areas.
Hence, the given statement is True.

Question 13

Areas of two circles are equal. Is it necessary that their circumferences are equal? Why?

Answer:

[True]
Solution
Use area of circle $\pi r^{2}$
Let two circles having radius r₁ and r₂
Here it is given that their areas are equal
$\pi r_{1}^{2}=\pi r_{2}^{2}\\ \\ r_{1}^{2}= r_{2}^{2}\\ \\ r_{1}= r_{2}\\$
We know that the circumference of the circle =$2\pi r$
Circumference of circle with radius r₁ = 2πr₁
Circumference of circle with radius r₂ = 2πr₂ …..(1)
Put r₂ = r₁ in (1) we get
2πr₁ = 2πr₂
Hence, the circumference of given circle are also equal because two circles with equal radii will also have equal circumference.
Therefore, the given statement is True.

Question 14

Is it true to say that the area of a square inscribed in a circle of diameter p cm is p²cm²? Why?

Answer:

[False]

In the figure, we see that the diameter of the circle is equal to the diagonal of a square
Hence, the diagonal of square = p cm
Let side of the square = a cm Using Pythagoras' theorem we get
$\\p^{2}=a^{2}+a^{2}\\ p^{2}=2a^{2}\\ \frac{p^{2}}{2}=a^{2}\\ a=\frac{p}{\sqrt{2}}$
Area of square = side × side
$=\frac{p}{\sqrt{2}} \times \frac{p}{\sqrt{2}}=\frac{p^{2}}{2}cm^{2}$
Here we found that area of square is not equal to p² cm².
Hence, the given statement is False

Question 1

Find the radius of a circle whose circumference is equal to the sum of the circumferences of two circles of radii 15 cm and 18 cm.

Answer:

Radius = 33 cm
Solution
$\because$ Circumference of circle =$2 \pi r$
The radii of the two circles are $r_{1}=15 cm$ and $r_{2}=18 cm$
Let the circumference of these two are $C_{1}$ and $C_{2}$ respectively.
Let the circumference of required circle is C with radius R.
So according to question $C=C_{1}+C_{2}$
$2 \pi R=2 \pi r_{1}+2 \pi r_{2}\; \; \; \; \; \; \; \; (\because C=2 \pi r)\\ 2 \pi R=2 \pi(15+18)\\ R=33 cm$
Hence, the radius of the required circle is 33cm.

Question 2

In Figure, a square of diagonal 8 cm is inscribed in acircle. Find the area of the shaded region.

Answer:

[18.24 cm²]
Area of square =(side)² ,
Area of circle =$\pi r^{2}$

Diagonal of square = Diameter of circle = 8 cm
Using Pythagoras' theorem in $\triangle$ABC
$(AB)^{2}+(BC)^{2}=(8)^{2}\\ a^{2}+a^{2}=(8)^{2}\\ 2a^{2}=(8)^{2}\\ a^{2}=\frac{64}{2}\\ a=\sqrt{32}=4\sqrt{2}$
Area of square ABCD =a²
$=4\sqrt{2}\\ =32 cm^{2}$
Diameter of circle = 8 cm
Radius (r) =$\frac{8}{2}=4cm$
Area of circle = $\pi r^{2}\\$
$= 3.14 \times 4 \times 4=50.24 cm^{2}$
Area of shaded region = Area of circle – Area of square
=50.24 - 32
=18.24 cm²

Question 3

Find the area of a sector of a circle of radius 28 cm and central angle 45°.

Answer:

308 cm²
Solution
Area of sector =$\frac{\pi r^{2} \theta }{360^{\circ}}$

The radius of the circle r = 28 cm
Angle (q) = 45°
Area of sector = $\frac{\pi r^{2} \theta }{360^{\circ}}$
$\\=\frac{22 \times 28 \times 28 \times 45 }{7 \times 360}\\ =308 cm^{2}$

Question 4

The wheel of a motor cycle is of radius 35 cm. How many revolutions per minute must the wheel make so as to keep a speed of 66 km/h?

Answer:

500 revolutions
Circumference of circle =$2 \pi r$
The speed of wheel = 66 km per hour =$\frac{66 \times 1000}{60}$
= 1100 m/min
Radius =$= 35m=\frac{35}{100}=0.35m$ (because 1m = 100cm)
Circumference of wheel =$2 \pi r=2 \times \frac{22}{7} \times 0.35$
=2.2 m
The distance covered by the wheel in one revolution = 2.2m
The number of revolutions per minute to keep a speed of 66 km per hour =$\frac{1100}{2.2} =500$ revolutions

Question 5

A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20m × 16m. Find the area of the field in which the cow can graze.

Answer:

154 m²
Solution
$\because$ Area of circle =$\pi r^{2}$
According to the question

In the figure, we see that the area grazed by the cow is in the form of the fourth part of a circle

Hence, area grazes by cow =$\frac{\theta}{360^{\circ}}\times \pi r^{2}$ here $\theta=90^{\circ}$
$=\frac{1}{4} \times \pi r^{2}$ (Because Area of circle $\pi r^{2}$
$=\frac{1}{4} \times \frac{22}{7}\times 14 \times 14$
$=11 \times 14=154 m^{2}$

Question 6

Find the area of the flower bed (with semi-circular ends) shown in Figure.

Answer:

458.5 cm²
Solution
If we observe the figure

We found that there is a rectangle and two semicircles in it.

The length and breadth of the rectangle are 38cm and 10cm, respectively.
Area of rectangle =$l \times b =38 \times 10=380 cm^{2}$
Diameter of semi-circle = 10cm
Radius of semi circle=$\frac{10}{2}=5 cm$
Area of semi circle=$\frac{1}{2}\pi r^{2}=\frac{1}{2} \times 3.14 \times 25=39.25 cm^{2}$
Hence, the total Required area = Area of rectangle + 2(Area of semi-circle)
$\\=380+2 \times 39.25\\ =380+78.5\\ =458.5 cm^{2}$

Question 7

In Figure, AB is the diameter of the circle, AC = 6 cm and BC = 8 cm. Find the area of the shaded region (Use p = 3.14).

Answer:

[54.5 cm²]
Given: AC = 6cm and BC = 8cm
In the figure $\triangle$ ABC is a right-angle triangle.
Hence, using Pythagoras' theorem
$\\(AB)^{2}=(AC)^{2}+(BC)^{2}\\ =(6)^{2}+(8)^{2}\\ =36+64\\ =100\\ AB=\sqrt{100}=10\\ AB=10 cm$
Diameter of circle = AB = 10 cm
Radius =$\frac{10}{2}=5 cm$
Area of circle =$\pi r^{2}$
$=3.14 \times (5)^{2}=78.5 cm^{2}$
Area of $\triangle ABC=\frac{1}{2}\times AC \times BC\\$
$=\frac{1}{2} \times 6 \times 8=24m^{2}$
Area of shaded region = Area of circle – Area of DABC
=78.5-24=54.5 cm²

Question 8

Find the area of the shaded field shown in Figure.

Answer:

[38.28 m²]

Here length and breadth of the rectangle ABCD are 8m and 4m, respectively.
Are of rectangle $ABCD=l \times b=8 \times 4 = 32 m^{2}$
Radius of semi-circle = 2m
Area of semi circle=$\frac{1}{2}\pi r^{2}$
=$\frac{1}{2}\times 3.14 \times (2)^{2}=6.28 m^{2}$
Area of shaded field = Area of rectangle ABCD + Area of semi-circle
= 32+6.28
= 38.28 m²

Question 9

Find the area of the shaded region in Figure.

Answer:

235.44 m²

There are two semi-circle with diameter (d) 4 cm.
Radius(r) =$\frac{d}{2}=\frac{4}{2}=2m$
Area of semi-circle =$\frac{1}{2}\times \pi \times (r)^{2}=\frac{1}{2}\times \pi \times (2)^{2}=2 \pi$
The length and breadth of rectangle ABCD is 16m and 4m respectively
Area of ABCD=16 x 4=64 m² ($\because$ Area of rectangle = length× breadth)
The length and breadth of rectangle UVWX is 26m and 12m respectively
Area of UVWX=26 x 12 =312 m² ($\because$ Area of rectangle = length× breadth)
Area of shaded region = Area of UVWX – Area of ABCD – 2 × Area of semi-circle
$\\=312 -64-2(2 \pi)\;\;\;\;\;\;\;\;\;\;(here \; \pi=3.14)\\ =312-64-12.56\\ =235.44 m^{2}$

Question 10

Find the area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60°.

Answer:

$\left [ \frac{308-147\sqrt{3}}{3} \right ]$
Solution
Here $\theta=60^{\circ}$
r=14 cm
Area of segment = $\frac{\pi r^{2} \theta}{360}-\frac{1}{2}r^{2}\sin \theta$

$\\=\frac{\frac{22}{7}\times 14 \times 14 \times 60}{360}-\frac{1}{2}\times 14 \times 14 \times \sin 60\\ =\frac{22 \times 28 \times 60}{360}-\frac{1}{2}\times 14 \times 14 \times \frac{\sqrt{3}}{2}\\ =\frac{308}{3}-49\sqrt{3}\\ =\frac{308-147\sqrt{3}}{3}cm^{2}$

Question 11

Find the area of the shaded region in Figure, where arcs drawn with centres A, B, C and D intersect in pairs at mid-points P, Q, R and S of the sides AB, BC, CD and DA, respectively of a square ABCD (Use $\pi$ = 3.14).

Answer:

[30.96 cm²]
Solution
Here ABCD is a square of side 12 cm
Area of ABCD= (side)²=(12)²=144 cm²
Area of sector =$\frac{\theta }{360^{\circ}} \times \pi r^{2}$ here $\theta=90^{\circ}$

Here PSAP, PQBP, QRCQ, RSDR are all equal sectors
Area of 4 sectors =$4 \times \frac{\theta }{360^{\circ}} \times \pi r^{2}$
$=4 \times \frac{1 }{4} \times \pi r^{2}\\ =3.14 \times 36\\ =113.04 cm^{2}$
Area of shaded region = Area of square – Area of 4 sectors
= 144-113.04
=30.96 cm²

Question 12

In Figure, arcs are drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm. to intesect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of the shaded region(Use $\pi$ = 3.14).

Answer:

39.25 cm²
Solution
Angle made by vertices A, B and C = 60° { In equilateral triangle all angles = 60°}
Diameter of circle = 10
Radius =$\frac{10}{2}=5 cm$
Area of shaded region = 3 × Area of sector
$\\=3 \times \frac{\pi r ^{2} \theta}{360}\\ =\frac{3 \times (5)^{2} \times 3.14 \times 60 }{360}\\ =\frac{25 \times 3.14}{2}\\ =\frac{25 \times 314}{2 \times 100}\\ =\frac{25 \times 157}{100}\\ =39.25 cm^{2}$

Question 13

In Figure, arcs have been drawn with radii 14 cm each and with centres P, Q and R. Find the area of the shaded region.

Answer:

[ 308 cm²]
Solution
Area of sector with angle $\theta = \frac{\pi r^{2} \theta}{360^{\circ}}$
Here $\angle p, \angle Q, \angle R=60^{\circ}$
The radius of each circle = 14 cm
There are three sectors
Area of each sector = $\frac{\pi \times (14)^{2} \times 60}{360}\\$
$\\=\frac{\frac{22}{7} \times 196}{6}\\ =\frac{616}{6}cm^{2}$
Area of shaded region = 3 x (Area of one sector)
$\\=3 \times \frac{616}{6}\\ =308cm^{2}$

Question 14

A circular park is surrounded by a road 21 m wide. If the radius of the park is 105 m, find the area of the road.

Answer:

15246 m²
Area of circle =$\pi r^{2}$

Given that AB = 105m, BC = 21m
Where AB is the radius of the park, and BC is the wide of road
AC=AB+BC
AC=105+21=126 m
Area of big circle=$\pi r^{2}$
$\\=\pi (126)^{2}\\ =49896 m^{2}$
Area of small circle =$\pi r^{2}$
$\\=\pi (105)^{2}\\ =34650 m^{2}$
Area of road =Area of big circle - Area of small circle
=49896-34650=15246 m²

Question 15

In Figure, arcs have been drawn of radius 21 cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region.

Answer:

[1386cm²]
Area of sector =$\frac{\pi r^{2}\theta}{360^{\circ}}$
Here $\theta=90^{\circ}$
Radius = 21 cm
There are four sectors in the figure
Area of sector =$\frac{\pi \times (21)^{2} \times 90}{360}$
=$\frac{\frac{22}{7} \times 441}{4}=346.5 cm^{2}$
Area of shaded region = 4 × Area of one sector
= 4 × 346.5
= 1386 cm²

Question 16

A piece of wire 20 cm long is bent into the form of an arc of a circle subtending an angle of 60° at its centre. Find the radius of the circle.

Answer:

$\frac{60}{\pi}cm$
Solution
Given $\theta=60^{\circ}$
Length of arc = 20 cm
We know that
Length of arc =$\frac{\theta}{360}\times 2 \pi r$
$\\20=\frac{\theta}{360}\times 2 \pi r\\\\ \frac{20 \times 360}{60 \times 2 \pi}=r\\ \\r=\frac{60}{\pi}cm$

Question 1

The area of a circular playground is 22176 m². Find the cost of fencing this ground at the rate of Rs 50 per metre.

Answer:

[26400]
Given:- Area of circular playground = 22176 m²Rate of fencing = 50 Rs. per meter
Circumference of circle =$2 \pi r$
We have to find the radius (r) of the playground
Area of playground =$\pi r^{2}$
$\\22176=\frac{22}{7}\times r^{2}$
$ r^{2}=\frac{22176 \times 7}{22}$
$ r^{2}=\frac{155232}{22}$
$r^{2}=7056$
$r=\sqrt{7056}$
$r=84m$
Circumference of playground =$2 \pi r$
$\\=2 \times \frac{22}{7}\times 84\\ =528 m$
Cost of fencing the playground = 528 $\times$ 50
= 26,400 Rs.
Hence, the cost of fencing the playground at the rate of ₹ 50 per meter is ₹ 26,400.

Question 2

The diameters of the front and rear wheels of a tractor are 80 cm and 2 m, respectively. Find the number of revolutions that the rear wheel will make in covering a distance in which the front wheel makes 1400 revolutions.

Answer:

560 revolutions.
Solution:
Circumference of circle =$2 \pi r$
The diameter of the front wheel = 80 cm
Radius =$\frac{80}{2}=40cm$
Diameter of rear wheel = 2m = 200 cm
Radius = $\frac{200}{2}=100cm$
We know that the distance covered in one revolution =$2 \pi r$
distance covered by front wheel in 1400 revolution =$1400 \times 2 \pi(40)$
Let rear wheel take x revolutions = x × 2π(100)
According to the question
$\\1400 \times 2\pi(40)= x \times 2\pi(100)\\\\ x=\frac{1400 \times 40}{100}=560$
The rear wheel will make 560 revolutions.

Question 3

Sides of a triangular field are 15 m, 16 m and 17 m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field. Find the area of the field which cannot be grazed by the three animals.

Answer:

$\left ( 24 \sqrt{21}-72 \right )m^{2}$
Solution

Let their angles of triangle are $\angle A,\angle B$ and $\angle C$ rope’s length (radius) = 7 cm
Area of sector with angle A = $\frac{\pi r^{2} \times \angle A }{360}=\frac{\pi \times (7)^{2} \times \angle A}{360}$
Area of sector with angle B= $\frac{\pi r^{2} \times \angle B }{360}=\frac{\pi \times (7)^{2} \times \angle B}{360}$
Area of sector with angle C= $\frac{\pi r^{2} \times \angle C }{360}=\frac{\pi \times (7)^{2} \times \angle C}{360}$
$\therefore$Sum of the areas is
$=\frac{\left ( \angle A +\angle B+\angle C \right )\times \pi \times (7)^{2}}{360}$
$=\frac{180}{360}\times\frac{22}{7}\times 49$ {$\because$ Sum of angles of a triangle = $180^{\circ}$ }
=77 m²
Sides of triangular field are 15m, 16m and 17m
Let a =15m, b =16m, c = 17m
$\\S=\frac{(a+b+c)}{2}\\ =\frac{(15+16+17)}{2}\\ =\frac{48}{2}\\ =24m$
Area of triangular field
$\\=\sqrt{s(s-a)(s-b)(s-c)}\\ =\sqrt{24(24-15)(24-16)(24-17)}\\ =\sqrt{24 \times 9 \times 8 \times 7}\\ =\sqrt{8 \times 3 \times 9 \times 8 \times 7}\\ =8\sqrt{3 \times 3 \times 3 \times 7}\\ =24 \sqrt{21}m^{2}$
So area of the field which cannot be grazed by the animals = Area of triangular field – Area of sectors of field
$=\left (24 \sqrt{21}-77 \right )m^{2}$

Question 4

Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a central angle of 60° (Use $\pi$ = 3.14).

Answer:

$(75.36-36\sqrt{3})cm^{3}$
Solution

Area of sector – Area of triangle
Radius = 12 cm
Angle = 60°
Area of sector OAB =$\frac{\pi r^{2} \theta}{360}$
$\\=\frac{3.14 \times 12 \times 12}{360}\times 60\\\\ =75.36 cm^{2}$
DAOB is isosceles triangles
Let $\angle OAB=OBA=X$
$\\OA=OB=12 cm\\ \angle AOB=60^{0}$
$\angle OAB +\angle OBA+\angle AOB=180^{0}$ {$\because$ Sum of all interior angles of a triangle is 180°}
x + x +60=180
2x =120
x=60
Here all the three angles are 60° $\therefore$given triangle is an equilateral triangle.
Area of $\triangle AOB=\frac{\sqrt{3}}{4}(side)^{2}$ {$\because$ Area of equilateral triangle$=\frac{\sqrt{3}}{4}(side)^{2}$ }
$=\frac{\sqrt{3}}{4}(12 \times 12)$
$=36\sqrt{3}cm^{2}$
Area of segment = Area of sector OBCA – Area of DAOB
= $\left (75.36-36\sqrt{3} \right )cm^{2}$

Question 5

A circular pond is 17.5 m is of diameter. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of Rs 25 per m²

Answer:

[3064.28 RS ]

Given the Diameter of the circular pond = 17.5 m
Radius (r)=$\frac{17.5}{2}m$
Width of Path = 2m
R=r+width of path
$\\=\frac{17.5}{2}m+2m\\\\=\frac{17.5+4}{2}=10.75m$
Area of path =$\pi(R^{2}-r^{2})$
$\\=\frac{22}{7}\times(10.75^{2}-8.75^{2})\\\\ =\frac{22}{7}\times(115.5625-76.5625)\\\\ =\frac{22}{7}\times(39)\\\\ =122.57m^{2}$
Area of path = 122.57m²
Rate of construction = Rs 25 per m²
Cost of construction $=25\times 122.57=3064.28Rs$

Question 6

In Figure, ABCD is a trapezium with AB || DC, AB = 18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If arcs of equal radii 7 cm with centres A, B, C and D have been drawn, then find the area of the shaded region of the figure.

Answer:

[196 m²]
Given AB = 18 cm
DC = 32 cm
Radius of circle = 7 cm
Area of trapezium =$\frac{1}{2}$ × sum of the parallel sides × distance between parallel sides.
= $\frac{1}{2}$ ×(AB+CD) × 14
=$\frac{1}{2}$ ×(18+32) × 14
=(50) × 7
=350 cm²
$\theta=\angle A+\angle B+\angle C+\angle D$ = 360° (sum of interior angles of quadrilateral)
Radius = 7 cm
Area of all sectors =$\frac{\theta}{360^{\circ}}\times \pi r^{2}$ here $\theta=360^{\circ}$
=$\pi r^{2}$
=$\frac{22}{7}\times 7 \times 7$
=154 cm²
Area of shaded region = Area of trapezium – Area of all sectors
=350-154=196 cm²

Question 7

Three circles each of radius 3.5 cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles.

Answer:

1.966 cm²
Radius = 3.5
Diameter = 3.5 + 3.5 = 7 cm

Here ABC is an equilateral triangle because AB = BC = CA = 7 cm
$\therefore$Area of $\triangle ABC=\frac{\sqrt{3}}{4}a^{2}$
$\\=\frac{\sqrt{3}}{4} \times 7 \times 7\\ =\frac{49\sqrt{3}}{4}\\ =21.217 cm^{2}$
In an equilateral triangle each angle = 60°
All the sectors are the same
$\therefore$ Area of all there sectors $=3 \times \frac{\theta}{360^{\circ}} \times \pi r^{2}$
$\\=3 \times \frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (3.5)^{2}\\ =19.251 cm^{2}$
Area of enclosed region = Area of DABC – Area of their sectors
=21.217-19.251
= 1.966cm²

Question 8

Find the area of the sector of a circle of radius 5 cm, if the corresponding arc length is 3.5 cm.

Answer:

[8.75 cm²]
Length of the Arc = 3.5 cm
Radius (r)= 5 cm
Area of sector=$\frac{1}{2}$ $\times$ radius $\times$ length of arc
$\\=\frac{1}{2} \times r \times \frac{\theta}{360^{\circ}}\times 2\pi r\\\\ =\frac{1}{2} \times 5\times3.5=8.75cm^{2}$

Question 9

Four circular cardboard pieces of radii 7 cm are placed on a piece of paper in such a way that each piece touches the other two pieces. Find the area of the portion enclosed Between these pieces.

Answer:

[42 cm²]
Area of shaded portion = Area of square – Area of 4 sectors

Radius = 7 cm
Side of square = 14 cm
AB=BC=CD=DA=14 cm
$\therefore$ ABCD is a square
$\therefore$Area of square= 14 $\times$ 14 (Area of square = (side)² )
=196 cm²
We know that each angle of square = $90^{\circ}$
Area of 4 sectors =$4 \times \frac{\pi r^{2} \theta}{360}\\$
$\\=\frac{4 \times 7 \times 7 \times 90}{360} \times \frac{22}{7}\\\\ =154 cm^{2}$
Area of shaded portion = Area of square – Area of 4 sectors
=196-154=42cm²

Question 10

On a square cardboard sheet of area 784 cm², four congruent circular plates of Maximum size are placed such that each circular plate touches the other two Plates and each side of the square sheet is tangent to two circular plates. Find The area of the square sheet not covered by the circular plates.

Answer:

[168 cm²]

Given area of sheet =784 cm²
Let the side of the sheet = a
a²=784² (Area of square = (side) ²)
a=$\sqrt{784}=28cm$
Diameter of each circular plate =$\frac{a}{2}=\frac{28}{2}=14cm$
Radius =$\frac{d}{2}=\frac{14}{2}=7cm$
Area of 4 circular plates =$4 \times \pi r^{2}$
$\\=4 \times \frac{22}{7} \times 7 \times 7\\ =616cm^{2}$
Area of sheet not covered with circular plates = Area of sheet – Area of 4 circular plates.
=784-616=168cm²

Question 11

The Floor of a room is of dimensions 5 m × 4 m and it is covered with circular tiles of diameters 50 cm each as shown in Figure. Find the area of the floor that remains uncovered with tiles. (Use $\pi$ = 3.14)

Answer:

[4.3 m²]
Diameter of tile =50cm=0.5m (1m = 100cm)
Radius =$\frac{50}{2}=25=0.25m$
Number of tiles lengthwise =$\frac{5}{0.5}=10$ tiles
Number of tiles widthwise =$\frac{4}{0.5}=8$ tiles
Total tiles =$10 \times 8=80$
Area of floor not covered by tiles = Area of rectangular floor – Area of 80 tiles
$\\=5 \times 4-80\pi r^{2}\\ $

$=20-80 \times \pi \times 0.25 \times 0.25\\ $

$=20-\frac{8 \times 314 \times 25 \times 25}{100 \times100\times100}\\$

$ =20-\frac{157}{10}\\$

$ =20-15.7\\ =4.3m^{2}$

Question 12

All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if area of the circle is 1256 cm². (Use $\pi$ = 3.14).

Answer: [800 cm²]
Solution
Given that area of circle =1256cm²
$\\\pi r^{2}=1256$
$ r^{2}=\frac{1256}{314}\times 100$
$ r^{2}=400 \; \; \; \; \; \; \; \; \left ( \pi=3.14 \right )$
$ r=\sqrt{400}$
$ r=20\ cm$
Diameter of circle = 40 cm
As we know that the diameter of circle is equal
Diagonals of rhombus = Diameters of circle = 40 cm
Each diagonals of rhombus = 40 cm
Area of rhombus =$\frac{1}{2}$ $\times$ product of digonals
= $\frac{1}{2}$ $\times$ 40 $\times$ 40
= 800 cm²
Hence, the required area of rhombus is =800 cm²

Question 13

An archery target has three regions formed by three concentric circles as shown in Figure. If the diameters of the concentric circles are in the ratio 1 : 2 : 3, then find the ratio of the areas of three regions

Answer:

[1 : 3 : 5]
Solution

d₁:d₂:d₃ = 1: 2 : 3 [multiplying by s]
= s : 2s : 3s
Radius of inner circle (r₁)=$\frac{s}{2}$
Radius of middle circle (r₂)=$\frac{2s}{2}=s$
Radius of outer circle (r₃)=$\frac{3s}{2}$
Area of region enclosed between second and first circle
$\\=\pi r_{2}^{2}-\pi r_{1}^{2}\\ =\pi s^{2}-\frac{\pi s^{2}}{4}\\ =\frac{3 \pi s^{2}}{4}$
Area of region enclosed between third and second circle
$\\=\pi r_{3}^{2}-=\pi r_{2}^{2}\\ =\frac{\pi 9s^{2}}{4}-\pi s^{2}\\ =\frac{ 5\pi s^{2}}{4}$
Area of first circle $=\pi r_{1}^{2}=\frac{\pi s^{2}}{4}$
Ratio of area of three regions
$\\=\frac{\pi s^{2}}{4}:\frac{3\pi s^{2}}{4}:\frac{5\pi s^{2}}{4}\\ =\pi s^{2}:3 \pi s^{2}:5\pi s^{2}\\ =1:3:5$

Question 14

The length of the minute hand of a clock is 5 cm. Find the area swept by the minute hand during the time period 6 : 05 am and 6 : 40 a m.

Answer:

$\left [ 45\frac{5}{6}cm^{2} \right ]$
Solution
We know that minute hand revolving in 60 min =$360^{\circ}$
In 1 minute it is revolving =$\frac{360}{60}=6^{\circ}$
Time difference =(6:40am -6:05am) =35 min
At 6:05 am and 6.40 am, there is 35 minutes
In 35 minutes angle between min hand and hour hand =$\left (6\times 35 \right )^{\circ} =210^{\circ}$
Length of minute hand (r)=5cm
Area of sector =$\frac{\pi r^{2}\theta}{360}$
$\\=\frac{22}{7}\times \frac{5 \times 5 \times 210^{\circ}}{360}\;\;\;\left \{\because \theta = 210^{\circ} \right \}\\ =\frac{11\times 5\times 5}{6}\\ =\frac{275}{6}\\ =45\frac{5}{6}cm^{2}$
Hence, required area is $45\frac{5}{6}cm^{2}$

Question 15

Area of a sector of central angle 200° of a circle is 770 cm². Find the length of the corresponding arc of this sector.

Answer:

$\left [ 73\frac{1}{3}cm \right ]$
Area of sector=$\frac{\pi r^{2}\theta}{360}$
Angle = 200°
Area of sector = 770 cm²
$\\\frac{\pi r^{2}\theta}{360}=770$
$\pi r^{2} \times 200^{\circ}=770 \times 360 ^{\circ}$
$r^{2}=\frac{770 \times 360 ^{\circ} \times 7}{200^{\circ} \times 22} \;\;\; \left [ here \ \pi=\frac{22}{7} \right ]$
$ r^{2}=49 \times 9$
$ r=\sqrt{49 \times 9}=7 \times 3=21cm$
Length of the corresponding arc
= $\frac{\theta \times 2\pi r}{360}$
$\\=\frac{200^{\circ} \times 2 \times \pi \times 21}{360^{\circ}}\\\\ =\frac{10 \times 7}{3}\times\frac{22}{7}\\\\ =\frac{220}{3}\\\\ =73\frac{1}{3}\ cm$

Question 16

The central angles of two sectors of circles of radii 7 cm and 21 cm are respectively 120° and 40°. Find the areas of the two sectors as well as the lengths of the corresponding arcs. What do you observe?

Answer:

$\because$ Area of sector=$\frac{\pi r^{2} \theta}{360}$
Radius of first sector(r₁) = 7 cm
Angle ($\theta_{1}$ ) = 120°
Area of first sector(A₁) =$\frac{\pi r_{1}^{2} \theta}{360}$
$\\=\frac{22 \times 7 \times 7 \times 120^{\circ} }{360^{\circ}}\\ =\frac{154}{3}cm^{2}$
Radius of second sector(r₂) = 21 cm
Angle ($\theta_{2}$ ) = 40°
Area of sector of second circle (A₂)=$\frac{\pi r_{2}^{2} \theta}{360}$
$\\=\frac{22}{7}\times \frac{21\times 21}{360}\times40\\\\ =154 cm^{2}$
Corresponding arc length of first circle =$\frac{2\pi r_{1} \theta}{360}$
=$\frac{\pi r_{1} \theta}{180}$
$\\=\frac{22}{7}\times \frac{7\times 120}{180}\\=\frac{44}{3}cm$
Corresponding arc length of second circle =$\frac{2\pi r_{2}\theta}{360}$
=$\frac{\pi r_{2}\theta}{180}$
$\\=\frac{22}{7}\times \frac{21\times 40}{180}\\=\frac{44}{3}cm$
We observe that the length of arc of both circle are equal.

Question 17

Find the area of the shaded region given in Figure.

Answer:

[Area of shaded area=154.88cm² ]

Area of square PQRS =(side)²=(14)²
=196 cm²
Area of ABCD (let side a) =(side)²=(a)²
Area of 4 semi-circle $\left ( r=\frac{a}{2} \right )=4 \times \frac{1}{2}\pi\left ( \frac{a}{2} \right )^{2}$
Area of semi-circle=$\frac{1}{2}\times \pi \times r^{2}$
$\\=\frac{2\pi a^{2}}{4}\\ =\frac{\pi a^{2}}{2}$
Total inner area = Area of ABCD + Area of 4 semi-circles
$\\=a^{2}+\frac{\pi a^{2}}{2}\\$
$\\EF=8cm\\ EF=\frac{a}{2}+a+\frac{a}{2}\\ 8=\frac{a+2a+a}{2}\\ 4a=16\\ a=4cm$
Area of inner region =$4^{2}+\frac{\pi 4^{2}}{2}\\$
$\\=16+\frac{ 16 \pi}{2}\\ =16+8 \pi$
Area of shaded area = Area of PQRS – inner region area
$\\=196-16-8 \pi\\ =180-8 \times 3.14\\ =180-25.12\\ =154.88 cm^{2}$

Question 18

Find the number of revolutions made by a circular wheel of area 1.54 m² in Rolling a distance of 176 m.

Answer:

[40] revolutions
Circumference of circle =$2\pi r$
Area of wheel = 1.54m²
Distance = 176 m

$\\\pi r^{2}=1.54$
$ r^{2}=\frac{154}{100}\times \frac{7}{22}$
​​​​​​​$ r=\sqrt{\frac{539}{1100}}$
r = 0.7m
Circumference =$2\pi r$
$\\=2 \times \frac{22}{7}\times0.7\\ =2 \times \frac{22}{7}\times \frac{7}{10}\\ =\frac{44}{10}\\ =4.4m$
Number of revolution $=\frac{\text{distance}}{\text{circumference}}\\$

$=\frac{176}{44} \times 10=40$ revolutions

Question 19

Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending an angle of 90° at the centre.

Answer [32.16cm²]
Solution

By using Pythagoras in $\triangle$ABC
$\\(AB)^{2}+(BC)^{2}=(AC)^{2}\\ r^{2}+r^{2}=25\\ 2r^{2}=25\\ r=\frac{5}{\sqrt{2}}cm$
Area of circle =$\pi r^{2}=\frac{22}{7} \times \frac{5}{\sqrt{2}}\times\frac{5}{\sqrt{2}}=39.28cm^{2}$
Area of sector =$\frac{\pi r^{2} \theta}{360}$
$\\=3.14 \times \frac{25}{2}\times\frac{90}{360}\\ =9.81cm^{2}$
Area of $\triangle ABC=\frac{1}{2}\times base \times height$
$\\=\frac{1}{2}\times \frac{5}{\sqrt{2}} \times \frac{5}{\sqrt{2}}\\ =\frac{25}{4} =6.25cm^{2}$
Area of minor segment = Area of sector – Area of DABC
=9.81-6.25
=3.56cm²
Area of major segment = Area of circle – Area of minor segment
=39.28-3.56=35.72cm²
Required difference = Area of major segment – Area of minor segment
=35.72-3.56=32.16cm²

Question 20

Find the difference of the areas of a sector of angle 120° and its corresponding major sector of a circle of radius 21 cm.

Answer: [462 cm²]
Solution

Radius = 21 cm
Angle = 120°
Area of circle = $\pi r^{2}$
$\\=\frac{22}{7}\times 21\times 21\\ =1386 cm^{2}$
Area of minor sector with angle 120° OABO =$\frac{\pi r^{2} \theta}{360}$ $\left [ \theta =120^{\circ} \right ]$
$\\=\frac{22}{7}\times \frac{21\times 21}{360}\times120\\ =462cm^{2}$
Area of major sector AOBA= Area of circle – area of minor sector
= 1386-462=924cm²
Required area =924-462=462cm²