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# Important Questions for CBSE Class 10 Maths Chapter 12 - Areas Related to Circles

Last updated date: 06th Sep 2024
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## Crucial Practice Problems for CBSE Class 10 Maths Chapter 12: Areas Related to Circles

Class 10 Maths Chapter 12 Important Questions Areas Related to Circles are given here based on the latest pattern of CBSE for 2024-2025. Students who are preparing for the board exams can practice Areas Related to Circles Important Questions to score full marks for the questions from this chapter. Along with the Important Questions for Class 10 Maths Chapter 12 related to Circles, we have also provided Extra Questions of Chapter 12 Class 10 Maths. Students can refer to the solutions whenever they get stuck while solving a problem. Also, in the end, some practice questions are provided for students to boost their preparation for the exam.

Areas Related to Circles chapter contains many formulas and concepts thus, it is important from the examination perspective, since most of the questions and objective-based questions will come in the exam from this chapter. Students can also refer to Chapter 12 Maths Class 10 Important Questions while preparing for their exam. NCERT Solution is also available at Vedantu to help the students in scoring full marks in the board exam.

Area related to circles includes the area of a circle, segment, sector, angle and length of a circle are provided in this pdf. In this chapter, we will discuss the concepts of the perimeter (also known as circumference) and area of a circle and apply this knowledge in finding the areas of two special parts of a circular region (or a circle) known as sector and segment. You can download Maths NCERT Solutions Class 10 and NCERT Solution for Class 10 Science to help you to revise the complete Syllabus and score more marks in your examinations.

Also, check CBSE Class 10 Maths Important Questions for other chapters:

 CBSE Class 10 Maths Important Questions Sl.No Chapter No Chapter Name 1 Chapter 1 Real Numbers 2 Chapter 2 Polynomials 3 Chapter 3 Pair of Linear Equations in Two Variables 4 Chapter 4 Quadratic Equations 5 Chapter 5 Arithmetic Progressions 6 Chapter 6 Triangles 7 Chapter 7 Coordinate Geometry 8 Chapter 8 Introduction to Trigonometry 9 Chapter 9 Some Applications of Trigonometry 10 Chapter 10 Circles 11 Chapter 11 Constructions 12 Chapter 12 Areas Related to Circles 13 Chapter 13 Surface Areas and Volumes 14 Chapter 14 Statistics 15 Chapter 15 Probability
Competitive Exams after 12th Science

## Study Important Questions for Class 10 Maths Chapter 12- Area Related to Circles

Very Short Answer Questions (1 Mark)

Unless stated otherwise, take $\pi =\dfrac{22}{7}$.

1. The radii of two circles are $19$ cm and $9$ cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Ans: Let us suppose R, radius of the circle whose circumference is equal to the circumference of the two circles with radius $19$ cm and $9$ cm respectively.

Thus, $2\pi R=2\pi \left( 19 \right)+2\pi \left( 9 \right)$

$\Rightarrow R=19+9$

$\Rightarrow R=28$ cm

$\therefore$ Radius of the circle, $R=28$ cm.

2.   The circumference of a circular field is $528$ cm. Then its radius is

a.   $42$ cm

b.   $84$ cm

c.   $72$ cm

d.   $56$ cm

Ans: b. $84$ cm

Circumference of a circle is given by, $2\pi R$.

$\therefore 2\pi R=528$

$\Rightarrow R=84$ cm

3. The circumference of a circle exceeds its diameter by $180$ cm. Then its radius is

a.   $32$ cm

b.   $36$ cm

c.   $40$ cm

d.   $42$ cm

Ans: d. $42$ cm

Let the diameter of the circle be $D$.

$\Rightarrow D=2R$.

Circumference of a circle is given by, $2\pi R$.

$\therefore 2\pi R=D+180$

$\Rightarrow 2\pi R=2R+180$

$\Rightarrow R=42$ cm

4. Area of the sector of angle ${{60}^{\circ }}$ of a circle with radius $10$ cm is

a.   $52\dfrac{5}{21}\text{ }c{{m}^{2}}$

b.   $52\dfrac{8}{21}\text{ }c{{m}^{2}}$

c.   $52\dfrac{4}{21}\text{ }c{{m}^{2}}$

d.   None of these

Ans: b. $52\dfrac{8}{21}\text{ }c{{m}^{2}}$

Area of the sector of a circle $=\left( \dfrac{\theta }{{{360}^{\circ }}} \right)\times \pi {{r}^{2}}$.

$\Rightarrow$ Area $=\left( \dfrac{{{60}^{\circ }}}{{{360}^{\circ }}} \right)\times \pi {{\left( 10 \right)}^{2}}$

$\Rightarrow$ Area $=52\dfrac{8}{21}\text{ }c{{m}^{2}}$.

5.   Area of a sector of angle ${{P}^{\circ }}$ of a circle of radius R is

a.   $\dfrac{{{P}^{\circ }}}{{{180}^{\circ }}}\times 2\pi R$

b.   $\dfrac{{{P}^{\circ }}}{{{180}^{\circ }}}\times \pi {{R}^{2}}$

c.   $\dfrac{{{P}^{\circ }}}{{{360}^{\circ }}}\times 2\pi R$

d.   $\dfrac{{{P}^{\circ }}}{{{720}^{\circ }}}\times 2\pi {{R}^{2}}$

Ans: d. $\dfrac{{{P}^{\circ }}}{{{720}^{\circ }}}\times 2\pi {{R}^{2}}$

Area of the sector of a circle $=\left( \dfrac{\theta }{{{360}^{\circ }}} \right)\times \pi {{r}^{2}}$.

$\Rightarrow$ Area $=\left( \dfrac{{{P}^{\circ }}}{{{360}^{\circ }}} \right)\times \pi {{\left( R \right)}^{2}}$

$\Rightarrow$ Area $=\dfrac{{{P}^{\circ }}}{{{720}^{\circ }}}\times 2\pi {{R}^{2}}$.

6. If the sum of the circumferences of two circles with radii $R_1$ and $R_2$ is equal to the circumference of a circle of radius $R$, then

a.   $R_1+R_2=R$

b.   $R_1+R_2 > R$

c.   $R_1+R_2 < R$

d.   None of these

Ans: a. $R_1+R_2=R$

Circumference of a circle is given by, $2\pi R$.

Thus, according to the given condition, $2\pi R_1+2\pi R_2=2\pi R$

$\Rightarrow R_1+R_2=R$.

7. 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.   $11:14$

Ans. b. $14:11$

The perimeter of a circle is given by, $2\pi R$.

The perimeter of a square is given by, $4a$.

Given, $2\pi R=4a$.

Ratio of area of circle to area of square is $\dfrac{\pi {{R}^{2}}}{{{a}^{2}}}$

$\Rightarrow \dfrac{\pi {{R}^{2}}}{{{a}^{2}}}=\dfrac{4\pi {{R}^{2}}}{{{\pi }^{2}}{{R}^{2}}}$

Hence, the ratio of their areas is, $14:11$.

8. The circumference of a circular field is $154$ m. Then its radius is

a.   $7$ m

b.   $14$ m

c.   $7.5$ m

d.   $28$ cm

Ans. Circumference of a circle is given by, $2\pi R$.

$\therefore 2\pi R=154$

$\Rightarrow R=24.5\text{ }m$.

9. The area of a circle is $394.24\text{ c}{{\text{m}}^{2}}$. Then the radius of the circle is

a.   $11.4$ cm

b.   $11.3$ cm

c.   $11.2$ cm

d.   $11.1$ cm

Ans. c. $11.2$ cm

The area of a circle is given by, $\pi {{R}^{2}}$.

$\therefore \pi {{R}^{2}}=394.24$

$\Rightarrow R=11.2$ cm.

10. If the perimeter and area of circle are numerically equal, then the radius of the circle is

a.   $2$ units

b.   $\pi$ units

c.   $4$ units

d.   $7$ units

Ans. a. $2$ units

Perimeter / Circumference of a circle is given by, $2\pi R$.

Area of a circle is given by, $\pi {{R}^{2}}$.

Given, $2\pi R=\pi {{R}^{2}}$.

$\Rightarrow R=2$ units.

11. The radius of a circle is $\dfrac{7}{\sqrt{\pi }}$ cm, then the area of the circle is

a.   $154\text{ c}{{\text{m}}^{2}}$

b.   $\dfrac{49}{\pi }\text{ c}{{\text{m}}^{2}}$

c.   $\text{22 c}{{\text{m}}^{2}}$

d.   $\text{49 c}{{\text{m}}^{2}}$

Ans. d. $49c{{m}^{2}}$

Area of a circle is given by, $\pi {{R}^{2}}$.

$\therefore$ Area $=\pi \left( \dfrac{49}{\pi } \right)$

$\Rightarrow$ Area $=49c{{m}^{2}}$.

12. 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

Ans. d. $25$ cm

Let $D$ ($D=2R$) be the diameter of the circle whose area is equal to the sum of the areas of the two circles of radii $24$ cm and $7$ cm.

$\therefore \pi {{R}^{2}}=\pi {{\left( 24 \right)}^{2}}+\pi {{\left( 7 \right)}^{2}}$

$\Rightarrow {{R}^{2}}=576+49$

$\Rightarrow R=25$ cm

13. The circumference of a circle is $528$ cm. Then its area is

a.   $22,176\text{ }c{{m}^{2}}$

b.   $22,576\text{ }c{{m}^{2}}$

c.   $23,176\text{ }c{{m}^{2}}$

d.   $24,576\text{ }c{{m}^{2}}$

Ans. a. $22,176\text{ }c{{m}^{2}}$

Circumference of a circle is given by, $2\pi R$.

$\therefore 2\pi R=528$ cm

$\Rightarrow R=84$ cm

Thus, area $=\pi {{\left( 84 \right)}^{2}}$

$\Rightarrow$Area $=22,176\text{ }c{{m}^{2}}$

Very Short Answer Questions (2 Marks)

1. The radii of two circles are $8$ cm and $6$ cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Ans: Area of a circle is given by, $\pi {{R}^{2}}$.

Let $R$ be the radius of the circle whose area is equal to the sum of the areas of the two circles of radii $8$ cm and $6$ cm, respectively.

Thus, we have: $\pi {{R}^{2}}=\pi {{\left( 8 \right)}^{2}}+\pi {{\left( 6 \right)}^{2}}$

$\Rightarrow {{R}^{2}}=64+36$

$\Rightarrow {{R}^{2}}=100$

$\Rightarrow R=10$ cm.

2. Figure depicts an archery target marked with its five scoring area from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is $21$ cm and each of the other bands is $10.5$ cm wide. Find the area of the five scoring regions.

Ans: Given, diameter of the region representing Gold score as $21$ cm and the width of other bands as $10.5$ cm.

Thus, radius of the Gold scoring region, $R=\dfrac{21}{2}$cm.

Area of a circle is given by, $\pi {{R}^{2}}$.

Area of the Gold scoring region, $Agold=\pi {{\left( \dfrac{21}{2} \right)}^{2}}$.

$\Rightarrow Agold=346.5\text{ }c{{m}^{2}}$

Area of scoring region = Area of previous circle – Area of next circle.

$\therefore$ Area of Red scoring region, $Ared=\pi {{\left( \dfrac{21}{2}+10.5 \right)}^{2}}-\pi {{\left( \dfrac{21}{2} \right)}^{2}}$

$\Rightarrow Ared=1386-346.5$

$\Rightarrow Ared=1039.5\text{ }c{{m}^{2}}$

Area of Blue scoring region, $Ablue=\pi {{\left( \dfrac{21}{2}+10.5+10.5 \right)}^{2}}-\left( 1039.5+346.5 \right)$

$\Rightarrow Ablue=3118.5-1386$

$\Rightarrow Ablue=1732.5\text{ }c{{m}^{2}}$

Area of Black scoring region, $Ablack=\pi {{\left( \dfrac{21}{2}+10.5+10.5+10.5 \right)}^{2}}-\left( 1039.5+346.5+1732.5 \right)$

$\Rightarrow Ablack=5544-3118.5$

$\Rightarrow Ablack=2425.5\text{ }c{{m}^{2}}$

Area of White scoring region, $Awhite=\pi {{\left( \dfrac{21}{2}+10.5+10.5+10.5 \right)}^{2}}-\left( 2425.5+1039.5+346.5+1732.5 \right)$

$\Rightarrow Awhite=8662.5-5544$

$\Rightarrow Awhite=3118.5\text{ }c{{m}^{2}}$

3. The wheels of a car are of diameter $80$ cm each. How many complete revolutions does each wheel make in $10$ minutes when the car is travelling at a speed of $66$ km per hour?

Ans: Given, diameter of car wheel $=80$ cm.

$\therefore$ Radius of the car wheel $=40$ cm.

Distance covered by the wheel in one revolution is equal to the circumference of the wheel.

$\therefore 2\pi R=2\times \dfrac{22}{7}\times 40$

Thus, distance covered by the wheel in one revolution is equal to $\dfrac{1760}{7}$ cm.

Given, the distance covered in $1$ hour $=66km=6600000cm$

Hence, distance covered by the wheel in $10$ minutes $=\dfrac{6600000}{60}\times 10=1100000cm$.

Number of revolutions $=\dfrac{\text{Total distance}}{\text{Distance covered in one revolution}}$

Number of revolutions $=\dfrac{1100000\times 7}{1760}$

Hence, number of revolutions $=4375$.

4. Find the area of a sector of a circle with radius $6$ cm, if angle of the sector is ${{60}^{\circ }}$.

Ans: Given, Angle of the sector is ${{60}^{\circ }}$.

Radius, $r=6$ cm

Area of the sector of a circle $=\left( \dfrac{\theta }{{{360}^{\circ }}} \right)\times \pi {{r}^{2}}$.

$\Rightarrow$ Area $=\left( \dfrac{{{60}^{\circ }}}{{{360}^{\circ }}} \right)\times \pi {{\left( 6 \right)}^{2}}$

$\Rightarrow$ Area $=\dfrac{132}{7}\text{ }c{{m}^{2}}$.

5.   Find the area of a quadrant of a circle whose circumference is $22$ cm.

Ans: Given, the circumference of the circle as $22$ cm.

We know, the circumference of a circle is given by, $2\pi R$.

Therefore, $2\pi R=22$

$\Rightarrow R=3.5$ cm

For the quadrant of a circle, $\theta ={{90}^{0}}$

Area of quadrant is given by, $A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$

$\Rightarrow \dfrac{{{90}^{0}}}{{{360}^{0}}}\times \dfrac{22}{7}\times \dfrac{7}{2}\times \dfrac{7}{2}=\dfrac{77}{8}c{{m}^{2}}$.

$\therefore$ Area of the quadrant of the circle $=9.625\text{ }c{{m}^{2}}$.

6. The length of the minute hand of a clock is $14$ cm. Find the area swept by the minute hand in $5$ minutes.

Ans: Given, length of the minute hand (radius) , $r=14$ cm and from the given data we interpret the angle in the clock, $\theta ={{30}^{{}^\circ }}$.

Area swept by the minutes hand in $5$ minutes, $A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$

$\Rightarrow A=\dfrac{{{30}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times 14\times 14=\dfrac{154}{3}c{{m}^{2}}$

$\Rightarrow A=51.33\text{ c}{{\text{m}}^{2}}$.

7. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:  (Use $\pi =3.14$)

i. minor segment

Ans: Given, Radius, $r~=10\text{ }cm$ and the angle, $\theta ={{90}^{{}^\circ }}$.

Area of minor sector, $A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$

$\Rightarrow A=~\dfrac{{{90}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 3.14\times 10\times 10~\text{ }=\text{ }78.5\text{ }c{{m}^{2}}~$

Area of $\Delta OAB$, = $\dfrac{1}{2}$ Base $\times$ Height.

$\dfrac{1}{2}bh=~\dfrac{1}{2}\times 10\times 10~~$

$\Rightarrow 50\text{ }c{{m}^{2}}$.

Area of minor segment can be obtained as,

Area of minor segment $=$  Area of minor sector $-$  Area of $\vartriangle OAB$ .

$\Rightarrow A=78.550~$

$\Rightarrow A=28.5\text{ }c{{m}^{2}}$

ii. Major Segment

Ans: For major sector, radius = $10$ cm and $\theta ={{360}^{{}^\circ }}-{{90}^{{}^\circ }}={{270}^{{}^\circ }}$

$\therefore$Area of major sector $A=~\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}~$

$\Rightarrow ~A=\dfrac{{{270}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \text{3}\text{.14}\times \text{10}\times \text{10}$

$\Rightarrow \text{A}=\text{ }235.5\text{ }c{{m}^{2}}~$.

8. A horse is tied to a peg at one corner of a square shape grass field of side $15$ m by means of a $5$ m long rope (see figure). Find:

i. the area of that part of the field in which the horse can graze.

Ans: Area of a sector of a circle is given by, $A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$

Area of the part of the field where the horse can graze is in the form of a quadrant with radius, $r=5m$.

For the quadrant of a circle, $\theta ={{90}^{0}}$

Thus, the area is given as, $A=\dfrac{{{90}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \text{3}\text{.14}\times \text{5}\times \text{5}$

$\Rightarrow ~A=19.625\text{ }{{m}^{2}}$.

ii. the increase in the grazing area if the rope were $10$ m long instead of $5$ m.

Ans: Area of quadrant with $10$m rope, the radius would be, $r=10m$.

Thus, the grazing area for $10$ m long rope is given by, $~A=\dfrac{{{90}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 3.14\times 10\times 10~$.

$\Rightarrow A=78.5\text{ }{{m}^{2}}$.

$\therefore$The increase in grazing area can be calculated by subtracting area of $5$m rope from the area of $10$m rope.

Hence, Increase in area $=78.519.625=58.875\text{ }{{m}^{2}}$

9. A brooch is made with silver wire in the form of a circle with diameter $35$ mm. The wire is also used in making $5$ diameters which divide the circle into $10$ equal sectors as shown in figure. Find:

i. The total length of the silver wire required.

Ans: Given, diameter of silver wire circle as $35$ mm

Thus, radius is $\dfrac{35}{2}$ mm.

Circumference of a circle is given by, $~2\pi r$.

Therefore, the circumference of the circle made by silver wire is,

$~\Rightarrow 2\times \dfrac{22}{7}~\times \dfrac{35}{2}~=110$ mm   ……(1)

As the wire is used in making $5$ diameters,

$\therefore$ length of $5$  diameters, $L=35\times 5$

$\Rightarrow L=175$ mm   ……(2)

Total length of the silver wire required would be the sum of circumference of the circle and the length of $5$ diameters.

Hence, adding equation (1) and (2), we get

$\Rightarrow 110\text{ }+\text{ }175\text{ }=\text{ }285\text{ }$ mm

ii. The area of each sector of the brooch.

Ans: Given, diameter of silver wire circle as $35$ mm

Thus, radius is $\dfrac{35}{2}$ mm.

Angle of each sector,$\theta =\dfrac{{{360}^{{}^\circ }}}{10}={{36}^{{}^\circ }}$.

$\therefore$The area of each sector of the brooch is given by, $A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$

$\Rightarrow A=\dfrac{{{36}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times \dfrac{35}{2}\times \dfrac{35}{2}~$

$\Rightarrow A=~\text{ }\dfrac{385}{4}m{{m}^{2}}~$

10. An umbrella has $8$  ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of  radius $45$ cm,  find  the area between  the  two  consecutive  ribs of  the umbrella.

Ans: Let umbrella be a flat circle with radius,$r=\text{ }45\text{ }cm$.

Given, the umbrella has $8$ ribs, thus the angle $\theta$ would be, $\theta =\dfrac{{{360}^{{}^\circ }}}{8}={{45}^{{}^\circ }}$

Area between two consecutive ribs of the umbrella can be obtained using the formula, $A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$.

$\Rightarrow A=\dfrac{{{45}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times 45\times 45~$

$\Rightarrow A=~\dfrac{22275}{28}\text{ }c{{m}^{2}}$

$\Rightarrow A=795.54\text{ }c{{m}^{2}}$.

11. A car has two wipers which do not overlap. Each wiper has a blade of length $25$ cm sweeping an angle of ${{115}^{\circ }}$. Find the total area cleaned at each sweep of the blades.

Ans: Given, a car has two wipers with a blade length of $25$ cm sweeping an angle of ${{115}^{\circ }}$.

Let us suppose radius, $~r~=\text{ }25\text{ }cm$ and

Angle, $\theta ={{115}^{{}^\circ }}$.

The two wipers do not overlap.

$\therefore$ The total area cleaned at each sweep of the blade will be,  $2\times \left( \dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}} \right)$

$\Rightarrow 2\times \left( \dfrac{{{115}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times 25\times 25 \right)=\dfrac{158125}{126}c{{m}^{2}}$

Hence, area cleaned at each sweep is $1254.96\text{ }c{{m}^{2}}$.

12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle ${{80}^{{}^\circ }}$ to a distance of $16.5$ km. Find the area of the sea over which the ships are warned. (Use$\pi =3.14$)

Ans: Given, a lighthouse spreads a red light over a sector of angle ${{80}^{\circ }}$ to a distance of $16.5$ km.

Let us suppose radius, $r~=16.5\text{ }km$ and

Angle, $\theta ={{80}^{{}^\circ }}$.

$\therefore$The area of sea over which the ships are warned is, $A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$$\Rightarrow A=\dfrac{{{80}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 3.14\times 16.5\times 16.5~$

$\Rightarrow A=189.97\text{ }k{{m}^{2}}~$

13.Find the area of the shaded region in figure, if radii of the two concentric circles with centre O are $7$ cm and $14$cm respectively and $\angle$AOC = ${{40}^{\circ }}$.

Ans: Area of shaded region $ABDC$ can be determined by subtracting area of sector $OBD$ from the area of sector $OAC$, i.e.,

Area of shaded region $ABDC=$ Area of sector $OAC-$ Area of sector $OBD$

Area of sector is given by, $A=~\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}~$.

Thus, Area of sector $OAC$$=\dfrac{{{40}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times {{(14)}^{2}} Area of sector OBD$$=\dfrac{{{40}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times {{(7)}^{2}}$

$\therefore$ Area of shaded region $ABDC=\dfrac{{{40}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times {{(14)}^{2}}-\dfrac{{{40}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times {{(7)}^{2}}$

$\Rightarrow Area=\dfrac{{{40}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\left[ {{(14)}^{2}}-{{(7)}^{2}} \right]=\dfrac{{{40}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times 7\times 21$

$\Rightarrow$ Area $=\dfrac{154}{3}~c{{m}^{2}}$.

Hence, area of the shaded region is $51.33\text{ }c{{m}^{2}}$.

14. Find the area of the shaded region in figure, if ABCD is a square of side $14$ cm and APD and BPC are semicircles.

Ans: From the given figure,

Area of shaded region = Area of square ABCD – (Area of semicircle APD + Area of semicircle BPC)

$\Rightarrow$Area of the shaded region $=14\times 14-\left[ \dfrac{1}{2}\times \dfrac{22}{7}{{\left( \dfrac{14}{2} \right)}^{2}}+\dfrac{1}{2}\times \dfrac{22}{7}{{\left( \dfrac{14}{2} \right)}^{2}} \right]$

$\Rightarrow$Area of the shaded region $=196-\dfrac{22}{7}\times 7\times 7$

$\therefore$ Area of the shaded region is $42c{{m}^{2}}$.

15. Find the area of the shaded region in figure, where a circular arc of radius $6$ cm has been drawn with vertex O of an equilateral triangle OAB of side $12$ cm as centre.

Ans: Area of the shaded region = Area of circle + Area of equilateral triangle $OAB$ - Area of the region common to the circle and the triangle.

Area of a circle is given by, $\pi {{R}^{2}}$.

Area of equilateral triangle is given by, $\dfrac{\sqrt{3}}{4}{{\left( a \right)}^{2}}$ ($a$ is the side)

$\Rightarrow$Area of the shaded region $=\pi {{(6)}^{2}}+\dfrac{\sqrt{3}}{4}{{(12)}^{2}}-\dfrac{{{60}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \pi {{(6)}^{2}}$

$\Rightarrow$Area of the shaded region $=30\pi +36\sqrt{3}=\left( \dfrac{660}{7}+36\sqrt{3} \right)\text{c}{{\text{m}}^{2}}$

Hence, the area of the shaded region is $156.63\text{ }c{{m}^{2}}$.

16. From each corner of a square of side $4$ cm a quadrant of a circle of radius $1$ cm is cut and also a circle of diameter of $2$ cm is cut as shown in figure. Find the area of the remaining portion of the figure:

Ans: Given, a square of side $4$ cm. Four quadrants of a circle of radius $1$ cm are cut from each corners of the square.

Also, a circle of diameter $2$ cm is cut.

$\therefore$ Area of the remaining portion = Area of square – ($4\times$ Area of quadrant of radius $1$ cm + Area of circle of diameter $2$ cm).

Area of a square is given as ${{\left( side \right)}^{2}}$.

Area of a circle is given by, $\pi {{R}^{2}}$.

Similarly, area of quadrant is given by $\left( \dfrac{\theta }{{{360}^{\circ }}} \right)\times \pi {{r}^{2}}$.

Thus,

Area of the remaining portion $=4\times 4-\left[ 4\times \dfrac{{{90}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times {{(1)}^{2}}+\dfrac{22}{7}\times {{\left( \dfrac{2}{2} \right)}^{2}} \right]$

$\Rightarrow$ Area of the remaining portion $=16-2\times \dfrac{22}{7}=\dfrac{68}{7}\text{ c}{{\text{m}}^{2}}$.

Hence, area of the remaining portion is $9.71\text{ }c{{m}^{2}}$.

17. In a circular table cover of radius $32$cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in figure. Find the area of the design (shaded region).

Ans: Given, circular table cover of radius, $r=32cm$ and an equilateral triangle $ABC$ is drawn in the middle.

Area of the design can be obtained by,

Area of design = Area of circle – Area of the equilateral triangle $ABC$.

Area of a circle is given by, $\pi {{R}^{2}}$.

Area of equilateral triangle is given by, $\dfrac{\sqrt{3}}{4}{{\left( a \right)}^{2}}$ ($a$ is the side).

Thus, area of design $=\pi {{\left( 32 \right)}^{2}}-\dfrac{\sqrt{3}}{4}{{\left( a \right)}^{2}}$   ……(1)

$\because$G is the centroid of the equilateral triangle.

$\therefore$radius of the circumscribed circle =  $\dfrac{2}{3}h$ cm

So, we have $32=\dfrac{2}{3}h$

$\Rightarrow h=48$ cm

From the figure of equilateral triangle, ${{a}^{2}}={{h}^{2}}+{{\left( \dfrac{a}{2} \right)}^{2}}$.

$\Rightarrow \dfrac{3{{a}^{2}}}{4}={{h}^{2}}$

$\Rightarrow {{a}^{2}}=\dfrac{4\times {{\left( 48 \right)}^{2}}}{3}$

$\Rightarrow a=\sqrt{3072}$

Substituting for $a$ in equation (1), we get:

Area of design $=\pi {{\left( 32 \right)}^{2}}-\dfrac{\sqrt{3}}{4}{{\left( \sqrt{3072} \right)}^{2}}$

$\Rightarrow \dfrac{22}{7}\times 1024-768\sqrt{3}=\left( \dfrac{22528}{7}-768\sqrt{3} \right)c{{m}^{2}}$

Thus, area of the design is $1888.08\text{ }c{{m}^{2}}$.

18. In figure ABCD is a square of side $14$cm. With centres A, B, C and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region.

Ans: Given, a square of side $14$ cm. With the corners of the square as centres four circles are drawn.

Thus, area of the shaded region = Area of the square – ($4\times$ Area of sector)

Area of a circle is given by, $\pi {{R}^{2}}$ and area of sector is given by, $\left( \dfrac{\theta }{{{360}^{\circ }}} \right)\times \pi {{r}^{2}}$Area of the shaded region $=14\times 14-4\times \dfrac{{{90}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times {{\left( \dfrac{14}{2} \right)}^{2}}=196-\dfrac{22}{7}\times 7\times 7$

$Hence, area of the shaded region is$42\text{ }c{{m}^{2}}$. 19. The given figure depicts a racing track whose left and right ends are semi-circular. The difference between the two inner parallel line segments is $60\text{ m}$ and they are each $106\text{ }m$ long. If the track is $10m$ wide, find: i. The distance around the track along its inner edge, Ans: Given that, difference between the two inner parallel line segments is $60m$ and they are each $106m$ long and the track is $10m$ wide. Distance around the track along its inner edge$\Rightarrow 106+106+2\times \left[ \dfrac{{{180}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 2\times \dfrac{22}{7}\times \left( \dfrac{60}{2} \right) \right]=212+60\times \dfrac{22}{7}\Rightarrow 212+\dfrac{1320}{7}=\dfrac{2804}{7}m$ii. The area of the track. Ans: Area of track can be calculated as,$ \Rightarrow 106\times 10+106\times 10+2\times \left[ \dfrac{1}{2}\times \dfrac{22}{7}{{(30+10)}^{2}}-\dfrac{1}{2}\times \dfrac{22}{7}{{(30)}^{2}} \right]  =1060+1060+\dfrac{22}{7}\left[ {{(40)}^{2}}-{{(30)}^{2}} \right] \Rightarrow 2120+\dfrac{22}{7}\times 700=4320{{m}^{2}}$20. In figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA =$7$cm, find the area of the shaded region. Ans: Given, radius of the bigger circle, OA or$R=7$cm. Diameter of the smaller circle is the radius of the bigger circle. Thus, Radius of the smaller circle,$r=\dfrac{7}{2}$cm. Area of the circle is given by,$\pi {{R}^{2}}$and Area of the semicircle is given by,$\dfrac{\pi {{r}^{2}}}{2}$.$\Delta ACB$is made up of two right- angled triangles and we know, area of a right-angled triangle is given as$\dfrac{1}{2}\times base\times height$. Area of the shaded region can be obtained as: Area of shaded region = Area of smaller circle + Area of semicircle$ACB\text{ }-$Area of$\Delta ACB$. Area of shaded region$=\dfrac{22}{7}\times {{\left( \dfrac{7}{2} \right)}^{2}}+\dfrac{1}{2}\times \dfrac{22}{7}\times {{(7)}^{2}}-\left( \dfrac{7\times 7}{2}+\dfrac{7\times 7}{2} \right)=\dfrac{133}{2}$Thus, area of the shaded region is$66.5\text{ }c{{m}^{2}}$. 21. The area of an equilateral triangle ABC is$17320.5\text{ }c{{m}^{2}}$. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region. (Use$\pi =3.14$and$\sqrt{3}=1.73205$) Ans: Given, area of the equilateral triangle$ABC=17320.5\text{ }c{{m}^{2}}$. Also, radius of each circle is half the length of the side of the equilateral triangle. Area of the equilateral triangle is given by,$\dfrac{\sqrt{3}}{4}\times {{\left( side \right)}^{2}}$. Let side of the equilateral triangle$ABC$be$a$.$\therefore $$\dfrac{\sqrt{3}}{4}\times {{a}^{2}}=17320.5 \Rightarrow {{a}^{2}}=40000 \Rightarrow a=200 cm \therefore  radius of circle, r=100 cm Area of the shaded region can be obtained as: Area of shaded region = Area of the equilateral triangle ABC\text{ }- 3\times  Area of the sector. We know, area of the sector of a circle =\left( \dfrac{\theta }{{{360}^{\circ }}} \right)\times \pi {{r}^{2}}. Thus, Area of shaded region =17320.5-3\times \left( \dfrac{{{60}^{\circ }}}{{{360}^{\circ }}} \right)\times 3.14\times {{\left( 100 \right)}^{2}} $\Rightarrow 17320.5-15700=1620.5cm{}^\text{2}$ Hence, area of the shaded region is 1620.5\text{ }c{{m}^{2}}. 22. On a square handkerchief, nine circular designs each of radius 7cm are made (see figure). Find the area of the remaining portion of the handkerchief. Ans: Given, radius of circular design is 7 cm. Observing the figure, the side of the square is 6\times 7=42 cm. The handkerchief consists of 9 circular designs. Thus, area of the shaded region = Area of square ABCD\text{ }-\text{ 9}\times  Area of circular design. Area of shaded region =\left( 42\times 42 \right)-9\times \pi \times {{\left( 7 \right)}^{2}} \Rightarrow 1764-1386=378\text{ }c{{m}^{2}} Hence, area of the shaded region is 378\text{ }c{{m}^{2}}. 23. In figure, OACB is a quadrant of a circle with centre O and radius 3.5cm. If OD = 2 cm, find the area of the: i. quadrant OACB Ans: Given radius, r=3.5\text{ }cm and OD=2 cm. Area of the quadrant of a circle =\left( \dfrac{\theta }{{{360}^{\circ }}} \right)\times \pi {{r}^{2}}. \therefore  Area of the quadrant OACB=\dfrac{{{90}^{\circ }}}{{{360}^{\circ }}}\times \dfrac{22}{7}\times {{\left( 3.5 \right)}^{2}} $\Rightarrow ~\dfrac{1}{4}\times \dfrac{22}{7}\times \dfrac{35}{10}\times \dfrac{35}{10}=~\dfrac{77}{8}c{{m}^{2}}~$ Area of the quadrant OACB is 9.625\text{ }c{{m}^{2}}. ii. shaded region Ans: Given radius, r=3.5\text{ }cm and OD=2 cm. Area of the shaded region = Area of the quadrant OACB\text{ }- Area of the triangle BOD. Area of the shaded region =9.625-\dfrac{1}{2}\times 3.5\times 2=6.125\text{ }c{{m}^{2}} Hence, area of the shaded region is 6.125\text{ }c{{m}^{2}}. 24.In figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use \pi =3.14) Ans: Given, a square OABC inscribed in a quadrant OPBQ and OA=20 cm. \therefore OA=AB=BC=OC. Taking the \Delta OAB into consideration and applying Pythagoras Theorem, O{{B}^{2}}=O{{A}^{2}}+A{{B}^{2}} \Rightarrow O{{B}^{2}}={{\left( 20 \right)}^{2}}+{{\left( 20 \right)}^{2}} \Rightarrow OB=\sqrt{800} \Rightarrow OB=20\sqrt{2} cm Area of the shaded region = Area of quadrant OPBQ - Area of square OABC. We know, area of the quadrant of a circle =\left( \dfrac{\theta }{{{360}^{\circ }}} \right)\times \pi {{r}^{2}}. \therefore  Area of shaded region =\dfrac{{{90}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 3.14{{\left( 20\sqrt{2} \right)}^{2}}-20\times 20 \Rightarrow \dfrac{1}{4}\times 3.14\times 800-400=228\text{ }c{{m}^{2}} Hence, area of the shaded region is 228\text{ }c{{m}^{2}}. 25. AB and CD respectively are the arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If \angle AOB={{30}^{\circ }}. Find the area of the shaded region. Ans: Given, two arcs AB and CD of two concentric circles with radii 21 cm and 7 cm respectively. Also, \angle AOB={{30}^{\circ }} We know, area of the quadrant of a circle =\left( \dfrac{\theta }{{{360}^{\circ }}} \right)\times \pi {{r}^{2}}. Area of the shaded region = Area of arc AB - Area of arc CD. \therefore  Area of shaded region =\left( \dfrac{{{30}^{{}^\circ }}}{{{360}^{{}^\circ }}} \right)\times \dfrac{22}{7}\times {{\left( 21 \right)}^{2}}-\left( \dfrac{{{30}^{{}^\circ }}}{{{360}^{{}^\circ }}} \right)\times \dfrac{22}{7}\times {{\left( 7 \right)}^{2}} \Rightarrow \left( \dfrac{{{30}^{{}^\circ }}}{{{360}^{{}^\circ }}} \right)\times \dfrac{22}{7}\times \left[ {{\left( 21 \right)}^{2}}-{{\left( 7 \right)}^{2}} \right]=\dfrac{308}{3}\text{ }c{{m}^{2}} Hence, area of the shaded region is \text{102}\text{.67 }c{{m}^{2}}. 26. In figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region. Ans: Given, a quadrant ABC of a circle of radius 14\text{ }cm. Let us take, \Delta ABC and apply Pythagoras Theorem. \therefore A{{B}^{2}}+A{{C}^{2}}=B{{C}^{2}} \Rightarrow {{\left( 14 \right)}^{2}}+{{\left( 14 \right)}^{2}}=B{{C}^{2}} \Rightarrow BC=\sqrt{392} \Rightarrow BC=14\sqrt{2}\text{ }cm. \therefore  Radius of the semicircle =\dfrac{14\sqrt{2}}{2}\text{=7}\sqrt{2}\text{ }cm. We know, area of the quadrant of a circle =\left( \dfrac{\theta }{{{360}^{\circ }}} \right)\times \pi {{r}^{2}}. Area of the shaded region BPCQ= Area of BCQB - (Area of BACP - Area of \Delta ABC). \therefore  Area of shaded region $=\dfrac{{{180}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}{{\left( 7\sqrt{2} \right)}^{2}}-\left[ \dfrac{{{90}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times {{\left( 14 \right)}^{2}}-\dfrac{14\times 14}{2} \right]~$ \Rightarrow \dfrac{1}{2}\times \dfrac{22}{7}\times 98-\left( \dfrac{1}{4}\times \dfrac{22}{7}\times 196-98 \right)=98\text{ }c{{m}^{2}} Hence, area of the shaded region is \text{98 }c{{m}^{2}}. 27.Calculate the area of the designed region in figure common between the two quadrants of circles of radius 8 cm each. Ans: Given, two quadrants each of radii 8 cm which has a design in the common region. Let us consider the figure as: Applying Pythagoras Theorem in the right- angled triangle ABC, $A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}$ $~\Rightarrow A{{C}^{2}}=\text{ }{{\left( 8 \right)}^{2}}+\text{ }{{\left( 8 \right)}^{2}}=\text{ }2{{\left( 8 \right)}^{2}}~$ $~\Rightarrow ~AC=\sqrt{128}=8\sqrt{2}~~cm~$ Draw a perpendicular, $BM\bot AC.~$ Then $AM=MC=~\text{ }\dfrac{1}{2}AC~$ $\Rightarrow ~\dfrac{1}{2}\times 8\sqrt{2}~=~4\sqrt{2}~cm~$ From the figure the right triangle AMB, (AB)^2 = (AM)^2 +(BM)^2 [Pythagoras theorem] $\Rightarrow {{\left( 8 \right)}^{2}}={{\left( 4\sqrt{2} \right)}^{2}}+B{{M}^{2}}$ $\Rightarrow B{{M}^{2}}=6432=32$ $\Rightarrow BM\text{ }=4\sqrt{2}~~cm$ Area of $\Delta ABC\text{ }=\dfrac{1}{2}~~\times \text{ }AC\text{ }\times \text{ }BM~$ $\Rightarrow ~\dfrac{8\sqrt{2}\times 4\sqrt{2}}{2}~=\text{ }32\text{ }c{{m}^{2}}~$ We know, area of the quadrant of a circle =\left( \dfrac{\theta }{{{360}^{\circ }}} \right)\times \pi {{r}^{2}}. Area of half shaded region = Area of quadrant ABC - Area of \Delta ABC. \therefore Area of half shaded region =\dfrac{{{90}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times {{\left( 8 \right)}^{2}}-32 $\Rightarrow ~16\times \dfrac{22}{7}-32~=~\dfrac{128}{7}\text{ }c{{m}^{2}}~$ Thus, area of designed region can be obtained as, twice the area of half shaded region. $\Rightarrow 2\text{ }\times \dfrac{128}{7}~~=~\dfrac{256}{7}\text{ }c{{m}^{2}}~$ Hence, area of the designed region is 36.57\text{ }c{{m}^{2}}. 28. Find the circumference of a circle of diameter 14cm. Ans: Given, diameter of circle, d=14 cm Circumference of a circle is given by, 2\pi r Since, d=14 cm \Rightarrow r=\dfrac{14}{2}=7 cm \therefore r=7cm Thus, the circumference of the circle is, C=2\pi r \Rightarrow C=2\times \dfrac{22}{7}\times 7 \Rightarrow C=44 cm The circumference of the circle is 44 cm. 29. The diameter of a circular pond is 17.5m. It is surrounded by a path of width 3.5m. Find the area of the path. Ans: Given, diameter of circular pond, d=17.5 m. \therefore  Radius, $r=\dfrac{17.5}{2}=8.25$ m It is surrounded by a path of width 3.5 m. \therefore Outer radius $R$ is the sum of circular pond radius with the width of the surrounded path. $\Rightarrow R=8.75+3.5=12.25$ m Equation for area of path A is given by \pi \left[ {{R}^{2}}-{{r}^{2}} \right]. \Rightarrow A=\pi \left[ \left( R+r \right)\left( R-r \right) \right] \Rightarrow A=\dfrac{22}{7}\left[ \left( 12.25+8.75 \right)\left( 12.25-8.75 \right) \right] \Rightarrow A=\dfrac{22}{7}\times 21\times 3.50 \Rightarrow A=231~{{\text{m}}^{2}} Hence, area of the path is 231~{{\text{m}}^{2}}. 30. Find the area of the shaded region where ABCD is a square of side 14 cm. Ans: Given, side of square, a=14 cm. Area of the shaded region can be calculated by subtracting area of 4 circles from the area of square ABCD, i.e., Area of shaded region = Area of square ABCD -\text{ }4\times Area of circle. Area of the given square ABCD is given by, A=14cm\times 14cm \Rightarrow A=196c{{m}^{2}} From the figure, we observe that the diameter of each circle is half the side of the square. Thus, diameter of circle =\dfrac{14}{2}=7 cm \therefore  Radius of each circle =\dfrac{7}{2}cm Area of the circle =\pi {{r}^{2}} \Rightarrow \dfrac{22}{7}\times {{\left( \dfrac{7}{2} \right)}^{2}}c{{m}^{2}} \therefore  Area of shaded region $=196\text{ }c{{m}^{2}}\left( 4\times 38.5 \right)\text{ }c{{m}^{2}}~$ \Rightarrow \left( 196-154 \right)c{{m}^{2}}=42\text{ }c{{m}^{2}} Hence, area of the shaded region is $42\text{ }c{{m}^{2}}$. 31. The radius of a circle is 20cm. Three more concentric circles are drawn inside it in such a manner that it is divided into four parts of equal area. Find the radius of the largest of the three concentric circles. Ans: Given radius of the circle, r=20 cm Area of a circle is given by, A=\pi {{r}^{2}} Substituting for radius, r. \therefore A=\pi {{\left( 20 \right)}^{2}} \Rightarrow A=400\pi c{{m}^{2}} As the circle is divided into four parts of equal area, area of each circle would be, \Rightarrow \dfrac{1}{4}400\pi c{{m}^{2}}=100\pi c{{m}^{2}} Let R be the radius of the largest of the three circles and its area is 400\pi -100\pi =300\pi cm. Thus the radius, R can be calculated as \Rightarrow \pi {{R}^{2}}=300\pi  \text{ }{{\text{R}}^{2}}=300 \therefore R=\sqrt{300}\text{ =}10\sqrt{3} cm Radius of the largest of the three concentric circles is 10\sqrt{3} cm. 32. OACB is a quadrant of a circle with centre O and radius 7 cm. If OD = 4cm, then find area of shaded region. Ans: Given radius r=7 cm and OD=4 cm Area of quadrant is given by, A=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}} From the figure it is clear that, angle \theta ={{90}^{\circ }} \therefore  Area of quadrant OACB=\dfrac{{{90}^{\circ }}}{{{360}^{\circ }}}\pi {{\left( 7 \right)}^{2}} \Rightarrow \dfrac{1}{4}\times \dfrac{22}{7}\times 49=\dfrac{77}{2}c{{m}^{2}} Area of \Delta OAD=\dfrac{1}{2}\left( 7\times 4 \right) Area of shaded region = Area of quadrant OACB$$-$Area of$\Delta AOB\Rightarrow \dfrac{77}{2}-\dfrac{1}{2}\left( 7\times 4 \right)=\dfrac{49}{2}c{{m}^{2}}$Thus, area of the shaded region is$24.5\text{ }c{{m}^{2}}$. 33. A pendulum swings through on angle of${{30}^{{}^\circ }}$and describes an arc$8.8$cm in length. Find the length of pendulum. Ans: Given, pendulum swings at angle${{30}^{\circ }}$, with an arc of length$8.8$cm. Sketching the figure, we have: In the below given figure let$r$be the length of pendulum and$\angle