## 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.

## Download CBSE Class 10 Maths Important Questions 2024-25 PDF

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 | |

2 | Chapter 2 | |

3 | Chapter 3 | |

4 | Chapter 4 | |

5 | Chapter 5 | |

6 | Chapter 6 | |

7 | Chapter 7 | |

8 | Chapter 8 | |

9 | Chapter 9 | |

10 | Chapter 10 | |

11 | Chapter 11 | |

12 | Chapter 12 | Areas Related to Circles |

13 | Chapter 13 | |

14 | Chapter 14 | |

15 | Chapter 15 |

## 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 $7$cm 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 $3.5$cm. 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 \[\Delta ABC\text{ }=\dfrac{1}{2}~~\times \text{ }AC\text{ }\times \text{ }BM~\]

\[\Rightarrow ~\dfrac{8\sqrt{2}\times 4\sqrt{2}}{2}~=\text{ }32\text{ }c{{m}^{2}}~\]

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 $14$cm.

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 $17.5$m. It is surrounded by a path of width $3.5$m. 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 $20$cm. 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 = $4$cm, 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

AOB={{30}^{{}^\circ }}$ then,$\angle AOB=\dfrac{\pi }{{{180}^{{}^\circ }}}\times {{30}^{{}^\circ }}$

$\Rightarrow \angle AOB=\dfrac{\pi }{6}$

Use $\theta =\dfrac{l}{r}$ to determine the length of the pendulum.

$\Rightarrow \dfrac{\pi }{6}=\dfrac{8.8}{r}$

$\therefore r=\dfrac{8.8\times 6}{\pi }$

$\Rightarrow r=16.8cm$

The length of the pendulum is $16.8$ cm.

34. The cost of fencing a circular field at the rate of Rs. $24$ per metre is Rs. $5280$. The field is to be ploughed at the rate of Rs.$0.50$per $m^2$. Find the cost of ploughing the field. (Take $\pi =\dfrac{22}{7}$)

Ans: As per the question the cost of fencing a circular field at the rate of Rs$24$, per meter is Rs \[5280.\]

With Rs. $24$, the length of fencing $=1$ metre

$\therefore $ for Rs.$5280$, the length fencing$=\dfrac{1}{24}\times 5280=220$meters

$\therefore $Perimeter i.e., circumference of the field $=220$meters

Let $r$ be the radius of the field.

Circumference is given by, $2\pi R$

$\therefore 2\pi r=220$

$\Rightarrow r=\dfrac{220\times 7}{2\times 22}=35m$

Area of the field, $A=\pi {{r}^{2}}$.

$\Rightarrow \pi {{\left( 35 \right)}^{2}}=1225\pi {{m}^{2}}$

Rate = Rs.\[0.50\] per m2

Total cost of ploughing the field

$\Rightarrow Rs.\left( 1225\pi \times 0.50 \right)=Rs.\dfrac{1225\times 22\times 1}{7\times 2}$

$\Rightarrow Rs.\left( 175\times 11 \right)=Rs.1925$

Hence, the cost of ploughing the field is Rs.$1925$.

35. Find the difference between the area of regular hexagonal plot each of whose side $72$ m and the area of the circular swimming take in scribed in it. (Take $\pi =\dfrac{22}{7}$)

Ans: Given, side of hexagonal plot \[=72\]m

Area of equilateral triangle is given by$A=\dfrac{\sqrt{3}}{4}{{(side)}^{2}}$

$\therefore A\Delta AOB=\dfrac{\sqrt{3}}{4}{{(72)}^{2}}=1296\sqrt{3}{{m}^{2}}$

Area of hexagonal plot $=6\times $ Area of triangle $AOB$

$\Rightarrow 6\times 1296\sqrt{3}=13468.032~{{\text{m}}^{2}}$

Now,$O{{C}^{2}}=O{{A}^{2}}-A{{C}^{2}}$

$\Rightarrow {{(72)}^{2}}-{{\left( \dfrac{72}{2} \right)}^{2}}~=5184-1296=3888$

$\Rightarrow O{{C}^{2}}=3888$

$\Rightarrow OC=\sqrt{3888}=62.35~\text{m}$

Area of circular region, $A=\pi {{r}^{2}}$

$\Rightarrow \dfrac{22}{7}\times {{\left( 62.35 \right)}^{2}}=12,219.42\text{ }{{m}^{2}}$.

Then the difference is, $13468.032~{{\text{m}}^{2}}-12,219.42\text{ }{{m}^{2}}=1248.603\text{ }{{m}^{2}}$.

36. In the given figure areas 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.

Ans: Given, areas drawn with radius $21$ cm at vertices, A,B,C and D.

Thus, required area is the area of the circle with radius $21$ cm.

$\Rightarrow A=\pi {{\left( 21 \right)}^{2}}=\dfrac{22}{7}\times 21\times 21$

$\Rightarrow A=1386c{{m}^{2}}$

37. A wheel has diameter \[84cm\], find how many complete revolutions it must make to complete $792$ meters.

Ans: Given, diameter of wheel, \[d=2r\text{ }=\text{ }84\] cm.

Distance covered in one revolution = circumference

Circumference of the circle is given by $2\pi r$.

Distance covered in one revolution $=2\pi r=\pi \left( 2r \right)=\dfrac{22}{7}\times 84$

$\Rightarrow 22\times 12=264$ cm

Thus, for distance covered\[264cm\], number of revolutions$=1$

$\therefore $For distance covered $792$m/ \[79200\] cm, number of revolutions$=\dfrac{1}{264}\times 79200=300$ .

Number of revolutions to complete $792$ m is $300$.

38. The given figure is a sector of a circle of radius \[10.5cm\]. Find the perimeter of the sector. (Take $\pi =\dfrac{22}{7}$ )

Ans: Given radius, $r=10.5$ cm

We know that circumference, i.e., perimeter of a sector of angle ${{P}^{{}^\circ }}$ of a circle with radius $R$ is given by, $\dfrac{{{P}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 2\pi R+2R$.

$\therefore $Required perimeter can be calculated as,

$\Rightarrow \dfrac{{{60}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{2\times 22}{7}\times \left( 10.5 \right)+2\left( 10.5 \right)=\dfrac{1}{6}\times \dfrac{44}{7}\times \dfrac{21}{2}+21$

$=32cm$

Thus, the perimeter of a sector is $32$ cm.

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

Ans: Given, radius of each wiper\[=\text{ }25\] cm and sweeping angle, $\theta ={{115}^{{}^\circ }}$.

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

Total area cleaned at each sweep of the blades $=2\left( \dfrac{115}{360}\times \dfrac{22}{7}\times 25\times 25 \right)$

$\Rightarrow \dfrac{230\times 22\times 5\times 25}{72\times 7}=\dfrac{230\times 11\times 125}{36\times 7}$

$\Rightarrow \dfrac{115\times 11\times 125}{18\times 7}=1254.96~\text{c}{{\text{m}}^{2}}$

Thus, the total area cleaned at each sweep of the blades is $1254.96~\text{c}{{\text{m}}^{2}}$

40. In the given figure arcs have been drawn with radii \[14\text{ }cm\] each and with centres P, Q and R. Find the area of the shaped region.

Ans: Given, arc with radius $14$ cm each.

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

From the figure area of shaded region is,$A=3\times \dfrac{{{60}^{\circ }}}{{{360}^{{}^\circ }}}\times \pi {{\left( 14 \right)}^{2}}=\dfrac{22}{7}\times \dfrac{{{\left( 14 \right)}^{2}}}{2}$

$\Rightarrow A=11\times 2\times 14=308c{{m}^{2}}$

Thus, area of shaded region is $308\text{ }c{{m}^{2}}$.

41. The radii of two circles are \[19cm\] and \[9cm\] respectively. Find the radius of the circle which has its circumference equal to the sum of the circumference of the two circles.

Ans: Given, radii of two circles as $19$ cm and $9$ cm.

Circumference of the circle is given by \[C=2\pi r\].

C1 = circumference of the 1st circle

\[\Rightarrow 2\pi \left( 19 \right)=\text{ }38\pi \text{ }cm~\]

C2 = circumference of the 2nd circle

\[\Rightarrow 2\pi \left( 19 \right)=18\pi \text{ }cm~\]

Circumference of the required circle \[=\] sum of circumferences of two circles

\[\Rightarrow 2\pi r=38\pi +18\pi \text{ }cm~\]

\[\Rightarrow 2\pi r=56\pi \]

\[\Rightarrow r=28cm\]

Thus, the radius of the required circle is $28$ cm.

42. A car travels \[0.99\,km\] distance in which each wheel makes \[450\] complete revolutions. Find the radius of its wheel

Ans: Given, distance travelled by a wheel in 450 complete revolutions \[=0.99\] km \[=\text{ }990m\]

Distance travelled in one revolution $=\dfrac{990}{450}=\dfrac{11}{5}$.

Let $r$ be the radius of the wheel.

Circumference of the circular field is given by, $2\pi r$.

$\therefore 2\pi r=\dfrac{11}{5}$

$\Rightarrow 2\times \dfrac{22}{7}r=\dfrac{11}{5}$

$\Rightarrow r=\dfrac{7\times 100}{20}=35m$.

43. A sector is cut from a circle of diameter \[21cm\]. If the angle of the sector is \[\mathbf{15}{{\mathbf{0}}^{{}^\circ }}\] find its area.

Ans: Given, diameter, \[d=\text{ }21cm~\]

Thus, radius can be obtained as, \[r=~\text{ }\dfrac{21}{2}cm~\]

Angle of sector $={{150}^{{}^\circ }}$

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

Hence, area of the sector is, $A=\dfrac{5}{12}\times \dfrac{22}{7}\times \dfrac{21}{2}\times \dfrac{21}{2}=\dfrac{5\times 11\times 21}{4\times 2}$

$\Rightarrow A=144.38\text{ }c{{m}^{2}}$.

44. In the given figure AOBCA represent a quadrant of area \[~9.625c{{m}^{2}}\]. Calculate the area of the shaded portion.

Ans: Given, area of quadrant $9.625\text{ }c{{m}^{2}}$.

From the figure, $OD=2$ cm and $OA=3.5$ cm.

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

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, the area of the shaded region is $6.125\text{ }c{{m}^{2}}$.

Short Answer Questions (3 Marks)

1. Find the area of the shaded region if \[PQ=24cm,\text{ }PR=7cm\] and O is the centre of the circle.

Ans: Given \[PQ=24\text{ }cm\] and \[PR=7\text{ }cm\]

Since QR is a diameter passing through the centre O of the circle

Angle of semi-circle, $\angle RPQ={{90}^{{}^\circ }}$

$\therefore Q{{R}^{2}}=P{{R}^{2}}+P{{Q}^{2}}$

$\Rightarrow {{7}^{2}}+{{24}^{2}}=625={{25}^{2}}$

$QR=25$ cm

Diameter of the circle \[=\text{ }25cm~\]

Thus radius, $r=\dfrac{25}{2}cm$

Area of the semi-circle$=\dfrac{1}{2}\pi {{r}^{2}}$

$\therefore $Area $=\dfrac{1}{2}\times \dfrac{22}{7}\times \dfrac{25}{2}\times \dfrac{25}{2}=\dfrac{11\times 625}{28}c{{m}^{2}}$

Also, area of $\Delta PQR=\dfrac{1}{2}PR\times PQ$

$\Rightarrow \dfrac{1}{2}\times 7\times 24=84c{{m}^{2}}$

From the given figure,

Area of the shaded region = Area of the semi-circle with O as centre and OQ as radius – area of $\Delta PQR$

Hence, area of the shaded region can be calculated as,

Area of the shaded region $=\dfrac{11\times 625}{28}-84=\dfrac{11\times 625-84\times 28}{28}$

$\Rightarrow \dfrac{6875-2352}{28}=\dfrac{4523}{28}$

Area of the shaded region $=161.54 c{{m}^{2}}$.

2. The perimeter of a sector of a circle of radius\[~5.7\text{ }m\] is \[27.2\text{ }m\].Calculate:

(i). The length of arc of the sector in cm.

Ans: Given radius, $r=5.7$ m and perimeter $=27.2$ m.

Then \[OA=OB=5.7m~\]

Now,

$OA+OB+arcAB=27.2m$

$\Rightarrow 5.7m+5.7m+arcAB=27.2$ m

$\Rightarrow arcAB=27.2-11.4=15.8$ m

The length of the arc of the sector is $15.8$ m.

ii. The area of the sector in cm2 correct to the nearest $cm^2$

Ans: Area of sector OACB$=\left( \dfrac{1}{2}\times radius\times arc \right)$

$\Rightarrow \left( \dfrac{1}{2}\times 5.7\times 15.8 \right)=45.03c{{m}^{2}}$

Area of sector correct to nearest $c{{m}^{2}}=45c{{m}^{2}}$

3. Find the area of the shaded region in the given figure where ABCD is a square of side \[10cm\] and semi-circles are drawn with each side of the square as diameter. $\left[ \pi =3.14 \right]$

Ans: Given, side of square ABCD $=10\text{ cm}$

Area of square ABCD $={{\left( side \right)}^{2}}={{\left( 10 \right)}^{2}}=100\text{ }c{{m}^{2}}$

Given that semicircle is drawn with side of square as diameter.

Diameter of semicircle $=10\text{ }cm$.

Radius of semicircle $=5\text{ }cm$

Area of semi-circle $=\dfrac{1}{2}\pi {{r}^{2}}$

Area $=\dfrac{1}{2}\times 3.14\times {{\left( 5 \right)}^{2}}=39.25\text{ }c{{m}^{2}}$

Area of 4 semi-circles – Area of shaded region = Area of square ABCD

$\therefore $Area of shaded region = Area of 4 semi-circles - Area of square ABCD

Area of shaded region $=4\times 39.25-100=57c{{m}^{2}}$

4. In the given figure $\Delta ABC$ is an equilateral triangle inscribed in a circle of radius \[4\text{ }cm\] and centre O. Show that the area of the shaded region is $\dfrac{4}{3}\left( 4\pi -3\sqrt{3} \right)c{{m}^{2}}$.

Ans: Given, $\Delta ABC$ as an equilateral triangle inscribed in circle with radius, $r=4$ cm.

In $\Delta OBD$, Let $BD=a,$$OB=4cm$

$\Rightarrow \dfrac{\sqrt{3}}{2}=\dfrac{a}{4}$

$\Rightarrow a=\dfrac{4\sqrt{3}}{2}$

$\Rightarrow a=2\sqrt{3}~\text{cm}$

$\Rightarrow BC=2a=2\times 2\sqrt{3}=4\sqrt{3}~\text{cm}$

$\dfrac{OD}{4}=\cos {{60}^{{}^\circ }}\Rightarrow OD=4\times \dfrac{1}{2}=2~\text{cm}$

Area of shaded region = Area of sector OBPC – Area of $\Delta OBC$

$\Rightarrow \dfrac{120}{{{360}^{{}^\circ }}}\times \pi \times {{4}^{2}}-\dfrac{1}{2}\times 4\sqrt{3}\times 2=\dfrac{4}{3}[4\pi -3\sqrt{3}]\text{c}{{\text{m}}^{2}}$

5. The radii of two circles are \[8cm\]and \[6cm\]respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles.

Ans: Given, radii of two circles, $r1=8\text{ cm}$ and $r2=6\text{ cm}$.

Area of a circle is given by $\pi {{r}^{2}}$

Let $A1$ = Area of the first circle

$\Rightarrow A1=\pi {{\left( 8 \right)}^{2}}=64\pi c{{m}^{2}}$

$A2=$Area of the second circle

$\Rightarrow A2=\pi {{\left( 6 \right)}^{2}}=36\pi c{{m}^{2}}$

Total area $=A1+A2$,

$\Rightarrow 64\pi +36\pi =100\pi c{{m}^{2}}$

Let R be the radius of the circle with area \[{{A}_{1}}+\text{ }{{A}_{2}}~\],

$\therefore \pi {{R}^{2}}=100\pi $

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

$\Rightarrow R=10cm$

Hence, required radius \[=\text{ }10cm~\]

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

i. minor sector

Ans: Given, radius of circle, $r=10$ cm subtending an angle ${{90}^{\circ }}$ at the centre. $\therefore \angle AOB={{90}^{\circ }}$.

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

$\Rightarrow \dfrac{90}{360}\times 3.14\times {{\left( 10 \right)}^{2}}=\dfrac{1}{4}\left( 3.14 \right)\left( 100 \right)$

$\Rightarrow \dfrac{314}{4}=78.5c{{m}^{2}}$

Area of minor sector is $78.5c{{m}^{2}}$.

ii. major sector

Ans: Area of circle $=\pi {{r}^{2}}=\pi {{\left( 10 \right)}^{2}}$

Area of major sector = Area of circle – Area of minor sector

$\Rightarrow \pi {{\left( 10 \right)}^{2}}-\dfrac{90}{360}\pi {{\left( 10 \right)}^{2}}=\left( 3.14 \right)\left( 100 \right)-\dfrac{1}{4}\left( 3.14 \right)\left( 100 \right)$

$\Rightarrow 314-78.50=235.5c{{m}^{2}}$

Area of major sector is \[235.5\text{ }c{{m}^{2}}\].

iii. minor segment

Ans: Area of minor segment = Area of minor sector OAB – Area of $\Delta OAB$

$\because $Area of $\Delta OAB=\dfrac{1}{2}\left( OA \right)\left( OB \right)\sin \angle AOB$

$=\dfrac{1}{2}\left( OA \right)\left( OB \right)\left( \because \angle AOB={{90}^{{}^\circ }} \right)$

Area of sector $=\dfrac{\theta }{360}\pi {{r}^{2}}$

$\Rightarrow \dfrac{1}{4}\left( 3.14 \right)\left( 100 \right)-50=25\left( 3.14 \right)-50$

$\Rightarrow 78.50-50=28.5c{{m}^{2}}$

Area of minor segment is \[28.5\text{ }c{{m}^{2}}\].

iv. major segment

Ans: Area of major segment = Area of the circle – Area of minor segment

$\Rightarrow \pi {{\left( 10 \right)}^{2}}-25.5=100\left( 3.14 \right)-28.5$

$\Rightarrow 314-18.5=285.5c{{m}^{2}}$

Area of major segment is \[285.5\text{ }c{{m}^{2}}\].

7. In Akshita’s house, there is a flower pot. The sum of radii of circular top and bottom of a flowerpot is 140 cm and the difference of their circumference is\[88cm\], find the diameter of the circular top and bottom.

Ans: Given, sum of radii of circular top and bottom \[=\text{ }140cm~\]

Let radius of top \[=r\text{ }cm~\]

$\therefore $ Radius of bottom$=\left( 140-r \right)cm$

Circumference of top$=2\pi r$ cm

Circumference of bottom$=2\pi \left( 140-r \right)cm$

Difference of circumference $=\left[ 2\pi r-2\pi \left( 140-r \right) \right]cm$

By the given conditions in the problem,

$2\pi r-2\pi \left( 140-r \right)=88$

$\Rightarrow 2\pi \left[ r-140+r \right]=88$

$\Rightarrow 2r-140=\dfrac{88}{2\left( \dfrac{22}{7} \right)}$

$\Rightarrow 2r=140+14-154$

$\Rightarrow r=\dfrac{154}{2}$

$\therefore $Radius of top is obtained as \[77cm~\]

Diameter of top can be calculated as,

$\Rightarrow 2\times 77=154cm$

Radius of bottom can be obtained as,

\[\Rightarrow 140\text{ }\text{ }r=140\text{ }\text{ }77\]

\[=63cm~\]

$\therefore $ Diameter of bottom $2\times 63=126cm~$.

8. A chord of a circle of radius \[15cm\] subtends an angle of ${{60}^{{}^\circ }}$ at the centre. Find the area of the corresponding minor and major segments of the circle. (use $\pi =3.14$ and $\sqrt{3}=1.73$ )

Ans: Given, radius of circle,$r=15\text{ }cm$ subtending an angle ${{60}^{\circ }}$ at centre.

$\therefore \angle AOB={{60}^{\circ }}$.

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

Area of minor segment $=$ Area of minor sector OAB $-$ area of $\Delta OAB$ $\Rightarrow \dfrac{60}{360}\pi {{(15)}^{2}}-\dfrac{\sqrt{3}}{4}{{(15)}^{2}}=\dfrac{1}{6}(3.14)(225)-\dfrac{(1.73)}{4}(225)$

$\Rightarrow 117.75-97.3125=20.4375~\text{c}{{\text{m}}^{2}}$

Area of major segment = Area of the circle – area of minor segment

$\Rightarrow \pi {{(15)}^{2}}-20.4375=(3.14)(225)-20.4375$

$\Rightarrow 706.5-20.4375=686.0625~\text{c}{{\text{m}}^{2}}$

Area of minor segment is $20.4375~\text{c}{{\text{m}}^{2}}$ and area of major segment is $686.0625\,~\text{c}{{\text{m}}^{2}}$.

9. A round table cover has six equal designs as shown in the figure. If the radius of the cover is\[28cm\] , find the cost of making the design at the rate of Rs.\[0.35\] per $c{{m}^{2}}$ . (Use $\sqrt{3}=1.7$ )

Ans: Given, radius of cover, $r=28\text{ }cm$. Table has six equal designs and the cost of making the design is Rs. $0.35$ per $c{{m}^{2}}$.

Area of equilateral triangle is given by,$A=\dfrac{\sqrt{3}}{4}{{a}^{2}}$ .

From the figure area of one design $=\dfrac{60}{360}\times \pi {{\left( 28 \right)}^{2}}$– Area of equilateral $\Delta OAB$

$\Rightarrow \dfrac{\pi }{6}\times {{\left( 28 \right)}^{2}}-\dfrac{\sqrt{3}}{4}{{\left( 28 \right)}^{2}}={{\left( 28 \right)}^{2}}\left( \dfrac{\pi }{6}-\dfrac{\sqrt{3}}{4} \right)$

$\Rightarrow {{\left( 28 \right)}^{2}}\left( \dfrac{22}{7\times 6}-\dfrac{1.7}{4} \right)$

Total cost can be calculated as follows,

$\Rightarrow 6\times {{(28)}^{2}}\left( \dfrac{22}{42}-\dfrac{17}{40} \right)\times 0.35\times \dfrac{440-357}{21\times 40}$

$\Rightarrow 21\times {{(28)}^{2}}\left( \dfrac{11}{21}-\dfrac{17}{40} \right)=\dfrac{21\times {{(28)}^{2}}}{100}\times \dfrac{440-357}{21\times 40}$

\[\Rightarrow \dfrac{28\times 28\times 83}{40}=\dfrac{7\times 28\times 83}{1000}\]

$\Rightarrow Rs.16.268$

The total cost of making the design is $Rs.16.268$.

10. ABCD is a flower bed. If \[OA\text{ }=\text{ }21m\] and \[DC\text{ }=\text{ }14m\]. Find the area of the bed.

Ans: Given,

\[OA\text{ }=\text{ }R\text{ }=\text{ }21m\]

\[OC\text{ }=\text{ }r\text{ }=\text{ }14m~\]

Area of the flower bed (i.e., shaded portion) $=$ area of quadrant of a circle of radius $R$ of the quadrant of a circle of radius $r$

$\Rightarrow \dfrac{1}{4}\pi {{R}^{2}}-\dfrac{1}{4}\pi {{r}^{2}}=\dfrac{\pi }{4}\left( {{R}^{2}}-{{r}^{2}} \right)$

$\Rightarrow \dfrac{1}{4}\times \dfrac{22}{7}\left[ {{(21)}^{2}}-{{(14)}^{2}} \right]{{m}^{2}}$

$\Rightarrow \dfrac{1}{4}\times \dfrac{22}{7}\times (21\times 14)(21-14){{m}^{2}}$

$\Rightarrow \dfrac{1}{4}\times \dfrac{22}{7}\times 35\times 7~{{\text{m}}^{2}}$

$\Rightarrow 192.5~{{\text{m}}^{2}}$

Hence, area of the flower bed is $192.5~{{\text{m}}^{2}}$.

11. In a circle of radius \[21\text{ }cm\], an arc subtends an angle of ${{60}^{{}^\circ }}$ at the centre. Find:

i. the length of the arc.

Ans: Given radius, \[r=\text{ }21\text{ }cm\] and angle $\theta ={{60}^{{}^\circ }}$.

Length of arc \[=\dfrac{\theta }{360{}^\circ }\times 2\pi r~\]

\[\Rightarrow \dfrac{{{60}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 2\times \dfrac{22}{7}\times 21~~=\text{ }22\text{ }cm~\]

ii. area of the sector formed by the arc.

Ans: Area of the sector \[=\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}\]

\[~\Rightarrow \dfrac{{{60}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times \dfrac{22}{7}\times 21\times 21~~=\text{ }231\text{ }c{{m}^{2}}\]

iii. area of the segment formed by the corresponding chord

Ans: Area of segment formed by corresponding chord \[=~\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}-~Area\text{ }of~\text{ }\Delta OAB~\]

$\Rightarrow $Area of segment $=231-$ Area of $\Delta OAB$ …(1)

In right triangles $OMA$ and $OMB$,

\[OA\text{ }=\text{ }OB\] [Radii of same circle]

\[OM\text{ }=\text{ }OM\] [Common]

\[\therefore \Delta OMA\cong \Delta OMB\] [RHS congruency]

\[~\therefore AM\text{ }=\text{ }BM\] [By CPCT]

Here $M$ is the midpoint.

\[\Rightarrow ~~\dfrac{1}{2}AB\text{ }and~\text{ }\angle AOM\text{ }=~\text{ }\angle BOM\]

\[\Rightarrow \dfrac{1}{2}\angle AOB\text{ }=\dfrac{1}{2}\times {{60}^{{}^\circ }}={{30}^{{}^\circ }}\]

In right angled \[\Delta OMA\],

$\cos {{30}^{{}^\circ }}=\dfrac{\text{OM}}{\text{OA}}$

$\Rightarrow \dfrac{\sqrt{3}}{2}=\dfrac{\text{OM}}{21}$

$\Rightarrow \text{OM}=\dfrac{21\sqrt{3}}{2}~cm$

Also $\sin {{30}^{{}^\circ }}=\dfrac{\text{AM}}{\text{OA}}$

$\Rightarrow \dfrac{1}{2}=\dfrac{\text{AM}}{21}$

$\Rightarrow 2AM=21cm$

$\Rightarrow AB=21cm$

$\therefore $ Area of $\Delta AOB=\dfrac{1}{2}\times AB\times OM$

\[\Rightarrow \dfrac{1}{2}\times 21\times \dfrac{21\sqrt{3}}{2}=\dfrac{441\sqrt{3}}{4}\]cm2

Using eq. $\left( 1 \right)$,

Area of segment formed by corresponding chord \[=~\left( 231-\dfrac{441\sqrt{3}}{4} \right)\text{ }c{{m}^{2}}~\].

12. A chord of a circle of radius \[15\text{ }cm\]subtends an angle of ${{60}^{{}^\circ }}$ at the centre. Find the area of the corresponding segment of the circle. (Use $\pi =3.14$ and $\sqrt{3}=1.73$)

Ans: Given radius,\[~r~=\text{ }15\text{ }cm\] and angle $\theta ={{60}^{{}^\circ }}$

Area of minor sector can be obtained from $\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$

\[\Rightarrow \dfrac{{{60}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 3.14\times 15\times 15=117.75c{{m}^{2}}\]

For, Area of $\Delta AOB$,

Draw $OM$, where $OM\bot AB$.

In right triangles $OMA$ and $OMB$,

\[OA\text{ }=\text{ }OB\] [Radii of same circle]

\[OM\text{ }=\text{ }OM\] [Common]

\[\therefore \Delta OMA\cong \Delta OMB\] [RHS congruency]

\[~\therefore AM\text{ }=\text{ }BM\] [By CPCT]

\[\Rightarrow ~~\dfrac{1}{2}AB\text{ }and~\text{ }\angle AOM\text{ }=~\text{ }\angle BOM\]

\[\Rightarrow \dfrac{1}{2}\angle AOB\text{ }=\dfrac{1}{2}\times {{60}^{{}^\circ }}={{30}^{{}^\circ }}\]

In right angled \[\Delta OMA\],

$\cos {{30}^{{}^\circ }}=\dfrac{\text{OM}}{\text{OA}}$

$\Rightarrow \dfrac{\sqrt{3}}{2}=\dfrac{\text{OM}}{15}$

$\Rightarrow \text{OM}=\dfrac{15\sqrt{3}}{2}~cm$

Also, $\sin {{30}^{{}^\circ }}=\dfrac{\text{AM}}{\text{OA}}$

$\Rightarrow \dfrac{1}{2}=\dfrac{\text{AM}}{15}$

$\Rightarrow 2AM=15cm$

$\Rightarrow AB=15cm$

$\therefore $ Area of $\Delta AOB=\dfrac{1}{2}\times AB\times OM$

$\Rightarrow \dfrac{1}{2}\times 15\times \dfrac{15\sqrt{3}}{2}=\dfrac{225\sqrt{3}}{3}$

$\Rightarrow \dfrac{225\times 1.73}{4}=97.3125c{{m}^{2}}$

$\therefore $Area of minor segment = Area of minor sector – Area of $\Delta AOB$

\[\Rightarrow 117.75\text{ }\text{ }97.3125\text{ }=\text{ }20.4375\text{ }c{{m}^{2}}~\]

And, Area of major segment $=\pi {{r}^{2}}-$ Area of minor segment

\[\Rightarrow 706.5\text{ }\text{ }20.4375\text{ }=\text{ }686.0625\text{ }c{{m}^{2}}~\]

13. A chord of a circle of radius 12 cm subtends an angle of ${{120}^{{}^\circ }}$ at the centre. Find the area of the corresponding segment of the circle. (Use $\pi =3.14$ and$\text{ }\sqrt{3}=1.73$ )

Ans: Given radius, \[r~~=\text{ }12cm\] and angle $\theta ={{120}^{{}^\circ }}$.

Area of corresponding sector can be obtained by using the formula $\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$

\[\Rightarrow \dfrac{{{120}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 3.14\times 12\times 12=\text{ }150.72\text{ }c{{m}^{2}}\]

For, Area of $\Delta AOB$,

Draw $OM$, where $OM\bot AB$.

In right triangles $OMA$ and $OMB$,

\[OA\text{ }=\text{ }OB\] [Radii of same circle]

\[OM\text{ }=\text{ }OM\] [Common]

\[\therefore \Delta OMA\cong \Delta OMB\] [RHS congruency]

\[~\therefore AM\text{ }=\text{ }BM\] [By CPCT]

\[\Rightarrow ~~\dfrac{1}{2}AB\text{ }and~\text{ }\angle AOM\text{ }=~\text{ }\angle BOM\]

\[\Rightarrow \dfrac{1}{2}\angle AOB\text{ }=\dfrac{1}{2}\times {{120}^{{}^\circ }}={{60}^{{}^\circ }}\]

In right angled \[\Delta OMA\],

$\cos {{60}^{{}^\circ }}=\dfrac{\text{OM}}{\text{OA}}$

$\Rightarrow \dfrac{1}{2}=\dfrac{\text{OM}}{12}$

$\Rightarrow \text{OM}=6~\text{cm}$

Also $\text{ }\sin {{60}^{{}^\circ }}=\dfrac{\text{AM}}{\text{OA}}$

$\Rightarrow \dfrac{\sqrt{3}}{2}=\dfrac{\text{AM}}{12}$

$\Rightarrow 2AM=12\sqrt{3}cm$

$\Rightarrow AB=12\sqrt{3}cm$

$\therefore $ Area of $\Delta AOB=\dfrac{1}{2}\times AB\times OM$

$\Rightarrow \dfrac{1}{2}\times 12\sqrt{3}\times 6=36\sqrt{3}$

$\Rightarrow 36\times 1.73=62.28c{{m}^{2}}$

Area of corresponding segment = Area of corresponding sector – Area of $\Delta AOB$

\[\Rightarrow 150.72\text{ }\text{ }62.28\text{ }=\text{ }88.44\text{ }c{{m}^{2}}~\]

14. Find the area of the shaded region in figure, if \[PQ\text{ }=\text{ }24\text{ }cm,\text{ }PR\text{ }=\text{ }7\text{ }cm\] and $O$is the centre of the circle.

Ans: Given radius, \[r~~=\text{ }12cm\] and angle $\theta ={{120}^{{}^\circ }}$.

Area of corresponding sector can be obtained by using the formula $\dfrac{\theta }{{{360}^{{}^\circ }}}\times \pi {{r}^{2}}$

\[\Rightarrow \dfrac{{{120}^{{}^\circ }}}{{{360}^{{}^\circ }}}\times 3.14\times 12\times 12=\text{ }150.72\text{ }c{{m}^{2}}\]

For, Area of $\Delta AOB$,

Draw $OM$, where $OM\bot AB$.

In right triangles $OMA$ and $OMB$,

\[OA\text{ }=\text{ }OB\] [Radii of same circle]

\[OM\text{ }=\text{ }OM\] [Common]

\[\therefore \Delta OMA\cong \Delta OMB\] [RHS congruency]

\[~\therefore AM\text{ }=\text{ }BM\] [By CPCT]

\[\Rightarrow ~~\dfrac{1}{2}AB\text{ }and~\text{ }\angle AOM\text{ }=~\text{ }\angle BOM\]

\[\Rightarrow \dfrac{1}{2}\angle AOB\text{ }=\dfrac{1}{2}\times {{120}^{{}^\circ }}={{60}^{{}^\circ }}\]

In right angled \[\Delta OMA\],

$\cos {{60}^{{}^\circ }}=\dfrac{\text{OM}}{\text{OA}}$

$\Rightarrow \dfrac{1}{2}=\dfrac{\text{OM}}{12}$

$\Rightarrow \text{OM}=6~\text{cm}$

Also $\text{ }\sin {{60}^{{}^\circ }}=\dfrac{\text{AM}}{\text{OA}}$

$\Rightarrow \dfrac{\sqrt{3}}{2}=\dfrac{\text{AM}}{12}$

$\Rightarrow 2AM=12\sqrt{3}cm$

$\Rightarrow AB=12\sqrt{3}cm$

$\therefore $ Area of $\Delta AOB=\dfrac{1}{2}\times AB\times OM$

$\Rightarrow \dfrac{1}{2}\times 12\sqrt{3}\times 6=36\sqrt{3}$

$\Rightarrow 36\times 1.73=62.28c{{m}^{2}}$

Area of corresponding segment = Area of corresponding sector – Area of $\Delta AOB$

\[\Rightarrow 150.72\text{ }\text{ }62.28\text{ }=\text{ }88.44\text{ }c{{m}^{2}}~\]

Long Answer Questions (4 Marks)

1. Three horses are tethered with \[7m\] long ropes at the three corners of a triangular field having sides \[20m,\text{ }34m\] and \[42\text{ }m\] . Find the area of the plot which can be grazed by the horses. Also, find the area of the plot which remains ungrazed.

Ans: Given, length of ropes at the three corners, $l=33$ m and the sides of the triangle as $20$ m, $34$ m and $42$ m.

From the figure let,

$\angle A=\theta {{1}^{{}^\circ }}$, $\angle B=\theta {{2}^{{}^\circ }}$ and $\angle C=\theta {{3}^{{}^\circ }}$

Area which can be grazed by three horses $=$ Area of sector with central angle $\theta {{1}^{{}^\circ }}$ and radius \[7\text{ }m\]$+$ Area of sector with central angle $\theta {{2}^{{}^\circ }}$ and radius \[7\text{ }cm\]$+$ Area of sector with central angle $\theta {{3}^{{}^\circ }}$ and radius \[7\text{ }cm~\]

$\Rightarrow \dfrac{\pi {{r}^{2}}\theta {{1}^{{}^\circ }}}{360}+\dfrac{\pi {{r}^{2}}\theta {{2}^{{}^\circ }}}{{{360}^{{}^\circ }}}+\dfrac{\pi {{r}^{2}}\theta {{3}^{{}^\circ }}}{{{360}^{{}^\circ }}}$

$\Rightarrow \dfrac{\pi {{r}^{2}}}{{{360}^{{}^\circ }}}\left( \theta {{1}^{{}^\circ }}+\theta {{2}^{{}^\circ }}+\theta {{3}^{{}^\circ }} \right)$

$\Rightarrow \dfrac{\pi {{r}^{2}}}{{{360}^{{}^\circ }}}\times {{180}^{\circ }}$

( $\because $Sum of three angles of a $\Delta ={{180}^{{}^\circ }}$)

$\Rightarrow \dfrac{22}{7}\times \dfrac{7\times 7\times {{180}^{{}^\circ }}}{{{360}^{{}^\circ }}}=77{{m}^{2}}$

Sides of plot \[ABC\] are given in the problem as,

\[a=\text{ }20m~\], \[b\text{ }=\text{ }34m\] and \[c=42m\]

$\because $Semi perimeter,$s=\dfrac{20+34+42}{2}=48m$

$\therefore $Area of triangular plot \[=\] Area of $\Delta ABC$

$\Rightarrow \sqrt{s(s-a)(s-b)(s-c)}=\sqrt{48\times 28\times 14\times 6}=336{{m}^{2}}$

Area grazed by the horses \[=\text{ }77{{m}^{2}}~\]

$\therefore $Ungrazed area can be calculated by subtracting area grazed by the horses from the area of the triangular plot.

\[\Rightarrow \] Ungrazed area \[=\left( 33677 \right)~=\text{ }259\text{ }{{m}^{2}}~\]

2. In the given figure, two circular flower beds have been shown on two sides of a square lawn ABCD of side $56$ m. If the corner of each circular flower bed is the point of enter section O of the diagonals of the square lawn. Find the sum of the areas of the lawn and the flower beds.

Ans: Given, side of square lawn $=56$ m.

Area of the square lawn $ABCD$, $A={{\left( 56 \right)}^{2}}$

Let $OA=OB=x$

$\therefore $By Pythagoras theorem,

$\Rightarrow {{x}^{2}}+{{x}^{2}}={{56}^{2}}$

$\Rightarrow 2{{x}^{2}}=56\times 56$

$\Rightarrow {{x}^{2}}=28\times 56$

Again, area of sector $OAB$ $=\dfrac{90}{360}\times \pi {{x}^{2}}=\dfrac{1}{4}\pi {{x}^{2}}$

Substituting for $x$, we get

$\Rightarrow $ Area of sector $OAB=\dfrac{1}{4}\times \dfrac{22}{7}\times 28\times 56{{m}^{2}}$

Area of $\Delta OAB=\dfrac{1}{4}56\times 56{{m}^{2}}$

(Here, $\angle AOB={{90}^{{}^\circ }}$ since square $ABCD$ is divided into 4 right triangles)

$\therefore $ Area of flowerbed$AB=\left( \dfrac{1}{2}\times \dfrac{22}{7}\times 28\times 56-\dfrac{1}{4}56\times 56 \right){{m}^{2}}$

$\Rightarrow \dfrac{1}{4}\times 28\times 56\left( \dfrac{22}{7}-2 \right)$

$\Rightarrow \dfrac{7\times 56\times 8}{7}$

Area of flower bed $AB=448{{m}^{2}}$

Similarly, area of the other flowerbed \[=\text{ }448{{m}^{2}}~\]

Total area of the lawn and the flowerbeds $=56\times 56+448+448=4032{{m}^{2}}$

3. An elastic belt is placed round the rein of a pulley of radius $5$ cm. One point on the belt is pulled directly away from the centre O of the pulley until it is at P, $10$ cm from O. Find the length of the best that is in contact with the rim of the pulley. Also find the shaded area.

Ans: Given, radius of rein of pulley, $OA=5$ cm.

$OP=10$ cm

Thus, $\cos \theta =\dfrac{OA}{OP}$

$\Rightarrow \cos \theta =\dfrac{5}{10}=\dfrac{1}{2}~$

$\Rightarrow \theta ={{60}^{{}^\circ }}$

$\Rightarrow \angle AOB={{120}^{{}^\circ }}$

$\therefore \operatorname{Arc}AB=\dfrac{120\times 2\times \pi \times 5}{360}~\text{cm}$

$\Rightarrow \dfrac{10\pi }{3}~\text{cm}\left[ \because l=\dfrac{\theta }{360}\times 2\pi r \right]$

Length of the belt that is in contact with the rim of the pulley can be calculated by subtracting length of arc AB from Circumference of the rim.

Length of the belt $=2\pi \times 5cm-\dfrac{10}{3}\pi cm$

Length of the belt $=\dfrac{20}{3}\pi cm$

Now, area of sector OAQB$=\dfrac{{{120}^{\circ }}}{{{360}^{\circ }}}\times \pi \times {{5}^{2}}cm=\dfrac{25}{3}\pi c{{m}^{2}}$ $\left[ \text{Since Area=}\dfrac{\theta }{{{360}^{\circ }}}\times \pi {{r}^{2}} \right]$

$\left[ \because AP=\sqrt{100-25}=\sqrt{75}=5\sqrt{3}cm \right]$

Area of quadrilateral OAPB is given by, $2(\text{Area of }\Delta OAP)=25\sqrt{3}c{{m}^{2}}$

Hence, shaded area can be obtained as,

Area of shaded region$=25\sqrt{3}-\dfrac{25}{3}\pi $

Hence, area of shaded region is $17.12\text{ }c{{m}^{2}}$.

## Download Class 10 Maths Chapter 12 Important Questions PDF

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.

### Area of a Circle

The area is the region enclosed by the circumference of the circle. The area covered by one complete cycle of the radius of the circle on a two-dimensional plane is known as the area of that circle.

Area of the circle = πr2 square units.

Where r is the radius of the circle.

### Area of Semicircle

To find the area of the semi-circle we need half of the area of the circle.

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The formula for the area of semicircle = \[\frac{\pi r^{2}}{2}\] square unit.

Where r is the radius of the circle.

### Area of a Ring

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We know the area of circle πr2. Here the ring is having two radii inner and outer radius. Area of the ring i.e. the coloured part shown in the above figure is calculated by subtracting the area of the inner circle from the area of the bigger circle.

Area of the ring = πR2 - πr2 = π(R2 - r2)

Where R = radius of outer circle

r = radius of the inner circle

### The Sector of a Circle

The sector of a circle is the region of a circle enclosed by an arc and two radii. The smaller area is called the minor sector and the larger area is called the major sector of the circle.

### Areas of Sectors of a Circle

The area formed by an arc and the two radii joining the endpoints of the arc is known as a sector of a circle.

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In the above diagram, OACB is the minor sector. OADB is the major sector.

### Minor Sector

The area including arc ∠AOB with point C is called Minor Sector. So, OACB is the minor sector. The area included in ∠AOB is the angle of the minor sector.

Area of the sector of angle \[\theta = \frac{\theta}{360} \times \pi r^{2}\]

### Major Sector

The area included in ∠AOB with point D is called the Major Sector. So OADB is the major sector. The angle of the major sector is 360° – ∠AOB.

If the degree measure of the angle at the center is 360, area of the sector = πr2.

Formula to find the area of Major Sector = πr2 - Area of the Minor Sector

Area of the circle can also be calculated by adding an area of a minor sector and area of a major sector.

Note: Area of Minor Sector + Area of Major Sector = Area of the Circle

To Calculate Length of an Arc of a Sector of Angle θ

An arc is the piece of the circumference of the circle so the length of an arc can be calculated as the θ part of the circumference.

Formula to calculate the length of an arc of a sector of angle \[\theta = \frac{\theta}{360} \times 2\pi r\].

### Areas of Segments of the Circle

The area made by an arc and a chord of a circle is called the segment of the Circle.

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### Minor Segment

The minor segment is defined as the region bounded by the chord and the minor arc.

In the above diagram, AB is a chord of the circle. The area made by chord AB and arc X is known as the minor segment.

Formulas to calculate Area of Minor Segment = Area of Minor Sector – Area of ∆ABO

\[= \frac{\theta}{360} \times \pi r^{2} - \frac{1}{2} r^{2} sin \theta\]

### Major Segment

The major segment is defined as the region bounded by the chord and the major arc.

Formulas to calculate Area of Major Segment = πr2 - Area of Minor Segment

Note: Area of major segment + Area of minor segment = Area of the circle.

### Areas of Combinations of Plane Figures

A plane figure is a geometric figure that has no thickness, it lies entirely on one plane. In this section, we will learn how to calculate the areas of some plane figures which are combinations of more than one plane figure.

To find the area of the composite area of the shapes, simply find the area of each shape and add them together. The order in which we calculate the areas does not matter and the commutative property states that it does not matter which order you add them in.

Following are the steps to find the area of combined figures:

Step I: First we divide the given combined figure into its simple geometrical shapes.

Step II: Then calculate the area of simple geometrical shapes separately.

Step III: Finally, to find the area of the combined figure we need to add or subtract these areas.

Below are some examples of combinations of plane figures:

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Above diagram is a combination of one rectangle and two semi-circles. So, to find an area of the figure, first, find an area of the rectangle and then add the area of two semi-circles.

Area of the given figure = Area of the rectangle + 2 x Area of semi-circles.

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Above diagram is the form of a ring.

Area of the required region = \[\frac{3}{4} (\pi r_{1}^{2} - \pi r_{2}^{2})\]

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Above diagram is a combination of circle and square. To find the area of a given figure, first, calculate the area of the circle then subtract the area of the square.

Area of the required region = πr2 - a2

Important formula to remember:

Area of circle = πr2

Length of an arc of a sector of a circle with radius r and angle with degree measure θ is \[\frac{\theta}{360} \times 2\pi r\].

Area of a sector of a circle with radius r and angle with degrees measure θ is \[\frac{\theta}{360} \times \pi r^{2}\]

Area of the segment of a circle = Area of the corresponding sector – Area of the corresponding triangle.

### Conclusion:

Refer to this** Important Questions For Class 10 Maths Chapter 12** for boards exam preparation. All the important concepts are available in the given PDF. The formula for the area of a circle also helps us calculate the area of circle sectors and segments as well. Download important questions of ch 12 Maths Class 10 from Vedantu site or app for more information.

## Important Related Links for CBSE Class 10 Maths

CBSE Class 10 Maths Study Materials |

## Conclusion

Vedantu's offering of "Important Questions for CBSE Class 10 Maths Chapter 12 - Areas Related to Circles" is a valuable resource for students seeking to excel in their mathematics studies. The comprehensive collection of questions and solutions provides a deep understanding of the intricate concepts related to circles and their areas. By using this platform, students can gain confidence in tackling complex problems, enhance their problem-solving skills, and prepare effectively for their CBSE Class 10 examinations. Vedantu's commitment to quality education is evident in the well-structured and engaging content, making it an indispensable tool for students aiming to achieve academic success in mathematics.

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

1. How many sums are there in class 10 Maths Chapter 12 Areas Related to Circles?

Chapter 12 of Maths for class 10 teaches about the application of the chapter ‘Circles’. There are a total of 35 sums in all the exercises combined together from the chapter. All of these problems are unique in their own way and offer excellent practice and knowledge to students. At times, however, these can get very confusing. The Important questions by Vedantu for this chapter can help students here because of their simple language structure.

2. What students will study in chapter 12 Areas Related to Circles of class 10 Maths?

In this chapter, students learn the different applications of circles in terms of their areas. Concepts like the area of circles, their circumference, and the various parts of it, segment, sector, angle, and length, everything is discussed in this chapter. Important Questions for chapter 12 is specifically designed to focus on the important areas by providing students with questions so that they can practice with them and score good results for questions from this chapter by Vedantu.

3. What are the benefits of using important questions for chapter 10 of class 12 Maths?

Important Questions provided by Vedantu are good for students to learn ‘Areas Related to Circles’. They can access the questions free of cost by visiting the link-Important Questions for class 10 maths, and they can be assured of their quality content as each question is kept up to date and follows the latest curriculum. Moreover, the easily understandable structure provides them with a good practice experience for good marks. Since these questions are from the perspective of the Board exam paper, these are excellent sources for studying.

4. State the important formulae in chapter 12 of class 10 Maths.

Chapter 12 is named 'Areas Related to Circles' and this chapter is practically an application of the previous chapter in class 10 named 'Circles'. So, this chapter has some important formulae to solve various problems:

- Area of circle: $\pi r^2$
Area of a semicircle: $\dfrac{1}{2}\pi r^2$

Area of quadrant: $\dfrac{\pi r^2}{4}$

Area of sector: $\dfrac{\theta}{360^\circ}\times \pi r^2$

Area of Ring: $\pi R^2 - \pi r^2$

The perimeter of the sector of Circle: 2 Radius + $\dfrac{\theta}{360^\circ} \times 2\pi r$

Length of arc = $\dfrac{2\pi r\theta}{360^\circ}

To know more students can download the vedantu app.

5. How to understand chapter 12 of class 10 Maths?

Circles are round figures without any sides or edges. Their applications are taught to students in chapter 12 of class 10 Maths where students learn how to solve different questions related to the areas of circles, the length and circumference, and a whole lot of other things. Since this chapter has a whole lot of things to learn, students can use Important Questions prepared for this chapter as it offers them an examination patterned structure, which is very beneficial.