**NCERT Solution For Class 10, Maths, Chapter 12, Areas Related To Circles Exercise 12.2** is based on **Areas of Sector** and **Segment of a Circle**. Exercise 12.2, Chapter 12, class 10 maths has total fourteen questions to discuss which are given below.

Table of Contents

Toggle## Class 10, Maths, Chapter 12, Exercise 12.2, Solutions

**Q.1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.**

**Ans: **we know that the area A of a sector of angles of a circle of radius r is given by

⇒ $\pi {{r}^{2}}\times \frac{\theta }{360}$

Here, r = 6 cm and θ = 60^{0}

= $\frac{22}{7}\times {{(6)}^{2}}\times \frac{60}{360}$

= $\frac{22}{7}\times 36\times \frac{1}{6}$

= $\frac{22}{7}\times 6=\frac{132}{7}$ cm^{2}

**Q.2. Find the area of a quadrant of a circle whose circumference is 22 cm.**

**Ans:**

**Q.3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.**

**Ans: **Here, minute hand of a clock is 14 cm in length. Therefore, it describes radius of circle (clock)

∴ r = 14 cm

∵ 60 min require to rotate through the circle = 360^{0}

∴ 1 min require to rotate through the circle = $\frac{360}{60}$

∴ 5 min require to rotate through the circle = $\frac{360}{60}\times 5$ = 30^{0}

Hence, the required area (Area of sector) i.e., the area swept in 5 minutes

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

Here, r = 14 cm and θ = 30^{0}

= $\frac{22}{7}\times {{(14)}^{2}}\times \frac{30}{360}$

= $\frac{22}{7}\times 14\times 14\times \frac{30}{360}$

= $22\times 2\times 14\times \frac{1}{12}=\frac{154}{3}$ cm^{2}

Hence, area of sector making an angle of 30^{0} is $\frac{154}{3}$ cm^{2}.

**Q.4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: **

**(i) minor segment **

**(ii) major sector. (Use π = 3.14)**

**Ans:**

**Q.5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:**

**(i) the length of the arc**

**(ii) area of the sector formed by the arc**

**(iii) area of the segment formed by the corresponding chord**

**Ans:**

**Q.6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and ****($\sqrt{3}=1.73$)**

**Ans:**

**Q.7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and **** $\sqrt{3}=1.73$**

** ****Ans:**

**Q.8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig. 12.11). Find**

**(i) The area of that part of the field in which the horse can graze.**

**(ii) The increase in the grazing area if the rope were 10 m long instead of 5 m. ****(Use π = 3.14)**

**Ans: **(i) The horse will graze over a quadrant of a circle with centre at the corner A of the field and radius AF = 5 cm

Then, the area of the quadrant of this circle

= $\frac{1}{4}$πr^{2 }= $\frac{1}{4}$ x 3.14 x 5 x 5

= $\frac{78.5}{4}$ = 19.625 m^{2}

(ii) if the rope were 10 m long instead of 5m then, the area of the quadrant of this circle and radius AH = 10 m

= $\frac{1}{4}$πr^{2} = $\frac{1}{4}$x 3.14 x 10 x 10

= $\frac{314}{4}$ = 78.5 m^{2}

Therefore, increase in the grazing area

= (78.5 – 19.625) = 58.875 m^{2}

**Q.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 Fig. 12.12. Find:**

**(i) the total length of the silver wire required.**

**(ii) the area of each sector of the brooch.**

**Ans:**

**Q.10. An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.**

** **

** ****Ans:**

**Ans:** Area between two consecutive ribs

= $\frac{1}{8}$x πr^{2} where r = 45 cm

= $\frac{1}{8}\times \frac{22}{7}$ x 45 x 45

= $\frac{22275}{28}$ cm^{2}

**Q.11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.**

**Ans:**

**Q.12. To warn ships for underwater rocks, a lighthouse spreads a red colored light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)**

**Ans:**

**Q.13. A round table cover has six equal designs as shown in Fig. 12.14. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ****₹**** 0.35 per cm ^{2}. (Use **

**$\sqrt{3}$= 1.72)**

** **

**Ans:**

**Q.14. Tick the correct answer in the following:**

**Area of a sector of angle p (in degrees) of a circle with radius R is**

**(i) $\frac{P}{180}\times 2\pi R$**

**(ii) $\frac{P}{180}\times \pi {{R}^{2}}$ **

**(iii) $\frac{P}{360}\times 2\pi R$**

**(iv) $\frac{P}{720}\times 2\pi {{R}^{2}}$**

** ****Ans: **(iv) ** $\frac{P}{720}\times 2\pi {{R}^{2}}$**

**Explanation: **We know that area A of a sector of angle 𝜃 in a circle of radius r is given by

A = πr^{2} × $\frac{\theta }{360}$ or

A = $\frac{\theta }{360}$× πr^{2}

But, r = R and 𝜃 = p

Therefore,

A = $\frac{\theta }{360}$× πr^{2} = $\frac{P}{720}$× 2πR^{2}