Class 10 Mathematics

# Area of Circle

## Exercise 12.2 (NCERT) Part 2

Question 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.

**Solution:** In ∆ OPR

`∠ROP = 60^o`

`∠OPR = 30^o`

OP = 12 cm

`text(cos)30°=(RP)/(OP)`

Or, `sqrt3/2=(RP)/(12)`

Or, `RP=6sqrt3`

So, `QP = 12sqrt3`

Area of ∆OQP

`=(1)/(2)xxtext(base)xxtext(height)`

`=(1)/(2)xx12xxsqrt3`

`=62.352 sq cm`

Area of minor sector

`=(120°)/(360°)xxπr^2`

`=(1)/(3)xxπxx12^2`

`=150.72 sq cm`

Area of minor segment `= 150.72 – 62.352= 88.368 sq cm`

Question 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. Find

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

**Solution:** Area of quadrant = ¼ πr^{2}

`= ¼ xx 25π = 19.625 sq cm`

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

**Solution:** Area of quadrant `= ¼ πr^2`

`= ¼ xx 100π = 78.5 sq cm`

As radius is doubled so area is quadrupled, so there is an increase of 300% in area

`78.625 – 19.625 = 58.875 sq cm`

Question 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. Find:

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

**Solution:** Circumference `= 2πr = 35π = 110 mm`

Wire required for 5 diameters `= 35 xx 5 = 175 mm`

Total length of wire `= 110 + 175 = 285 mm`

(ii) The area of each sector of the brooch

**Solution:** Since there are 10 equal sectors, so area of each sector will 1/10 the area of circle

`=(1)/(10)xxπr^2`

`=(1)/(10)xx(22)/(7)xx(35/2)^2`

`=96.25 sq mm`

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

**Solution:** Area of a sector between two consecutive ribs

`=(1)/(8)xxπr^2`

`=(1)/(8)xxπxx45^2`

`=794.8 sq cm`

Question 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.

**Solution:** Area swept by two wipers can be calculated using the area of sector formula as follows:

`=(θ)/(360°)xxπr^2`

`=(230°)/(360°)xxπxx25^2`

`=1254.9 sq cm`

Question 12: To warn ships for underwater rocks, a lighthouse spreads a red coloured 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.

**Solution:** Required area can be calculated by using the area of sector formula:

`=(θ)/(360°)xxπr^2`

`=(80°)/(360°)xxπxx16.5^2`

`=189.97 sq km`

Question 13: A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs 0.35 per cm^{2}.

**Solution:** The area of the hexagon will be equal to six equilateral triangles with each side equal to the radius.

Area of hexagon

`=6xxsqrt3/4xxtext(side)^2`

`=6xxsqrt3/4xx28^2`

`=2036.832 sq cm`

Area of circle `= πr^2 = π xx 28^2 = 2461.76 sq cm`

So, area of designed portion `= 2461.76 – 2036.832 = 424.928 sq cm`

Cost of making design `= 425 xx 0.35 = Rs. 148.75`

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Exercise 1

Exercise 2 (Part 1)

Exercise 3 (Part 1)

Exercise 3 (Part 2)