Class 10 Mathematics

# Area of Circle

## Exercise 12.3 (NCERT) Part 2

Question: 10. The area of an equilateral triangle ABC is 17320.5 cm^{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. Find the area of the shaded region. (Use `π = 3.14` and `sqrt3 = 1.73205`)

**Solution:** Area of equilateral triangle

`=(sqrt3)/(4)xxtext(side)^2`

Or, `a^2=17320.5xx(4)/(1.73205)`

`=10000xx4`

Or, `a=200 cm`

Radius of circle = 100 cm (half of a)

Area of circle `= πr^2 = π xx 100^2 = 31400 sq cm`

Area of three sectors = ½ Area of circle (Because all angles sum up to 180^{o})

`= ½ xx 31400 = 15700 sq cm`

So, area of shaded portion `= 17302.5 – 15700 = 1602.5 sq cm`

Question: 11. On a square handkerchief, nine circular designs each of radius 7 cm are made. Find the area of the remaining portion of the handkerchief.

**Solution:** Area of 9 circles `= 9 x πr^2`

`= 9 xx π xx 7^2 = 1386 sq cm`

Side of square `= 6 xx 7 = 42 cm`

Area of square `= text(Side)^2 = 42^2 = 1764 sq cm`

Area of remaining portion `= 1764 – 1386 = 378 sq cm`

Question: 12. In the given 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, (ii) shaded region.

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

`= ¼ xx π xx 3.5^2 = 9.625 sq cm`

Area of ∆BDO `= ½ xx BD xx OD`

`= ½ xx 3.5 xx 2 = 3.5 sq cm`

Hence, area of shaded portion `= 9.625 – 3.5 = 6.125 sq cm`

Question 13. In the given figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region.

**Solution:** Using Pythagoras Theorem;

`BO^2 = OA^2 + OC^2`

`= 20^2 + 20^2`

Or, `BO = 20sqrt2 cm` = Radius of circle

Area of quadrant `= ¼ xx πr^2`

`= ¼ xx π xx (20sqrt2)^2 = 628 sq cm`

Area of square `= text(Side)^2 = 20^2 = 400 sq cm`

Area of shaded portion `= 628 – 400 = 228 sq cm`

Question 14: AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O. If angle AOB = 30°, find the area of the shaded region.

**Solution:** Area of shaded region

= Area of sector of bigger circle – Area of sector of smaller circle

Area of sector of bigger circle

`=(30°)/(360°)πxx21^2`

Area of sector of smaller circle

`=(30°)/(360°)πxx7^2`

Area of shaded region

`=(30°)/(360°)π(21^2-7^2)`

`=102(2)/(3) sq cm`

Question: 15. In the given 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.

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

`= ¼ xx π xx 14^2 = 154 sq cm`

Area of triangle `= ½ xx b xx h`

`= ½ xx 14 xx 14 = 98 sq cm`

Area of segment made by hypotenuse of triangle

`= 154 – 98 = 56 sq cm`

Diameter of external semicircle `= 14sqrt2` cm (since other two sides of triangle are 14 cm)

So, area of semicircle `= ½ xx πr^2`

`= ½ xx π xx (14sqrt2)^2 = 154 sq cm`

Area of shaded portion `= 154 – 56 = 98 sq cm`

Question 16: Calculate the area of the designed region in the figure common between the two quadrants of circles of radius 8 cm each.

**Solution:** Area of square `= text(Side)^2 = 8^2 = 64 sq cm`

Area of shaded portion

= Double area of segment formed by diagonal of the square

Area of two quadrants `= ½ πr^2`

`= ½ xx π xx 8^2 = 100.48 sq cm`

Area of square = Area of two triangles formed by radii and the cord

So, Area of shaded portion `= 100.48 – 64 = 36.48 sq cm`

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Exercise 1

Exercise 2 (Part 1)

Exercise 2 (Part 2)

Exercise 3 (Part 1)