Area of Circle NCERT Exercise 12.3 solution part two Class Ten Mathematics

# Area of Circle

## Exercise 12.3 Part 2

Question: 10. The area of an equilateral triangle ABC is 17320.5 cm2. 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 180o)

= 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

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.

= 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

=(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

= 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