Area of Circle
NCERT 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`)
Answer: 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.
Answer: 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.
Answer: 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.
Answer: 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.
Answer: 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.
Answer: 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.
Answer: 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`