Ex 12.3, 9 - AB and CD are two diameters of a circle - Ex 12.3

Ex 12.3, 9 - Chapter 12 Class 10 Areas related to Circles - Part 2
Ex 12.3, 9 - Chapter 12 Class 10 Areas related to Circles - Part 3
Ex 12.3, 9 - Chapter 12 Class 10 Areas related to Circles - Part 4

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Transcript

Question 9 In figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region. Area of shaded region = Area of circle with diameter OD + Area of semicircle ACB Area of triangle ABC Area of circle with diameter OD Since OD & OA are radius of circle with centre O OD = OA = 7 cm Diameter = OD = 7 cm radius = r = /2 = 7/2 cm Area of circle with diameter OD = r2 = 22/7 (7/2)^2 = 22/7 7/2 7/2 = 77/2 cm2 Area of circle with diameter OD = 77/2 cm2 Area of semicircle ACB Diameter = AB So, radius = OA = 7 cm Area of semicircle ACB = 1/2 Area of circle = 1/2 r2 = 1/2 22/7 7 7 = 11 7 = 77 cm2 So, Area of semicircle with diameter AB = 77 cm2 Area ABC Given that diameters AB & CD are perpendicular So, AB DC BOC = AOC = 90 Area ABC = 1/2 Base Height Here, Base = AB = 2 radius = 2OA = 2 7 = 14 & Height = OC = 7 cm Putting values Area ABC = 1/2 Base Height = 1/2 AB OC = 1/2 14 7 = 49 cm2 Area of shaded region = Area of circle with diameter OD + Area of semicircle with diameter AB Area of triangle ABC = 77/2 + 77 49 = 77/2 + 28 = (77 + 2 28)/2 = 133/2 = 66.5

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo