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Ex 12.3

Ex 12.3, 1
Deleted for CBSE Board 2023 Exams

Ex 12.3, 2 Deleted for CBSE Board 2023 Exams

Ex 12.3, 3 Deleted for CBSE Board 2023 Exams

Ex 12.3, 4 Important Deleted for CBSE Board 2023 Exams

Ex 12.3, 5 Deleted for CBSE Board 2023 Exams

Ex 12.3, 6 Deleted for CBSE Board 2023 Exams

Ex 12.3, 7 Deleted for CBSE Board 2023 Exams

Ex 12.3, 8 Important Deleted for CBSE Board 2023 Exams

Ex 12.3, 9 Deleted for CBSE Board 2023 Exams You are here

Ex 12.3, 10 Deleted for CBSE Board 2023 Exams

Ex 12.3, 11 Deleted for CBSE Board 2023 Exams

Ex 12.3, 12 Important Deleted for CBSE Board 2023 Exams

Ex 12.3, 13 Important Deleted for CBSE Board 2023 Exams

Ex 12.3, 14 Deleted for CBSE Board 2023 Exams

Ex 12.3, 15 Important Deleted for CBSE Board 2023 Exams

Ex 12.3, 16 Important Deleted for CBSE Board 2023 Exams

Chapter 12 Class 10 Areas related to Circles

Serial order wise

Last updated at Dec. 28, 2018 by Teachoo

Ex 12.3, 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