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Ex 12.3

Ex 12.3, 1
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Ex 12.3, 2 Deleted for CBSE Board 2023 Exams

Ex 12.3, 3 Deleted for CBSE Board 2023 Exams You are here

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

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 May 29, 2018 by Teachoo

Ex 12.3, 3 Find the area of the shaded region in figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles. Area of shaded region = Area of square ABCD Area of semicircle APD Area of semicircle BPC Area of square ABCD Side of square = 14 cm Area of square = Side Side = 14 14 = 196 cm2 Area of semi circle APD Diameter = AD = 14 cm So, radius = /2 = 14/2 = 7 cm Area of semi circle APD = 1/2 r2 = 1/2 22/7 7^2 = 1/2 22/7 7 7 = 77 cm2 Similarly, For semicircle BPC Diameter = BC = 14 cm Since diameter is same Area of semi circle BPC = Area of semi circle APD = 77 cm2 Now, Area of shaded region = Area of square Area of semi circle APB Area of semi circle BPC = 196 77 77 = 196 (77 + 77) = 196 (154) = 42 cm2 Hence, area of shaded region = 42 cm2