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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

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 You are here

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 Aug. 5, 2021 by Teachoo

Ex 12.3, 13 In figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π= 3.14) Area of shaded region = Area of quadrant OPBQ – Area of square OABC Area of square Side of square = OA = 20 cm Area of square = (side)2 = (20)2 = 20×20 = 400 cm2 Area of quadrant, We need to find radius Joining OB. Also, all angles of square are 90° ∴ ∠ BAO = 90° Hence, Δ OBA is a right triangle Using Pythagoras theorem in Δ OBA (Hypotenuse)2 = (Height)2 + (Base)2 (OB)2 = (AB)2 + (OA)2 (OB)2 = 202 + 202 (OB)2 = 400 + 400 OB2 = 800 OB = √800 OB = √(10×10×2×2×2) OB = √(〖10〗^2×2^2×2) OB = 10×2 √2 OB = 20√2 Hence, radius = OB = 20√2 Now area of quadrant = 1/4×𝑎𝑟𝑒𝑎 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒 = 1/4×𝜋𝑟2 = 1/4×3.14×(20√2 )2 = 1/2×3.14×(20×20×√2×√2 ) = 1/4×3.14×(400×2) = 1/4×3.14×800 = 3.14 × 200 = 628 cm2 Now , Area of shaded region = Area of Quadrant OPBQ – area of square OABC = 628 – 400 = 228 cm2 Hence, area of shaded region = 288 cm2