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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 You are here

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 March 29, 2023 by Teachoo

Ex 12.3, 5 From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in figure. Find the area of the remaining portion of the square. Area of remaining portion = Area of square – Area of middle circle – Area of 4 quadrants For square ABCD Area of square = (side)2 = (4)2 = 4 × 4 = 16 For middle circle Diameter = 2 cm Hence, radius = r = 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟/2 = 2/2 = 1 cm Area of middle circle = 𝜋𝑟2 = 22/7×1×1 = 22/7 cm2 For quadrant radius = r = 1 cm Area of 1 quadrant = 1/4 of (area of circle ) = 1/4×πr2 = 1/4×22/7×12 = 22/28 Since, all quadrants have same radius So, area of 4 quadrant = 4 × Area of 1 quadrant = 4 × 22/28 = 22/7 cm2 Area of remaining portion = Area of square – Area of middle circle – Area of 4 quadrants = 16 – 22/7−22/7 = (16 × 7 − 22 − 22)/7 = (112 − 44)/7 = 68/7 cm2 Hence, area of remaining portion = 68/7 cm2