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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 You are here

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

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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 16, 2023 by Teachoo

Ex 12.3, 4 Find the area of the shaded region in figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre. Area of shaded region = Area of circle with radius 6 cm + Area of equilateral triangle with side 12 cm – Area of sector ODE Area of circle Radius of circle = r = 6 cm Area of circle = 𝜋r2 = 22/7×(6)2 = 22/7 × 36 = 792/7 cm2 Area of equilateral triangle Area of equilateral triangle = √3/4 (side)2 = √3/4×(12)^2 = √3/4×12×12 = √3×3×12 = 36√3 cm2 Area of sector ODE Radius = r = 6 cm , & θ = ∠ DOE = 60° Area of sector OCD = θ/360×𝜋𝑟2 = (∠DO𝐸)/360×𝜋𝑟2 = (60°)/(360°)×22/7×62 = 1/6×22/7×6×6 = 22/7×6 = 132/7 Now, Area of shaded region = Area of circle with radius 6 cm + Area of equilateral triangle with side 12 cm – Area of sector ODE = 792/7+36√3 −132/7 = (792 + 7 × 36√(3 )− 132)/7 = (792 + 252√3 − 132)/7 = (660 + 252√3)/7 = 660/7 + (252√3)/7 = (660/7 +36√3) cm2 Hence, area of shaded region = (660/7+36√3) cm2