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CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard

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Question 29 - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

Last updated at Oct. 27, 2020 by

In the figure, ABCD is a square of side 14 cm. Semi-circles are drawn with each side of square as diameter. Find the area of the shaded region.

In the figure, ABCD is a square of side 14 cm. Semi-circles are drawn

Question 29 - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard - Part 2
Question 29 - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard - Part 3
Question 29 - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard - Part 4

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Note : This is similar to Example 6 of NCERT – Chapter 12 Class 10

Check the answer here https://www.teachoo.com/1885/549/Example-6---Find-area-of-shaded-design--ABCD-is-a-square-10-cm/category/Examples/

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Transcript

Question 29 In the figure, ABCD is a square of side 14 cm. Semi-circles are drawn with each side of square as diameter. Find the area of the shaded region. First, let us find Area of Square Given Side of square ABCD = 14 cm Area of square ABCD = (Side)2 = (14)2 = 196 cm2 Let us mark the shaded region as I , II, III, and IV. Also, Diameter of semicircle = Side of square = 14 cm Radius of semicircle = 𝑆𝑖𝑑𝑒/2 = 14/2 = 7 cm Area of semi circle AD = 1/2Γ— area of circle = 1/2Γ—Ο€r2 = 1/2Γ—22/7 Γ— (7)2 = 11 Γ— 7 = 77 cm2 Since radius is same for semi-circle AD, BC, AB, CD Area of all semi circles = 77 cm2 Area of region I + Area of region III = Area of square ABCD – (Area of semicircle AD + area of semi circle BC) Area of region II + Area of region IV = Area of square ABCD – (Area of semicircle AB + area of semi circle CD) So, Area of region (I + II + III + IV) = 2(Area of square ABCD) – (Area of semicircle AD + BC + AB + CD) Putting values = 2(196) – (77 + 77 + 77 + 77) = 200 – 4 Γ— 77 = 392 βˆ’ 308 = 84 cm2

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