CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard

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Question 25 (Choice 2) - CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

Last updated at April 16, 2024 by

Find the area of the unshaded region shown in the given figure.

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Transcript

Area of unshaded region = 4 × (Area of semicircle with radius r) + Area of square with side a We note that r + r = a 2r = a a = 2r Also, 14 = 3 + r + a + r + 3 14 = 6 + 2r + a Putting a = 2r 14 = 6 + 2r + 2r 14 = 6 + 4r 14 − 6 = 4r 8 = 4r 4r = 8 r = 8/4 r = 2 Now, Area of semicircle Area of semicircle = 𝟏/𝟐 × π × 𝑟^2 = 𝟏/𝟐 × π × 22 = 2π cm2 Area of square Since a = 2r a = 2(2) = 4 cm Area of square = a2 = 42 = 16 cm2 Now, Area of unshaded region = 4 × (Area of semicircle with radius r) + Area of square with side a = 4 × 2π + 16 = (8π + 16) 〖𝒄𝒎〗^𝟐

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