SSC CPO (SI in Delhi Police & CAPFs) 2016 · Question 110 of 181

A string of length 24 cm is bent first into a square and then into a right-angled triangle by keeping one side of the square fixed as its base. Then the area of triangle equals to: x + = x 1 3 −5 (x + ) = x 1 3 ( ) 3 −5 3 x + 3 + x31 3(x)( )(x + x 1 ) = x 1 27 −125 x + 3 + x31 3(1)(x + ) = x 1 27 −125 x + 3 + x31 3( ) = 3 −5 27 −125 x + 3 = x31 + 27 −125 5 x + 3 = x31 27 −125+135 x + 3 = x31 27 10 π 3π π π 2a 2a = 2 2a a (2a) = 3 8a3 cm3 πr 3 4 3 π × 3 4 (a) = 3 3 4a π 3 cm3 3 4a π 3 8a3 = 4π 8×3 π 6 ∴ 6 : π 2 2

1. A24 cmCorrect
2. B60 cm
3. C40 cm
4. D28 cm