If the diagonal of a square is increased by 4 cm, its

Solution

Area of Square (A1) = a² ................... (1) Length of diagonal (d1) = √2a .............. (2) From eq(2) → a = d1/√2 Area of square (A1) = (d1/√2)² A1 = (d1)²/2 ........... (3) ∴ Diagonal of square increased by 4cm and its are increases by 56cm². Increased diagonal of square (d2) = d1 + 4 ........... (4) Increased area of square (A2) = A1 + 56 .......... (5) A2 = (d2)²/2 From eq (3) we get,  A2 = (d1 + 4)²/2 .......... (6) From eq (4) we get, A2 - A1 = 56 .......... (7) Substitute eq (3) and eq (6) in eq (7) ⇒ (d1 + 4)²/2 - (d1)²/2 = 56 ⇒ (d1 + 4)² - (d1)² = 56 × 2 ⇒ (d1 + 4)² - (d1)² = 112 ⇒ (d1)² + 8d1 + 16 - (d1)² = 112 ⇒  8d1 + 16 = 112 ⇒  8d1 = 112 - 16 ⇒  8d1 = 96 ⇒ d1 = 12 ∴ d2 = d1 + 4 d2 = 12 + 4 = 16 Ratio of the new area of the square to the initial area of the square = A2 / A1 A2 / A1 = ((d2)²/2) / ((d1)²/2 ) A2 / A1 = (16² / 2) / (12² / 2) A2 / A1 = (256 / 2) / (144 / 2) A2 / A1 = 128 / 72 A2 / A1 = 16 / 9 ∴ Here, Ratio of the new area of the square to the initial area of the square A2 : A1 = 16 : 9