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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
79.99% of (84.89 × 5.99) - (3.89)2 × 21.87 = ?
2 (1/4)% of 7999.78 + {49.77% of 899.71} + √144.14 - 20% of 1499.83 = ?
?% of 1200.22 + 319.82 = 3.99 × 295.64
45.22% of (71.9 x 5.01) + 69.97 =?
15.001% of 799.99 - 3/11% of 1099.99 + 111.002 = ?
(74.76 ÷ 12.11 X ?)% of 239.89 = 600.19
The speed of the boat in still water is 20% less than the speed of the boat in downstream. The time taken by the boat to cover 780 km distance in upstre...
(44/25) ÷ (154/199.5) × 419.91 = ? – (11.11)3
(2310.23 ÷ 32.98) + (1008.32 ÷ 23.9) + 1594.11 = ?