Question
If the diagonal of a square is increased by 4 cm, its
area increases by 56 cm2 . Find the ratio of the new area of the square to the initial area of the square.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
300% of (3341 – 471) = ? × (√4225/195)
1404 ÷ 26 x 3 + 7 = ?2
2(1/3) + 2(5/6) – 1(1/2) = ? – 6(1/6)
35% of 240 – 6 2 = ? 2 – √256
16 × 35 + 119 + 23 × 17 = ? + 370
{(? × 15) + (? × 45)} – 120 = 360
721 +21 x 9 - 118 = ? + 82
If a³ - b³ = (a - b)(a² + ab + b²), find a³ - b³ when a = 10 and b = 4.
`(450 -: ?)/(2.5 xx 1.2)` = 250
(750 / 15 × 15 + 152 + 20% of 125) = ?3
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