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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
I. 4x2 + 9x - 9 = 0
II. 4y2 - 19y + 12 = 0
I. x − √2401 = 0
II. y2 − 2401 = 0
I. p² - 10p +21 = 0
II. q² + q -12 = 0
Equation 1: x² - 220x + 12100 = 0
Equation 2: y² - 210y + 11025 = 0
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the...
What will be the product of smaller roots of both equations.
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 33x² - 186x + 240 = 0
Equation 2: 35y² - 200y + ...
How many values of x and y satisfy the equation 2x + 4y = 8 & 3x + 6y = 10.
I. 2x² - 9x + 10 = 0
II. 3y² + 11y + 6 = 0
I. x2 - 17x + 70 = 0
II. y2 - 11y + 28 = 0