Question
AB is parallel to DC in a trapezium ABCD. It is given
that AB > DC and the diagonals AC and BD intersect at O. If AO = 3x - 15 OB = x + 9, OC = x - 5 and OD = 5, and x has two values x1 and x2 then the value of ( x12Â + x22) is:Solution
ATQ, 
⇒ (3x - 15) × 5 = (x + 9) × (x - 5) ⇒ 15x - 75 = x2 + 9x - 5x - 45 ⇒ 15x - 75 = x2 + 4x - 45 ⇒ x2 - 11x + 30 = 0 Solving the quadratic equation: 
⇒ x1 = 6, x2 = 5 To find x12 + x22: ⇒ 62 + 52 = 36 + 25 = 61
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