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 (x₁2 + x22) is:
Show answer & explanation
Correct answer: B
Explanation
The correct answer is option 2 i.e. 61.
AO/OB = OC/OD
⇒ (3x - 15)/(x + 9) = (x - 5)/5
⇒ 15x - 75 = x2 + 9x - 5x - 45
⇒ x2 - 11x + 30 = 0
⇒ x = [11 ± √(121 - 120)]/2
⇒ x1 = 6, x2 = 5
So, x12 + x22 = 62 + 52 = 36 + 25 = 61
Therefore, the required value is 61.