Question

In the adjoining figure, ABCD is trapezium in which CD || AB and its diagonals intersect at O. If AO = (5x - 7) cm, OC = (2x + 1) cm, DO = (7x - 5) cm and OB = (7x + 1) cm, find the value of x.

Answer 1

Here is the answer.



We know that diagonals of a trapezium divide each other proportionally.So,
AO/OC = DO/OB
⇒5x−7/2x+1 = 7x−5/7x+1⇒$35x^2-49x+5x-7$ = $14x^2+7x-10x-5$
⇒$21x^2-41x-2$ = 0⇒$21x^2-42x+x-2$ = 0
⇒21x(x−2) + 1(x−2) = 0
⇒(21x + 1)(x−2) = 0
⇒x−2 = 0 or 21x+1 = 0
⇒x = 2 or x = −1/21 (rejected)