In the adjoining figure, ABCD is a 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.

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Solution

In trapezium ABCD, $A B ∥ C D$ and the diagonals AC and BD intersect at O.
Therefore,

$\frac{A O}{O C} = \frac{B O}{O D} \Rightarrow \frac{5 x - 7}{2 x + 1} = \frac{7 x - 5}{7 x + 1} \Rightarrow (5 x - 7) (7 x + 1) = (7 x - 5) (2 x + 1) \Rightarrow 35 x^{2} + 5 x - 49 x - 7 = 14 x^{2} - 10 x + 7 x - 5 \Rightarrow 21 x^{2} - 41 x - 2 = 0 \Rightarrow 21 x^{2} - 42 x + x - 2 = 0 \Rightarrow 21 x (x - 2) + 1 (x - 2) = 0 \Rightarrow (x - 2) (21 x + 1) = 0 \Rightarrow x = 2, - \frac{1}{21} ∵ x \neq - \frac{1}{21} ∴ x = 2$

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Similar questions

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.

Q. In the adjoining figure

A B | | D C

and diagonal AC & BD intersect at point O. If

A O = (3 x - 1) c m

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(a) 2
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A B C D

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