Question

In the given figure, ABCD is a trapezium whose diagonals AC and BD intersect at O such that OA = (3x −1) cm, OB = (2x + 1) cm, OC = (5x − 3) cm and OD = (6x − 5) cm. Then, x = ?


(a) 2
(b) 3
(c) 2.5
(d) 4

Answer 1

(a) 2

We know that the diagonals of a trapezium are proportional.
Therefore,

$$\begin{gathered} \frac{OA}{OC}=\frac{OB}{OD} \\ \Rightarrow \frac{3X-1}{5X-3}=\frac{2X+1}{6X-5} \\ \Rightarrow \left(3X-1\right)\left(6X-5\right)=\left(2X+1\right)\left(5X-3\right) \\ \Rightarrow 18X^2-15X-6X+5=10X^2-6X+5X-3 \\ \Rightarrow 18X^2-21X+5=10X^2-X-3 \\ \Rightarrow 18X^2-21X+5-10X^2+X+3=0 \\ \Rightarrow 8X^2-20X+8=0 \\ \Rightarrow 4\left(2X^2-5X+2\right)=0 \\ \Rightarrow 2X^2-5X+2=0 \\ \Rightarrow 2X^2-4X-X+2=0 \\ \Rightarrow 2X\left(X-2\right)-1\left(X-2\right)=0 \\ \Rightarrow \left(X-2\right)\left(2X-1\right)=0 \\ \Rightarrow \mathrm{Either}x-2=0\mathrm{or}2x-1=0 \\ \Rightarrow \mathrm{Either}x=2\mathrm{or}x=\frac{1}{2} \\ \mathrm{When}x=\frac{1}{2},6x-5=-2<0,\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{possible}. \\ \mathrm{Therefore},x=2 \end{gathered}$$