Question

In the parallelogram ABCD , the side AB is produced to the point X , so that BX=AB. The line DX cuts BC at E. Area of $\Delta AED$=

• A. 2 X area $\left(\Delta CEX\right)$
• B. $\frac{1}{2}\times area\left(\Delta CEX\right)$
• C. area$\left(\Delta CEX\right)$
• D. $\frac{1}{3}\times area\left(\Delta CEX\right)$

Answer 1

Answer: (A)

2 X area $\left(\Delta CEX\right)$

$ar\left(\Delta ADE\right)=\frac{1}{2}ar\left(ABCD\right)$ ............$\left(1\right)$

$ar\left(\Delta CEX\right)=\frac{1}{2}ar\left(\Delta CBX\right)$

$(\because CE=EB)$

$=\frac{1}{2}ar\left(\Delta CBA\right)$

$(\because BX=BA)$

$=\frac{1}{2}\times \frac{1}{2}ar\left(ABCD\right)$

$=\frac{1}{4}ar\left(ABCD\right)$ ..........$\left(2\right)$

From $\left(2\right)$, we can understand that

$2.ar\left(\Delta CEX\right)=\frac{1}{2}ar\left(ABCD\right)$

$=ar\left(\Delta ADE\right)$ (From $\left(1\right)$)

$\left(A\right)$ $2\times$area$\left(\Delta CEX\right)$.