Question

In the figure, $\square ABCD$ and $\square EFGD$ are two parallelograms and G is the mid-point of CD. Then $A\left(\Delta DPC\right)$ = $xA\left(EFGD\right)$. Find $x$.

• A. 1/2
• B. 1/4
• C. 1/8
• D. 1

Answer 1

Answer: (D)

1

Given-

$\square ABCD$ & $\square EFGD$ are two parallelograms.

Their bases $DC$ & $DG$ are on the same line DC and G is the mid point of DC.

Also, $\square ABCD$ & $\square EFGD$ are within the same parallels $AB$ & $DC.$

To find out- The statement, $ar\Delta DPC=\frac{1}{2}arEFGD$ is true or not.

Solution-

$\Delta DPC$ & $\square ABCD$ are on the same base DC

And within the same parallels $AB$ & $DC.$

$\therefore ar\Delta DPC=\frac{1}{2}ABCD$ .......(i)

Again, G is the midpoint of DC

$\therefore DG=\frac{1}{2}DC$ ......(ii)

Now, paralleograms $ABCD$ & $DGEF$ are on the same base line DC

And within the same parallels $AB$ & $DC.$

$\therefore arEFGD=\frac{1}{2}ABCD$ ...(iii)

Since base of $EFGD$ is $DG=\frac{1}{2}DC$

$\therefore$ From $\left(ii\right)$ & $\left(iii\right)$,

$ar\Delta DPC=arEFGD$

$⟹x$ = $1$