Question

In a parallelogram ABCD, P and Q are mid-points of BC and CD then area of $△APQ$ will be_____ of parallelogram.

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{8}$
• D. $\frac{3}{8}$

Answer 1

The correct option is D $\frac{3}{8}$

ABCD is a parallelogram in which P ,Q are midpoint of BC and CD

Join BD and CD

In $△ACD$,Q is the mid-point of DC

$\therefore AQ$ is a median

Since median divide the triangle in to two triangle equal in area.

$\therefore ar\left(△ADQ\right)=ar\left(△ACQ\right)$

$\Rightarrow ar\left(△ADQ\right)=\frac{1}{2}ar\left(△ACD\right)$

$\Rightarrow ar\left(△ADQ\right)=\frac{1}{2}\times \frac{1}{2}ar\left(parallologramABCD\right)$

$\Rightarrow ar\left(△ADQ\right)=\frac{1}{4}ar\left(parallologramABCD\right)$....(i)

Similarly in $△ABC$ P is the mid-point of BC

$\therefore AP$ is a median

Since median divide the triangle in to two triangle equal in area.

$\therefore ar\left(△ABP\right)=ar\left(△AcP\right)$

$\Rightarrow ar\left(△ABP\right)=\frac{1}{2}ar\left(△ABC\right)$

$\Rightarrow ar\left(△ABP\right)=\frac{1}{2}\times \frac{1}{2}ar\left(parallologramABCD\right)$

$\Rightarrow ar\left(△ABP\right)=\frac{1}{4}ar\left(parallologramABCD\right)$.....(ii)

In $△BCD$ P ,Q are the mid-point of BC and CD.

$\therefore PQ\parallel BDandPQ=\frac{1}{2}BD$

$\Rightarrow ar\left(△CPQ\right)=\frac{1}{4}ar\left(△BCD\right)$

$\Rightarrow \frac{1}{4}\times \frac{1}{2}ar(paral.ABCD)$

$\Rightarrow \frac{1}{8}ar(paral.ABCD)$.......(iii)

Adding (i),(ii) and (iii)

$ar\left(△ADQ\right)+ar\left(△ABP\right)+ar\left(△CPQ\right)$

$\Rightarrow \frac{1}{4}ar(paral.ABCD)+\frac{1}{4}ar(paral.ABCD)+\frac{1}{8}ar\left(paralABCD\right)$

$\Rightarrow \frac{5}{8}ar(paral.ABCD)$

$\therefore △APQ=ar(paral.ABCD)-\frac{5}{8}ar\left(paralABCD\right)$

$\Rightarrow (1-\frac{5}{8})ar\left(paralABCD\right)$

$\Rightarrow \frac{3}{8}ar(paral.ABCD)$