Question

In parallelogram ABCD shown below, the vertical distance between the lines AD and BC is 5 cm and length of BC is 4 cm. P and Q are the midpoints of AB and AD respectively. Area of triangle AQP is ____.

• A. 1.25 $c{m}^2$
• B. 2.5 $c{m}^2$
• C. 5 $c{m}^2$
• D. Insufficient data

Answer 1

The correct option is B 2.5 $c{m}^2$

Given:
ABCD is a parallelogram.
Base BC = 4 cm
Height of a parallelogram = 5 cm

Area of parallelogram ABCD = Base x Height
$=5cm\times 4cm$
$=20c{m}^2$

A diagonal divides a parallelogram in two congruent triangles which have equal areas.
$\therefore Area\Delta ABD=Area\Delta BDC$
$\therefore Area\Delta ABD$ =$\frac{1}{2}$ Area of ABCD
$=10c{m}^2$

Let R be the midpoint of the diagonal. Join P and Q to R.
P and Q are midpoints of AB and AD.
Therefore PQ is parallel to BD (Midpoint theorem)

Similarly, QR is parallel to AB and PR is parallel to AD
Therefore, APQR, DQPR and BPQR are all parallelograms
$\therefore Area\Delta AQP=Area\Delta RQP$
$=Area\Delta BPR=Area\Delta QPR$...(i)

$Area\Delta ABD=Area\Delta AQP+Area\Delta RQP+Area\Delta BPR+Area\Delta QPR$

$Area\Delta ABD=4\times Area\Delta AQP$$=\frac{1}{4}\times Area\Delta ABD$
$\therefore Area\Delta AQP=\frac{1}{4}\times 10$
$=2.5c{m}^2$