Question

ABCD is a trapezium. P is the midpoint of BC. AB = 14 cm, DC = 6 cm. The distance between AB and CD is 8 cm. Find the area of the triangle AQD.

• A. 80 square cm.
• B. 10 square cm.
• C. 50 square cm.
• D. 90 square cm.

Answer 1

The correct option is A

80 square cm.

In triangles DCP and PBQ,

CP = PB [Given]

$\angle DPC=\angle BPQ$ [Vertically opposite angles]

$\angle CDP=\angle PQB$ [Interior alternate angles] [DC is parallel to AB and AQ is the extension of AB so, DC is parallel to AQ and DQ acts as a transversal.]

As two angles and one side of triangle DCP are equal to two angles and one side of triangle PBQ, through AAS congruency rule, both the triangles are congruent.

DCP ≅ PBQ

Now, DC = BQ [Corresponding parts of congruent triangles]

DC = 6 cm [given]; Hence, BQ = 6 cm

AQ = AB + BQ = 14 + 6 = 20 cm.

The distance between AB and CD, which is also the height of the triangle= 8 cm.

Base of the triangle AQD = 20 cm; Height of the triangle AQD = 8 cm

Area of the triangle AQD = half the product of base and height = $\frac{1}{2}(20\times 8)$ = 80 square cm.