Question

ABCD is a trapezium. P is the midpoint of BC. AB = 12 cm, DC = 8 cm. The distance between AB and CD is 6 cm. Find the area of the triangle AQD.

• A. 100 square cm.
• B. 60 square cm.
• C. 90 square cm.
• D. 120 square cm.

Answer 1

The correct option is B

60 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 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 = 8 cm [given]

Hence, BQ = 8 cm

AQ = AB + BQ = 12 + 8 = 20 cm.

Distance between AB and CD, which is also the height of the triangle= 6 cm.

Base of the triangle AQD = 20 cm

Height of the triangle AQD = 6 cm

Area of the triangle AQD = half the product of base and height

= $\frac{1}{2}(20\times 6)$ = 60 square cm.