ABCD is a trapezium.P is the mid point ofBC,AB=12cm,DC=8cm. the distance between AB andCD is 6cm a)what is the area of the trapezium? b) what is the area of triangle AQD?
Answers
= 1/2 ×12+8×6
=1/2×20×6
=10×6
=60cm^2
2). area of triangle AQD=1/2 base × height
=1/2 ×2×6
=6 cm^2
Answer: a) 60 cm sq.
b) 60 cm sq.
Step-by-step explanation:
In triangles DCP and PBQ,
CP = PB [Given]
∠DPC=∠BPQ [Vertically opposite angles]
∠CDP=∠PQB
[Interior alternate angles] [DC is parallel to AB and AQ is an extension of AB so, DC is parallel to AQ and DQ acts as a transversal.]
As two angles and one side of the triangle, DCP is equal to two angles and one side of triangle PBQ, through the AAS congruency rule, both 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.
BaThe 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
= 1/2(20×6) = 60 square cm.