Question

# The following figure shows a trapezium ABCD in which AB || DC. P is the mid-point of AD and PR || AB. Then:

A
PR=12(AB+CD)
B
PR=(AB+CD)
C
PR=13(AB+CD)
D
PR=14(AB+CD)

Solution

## The correct option is A $$PR = \displaystyle \frac{1}{2} (AB + CD)$$In $$\triangle$$, BDC and BQR,$$\angle B = \angle B$$ (Common)$$\angle BDC = \angle BQR$$ (Corresponding angles of parallel lines)$$\angle BCD = \angle BRQ$$ (Corresponding angles of parallel lines)thus, $$\triangle BDC \sim \triangle BQR$$Thus, $$\dfrac{BD}{QB} = \dfrac{DC}{QR}$$$$2 = \dfrac{DC}{QR}$$ (Q is the mid point of BD)$$QR = \dfrac{1}{2} DC$$Similarly, $$QP = \dfrac{1}{2} AB$$Hence, $$QR + QP = \dfrac{1}{2} (AB + DC)$$$$PR = \dfrac{1}{2}(AB + CD)$$Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More