Question

The following figure shows a trapezium ABCD in which AB || DC.

P is the mid-point of AD and PR || AB. Then:

• A. $PR=\frac{1}{2}(AB+CD)$
• B. $PR=(AB+CD)$
• C. $PR=\frac{1}{3}(AB+CD)$
• D. $PR=\frac{1}{4}(AB+CD)$

Answer 1

The correct option is A $PR=\frac{1}{2}(AB+CD)$

In $△$, 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, $△BDC\sim △BQR$

Thus, $\frac{BD}{QB}=\frac{DC}{QR}$

$2=\frac{DC}{QR}$ (Q is the mid point of BD)

$QR=\frac{1}{2}DC$

Similarly, $QP=\frac{1}{2}AB$

Hence, $QR+QP=\frac{1}{2}(AB+DC)$

$PR=\frac{1}{2}(AB+CD)$