Question

In a trapezium ABCD, in which $AB\parallel CD.$ P&Q are mid points of AC & BD. If PQ = 5 cm then AB - CD =________

Answer 1

Draw CN passing through Q meeting on AB.

Now consider $△DQC$ and $△BQN,$

$DQ=QB$ (Q is the mid point of $BD$)

$\angle CDQ=\angle NBQ$ (Alternate angles as $CD\parallel AB$)

$\angle DQC=\angle BQN$ (Vertically oppposite angles)

$△DQC\cong △BQN$ (By ASA property)

$⟹CD=BN(C.P.C.T.)---\left(1\right)$

By mid point theorem in $△ACN,$

$MN=\frac{1}{2}AP$

$⟹PQ=\frac{1}{2}(AB-BN)$

$⟹5\ast 2=AB-CD\left(from1\right)$

$⟹AB-CD=10cm$ (Answer)