Question

In a trapezium ABCD with AB = 20 cm, CD = 10 cm AB ||CD, P and Q are the mid-points of AC and
is BD then the length of PQ is
​

Answer 1

Length of $PQ = 5$ $cm$.

Explanation:

Given that, in trapezium ABCD

Length of  

Length of $CD = 10$ $cm$

And both are parallel sides, AC and BD are the diagonal of trapezium and mid points are P and Q.

Length of  ?  

We draw DP and produce it to meet AB at point E  

∠ $APE =$ ∠ $CPD$            (vertically opposite angles)  

$AP = PC$                                         (P is midpoint of AC)  

∠ $PCD =$ ∠ $PAE$              (alternate angles)  

△$APE$ is similar to △$CPD$  

⇒  $DP = PE$

And $AE = CD$  

In △ $DEB$, P is the midpoint of DE

Q is the midpoint of BD  

Hence, PQ is parallel to EB  

⇒  PQ is parallel to AB  

⇒ PQ is parallel to AB and CD  

Also, $PQ = \frac{1}{2}\times EB$

⇒  $PQ = \frac{1}{2}(AB - AE)$  

⇒  $PQ = \frac{1}{2}(AB - CD)$  

Length of $PQ = 1/2 (20 - 10) = 5$ $cm$.