The mid points P,Q,R and S of the sides of a quadrilateral ABCD are joined. Can I compare the areas of both the quadrilaterals. I know that the new quadrilateral is a parallelogram by mid point theorem..I want little help to compare the areas

  • $\begingroup$ I edit my answer. And write your doubts in comments. Not in answers. And if any other doubt. Tell me plz. $\endgroup$ – Kanwaljit Singh Dec 18 '16 at 6:18

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Join PQ, then in △PAD and △PQS,

DA || SQ

The line joining the midpoints of two sides of a triangle i.e DA is parallel to third side SQ and half of it.

Hence DA = $\frac{1}{2}$SQ.

So, $\frac{(ar△PAD)}{(ar△PQS)}$ = $\frac{(side DA)^2}{(side SQ)^2}$

= $\frac{1/4 SQ^2}{SQ^2}$

$\frac{(ar△PAD)}{(ar△PQS)}$ = $\frac{1}{4}$

(ar△PAD) = $\frac{1}{4}$(ar△PQS)

Similarly, △QAB= $\frac{1}{4}$△QPR.

△RBC= $\frac{1}{4}$△RQS.

△SDC= $\frac{1}{4}$△SPR.

Adding the four equations, we get:

△PAD+△QAB+△RBC+△SDC= $\frac{1}{4}$(△PQS+△QPR+△RQS+△SPR)

Note that △PQS+△QPR+△RQS+△SPR = 2 × area of quadrilateral PQRS.

hence, △PAD+△QAB+△RBC+△SDC is half the area of PQRS.

But, area of PQRS −△PAD+△QAB+△RBC+△SDC also equals the area of parallelogram ABCD.

area of ABCD= $\frac {1}{2}$ × area of PQRS.

  • $\begingroup$ Nice work +1.... $\endgroup$ – Vidyanshu Mishra Dec 18 '16 at 6:34
  • $\begingroup$ @THE LONE WOLF. Thank you. $\endgroup$ – Kanwaljit Singh Dec 18 '16 at 7:26

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