1
$\begingroup$

enter image description here

I tried this question by constructing a line $PD$ therefore forming two triangles $ADP$ and QDP but couldn't establish the congruency relation between the triangles. My approach was that if I have those triangles congruent then, $ PQ = AD (CPCT)$ where $PQ$ and $AD$ is the individual side of parm. $ABCD$ and $PQRC$ respectively and we have $AD=BC$ & $ PQ=RC$ (since, opposite side of parm. is equal) by joing $BD$ (forms diagnol) of parm. $ABCD$ and triangles $ABD$ and $BCD$ be equal in area similarly for parm. $PQRC$.

(may be my approach isn't on a right track or their are other ways to solve this particular question, please give it a try! )

$\endgroup$
6
$\begingroup$

Referring to the following:-enter image description here

Area of the red triangle = (1/2)[area of ABCD]

Area of the red triangle = (1/2)[area of PQRC]

Therefore, area of ABCD = area of PQRC

$\endgroup$
1
$\begingroup$

A hint:

Extend the lines $A\vee B$ and $Q\vee R$ to the left util they intersect at $S$. Then bring the parallelogram $SDCP$ into the game.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.