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** **__Solve this :__

$1.\mathrm{PQRS}\mathrm{is}\mathrm{a}\mathrm{trapezium}\mathrm{in}\mathrm{which}\mathrm{PQ}\mathrm{is}\mathrm{parallel}\mathrm{to}\mathrm{SR}\mathrm{and}\mathrm{SR}\mathrm{PQ}.\mathrm{A}\mathrm{line}\mathrm{through}\mathrm{P}\mathrm{parallel}\mathrm{to}\mathrm{QR}\mathrm{cuts}\mathrm{the}\mathrm{diagonal}\mathrm{QS}\mathrm{at}\mathrm{X}.\mathrm{Prove}\mathrm{that}\mathrm{ar}(\u2206\mathrm{XQR})=\mathrm{ar}(\u2206\mathrm{PQS}).\phantom{\rule{0ex}{0ex}}[\mathrm{hint}:\mathrm{Join}\mathrm{PR}]$

__Solve this :__

Dear Student,

Please find below the solution to the asked query:

We have our diagram , As :

Here we have join PR .

We know area of triangle = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$

Given : A line through P parallel to QR cuts the diagonal QS at X. So **QR | | PX**

Here ,

Area of $\u2206$ XQR = Area of $\u2206$ PQR --- ( 1 ) ( As both lies in same parallel lines " QR | | PX " , So height is same and have same base = QR )

And

Area of $\u2206$ PQR = Area of $\u2206$ PQS --- ( 2 ) ( As both lies in same parallel lines " PQ | | SR "(Given ),So height is same and have same base = PQ )

We know " Euclid's first axiom : Things which are equal to the same thing are also equal to one another. " , So from equation 1 and 2 we get :

**Area of $\u2206$ XQR = Area of $\u2206$ PQS ( Hence proved )**

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