Solve this :

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

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

And

Area of $∆$ PQR =  Area of $∆$ 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 $∆$ XQR =  Area of $∆$ PQS                                         ( Hence proved )



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