Question

PQRS is a square E and F are points on the sides QR and RS respectively show that $ar\left(\Delta PQE\right)=ar\left(\Delta PSF\right)$

Answer 1

Given PQRS is a square $E$ and $F$ are points on the sides $QR$ and $RS.$

We have to prove that $ar\left(△PQE\right)=ar\left(△PSF\right)$

We know all four sides of a square are equal in length,

$\Rightarrow PQ=QR=RS=SP=a$

In $△PQE$

$\Rightarrow$ Area of triangle $=\frac{1}{2}bh$

$=\frac{1}{2}(PQ\times QE)$

$\therefore$ Area of $△PQE=\frac{1}{2}(a\times QE)⟶\left(1\right)$

$\Rightarrow$ In $△PSF=\frac{1}{2}(PS\times SF)$

$\therefore$ Area of $△PSF=\frac{1}{2}(a\times SF)$

$\Rightarrow$ here $SF=QE$ ( because both are the distance between two parallel line as $SQ\parallel EF$ )

$\therefore$ Area of $PSF=\frac{1}{2}(a\times QE)⟶\left(2\right)$

$\therefore ar\left(△PQE\right)=ar\left(△PSF\right).$

Hence, the answer is proved.