Question

Prove that $ar\left(△AXS\right)=\frac{1}{2}ar\left(PQRS\right)$ in the adjoining figure.

Answer 1

Since $PQRS$ is a parallelogram

$PQ||RS$ (opposite sides of parallelogram are parallelogram are parallel )

And $ABRS$ is also a parallelogram

So, $AB||RS$ (opposite sides of parallelogram are parallelogram are parallel)

Since $PQ||RS$ and $AB||RS$

we can say that $PB||RS$

Now, $PQRS$ and $ABRS$ are two parallelograms with same base $RS$ and between the same parallels $PB$ and $RS$

Therefore, $ar\left(PQRS\right)=ar\left(ABRS\right).......\left(i\right)$

Now, Parallelogram $ABRS$ and triangle $AXS$ are on the same base $AS$ and we know that if a triangle and a parallelogram lie on the same base and between two parallel lines then the area of the triangle will half the area of parallelogram, i.e.

$ar\left(AXS\right)=ar\left(ABRS\right)/\mathrm{2.........}\left(ii\right)$

Form (i) and (ii)

$ar\left(AXS\right)=ar\left(PQRS\right)/2$