Question

ABCD is parallelogram and ABEF is a rectangle and DG is perpendicular on AB.
Prove that (i) $ar\left(ABCD\right)=ar\left(ABEF\right)$
(ii) $ar\left(ABCD\right)=AB\times DG$

Answer 1

(i) A rectangle is also a parallelogram

The height of parllelogram and rectangle are same

Then Parallelograms lie on the same base and between the same parallels

So area of parllelogram (ABCD)=area of rectangle (ABEF) ....... (1)

(ii) area of parllelogram (ABCD)=area of rectangle (ABEF) (from (1)

The area of rectangle ABEF =$AB\times BE$

=$AB\times DG$

From (1)The area of pallelogram=$AB\times DG$(Where DG is parpendicula from D at AB)

From the result, we can say that "area of a parallelogram is the product of its any side

and the corresponding altitude".