Question

Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle. [4 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Proof : 2 Marks

Consider the parallelogram ABCD and rectangle ABEF as follows.

Here, it can be observed that parallelogram ABCD and rectangle ABEF are between the same parallels AB and CF.
ar(ABCD) = ar(ABEF)

We know that opposite sides of a parallelogram or rectangle are of equal lengths.
Therefore,
AB =EF (For rectangle)
AB =CD (For parallelogram)
$\Rightarrow CD=EF$

$\therefore AB+CD=AB+EF......\left(1\right)$ [Adding AB on both sides]

Of all the line segments that can be drawn to a given line from a point not lying on it, the perpendicular line segment is the shortest.
$\Rightarrow AF<AD$

And similarly, BE <BC
$AF+BE<AD+BC...\left(2\right)$

From equations (1) and( 2), we obtain
AB +EF +AF+ BE < AD + BC+ AB+ CD

i,e Perimeter of rectangle ABEF < Perimeter of parallelogram ABCD