Question

Question 1
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.

Answer 1

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