Question

Parallelogram $ABCD$ and rectangle $ABEF$ are on the same base and have equal areas. Show that perimeter of the parallelogram is greater than that of rectangle.

Answer 1

Since opposite sides of parallelogram and rectangle are equal, we have

$AB=DC$ since $ABCD$ is a parallelogram

And $AB=EF$ since $ABEF$ is a rectangle

$DC=EF$ .....$\left(1\right)$

$\Rightarrow AB+DC=AB+EF$ ......$\left(2\right)$

Since, of all the segments that can be drawn to a given line from a point not lying on it, the perpendicular segment is the shortest,

$BE<BC$ and $AF<AD$

$\Rightarrow BC>BE$ and $AD>AF$

$\Rightarrow BC+AD>BE+AF$ .....$\left(3\right)$

Adding $\left(2\right)$ and $\left(3\right)$,

$AB+DC+BC+AD>AB+EF+BE+AF$

$\Rightarrow AB+BC+CD+DA>AB+BE+EF+FA$

Perimeter of a parallelogram$>$ Perimeter of rectangle.