Question

Parallelogram $ABCD$ and rectangle $ABPQ$ have the same base $AB$ and also have equal areas (see figure). Show that the perimeter of the parallelogram is greater than that of the rectangle

Answer 1

Since opposite side of parallelogram are rectangle are equal,

$AB=DC$ ($ABCD$ is parallelogram)

$AB=PQ$ ($ABPQ$ is rectangle)

$\to CD=PQ$

$AB+CD=AB+PQ$ (On adding $AB$ in both sides)

Since all segment that can be drawn to a given line

$\therefore BP<BR$ & $AQ<AD$

$BC>BP$ & $AD>AQ$

$BC+AD>BP+AQ$

Adding $\left(2\right)+\left(3\right)$

$AB+DC+BC+AD>AB+PQ+BP+AQ$

$\to AB+BC+CD+AD>AB+BP+PQ+AQ$

Perimeter of parallelogram $ABCD>$ Perimeter of $ABPQ$

Perimeter of parallelogram $>$ Perimeter of rectangle.