Question

Find the area of a rhombus whose perimeter is 80m and one of whose diagonal is 24m.

Answer 1

Perimeter of rhombus ABCD = 80m
$\therefore$ Each side = $\frac{80}{4}=20m$
One diagonal = 24m
$\because$ Diagonal of a rhombus
Bisect each other right angles
$\therefore$ In right $\Delta AOB,AB=20m$,
AO= $\frac{1}{2}AC=\frac{1}{2}\times 24=12m$
$A{B}^2=A{O}^2+O{B}^2$
$\Rightarrow (20{)}^2=(12{)}^2+O{B}^2$
$\Rightarrow 400=144+O{B}^2$
$\Rightarrow O{B}^2=400-144=256=(16{)}^2$
$\therefore OB=16m$
and $BD=2\times OB=2\times 16=32m$
Now area of rhombus = $\frac{1}{2}\times d_1\times d_2$
= $\frac{1}{2}\times 24\times 32m^2=384m^2$