Question

If the perimeter of a rhombus is $20cm$ and one of its diagonals is $8cm$, then find the area of the rhombus.

Answer 1

Let first diagonal be $d$

Second diagonal be $d_2$

Giver perimeter of a rhombus $=20cm$

We know perimeter of a rhombus $=4$ sides

Hence, side $=\frac{20}{4}cm=5cm$

Now, we know that the diagonals of a rhombus bisect each other at right angles $\left(90°\right)$

Hence, a right angled triangle can be visualized with side as hypotenuse.

Diagonal length $=8cm$

Half the length ( since diagonal bisect each other ) $=\frac{8}{2}=4cm$

$\Rightarrow {\left(\frac{d}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2=5^2$

$\Rightarrow {\left(\frac{8}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2=25$

$\Rightarrow 4^2+{\left(\frac{d_2}{2}\right)}^2=25$

$\Rightarrow {\left(\frac{d_2}{2}\right)}^2=25-16$

$\Rightarrow {\left(\frac{d_2}{2}\right)}^2=9$

$\Rightarrow {\left(\frac{d_2}{2}\right)}^2=\sqrt{9}$

$\Rightarrow \frac{d_2}{2}=3$

$\Rightarrow d_2=6cm$

Hence other diagonal $=6cm$

Area $=\frac{1}{2}\times d\times d_2$

$=\frac{1}{2}\times 8\times 6$

$=24c{m}^2.$

Hence, the answer is $24c{m}^2.$