Question

The diagonals of a rhombus are $24$ cm and $10$ cm. Calculate the area of the rhombus

Answer 1

Given:

The diagonals of a rhombus are,

$d_1=24cm,d_2=10cm$

Area of the rhombus $=\frac{1}{2}\times d_1\times d_2$

$=\frac{1}{2}\times 24cm\times 10cm$

$=120c{m}^2$

(OR)

Lets draw a rough figure,

According to question,

$AC=24$ cm; $BD=10$ cm.

We know that, the diagonals of a rhombus bisect each other at right angles.

Let $O$ be the point of intersection of these diagonals.

Then we have,

$AO=CO=12$ cm

$BO=DO=5$ cm.

We also know that $AOD$ is a right angled triangle.

Hence the area of $AOD$ is
$\frac{1}{2}\times OA\times OD=\frac{1}{2}\times 12\times 5=30{\text{cm}}^2$.
Since a rhombus has four congruent right triangles, therefore the area of rhombus is $=4\times A\left(△AOD\right)=4\times 30=120{\text{cm}}^2$.

Hence, the answer is $120{\text{cm}}^2$.