Question

Find the area of a rhombus one side of which measures 20 cm and one of whose diagonals is 24 cm.

Answer 1

Let ABCD is a rhombus with the diagonals AC and BD which intersect each other at O.

Given AC = 24cm.

Then AO = 12cm.

Let BO = x and AB = 20cm.

By Pythagoras theorem, we know that

$c^2=a^2+b^2$

$\Rightarrow {20}^2={12}^2+x^2$

$\Rightarrow 400=144+x^2$

$\Rightarrow x^2=400-144$

$\Rightarrow x^2=256$

x = 16cm.

Diagonal BD = 2 * 16

= 32cm.

We know that Area of the rhombus = 1/2 * (product of diagonals)

= 1/2 * 24 * 32

= 12 * 32

= 384 sq.cm.