Question

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

Answer 1

The diagonals of a rhombus bisect at right angles
Consider $\Delta$ AOD
Using the Pythagoras theorem
$A{D}^2=O{D}^2+A{O}^2$
By substituting the values
${20}^2=O{D}^2+{12}^2$
On further calculation
$O{D}^2=400-144$
$O{D}^2=256$
By taking out the square root
OD = $\sqrt{256}$
So we get
OD $=16cm$
We know that BD $=2$OD
So we get
BC $=2\times \left(16\right)=32cm$

We know that

Area of rhombus $=\frac{1}{2}\times$ 1st diagonal $\times$ 2nd diagonal

Area of rhombus ABCD $=\frac{1}{2}\times AC\times BD$
By substituting the values
Area of rhombus ABCD $=\frac{1}{2}\times 24\times 32$
On further calculation
Area of rhombus ABCD $=384c{m}^2$
Therefore, the area of rhombus ABCD is $384c{m}^2.$