Question

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

Answer 1

It is given that,
The sides of rhombus = 20 cm.
One of the diagonal = 24 cm.



In ∆ABC,
The sides of the triangle are of length 20 cm, 20 cm and 24 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{20+20+24}{2}=\frac{64}{2}=32\mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{32\left(32-20\right)\left(32-20\right)\left(32-24\right)} \\ =\sqrt{32\left(12\right)\left(12\right)\left(8\right)} \\ =192{\mathrm{cm}}^2...\left(1\right) \end{gathered}$$

In ∆ACD,
The sides of the triangle are of length 20 cm, 20 cm and 24 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{20+20+24}{2}=\frac{64}{2}=32\mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ACD=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{32\left(32-20\right)\left(32-20\right)\left(32-24\right)} \\ =\sqrt{32\left(12\right)\left(12\right)\left(8\right)} \\ =192{\mathrm{cm}}^2...\left(2\right) \end{gathered}$$


∴ Area of the rhombus = Area of ∆ABC + Area of ∆ACD
= 192 + 192
= 384 cm²

Hence, the area of a rhombus is 384 cm².