Question

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

Answer 1

Let ABCD be the rhombus, whose diagonals intersect at O.

AB = 20 cm and AC = 24 cm
The diagonals of a rhombus bisect each other at right angles.

Therefore, ΔAOB is a right angled triangle, right angled at O.

Here, OA =$\frac{1}{2}\mathrm{AC}$ = 12 cm
AB = 20 cm

By Pythagoras theorem:
(AB)² = (OA)² + (OB)²
⇒ (20)² = (12)² + (OB)²
⇒ (OB)² = (20)² − (12)²
⇒ (OB)² = 400 − 144 = 256
⇒ (OB)² = (16)²
⇒ OB = 16 cm
∴ BD = 2 $\times$ OB = 2 $\times$ 16 cm = 32 cm

∴ Area of the rhombus ABCD = $\left(\frac{1}{2}\times \mathrm{AC}\times \mathrm{BD}\right)$ cm²
= $\left(\frac{1}{2}\times 24\times 32\right)$ cm²
= 384 cm²