Question

Find the area of a rhombus having each side equal to 13 cm and one of the diagonals equal to 24 cm.

Answer 1

Let ABCD be the rhombus whose diagonals intersect at O.

Then, AB = 13 cm
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, such that:
OA = $\frac{1}{2}$AC = 12 cm
AB = 13 cm

By Pythagoras theorem:
(AB)² = (OA)² + (OB)²
⇒ (13)² = (12)² + (OB)²
⇒ (OB)² = (13)² − (12)²
⇒ (OB)² = 169 − 144 = 25
⇒ (OB)² = (25)²
⇒ OB = 5 cm
∴ BD = 2 × OB = 2 × 5 cm = 10 cm

∴ Area of the rhombus ABCD = $\frac{1}{2}$ × AC × BD cm²
= $\left(\frac{1}{2}\times 24\times 10\right)$ cm² = 120 cm²