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

ABCD is a rhombus with diagonals BD and AC.

DC = 20 cm

BD = 24 cm

Diagonals of a rhombus bisect each other at 90°.

DE = $\dfrac{BD}{2} = \dfrac{24}{2}$ = 12 cm.

In a right triangle DEC,

Hypotenuse (DC) = 20 cm

DE = 12 cm.

By using pythagoras theorem for the right triangle DEC,

⇒ DC² = DE ² + EC²

⇒ 20² = 12² + EC²

⇒ EC² = 20² - 12²

⇒ EC² = 400 - 144

⇒ EC² = 256

⇒ EC = $\sqrt{256}$

⇒ EC = 16 cm.

AC = 2 × EC

= 2 × 16 = 32 cm.

Area of rhombus = $\dfrac{1}{2}$ × (product of diagonals)

= $\dfrac{1}{2}$ × 24 × 32

= 12 × 32

= 384 cm².

Hence, area of rhombus = 384 cm².