In a rhombus PQRS, side PQ = 17 cm and diagonal PR = 16 cm. Calculate the area of the rhombus.

We know that,

The diagonals of a rhombus bisect each other at right angles.

From figure,

OP = OR = 8 cm

PQ = 17 cm

In right angled △ POQ,

Hypotenuse² = Perpendicular² + Base²

⇒ PQ² = OP² + OQ²

⇒ 17² = 8² + OQ²

⇒ 289 = 64 + OQ²

⇒ OQ² = 289 - 64

⇒ OQ² = 225

⇒ OQ = $\sqrt{225}$

⇒ OQ = 15 cm.

Since, OQ = OS = 15 cm (the diagonals of a rhombus bisect each other)

QS = OQ + OS = 15 + 15 = 30 cm

∴ Diagonals are PR = 16 cm and QS = 30 cm.

As we know,

Area of rhombus = $\dfrac{1}{2}$ × Product of diagonals

= $\dfrac{1}{2} \times (30 × 16) = \dfrac{1}{2} \times 480$ = 240 cm².

Hence, the area of rhombus = 240 cm².