A rectangle with one side of length 4 cm is inscribed in a circle of diameter 5 cm. Find the area of rectangle.

Let side BC = 4 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ BM = MC = 2 cm.

Given,

Diameter = 5 cm, radius = $\dfrac{5}{2}$ = 2.5 cm.

In right ∆OBM,

⇒ OB² = BM² + OM²

⇒ (2.5)² = 2² + OM²

⇒ 6.25 = 4 + OM²

⇒ OM² = 2.25

⇒ OM = $\sqrt{2.25}$ = 1.5 cm.

Similarly in right ∆OAN,

⇒ OA² = AN² + ON²

⇒ (2.5)² = 2² + ON²

⇒ 6.25 = 4 + ON²

⇒ ON² = 2.25

⇒ ON = $\sqrt{2.25}$ = 1.5 cm.

From figure,

⇒ MN = OM + ON = 1.5 + 1.5 = 3 cm.

⇒ AB = DC = MN = 3 cm.

⇒ Area = length × breadth = 4 cm × 3 cm = 12 cm².

Hence area of rectangle = 12 cm².