Question

Find the area of a rhombus having each side equal to 15 cm and one of whose diagonals is 24 cm.

Answer 1

Let ABCD be the rhombus where diagonals intersect at 0 as given in the figure.

Then AB = 15 cm and AC = 24 cm.

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

Therefore, from the figure triangle A0B is a right-angled triangle, right angled at O such that

Therefore, OA = ½(AC) = ½ (24) = 12 cm and AB = 15 cm.

By applying Pythagoras theorem, we get,

(AB)² = (OA)² + (OB)²

(15)² = (12)² + (OB)²

(OB)²= (15)²– (12)²

(OB)² = 225 – 144

= 81

(OB)² = (9)²

OB = 9 cm

BD = 2 x OB

= 2 x 9

= 18 cm

Hence, Area of the rhombus ABCD = (½ x AC x BD)

= (1/2 x 24 x 18)

= 216 cm²