The perimeter of a rhombus is 52 cm. If one of its diagonals is 24 cm long, find:

(i) the length of the other diagonal,

(ii) the area of the rhombus.

(i) Let ABCD be a rhombus.

Perimeter of the rhombus = 52 cm

One diagonal BD = 24 cm

Let a be the length of a side of the rhombus.

Perimeter of a rhombus = 4 x Side

⇒ 4 x a = 52

⇒ a = $\dfrac{52}{4}$

⇒ a = 13 cm.

BD = 24 cm

Since the diagonals of a rhombus bisect at 90°.

Then, OB = OD = $\dfrac{24}{2}$ = 12 cm.

Applying pythagoras theorem for △AOB, we get:

⇒ AB² = OA² + OB²

⇒ (13)² = OA² + (12)²

⇒ 169 = OA² + 144

⇒ OA² = 169 - 144

⇒ OA² = 25

⇒ OA = $\sqrt{25}$

⇒ OA = 5 cm.

⇒ AC = 2 x OA = 2 x 5 cm = 10 cm.

Hence, the length of the other diagonal = 10 cm.

(ii) By formula,

Area of rhombus = $\dfrac{1}{2}$ x product of diagonals

= $\dfrac{1}{2}$ x 24 x 10

= 12 x 10 = 120 cm².

Hence, area of the rhombus = 120 cm².