The perimeter of a rhombus is 40 cm. If one diagonal is 16 cm, find :

(i) its other diagonal

(ii) its area

(i) Given:

The perimeter of a rhombus = 40 cm.

One diagonal = 16 cm.

As we know, the perimeter of the rhombus = 4 x side

⇒ 4 x side = 40 cm

⇒ Side = $\dfrac{40}{4}$ cm

⇒ Side = 10 cm

AB = 10 cm

AC = 16 cm

Then, OA = OC = $\dfrac{16}{2}$ = 8 cm

Let OB be a cm.

Since the diagonal of a rhombus bisect at 90°.

Applying pythagoras theorem in triangle AOB, we get:

AB² = OA² + OB²

⇒ (10)² = (8)² + a²

⇒ 100 = 64 + a²

⇒ a² = 100 - 64

⇒ a² = 36

⇒ a = $\sqrt{36}$

⇒ a = 6

So, OB = 6 cm

BD = 2 x 6 cm

= 12 cm

Hence, the length of other diagonal is 12 cm.

(ii) As we know that area of rhombus = $\dfrac{1}{2}$ x product of its diagonal

= $\dfrac{1}{2}$ x 16 x 12 cm²

= $\dfrac{1}{2}$ x 192 cm²

= 96 cm²

Hence, the area of the rhombus is 96 cm².