​Find the area of a rhombus whose side is 10 cm and the longest diagonal is 16 cm. ​

Correct option is C

Given:

Side of the rhombus = 10 cm

Longest diagonal = 16 cm

Concept Used:

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

​$$\text{Area of rhombus} = \frac{1}{2} \times d_1 \times d_2$$​​

Solution:

Let the diagonals be d₁ and d₂, where d₁ = 16 cm (longest diagonal).

​$$\frac{d_1}{2} = \frac{16}{2} = 8 \text{ cm}$$​​

Using Pythagoras' theorem,

​$$\left(\frac{d_2}{2}\right)^2 + 8^2 = 10^2$$​​

​$$\left(\frac{d_2}{2}\right)^2 + 64 = 100$$​​

​$$\left(\frac{d_2}{2}\right)^2 = 36$$​​

​​​​$$d_2 = 12 \text{ cm}$$​​

​​​​$$\text{Area} = \frac{1}{2} \times 16 \times 12$$​​​$$= \frac{192}{2} = 96 \text{ cm}^2$$

Option (C) is right.