The length of each side of a rhombus is 10 cm. If the length of one of its diagonals is 16 cm, then what is the area of the rhombus?

As we know that every rhombus is a parallelogram. So the law of parallelogram should be applicable in rhombus.

As per this law, the sum of the squares of the length of the four sides is equal to the sum of the squares of the length of the two diagonals.

Let's assume the length of each side of the rhombus is 'a' cm.

Let's assume the length of diagonals are $$d_{1}$$ and $$d_{2}$$

$$a^2+a^2+a^2+a^2\ =\ \left(d_1\right)^2\ +\ \left(d_2\right)^2$$

$$4a^2\ =\ \left(d_1\right)^2\ +\ \left(d_2\right)^2$$

The length of each side of a rhombus is 10 cm. If the length of one of its diagonals is 16 cm.

$$4\times\left(10\right)^2\ =\ \left(16\right)^2\ +\ \left(d_2\right)^2$$

$$400\ =\ 256\ +\ \left(d_2\right)^2$$
$$400-256\ =\ \ \ \left(d_2\right)^2$$
$$\ 144=\ \ \ \left(d_2\right)^2$$
$$\ 12^2=\ \ \ \left(d_2\right)^2$$

$$\ 12^{ }=\ \ \ d_2$$

the length of another diagonal = 12 cm

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

= $$\frac{1}{2}\times\ 16\times12$$

= $$\ 16\times6$$

= 96 $$cm^{2}$$