Question

If the side of a rhombus is 10 cm and one diagonal is 16 cm, then its area is _________.

Answer 1

Given:
The side of a rhombus is 10 cm
One diagonal is 16 cm

Let ABCD be a rhombus.
AB = 10 cm ...(1)
AC = 16 cm ...(2)

We know, the diagonals of a rhombus bisects each other at right angles.
Let the point of intersection of the diagonals be O.

Then,
AO = OC = $\frac{16}{2}=8$ cm ...(3)

$$\begin{gathered} \mathrm{In}∆AOB, \\ \mathrm{Using}\mathrm{pythagoras}\mathrm{theorem}, \\ A{B}^2=A{O}^2+B{O}^2 \\ \Rightarrow {10}^2=8^2+B{O}^2 \\ \Rightarrow 100=64+B{O}^2 \\ \Rightarrow B{O}^2=100-64 \\ \Rightarrow B{O}^2=36 \\ \Rightarrow BO=6 \\ \mathrm{Thus},BD=2\times BO \\ =2\times 6 \\ =12\mathrm{cm}...\left(4\right) \end{gathered}$$



$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{rhombus}=\frac{1}{2}\times \left(\mathrm{product}\mathrm{of}\mathrm{diagonals}\right) \\ =\frac{1}{2}\times 12\times 16\left(\mathrm{from}\left(2\right)\mathrm{and}\left(4\right)\right) \\ =6\times 16 \\ =96{\mathrm{cm}}^2 \end{gathered}$$


Hence, the area is $\underline{96{\mathrm{cm}}^2}$.