Question

Each side of a rhombus is 10 cm long and one of its diagonals measures 16 cm. Find the length of the other diagonal and hence find the area of the rhombus.

Answer 1

Let ABCD be a rhombus.
∴ AB = BC = CD = DA = 10 cm
Let AC and BD be the diagonals of ABCD. Let AC = x and BD = 16 cm and O be the intersection point of the diagonals.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴​ ∆AOB is a right angle triangle, in which OA = AC ^$÷$ 2 = x ^$÷$ 2 and OB = BD ^$÷$2 = 16 ^$÷$ 2 = 8 cm.

Now, AB²= OA² + OB² [Pythagoras theorem]
$$\begin{gathered} \Rightarrow {10}^2={\left(\frac{x}{2}\right)}^2+8^2 \\ \Rightarrow 100-64=\frac{x^2}{4} \\ \Rightarrow 36\times 4=x \end{gathered}$$²

⇒x² =144
∴ x = 12 cm

Hence, the other diagonal of the rhombus is 12 cm.
∴ Area of the rhombus = $\frac{1}{2}\times \left(12\times 16\right)=96{\mathrm{cm}}^2$