Question

If perimeter of a rhombus is 100 cm and length of one diagonal is 48 cm, what is the area of the quadrilateral ?

Answer 1

Perimeter of the rhombus = 100 cm
$$\begin{gathered} \Rightarrow 4\times \mathrm{side}=100 \\ \Rightarrow \mathrm{side}=\frac{100}{4}=25\mathrm{cm} \end{gathered}$$
Thus, each side of the rhombus = 25 cm.
Diagonals of a rhombus bisect each other at 90$°$.
So, AO = OC = $\frac{48}{2}=24\mathrm{cm}$
In $△$AOB
We apply Pythagoras theorem,
$$\begin{gathered} {\mathrm{AO}}^2+{\mathrm{OB}}^2={\mathrm{AB}}^2 \\ \Rightarrow {24}^2+{\mathrm{OB}}^2={25}^2 \\ \Rightarrow {\mathrm{OB}}^2=625-576=49 \\ \Rightarrow \mathrm{OB}=7\mathrm{cm} \end{gathered}$$
So, DB = $2\times \mathrm{OB}=2\times 7=14\mathrm{cm}$
Area of rhombus = $\frac{1}{2}\times \left(\mathrm{Product}\mathrm{of}\mathrm{diagonals}\right)=\frac{1}{2}\times 14\times 48=336{\mathrm{cm}}^2$