Question

Find the area of a rhombus if its vertices are $\left(3,0\right)$, $\left(4,5\right)$, $\left(-1,4\right)$ and $\left(-2,-1\right)$ taken in order. [Hint:$\mathrm{Area}\mathrm{of}\mathrm{Rhombus}=\frac{\mathrm{Product}\mathrm{of}\mathrm{Diagonals}}{2}$]

Answer 1

Step 1: Find the diagonals

Let assume $A(3,0),B(4,5),C(-1,4)$ and $D\left(-2,-1\right)$ are the vertices of the rhombus $ABCD$, taken in order.

Length of the diagonals

$$\begin{gathered} AC=\sqrt{{\left(3-\left(-1\right)\right)}^2+{\left(0-4\right)}^2} \\ =\sqrt{4^2+4^2} \\ =4\sqrt{2} \end{gathered}$$ [ Distance formula $\mathbf{}\mathbf{=}\mathbf{}\sqrt{{\mathbf{(}{\mathbf{x}}_{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}{\mathbf{x}}_{\mathbf{1}}\mathbf{)}}^{\mathbf{2}}\mathbf{}\mathbf{+}{\mathbf{(}\mathbf{}{\mathbf{y}}_{\mathbf{2}}\mathbf{}\mathbf{-}{\mathbf{y}}_{\mathbf{1}}\mathbf{)}}^{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}$]

$$\begin{gathered} BD=\sqrt{{\left(4-\left(-2\right)\right)}^2+{\left(5-\left(-1\right)\right)}^2} \\ =\sqrt{6^2+6^2} \\ =\sqrt{72} \\ =6\sqrt{2} \end{gathered}$$

Step 2: Find the area of the Rhombus

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{Rhombus}=\frac{\mathrm{Product}\mathrm{of}\mathrm{Diagonals}}{2} \\ =\frac{AC\times BD}{2} \\ =\frac{4\sqrt{2}\times 6\sqrt{2}}{2} \\ =24sq.\mathrm{units} \end{gathered}$$

Hence, the area of the Rhombus is $24sq.\mathrm{units}$