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

Given:

Given points are $(3, 0), (4, 5), (-1, 4)$ and $(-2, -1)$.

To do:

We have to find the area of the rhombus formed by the given points.

Solution:

Let $\mathrm{ABCD}$ is a rhombus whose vertices are $\mathrm{A}(3,0), \mathrm{B}(4,5), \mathrm{C}(-1,4)$ and $\mathrm{D}(-2,-1)$.

We know that,

The distance between two points $\mathrm{A}(x_{1}, y_{1})$ and $\mathrm{B}(x_{2}, y_{2})$ is $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$.

Therefore,

$AC^{2}=(-1-3)^{2}+(4-0)^{2}$

$=(-4)^{2}+(4)^{2}$

$=16+16$

$=32$

$\Rightarrow AC=\sqrt{32}$

$=4\sqrt2$

$B D^{2}=(-2-4)^{2}+(-1-5)^{2}$

$=(-6)^{2}+(-6)^{2}$

$=36+36$

$=72$

$\Rightarrow BD=\sqrt{72}$

$=6\sqrt2$

We know that,

Area of rhombus $=\frac{\text { Product of diagonals }}{2}$

$=\frac{4 \sqrt{2} \times 6 \sqrt{2}}{2}$

$=\frac{24\times2}{2}$

$=24$ sq. units

The area of the rhombus is $24$ sq. units. 

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Updated on: 10-Oct-2022

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