Find the area of a rhombus if its vertices are
Question.

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (–1, 4) and (–2, – 1) taken in order.

Solution:

Diagonals AC and BD bisect each other at right angle to each other at O.

$A C=\sqrt{(-1-3)^{2}+(4-0)^{2}}$

$=\sqrt{16+16}=\sqrt{32}=4 \sqrt{2}$

$B D=\sqrt{(4+2)^{2}+(5+1)^{2}}=\sqrt{36+36}=6 \sqrt{2}$

Then $\mathrm{OA}=\frac{1}{2} \mathrm{AC}=\frac{1}{2} \times 4 \sqrt{2}=2 \sqrt{2}$

$\mathrm{OB}=\frac{1}{2} \mathrm{BD}=\frac{1}{2} \times \mathbf{6} \sqrt{2}=3 \sqrt{2}$

Area of $\Delta \mathrm{AOB}=\frac{1}{2}(\mathrm{OA}) \times(\mathrm{OB})=\frac{1}{2} \times \mathbf{2} \sqrt{2} \times 3 \sqrt{2}=6$ sq. units

Hence, the area of the rhombus ABCD

$=4 \times$ area of $\Delta \mathrm{AOB}=4 \times 6=24$ sq. units.
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