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

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From the figure:

Rhombus figure

Let the vertices of the rhombus are: 

A(3,0),\ B(4,5),\ C(-1,4),\ D(-2,-1)

Area of the rhombus ABCD is given by;

= \frac{1}{2}\times(Product\ of\ lengths\ of\ diagonals)

Hence we have to find the lengths of the diagonals AC and BD of the rhombus.

The distance formula:

D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Length of the diagonal AC:

AC = \sqrt{(3-(-1))^2+(0-4)^2} = \sqrt{16+16} = 4\sqrt{2}

Length of the diagonal BD:

BD = \sqrt{(4-(-2))^2+(5-(-1))^2} = \sqrt{36+36} = 6\sqrt{2}

Thus, the area will be,

= \frac{1}{2}\times (AC)\times(BD)

= \frac{1}{2}\times (4\sqrt{2})\times(6\sqrt{2}) = 24\ square\ units.

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Divya Prakash Singh

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