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)


Answer :



Let A(3, 0), B (4, 5), C( – 1, 4) and D ( – 2, – 1) are the vertices of a rhombus ABCD.

Length of diagonal AC
\(=\sqrt{[3-(-1)]^2 + (0-4)^2} \)
\( = \sqrt{16+16} \)
\( = \sqrt{32} \)
\( = 4\sqrt{2} \)
Length of diagonal BD
\(=\sqrt{[(-2)-4]^2+ (-1-5)^2} \)
\( = \sqrt{(-6)^2+(-6)^2} \)
\( = \sqrt{36+36} \)
\( = \sqrt{72} \)
\( = 6\sqrt{2}\)

Area of rhombus = \(\frac{1}{2} \) × product of diagonals
= \(\frac{1}{2} × 4\sqrt{2}×6\sqrt{2} = 24 \)sq unit

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