Question:
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)]
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)]
Updated On: Jan 13, 2026
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Solution and Explanation
Let (3, 0), (4, 5), (−1, 4) and (−2, −1) are the vertices A, B, C, D of a rhombus ABCD.
Length of diagonal AC=\(\sqrt{[3-(-1)]^2+(0-4)}\)
= \(\sqrt{16+16}=4\sqrt2\)
Length of diagonal BD=\(\sqrt{[4-(-2)]^2+[5-(-1)]^2}\)
=\(\sqrt{36+36}=6\sqrt2\)
Therefore the area of rhombus ABCD = \(\frac{1}{2}\times4\sqrt2\times6\sqrt2\)
= 24 square units
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