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Find the area of a rhombus if its vertices are A(3, 0), B(4, 5), C(–1, 4) and D(–2, –1). |
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Answer» Given that, A(3, 0), B(4, 5), C(–1, 4)and D (–2, –1) are vertices of rhombus. We know that, Area of rhombus = \(\frac{1}{2}\) × product of diagonals = \(\frac{1}{2}\) × AC × BD Now, AC = \(\sqrt{(−1 − 3)^2 + (4 − 0)^2}\) (By distance formula) = \(\sqrt{16 + 16}\) = 4√2. And BD= \(\sqrt{(−2 − 4) ^2 + (−1 − 5) ^2}\) = \(\sqrt{36 + 36}\) = 6√2. ∴ Area of rhombus = \(\frac{1}{2}\) × AC × BD = \(\frac{1}{2}\) × 4√2 × 6√2 = 24 square units. Hence, Area of rhombus = 24 square units. |
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