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The vertices of a rhombus ABCD are A$(-3, -4)$, B$(5, -3)$, C$(1, 4)$ and D$(-7, 3)$. Find the length of both the diagonals. Hence, find area of the rhombus ABCD.
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Solution: (a) AC = $\sqrt{(1+3)^2 + (4+4)^2} = 4\sqrt{5}$ (1 Mark)
BD = $\sqrt{(-7-5)^2 + (3+3)^2} = 6\sqrt{5}$ (1 Mark)
Area of rhombus ABCD = $\frac{1}{2} \times 4\sqrt{5} \times 6\sqrt{5}$
$= 60$ (1 Mark)
BD = $\sqrt{(-7-5)^2 + (3+3)^2} = 6\sqrt{5}$ (1 Mark)
Area of rhombus ABCD = $\frac{1}{2} \times 4\sqrt{5} \times 6\sqrt{5}$
$= 60$ (1 Mark)