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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 = `1/2` (product of its diagonals)]
Solution
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)^2)`
= `sqrt((-4)^2 + (4)^2)`
= `sqrt (16 + 16)`
= `4sqrt2`
Length of diagonal BD = `sqrt([4-(-2)]^2+[5-(-1)]^2)`
= `sqrt((-6)^2 + (-6)^2)`
= `sqrt(36+36) `
= `6sqrt2`
Therefore, area of rhombus ABCD = `1/2xx("Product of diagonals")`
= `1/2xx"AC"xx"BD"`
= `1/2xx4sqrt2xx6sqrt2`
= `1/2xx2xx4xx6`
= 24 square units
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