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)]

Answer 1

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