Question

Find the area of rhombus if its vertices are $(3,0),(4,5),(-1,4)$ and $(-2,-1)$ taken in order.

[Hint: Area of rhombus $=\frac{1}{2}\times$ product of its diagonals.]

Answer 1

We know that,

Area of Rhombus $=\frac{1}{2}$ (Product of its diagonals)

Given co-ordinates of rhombus are $(3,0),(4,5),(-1,4)$ and $(-2,-1)$

Diagonal $1=\sqrt{(3+1{)}^2+(0-4{)}^2}=4\sqrt{2}$

Diagonal $2=\sqrt{(4+2{)}^2+(5+1{)}^2}=6\sqrt{2}$

$\therefore$ Area of rhombus $=\frac{1}{2}(4\sqrt{2}\times 6\sqrt{2})$

$=24$ unit${}^2$