Question

Find the area of the rhombus ABCD with vertices $A(2,0),B(5,-5),C(8,0)$and $D(5,5)$.

Answer 1

Consider the given vertices.

$A(2,0),B(5,-5),C(8,0)$ and $D(5,5)$

Diagonal $AC=\sqrt{(2-8{)}^2+(0-0{)}^2}$

$=\sqrt{36+0}=6unit$

Diagonal $BD=\sqrt{(5-5{)}^2+(-5-5{)}^2}$

$=\sqrt{0^2+(-10{)}^2}=10unit$

We know that the area of the rhombus

$=\frac{1}{2}\times productofdiagonal$

$=\frac{1}{2}\times AC\times BD$

$=\frac{1}{2}\times 6\times 10$

$=3\times 10=30unit{s}^2$

Hence, this is the answer.