Question

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

Answer 1

Let a rhombus is $ABCD$ then

Given that:

$A(3,0).B(4,5),C(-1,4),D(-2,-1)$

So, $AC$ and $BD$ be the diagonal then

Using distance formula we can calculate $AC$ and $BD$,

$AC=\sqrt{(3+1{)}^2+(0-4{)}^2}=\sqrt{32}$

$BD=\sqrt{(4+2{)}^2+(5+1{)}^2}=\sqrt{72}$

Area of rhombus $ABCD=\frac{1}{2}\times AC\times BD$

$=\frac{1}{2}\times \sqrt{16\times 2}\times \sqrt{36\times 2}$

$=6\times 4=24unit{s}^2$

hence, the area of rhombus is $24unit{s}^2$