Question

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

Answer 1

Let $A(3,0$),$B(4,5)$, $C(-1,4)$ and $D(-2,-1)$ are the vertices of rhombus $ABCD.$

$AC$ and $BD$ are the diagonals of rhombus.

$\therefore BD=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$

Here $x_1=4,x_2=-2$ and $y_1=5,y_2=-1$

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

$\Rightarrow BD=\sqrt{6^2+6^2}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2}$

$\therefore AC=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$

Here $x_1=3,x_2=-1$ and $y_1=0,y_2=4$

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

$\Rightarrow BD=\sqrt{4^2+(-4{)}^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$

Area of rhombus $=\frac{1}{2}\times$Product of diagonals

$=\frac{1}{2}\times 6\sqrt{2}\times 4\sqrt{2}$

$=\frac{24\times 2}{2}=24$ Sq.unit