Question

If $A(2,-1)$, $B(3,4)$, $C(-2,3)$ and $D(-3,-2)$ be four points in a coordinate plane, show that ABCD is a rhombus but not a square. Find the area of the rhombus.

• A. $28$ sq. units
• B. $32$ sq. units
• C. $24$ sq. units
• D. $20$ sq. units

Answer 1

The correct option is C $24$ sq. units

$AB=\sqrt{{(3-2)}^2+{(4+1)}^2}=\sqrt{1+25}=\sqrt{26}$
$BC=\sqrt{{(-2-3)}^2+{(3-4)}^2}=\sqrt{25+1}=\sqrt{26}$
$CD=\sqrt{{(-3+2)}^2+{(-2-3)}^2}=\sqrt{1+25}=\sqrt{26}$
$DA=\sqrt{{(-3-2)}^2+{(-2+1)}^2}=\sqrt{25+1}=\sqrt{26}$
$AC=\sqrt{{(-2-2)}^2+{(3+1)}^2}=\sqrt{16+16}=\sqrt{32}$

$BD=\sqrt{{(-3-3)}^2+{(-2-4)}^2}=\sqrt{36+36}=\sqrt{72}$
Now we can see that
$AB=BC=CD=DA=\sqrt{26}$ units but daigonals $AC\ne BD$
=> ABCD is a quadrilateral whose all sides are equal but diagonals are not equal.
=> ABCD is a rhombus, not a square.
Area of rhombus ABCD $=\frac{1}{2}\times$ (Product of diagonals)

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

$=\frac{1}{2}\times (\sqrt{3}2\times \sqrt{7}2)$
$=24$ sq.units