Question

Shows that the point $A(6,1)$,$B(8,2)$, $C(9,4)$ and $D(7,3)$ are the vertices of a rhombus. find its area.

• A. $4squnits$
• B. $3squnits$
• C. $5squnits$
• D. $6squnits$

Answer 1

The correct option is B $3squnits$

The given points are $A(6,1),B(8,2),C(9,4)$ and $D(7,3)$

$AB=\sqrt{(6-8{)}^2+(1-2{)}^2}=\sqrt{4+1}=\sqrt{5}$ units

$BC=\sqrt{(8-9{)}^2+(2-4{)}^2}=\sqrt{1-4}=\sqrt{5}$ units

$CD=\sqrt{(9-7{)}^2+(4-3{)}^2}=\sqrt{4+1}=\sqrt{5}$ units

$AD=\sqrt{(7-6{)}^2+(3-1{)}^2}=\sqrt{1+4}=\sqrt{5}$ units

$AC=\sqrt{(6-9{)}^2+(1-4{)}^2}=\sqrt{9+9}=3\sqrt{2}$ units

$BD=\sqrt{(8-7{)}^2+(2-3{)}^2}=\sqrt{1+1}=\sqrt{2}$ units

$\because AB=BC=CD=AD=\sqrt{5}$ and $AC\ne BD$

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

$=\frac{1}{2}\times 3\sqrt{2}\times \sqrt{2}$

$=3$ sq. units

Hence, Area of rhombus is $3$ sq. units.