Question

Find the area of a rectangle whose vertices are $A(-2,6),B(5,3),C(-1,-11)$ and $D(-8,-8)$.

• A. $4\sqrt{29}sq.units$
• B. $116sq.units$
• C. $29\sqrt{5}sq.units$
• D. $58\sqrt{2}sq.units$

Answer 1

The correct option is B $116sq.units$

Vertices of a rectangle are $A(-2,6),B(5,3),C(-1,-11)$ and $D(-8,-8)$.

$AB=\sqrt{(5+2{)}^2+(3-6{)}^2}$
$=\sqrt{7^2+(-3{)}^2}=\sqrt{49+9}=\sqrt{58}$
$BC=\sqrt{(-1-5{)}^2+(-11-3{)}^2}=\sqrt{232}$
$\therefore$ Area of the rectangle $ABCD=AB\times BC$
$=\sqrt{58}\times \sqrt{232}=\sqrt{29\times 2\times 29\times 8}$
$=29\times 4=116$.Sq. Units.