Question

If the vertices of quadrilateral are $(-4,5),(0,7)$, $(5,-5)$ and $(-4,-2)$, then area of quadrilateral is

• A. 70 sq units
• B. 60 sq units
• C. $75.5$ sq units
• D. $60.5$ sq units

Answer 1

Answer: (D)

Given vertices $A(-4,5),B(0,7),C(5,-5)$ and $D(-4,-2)$ are shown below in the $xy$-plane. Joining all the vertices a quadrilateral $ABCD$ is formed.

Area of quadrilateral $ABCD$
$=$ Area of $△ADC+$ Area of $△ABC.....$(i)
Now, area of $△ADC$
$=\frac{1}{2}|[-4(-2+5)-4(-5-5)+5(5+2)]$
$[\because$ Area $=\frac{1}{2}|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)$ and here,
$\left(x_1,y_1\right)=(-4,5),\left(x_2,y_2\right)=(-4,-2),\left(x_3,y_3\right)=(5,-5)]$
$=\frac{1}{2}|[-4\times 3-4(-10)+5\times 7]|$
$=-\frac{1}{2}|[-12+40+35]|=\frac{1}{2}|[75-12]|=\frac{63}{2}$
Area of $△ABC=\frac{1}{2}|[-4(7+5)+0(-5-5)+5(5-7)]|$
$[\text{ here, }\left(x_1,y_1\right)=(-4,5),\left(x_2,y_2\right)=(0,7)\mathrm{and}\left(x_3,y_3\right)=(5,-5)]$
$\frac{1}{2}|[-4\times 12+0\times (-10)+5(-2)]|$
$=\frac{1}{2}|-48-10|$
$=\frac{|-58|}{2}=29$
$\therefore$ From Eq. (i)
Area of quadrilateral $ABCD$
$=\frac{63}{2}+29=\frac{63+58}{2}=\frac{121}{2}$
$=60.5$ sq units