Question

The area of the quadrilateral with vertices at $(-4,-2),(-3,-5),(3,-2),(2,3)$ is

• A. $14$
• B. $28$
• C. $56$
• D. $Noneofthese$

Answer 1

The correct option is B $28$

Let the vertices of quadrilateral be $A(-4,-2),B(-3,-5),C(3,-2),D(2,3)$.

Joining $AC$, we have two triangles $\Delta ABC$ and $\Delta ACD$ as shown in the figure below.

Therefore,

Area of quadrilateral $ABCD=$ Area of $\Delta ABC+$ Area of $\Delta ACD$

Calculate the area of $\Delta ABC$.

$=\frac{1}{2}[-4(-5+2)-3(-2+2)+3(-2+5)]$

$=\frac{1}{2}[-4(-3)-3\left(0\right)+3\left(3\right)]$

$=\frac{1}{2}[12+9]$

$=\frac{21}{2}sq.units$

Calculate the area of $\Delta ACD$.

$=\frac{1}{2}[-4(3-(-2))+2(-2-(-2))+3(-2-3)]$

$=\frac{1}{2}[-4\left(5\right)+2\left(0\right)+3(-5)]$

$=\frac{1}{2}[-20-15]$

$=-\frac{35}{2}$

Area cannot be negative. Therefore,

Area of $\Delta ACD=\frac{35}{2}sq.units$

Therefore,

Area of quadrilateral $ABCD=\frac{21}{2}+\frac{35}{2}=28sq.units$

Hence, this is the required result.