Question

The area of a quadrilateral (-4, -2), (-3, -5), (3, -2) and (2, 3)is

• A. 20 sq. units
• B. 24 sq. units
• C. 28 sq. units
• D. 30 sq. units

Answer 1

Answer: (C)

28 sq. units

Let the points be $A(4,2),B(3,5),C(3,2)$ and $D(2,3)$
The quadrilateral ABCD can be divided into triangles ABC and ACD and hence the
area of the quadrilateral is the sum of the areas of the two triangles.
Area of a triangle with vertices $(x_1,y_1)$ ; $(x_2,y_2)$ and $(x_3,y_3)$ is $\left|\frac{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}{2}\right|$
Hence, substituting the points $(x_1,y_1)=(-4,-2)$ ; $(x_2,y_2)=(-3,-5)$ and $(x_3,y_3)=(3,-2)$ in the area formula, we get
Area of triangle ABC $=\left|\frac{(-4)(-5+2)+(-3)(-2+2)+3(-2+5)}{2}\right|=\left|\frac{12+0+9}{2}\right|=\frac{21}{2}=10.5squnits$
And, substituting the points $(x_1,y_1)=(-4,-2)$ ; $(x_2,y_2)=(3,-2)$ and $(x_3,y_3)=(2,3)$ in the area formula, we get
Area of triangle ACD $=\left|\frac{-4(-2-3)+\left(3\right)(3+2)+2(-2+2}{2}\right|=\left|\frac{20+15+0}{2}\right|=\frac{35}{2}=17.5squnits$
Hence, Area of quadrilateral $=10.5+17.5=28squnits$