Question

Area of the quadrilateral (in sq units) with vertices A(-1, 6), B(-3, -9), C(5, -8) and D(3, 9) is

• A. 48
• B. 96
• C. 24
• D. None of these

Answer 1

The correct option is B 96

Let the points be $A(-1,6),B(-3,-9),C(5,-8)$ and $D(3,9)$
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, Area of triangle ABC $=\left|\frac{(-1)(-9+8)+(-3)(-8-6)+5(6+9)}{2}\right|=\left|\frac{1+42+75}{2}\right|=\frac{118}{2}=59squnits$
And, Area of triangle ACD $=\left|\frac{(-1)(-8-9)+\left(5\right)(9-6)+3(6+8}{2}\right|=\left|\frac{17+15+42}{2}\right|=\frac{74}{2}=37squnits$
Hence, Area of quadrilateral $=59+37=96squnits$