Question

The area of the quadrilaterals, the coordinates of whose vertices are (1, 2), ( 6, 2), (5, 3) and (3, 4) are

• A. $\frac{9}{2}$
• B. 5
• C. $\frac{11}{2}$
• D. 11

Answer 1

The correct option is C $\frac{11}{2}$

Let the points be $A(1,2),B(6,2),C(5,3)$ and $D(3,4)$
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 $=\left|\frac{\left(1\right)(2-3)+\left(6\right)(3-2)+5(2-2)}{2}\right|=\left|\frac{-1+6}{2}\right|=\frac{5}{2}squnits$
And, Area of triangle ACD $=\left|\frac{\left(1\right)(3-4)+\left(5\right)(4-2)+3(2-3)}{2}\right|=\left|\frac{-1+10-3}{2}\right|=\frac{6}{2}=3squnits$
Hence, Area of quadrilateral $=\frac{5}{2}+3=\frac{11}{2}squnits$