Question

The area of the quadrilateral formed by the points $(2,6),(3,2),(5,6)$ and $(-3,2)$ is

• A. $9$
• B. $12$
• C. $18$
• D. $2$

Answer 1

Answer: (C)

$18$

Plotting the given points, as shown in the Figure as A, B, C, D we find AB is parallel to CD
$\therefore$ ABCD is a trapezium
Hence the area of ABCD
= $\frac{1}{2}(AB+CD)\times$ the distance between AB and CD ...(find distance using distance formula)
$=\frac{1}{2}\times (6+3)\times 4=18$ square units.