Question

Let $A(3,2),B(-4,1),C(-3,1)$ and $D(2,-4)$ be the vertices of a quadrilateral ABCD. Find the aea of the quadrilateral formed by the mid-points of the sides of the quadrilateral ABCD.

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

The correct option is C $9$

Given the vertices of quadrilateral ABCD

$A(3,2),B\left(\right(-4,1),C(-3,1)$ and $D(2,-4)$

Let the midpoint be denoted as P, Q, R and S of AB<BC<CD and DA respectively.

Mid point of AB,P $=(\frac{3-4}{2},\frac{2+1}{2})=(\frac{-1}{2},\frac{3}{2})$

Mid point of BC,Q $=(\frac{-4-3}{2},\frac{1+1}{2})=(\frac{-7}{2},1)$

Mid point of CD,R $=(\frac{-3+2}{2},\frac{1-4}{2})=(\frac{-1}{2},\frac{-3}{2})$

Mid point of DA, S $=(\frac{3+2}{2},\frac{2-4}{2})=(\frac{5}{2},-1)$

Area of a quadrilateral or polygon is given by the formula,

$\frac{1}{2}\left|\right(x_1{y}_2-y_1{x}_2)+(x_2{y}_3-y_2{x}_3)+...+(x_n{y}_1-y_n{x}_1\left)\right|=\frac{1}{2}|[(\frac{-1}{2}-\times 1)-(\frac{-7}{2}\times \frac{3}{2})]+[(\frac{-7}{2}\times \frac{-3}{2})-(\frac{-1}{2}\times 1)]+[(\frac{-1}{2}\times -1)-(\frac{5}{2}\times \frac{-3}{2})]+[(\frac{5}{2}\times \frac{3}{2})-(-1\times \frac{-1}{2})]|=\frac{1}{2}|[\frac{-1}{2}+\frac{21}{4}]+[\frac{21}{4}+\frac{1}{2}]+[\frac{1}{2}+\frac{15}{4}]+[\frac{15}{4}-\frac{1}{2}]|=\frac{1}{2}\left|\frac{-2+21+21+2+2+15+15+-2}{4}\right|=9$