Question

Find the area of the quadrilateral $\underset{–}{PQRS}$ whose vertices in order are $P(-5,-3),Q(-4,-6),R(2,-1)$ and $S(1,2)$.

Answer 1

Area of a quadrilateral$PQRS=$Area of $△PQS+$Area of $△QSR$

We have Area of triangle$=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

Let $P(-5,-3),Q(-4,-6),R(2,-1)$ and $S(1,2)$.

Area of $△PQS=\frac{1}{2}\left|\begin{array}{ccc}-5& -3& 1\\ -4& -6& 1\\ 1& 2& 1\end{array}\right|$

$=\frac{1}{2}[-5(-6-2)+3(-4-1)+1(-8+6)]$

$=\frac{1}{2}[40-15-2]=\frac{23}{2}$sq.units

Area of $△QSR=\frac{1}{2}\left|\begin{array}{ccc}-4& -6& 1\\ 2& -1& 1\\ 1& 2& 1\end{array}\right|$

$=\frac{1}{2}[-4(-1-2)+6(2-1)+1(4+1)]$

$=\frac{1}{2}[12+6+5]=\frac{23}{2}$sq.units

$\therefore$ Area of quadrilateral $PQRS=$Area of $△PQS+$Area of $△QSR$

$=\frac{23}{2}+\frac{23}{2}=\frac{46}{2}=23$sq.units