Question

Find the area of quadrilateral, the coordinates of whose angular points taken in order are \[(1,1),(3,4),(5,-2)\] and \[(4,-7)\].
\[\begin{align}
  & (A)\text{ 20}\text{.5 sq}\text{.units} \\
 & \text{(B) 41 sq}\text{.units} \\
 & \text{(C) 82 sq}\text{.units} \\
 & \text{(D) 61}\text{.5 sq}\text{.units} \\
\end{align}\]

Answer 1

Hint: First of all, we will illustrate the given question in the form of a diagram. Now we will assume this as quadrilateral ABCD. Now let is draw a line AC. Now the quadrilateral is divided into two parts. Now we will find the area of two parts. By adding these two areas, we will find the area of quadrilateral ABCD.

Complete step by step answer:
Before solving the question, we should illustrate the given quadrilateral of vertices \[A(1,1),B(3,4),C(5,-2)\]and \[D(4,-7)\].

Now let us divide quadrilateral ABCD into two parts by constructing a line AC.

Now quadrilateral ABCD is divided into triangle ABC and triangle ACD.
So, the area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
We know that the area of triangle ABC whose vertices are \[A({{x}_{1}},{{y}_{1}}),B({{x}_{2}},{{y}_{2}}),C({{x}_{3}},{{y}_{3}})\] is equal to\[\dfrac{1}{2}\left| \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & 1 \\
   {{x}_{2}} & {{y}_{2}} & 1 \\
   {{x}_{3}} & {{y}_{3}} & 1 \\
\end{matrix} \right|\].
So, the area of triangle ABC whose vertices are \[A(1,1),B(3,4),C(5,-2)\].
So, the area of triangle ABC
\[=\dfrac{1}{2}\left| \begin{matrix}
   1 & 1 & 1 \\
   3 & 4 & 1 \\
   5 & -2 & 1 \\
\end{matrix} \right|=\left| \dfrac{1}{2}\left( 1(4-(-2)-1(3-5)+1(-6-20) \right) \right|=9sq.units\]
In the similar manner, we can find the find the area of triangle ACD.
So, the area of triangle ACD \[=\dfrac{1}{2}\left| \begin{matrix}
   1 & 1 & 1 \\
   4 & -7 & 1 \\
   5 & -2 & 1 \\
\end{matrix} \right|=\left| \dfrac{1}{2}\left( 1(-7-(-2)-1(4-5)+1(-8+35) \right) \right|=11.5sq.units\]
We know that the sum of area of triangle ABC and area of triangle ACD is equal to area of quadrilateral ABCD.
Area of quadrilateral ABCD \[=9+11.5=20.5sq.units\]
Hence, option (A) is correct.
Note:
We should know that the area of n-sided polygon whose vertices are \[({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}}),({{x}_{3}},{{y}_{3}})......({{x}_{n}},{{y}_{n}})\] is equal to \[\dfrac{1}{2}\left| ({{x}_{1}}{{y}_{2}}-{{y}_{1}}{{x}_{2}})+({{x}_{2}}{{y}_{3}}-{{y}_{2}}{{x}_{3}})+....+({{x}_{n}}{{y}_{1}}-{{y}_{1}}{{x}_{n}}) \right|\].
So, the area of quadrilateral of vertices \[(1,1),(3,4),(5,-2)\] and \[(4,-7)\]
\[\begin{align}
  & =\dfrac{1}{2}\left| (4-3)+(-6-20)+(-35+8)+(4+7) \right| \\
 & =\dfrac{1}{2}|1-26-27+11|=\dfrac{1}{2}|-41|=20.5sq.units \\
\end{align}\]