Question

Find the area of the quadrilateral ABCD. If \[A\left( { - 5,7} \right), B\left( { - 4, - 5} \right), C\left( { - 1, - 6} \right){\text{ and }}D\left( {4,5} \right)\] are the vertices of a quadrilateral.

Answer 1

Hint: In this question first of all, draw the figure of the quadrilateral which will give us a clear idea of what we have to find, divide it into two triangles. The area of the quadrilateral is the sum of areas of the two triangles. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Given points are \[A\left( { - 5,7} \right), B\left( { - 4, - 5} \right), C\left( { - 1, - 6} \right){\text{ and }}D\left( {4,5} \right)\]
Joining \[AC\], two triangles are formed i.e., \[\Delta ABC \& \Delta ACD\] as shown in the below figure:

Hence area of quadrilateral \[ABCD\]= area of \[\Delta ABC\] + area of \[\Delta ACD\]
We know that for the given three points \[P\left( {{x_1},{y_1}} \right),Q\left( {{x_2},{y_2}} \right){\text{ and }}R\left( {{x_3},{y_3}} \right)\] the area of the \[\Delta PQR\] is given by \[\Delta = \dfrac{1}{2}\left| {\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]} \right|\].
Now the area of the \[\Delta ABC\] where \[A\left( { - 5,7} \right),B\left( { - 4, - 5} \right){\text{ and }}C\left( { - 1, - 6} \right)\] is given by
\[
   \Rightarrow {\Delta _1} = \dfrac{1}{2}\left| {\left[ { - 5\left( { - 5 - \left( { - 6} \right)} \right) + \left( { - 4} \right)\left( { - 6 - 7} \right) + \left( { - 1} \right)\left( {7 - \left( { - 5} \right)} \right)} \right]} \right| \\
   \Rightarrow {\Delta _1} = \dfrac{1}{2}\left| {\left[ { - 5\left( 1 \right) - 4\left( { - 13} \right) + \left( { - 1} \right)\left( {12} \right)} \right]} \right| \\
   \Rightarrow {\Delta _1} = \dfrac{1}{2}\left| {\left[ { - 5 + 52 - 12} \right]} \right| \\
  \therefore {\Delta _1} = \dfrac{{35}}{2}{\text{ sq}}{\text{.units}} \\
\]
And the area of the \[\Delta ADC\] where \[A\left( { - 5,7} \right),D\left( {4,5} \right){\text{and C}}\left( { - 1, - 6} \right)\] is given by
\[
   \Rightarrow {\Delta _2} = \dfrac{1}{2}\left| {\left[ { - 5\left( {5 - \left( { - 6} \right)} \right) + 4\left( { - 6 - 7} \right) + \left( { - 1} \right)\left( {7 - 5} \right)} \right]} \right| \\
   \Rightarrow {\Delta _2} = \dfrac{1}{2}\left| {\left[ { - 5\left( {11} \right) + 4\left( { - 13} \right) + \left( { - 1} \right)\left( 2 \right)} \right]} \right| \\
   \Rightarrow {\Delta _2} = \dfrac{1}{2}\left| {\left[ { - 109} \right]} \right| \\
  \therefore {\Delta _2} = \dfrac{{109}}{2}{\text{ sq}}{\text{.units}} \\
\]
Therefore, area of the quadrilateral \[ABCD\]\[ = {\Delta _1} + {\Delta _2} = \dfrac{1}{2}\left[ {35 + 109} \right] = \dfrac{{144}}{2} = 72{\text{ sq}}{\text{.units}}\]
Thus, the area of the quadrilateral \[ABCD\] is 72 sq. units.

Note: For the given three points \[P\left( {{x_1},{y_1}} \right), Q\left( {{x_2},{y_2}} \right){\text{ and }}R\left( {{x_3},{y_3}} \right)\] the area of the \[\Delta PQR\] is given by \[\Delta = \dfrac{1}{2}\left| {\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]} \right|\]. The area of the quadrilateral and triangle are always positive.