Question

Find the area of quadrilateral math with vertices M(7, 6), A(3, - 2), T(- 7, 1) and H(- 1, 9)

Answer 1

Lets draw a picture of our quadrilateral:

In order to find the area, we can divide our parallelogram in 2 triangles:

The area of triangle AHT is given by

`\text{Area }\Delta AHT=(1)/(2)(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))`

where

`\begin{gathered} (x_1,y_1)=(3,-2)=A \\ (x_2,y_2)=(-1,9)=H \\ (x_3,y_3)=(-7,1)=T \end{gathered}`

By substituting these points into the given formula, we get

`\text{Area }\Delta AHT=(1)/(2)(3_{}(9_{}-(-7))-1((-7)-(-2))-7((-2)-9))`

which gives

`\begin{gathered} \text{Area }\Delta AHT=(1)/(2)(3_{}(16)-1(-5)-7(-11)) \\ \text{Area }\Delta AHT=(1)/(2)(48+5+77) \\ \text{Area }\Delta AHT=(130)/(2) \\ \text{Area }\Delta AHT=65 \end{gathered}`

Similarly, for the area of triangle AHM, we can choose

`\begin{gathered} (x_1,y_1)=(3,-2)=A \\ (x_2,y_2)=(-1,9)=H \\ (x_3,y_3)=(7,6)=M \end{gathered}`

By substuting in our area formula, we get

`\text{Area }\Delta AHM=(1)/(2)(3_{}(9_{}-6)-1(6-(-2))+7((-2)-9))`

which gives

`\begin{gathered} \text{Area }\Delta AHM=(1)/(2)(3_{}(3)-1(8)+7(-11) \\ \text{Area }\Delta AHM=(1)/(2)(9-8-77) \\ \text{Area }\Delta AHM=(76)/(2) \\ \text{Area }\Delta AHM=38 \end{gathered}`

Then, the total area is given by

`\begin{gathered} A=\text{Area }\Delta AHT+\text{Area }\Delta\text{AHM} \\ A=65+38 \\ A=103 \end{gathered}`

then, the answer is 103 units squared.