Question

Find the area of quadrilateral EFGH whose vertices are (4, 8), ($-$6, $-$2), ($-$3, $-$4), and (3, 1)

Answer 1

$$\begin{gathered} \mathrm{Area}\mathrm{of}△\mathrm{EHG}=\frac{1}{2}\left[4\left\{1-\left(-4\right)\right\}+3\left(-4-8\right)-3\left(8-1\right)\right] \\ =\frac{1}{2}\left[4\times 5+3\left(-12\right)-3\left(7\right)\right] \\ =\frac{1}{2}\left(20-36-21\right) \\ =\frac{1}{2}\times \left(-37\right)=-18.5 \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{area}\mathrm{is}\mathrm{always}\mathrm{positive}. \\ \therefore \mathrm{Area}\mathrm{of}△EHG=18.5\mathrm{square}\mathrm{units}. \end{gathered}$$

$$\begin{gathered} \mathrm{Are}a\mathrm{of}△\mathrm{EFG}=\frac{1}{2}\left[4\left\{-2-\left(-4\right)\right\}-6\left(-4-8\right)-3\left\{8-\left(-2\right)\right\}\right] \\ =\frac{1}{2}\left[4\times 2-6\left(-12\right)-3\left(10\right)\right] \\ =\frac{1}{2}\left(8+72-30\right) \\ =\frac{1}{2}\times \left(50\right)=25 \\ \therefore \mathrm{Area}\mathrm{of}△\mathrm{EFG}=25\mathrm{square}\mathrm{units} \end{gathered}$$

$$\begin{gathered} \mathrm{Area}\left(\square \mathrm{EFGH}\right)=\left(△\mathrm{EHG}\right)+\mathrm{A}\left(△\mathrm{EFG}\right) \\ =18.5+25=43.5\mathrm{square}\mathrm{units} \end{gathered}$$