Question

right triangle egf is shown.
what is the exact area, in square units, of triangle egf?

Explanation:

Step1: Calculate the length of \(EF\)

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For \(E(-6,3)\) and \(F(5,0)\), \(x_1=-6,y_1 = 3,x_2=5,y_2 = 0\).
\(EF=\sqrt{(5+6)^2+(0 - 3)^2}=\sqrt{121 + 9}=\sqrt{130}\)

Step2: Calculate the length of \(EG\)

For \(E(-6,3)\) and \(G(-4,-3)\), \(x_1=-6,y_1 = 3,x_2=-4,y_2=-3\).
\(EG=\sqrt{(-4 + 6)^2+(-3 - 3)^2}=\sqrt{4+36}=\sqrt{40}=2\sqrt{10}\)

Step3: Calculate the length of \(GF\)

For \(G(-4,-3)\) and \(F(5,0)\), \(x_1=-4,y_1=-3,x_2=5,y_2 = 0\).
\(GF=\sqrt{(5 + 4)^2+(0+3)^2}=\sqrt{81 + 9}=\sqrt{90}=3\sqrt{10}\)
Since \(EG^{2}+GF^{2}=(2\sqrt{10})^{2}+(3\sqrt{10})^{2}=40 + 90=130\) and \(EF^{2}=(\sqrt{130})^{2}=130\), by Pythagorean theorem \(EG\perp GF\)

Step4: Calculate the area of \(\triangle EGF\)

The area of a right - triangle \(A=\frac{1}{2}\times base\times height\). Here, base \(b = EG\), height \(h = GF\)
\(A=\frac{1}{2}\times2\sqrt{10}\times3\sqrt{10}\)

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Answer:

\(30\)