Question

type the correct answer in each box. if necessary, round your answer(s) to the nearest hundredth.
the vertices of (\triangle abc) are (a(-2,2)), (b(6,2)), and (c(0,8)). the perimeter of (\triangle abc) is (square) units, and its area is (square) square units.

Explanation:

Step1: Calculate the length of \(AB\)

Use the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For \(A(-2,2)\) and \(B(6,2)\), \(x_1=-2,y_1 = 2,x_2=6,y_2=2\).
\(AB=\sqrt{(6-(-2))^2+(2 - 2)^2}=\sqrt{(8)^2+0^2}=8\)

Step2: Calculate the length of \(AC\)

For \(A(-2,2)\) and \(C(0,8)\), \(x_1=-2,y_1 = 2,x_2=0,y_2=8\).
\(AC=\sqrt{(0-(-2))^2+(8 - 2)^2}=\sqrt{(2)^2+(6)^2}=\sqrt{4 + 36}=\sqrt{40}=2\sqrt{10}\approx6.32\)

Step3: Calculate the length of \(BC\)

For \(B(6,2)\) and \(C(0,8)\), \(x_1=6,y_1 = 2,x_2=0,y_2=8\).
\(BC=\sqrt{(0 - 6)^2+(8 - 2)^2}=\sqrt{(-6)^2+(6)^2}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2}\approx8.49\)

Step4: Calculate the perimeter

Perimeter \(P=AB + AC+BC\).
\(P=8+6.32 + 8.49=22.81\)

Step5: Calculate the area

Base \(b = AB=8\), height \(h\) is the vertical distance from \(C\) to the line \(y = 2\) (since \(AB\) is on \(y = 2\)), \(h=8 - 2=6\).
Area \(A=\frac{1}{2}\times b\times h=\frac{1}{2}\times8\times6 = 24\)

Answer:

The perimeter is \(22.81\) units and the area is \(24\) square units.