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 units, and its area is square units. reset next

Explanation:

Step1: Calculate the length of \(AB\)

Using 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 \(BC\)

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

Step3: Calculate the length of \(AC\)

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

Step4: Calculate the perimeter \(P\)

\(P=AB + BC+AC\)
\(P=8+6\sqrt{2}+2\sqrt{10}\approx8 + 8.49+6.32=22.81\)

Step5: Calculate the area \(A\)

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

Answer:

Perimeter: \(22.81\), Area: \(24\)