Question

the vertices of △abc are a(2, 8), b(16, 2), and c(6, 2). the perimeter of △abc is blank units, and its area is blank square units. type the correct answer in each box. if necessary, round your answers to the nearest hundredth.

Explanation:

Step1: Calculate the length of side AB

Use the distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For points $A(2,8)$ and $B(16,2)$, we have $AB=\sqrt{(16 - 2)^2+(2 - 8)^2}=\sqrt{14^2+( - 6)^2}=\sqrt{196 + 36}=\sqrt{232}\approx15.23$.

Step2: Calculate the length of side BC

For points $B(16,2)$ and $C(6,2)$, $BC=\sqrt{(6 - 16)^2+(2 - 2)^2}=\sqrt{( - 10)^2+0^2}=10$.

Step3: Calculate the length of side AC

For points $A(2,8)$ and $C(6,2)$, $AC=\sqrt{(6 - 2)^2+(2 - 8)^2}=\sqrt{4^2+( - 6)^2}=\sqrt{16 + 36}=\sqrt{52}\approx7.21$.

Step4: Calculate the perimeter

Perimeter $P=AB + BC+AC\approx15.23+10 + 7.21=32.44$.

Step5: Calculate the area

Since $B(16,2)$ and $C(6,2)$ have the same $y$ - coordinate, the base $BC = 10$. The height from $A$ to the line $BC$ is the difference in the $y$ - coordinates of $A$ and the $y$ - coordinate of $B$ (or $C$), so $h=8 - 2 = 6$. Area $A=\frac{1}{2}\times base\times height=\frac{1}{2}\times10\times6 = 30$.

Answer:

Perimeter: $32.44$
Area: $30$