Question

this question has two parts. first, answer part a. then, answer part b. part a identify the figure with the vertices a(0,5), b(0,1) and c(6,1). part b find the perimeter and area of the figure. perimeter units area units²

Explanation:

Step1: Calculate side - lengths using distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
For side $AB$ with $A(0,5)$ and $B(0,1)$:
$AB=\sqrt{(0 - 0)^2+(1 - 5)^2}=\sqrt{0+( - 4)^2}=4$.
For side $BC$ with $B(0,1)$ and $C(6,1)$:
$BC=\sqrt{(6 - 0)^2+(1 - 1)^2}=\sqrt{6^2+0}=6$.
For side $AC$ with $A(0,5)$ and $C(6,1)$:
$AC=\sqrt{(6 - 0)^2+(1 - 5)^2}=\sqrt{6^2+( - 4)^2}=\sqrt{36 + 16}=\sqrt{52}=2\sqrt{13}$.

Step2: Calculate the perimeter

The perimeter $P$ of a triangle is $P=AB + BC+AC$.
$P=4 + 6+2\sqrt{13}=10 + 2\sqrt{13}$.

Step3: Calculate the area

Since $AB$ is vertical ($x$-coordinates are the same) and $BC$ is horizontal ($y$-coordinates are the same), the area $A$ of a right - triangle (as $\angle B = 90^{\circ}$) is $A=\frac{1}{2}\times base\times height$. Here, base $BC = 6$ and height $AB = 4$.
$A=\frac{1}{2}\times4\times6 = 12$.

Answer:

perimeter: $10 + 2\sqrt{13}$ units
area: $12$ units$^2$