Question

this question has two parts. first, answer part a. then, answer part b. part a identify the figure with the vertices g(-4,1), h(4,1), and i(0,-2). select choice 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 $GH$ with $G(-4,1)$ and $H(4,1)$:
$d_{GH}=\sqrt{(4 - (-4))^2+(1 - 1)^2}=\sqrt{(8)^2+0^2}=8$.
For side $GI$ with $G(-4,1)$ and $I(0,-2)$:
$d_{GI}=\sqrt{(0 - (-4))^2+(-2 - 1)^2}=\sqrt{(4)^2+(-3)^2}=\sqrt{16 + 9}=\sqrt{25}=5$.
For side $HI$ with $H(4,1)$ and $I(0,-2)$:
$d_{HI}=\sqrt{(0 - 4)^2+(-2 - 1)^2}=\sqrt{(-4)^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5$.
Since two sides have equal length ($GI = HI = 5$), the figure is an isosceles triangle.

Step2: Calculate the perimeter

The perimeter $P$ of a triangle is the sum of the lengths of its sides.
$P=d_{GH}+d_{GI}+d_{HI}=8 + 5+5=18$ units.

Step3: Calculate the area

The area $A$ of a triangle with base $b$ and height $h$. Here, base $b = GH = 8$ and height $h$ is the perpendicular distance from $I$ to $GH$. Since $GH$ is a horizontal line ($y = 1$), the height $h$ is the vertical distance from $y=-2$ to $y = 1$, so $h=3$.
$A=\frac{1}{2}\times b\times h=\frac{1}{2}\times8\times3 = 12$ square units.

Answer:

Part A: Isosceles triangle
Part B:
perimeter: 18 units
area: 12 units²