Question

Find the perimeter and the area of right triangle ABC with vertices A(-3,-4), B(13,8) and C(22,-4).what type of triangle is it?Show your work.#2PerimeterArea

Answer 1

Perimeter of the triangle = 60

Area of the triangle = 150

The triangle is a scalene triangle

Step-by-step explanation:

The given vertices: A(-3,-4), B(13,8) and C(22,-4)

To get the perimeter and the area, we need to find the distance of the three sides of the triangle.

We will find the distance between A and B, B and C and A and C

The distance formula is given by:

`dis\tan ce\text{ = }\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2}`
`\begin{gathered} \text{For distance AB:} \\ x_1=-3,y_1=-4,x_2=13,y_2\text{ = }8 \\ \text{distance AB = }\sqrt[]{(8-(-4))^2+(13-(-3))^2} \\ \text{distance AB =}\sqrt[]{(8+4)^2+(13+3)^2}\text{ = }\sqrt[]{12^2+16^2} \\ \text{distance AB = }\sqrt[]{400}\text{ = 20} \end{gathered}`
`\begin{gathered} \text{For distance BC:} \\ x_1=13,y_1=8,x_2=22,y_2\text{ = }-4 \\ dis\tan ce\text{ BC= }\sqrt[]{(-4-8)^2+(22-13)^2} \\ \text{distance BC = }\sqrt[]{(-12)^2\text{ }+9^2}\text{ = }\sqrt[]{225} \\ \text{distance BC = 15} \end{gathered}`
`\begin{gathered} \text{distance AC:} \\ x_1=-3,y_1=-4,x_2=22,y_2\text{ = }-4 \\ dis\tan ce\text{ AC= }\sqrt[]{(-4-(-4))^2+(22-(-3))^2} \\ \text{distance AC = }\sqrt[]{(-4+4)^2+(22+3)^2}\text{ = }\sqrt[]{0^2+25^2} \\ \text{distance AC = }\sqrt[]{625}\text{ = 25} \end{gathered}`

We have gotten the three sides:

AB = 20, BC = 15, AC = 25

Perimeter of a triangle = sum of all three sides

Perimter = AB + BC + AC

Perimeter = 20 + 15 + 25

Perimeter of the triangle = 60

To get the area of the triangle, we will use Heron's formula:

`\begin{gathered} \text{Area = }\sqrt[]{s(s-a)(s-b)(s-c)} \\ \text{where s = }\frac{a\text{ + b + c}}{2} \end{gathered}`
`\begin{gathered} u\sin g\text{ the variables we have:} \\ \text{s = }\frac{AB\text{ + BC + AC}}{2} \\ s\text{ = }(20+15+25)/(2)\text{ = 60/2} \\ s\text{ = 30} \\ \\ \text{Area = }\sqrt[]{s(s-AB)(s-BC)(s-AC)} \\ \text{Area = }\sqrt[]{30(30-20)(30-15)(30-25)} \end{gathered}`
`\begin{gathered} \text{Area = }\sqrt[]{30(10)(15)(5)} \\ \text{Area = }\sqrt[]{22500} \\ Area\text{ of the triangle = }150 \end{gathered}`

None of the three sides of the triangle are equal.

A triangle with three sides that are not equal is a scalene triangle.

Hence, the triangle is a scalene triangle