Question

What is the perimeter of the triangle shown on the coordinate plane, to the nearest tenth of a unit? 20.6 units 22.7 units 25.6 units 27.6 units

Answer 1

we know that

the perimeter of a polygon is the sum of the length sides

in this problem we have a triangle

so

the polygon has three sides

Let

`A(-5,4)\\B(1,4)\\C(3,-4)`

the perimeter is equal to

`P=AB+BC+AC`

The formula to calculate the distance between two points is equal to

`d=\sqrt{(y2-y1)^(2)+(x2-x1)^(2)}`

Step 1

Find the distance AB

`A(-5,4)\\B(1,4)`

substitutes the values in the formula

`d=\sqrt{(4-4)^(2)+(1+5)^(2)}`

`d=\sqrt{(0)^(2)+(6)^(2)}`

`dAB=6\ units`

Step 2

Find the distance BC

`B(1,4)\\C(3,-4)`

substitutes the values in the formula

`d=\sqrt{(-4-4)^(2)+(3-1)^(2)}`

`d=\sqrt{(-8)^(2)+(2)^(2)}`

`d=√(68)`

`dBC=8.25\ units`

Step 3

Find the distance AC

`A(-5,4)\\C(3,-4)`

substitutes the values in the formula

`d=\sqrt{(-4-4)^(2)+(3+5)^(2)}`

`d=\sqrt{(-8)^(2)+(8)^(2)}`

`d=√(128)`

`dAC=11.31\ units`

Step 4

Find the perimeter

the perimeter is equal to

`P=AB+BC+AC`

substitutes the values

`P=6+8.25+11.31=25.56\ units=25.6\ units`

therefore

the answer is

`25.6\ units`