Question

What is the perimeter of the triangle shown on the coordinate plane, to the nearest tenth of a unit?

14.6 units

15.5 units

21.0 units

21.6 units

Answer 1

see the attached figure to better understand the problem

we know that the distance between two points is equal to

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

Let

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

Step 1

Find the distance AB

`A(-3,3)\ B(3,4)`

substitute in the formula

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

`d=\sqrt{(1)^(2)+(6)^(2)}`

`dAB=√(37)\ units`

Step 2

Find the distance BC

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

substitute in the formula

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

`d=\sqrt{(-7)^(2)+(0)^(2)}`

`dBC=7\ units`

Step 3

Find the distance AC

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

substitute in the formula

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

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

`dAC=√(72)\ units`

Step 4

Find the perimeter of the triangle

we know that

the perimeter of the triangle is the sum of the length sides of the triangle

`P=AB+BC+AC`

substitute the values

`P=√(37)\ units+7\ units+√(72)\ units=21.6\ units`

therefore

the answer is

the perimeter of the triangle is
`21.6\ units`