Question

Triangle ABC has vertices at (-4,0) , (-1, 6) and (3,-1) perimeter of triangle ABC, rounded to the nearest tenth ?

Answer 1

The perimeter of triangle ABC is
`P=21.8\ units`

Explanation:

The perimeter of triangle ABC 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)}`

we have

`A(-4,0),B(-1, 6),C(3,-1)`

step 1

Find the distance AB

`A(-4,0),B(-1, 6)`

substitute in the formula

`d_A_B=\sqrt{(6-0)^(2)+(-1+4)^(2)}`

`d_A_B=\sqrt{(6)^(2)+(3)^(2)}`

`d_A_B=√(45)\ units`

step 2

Find the distance BC

`B(-1, 6),C(3,-1)`

substitute in the formula

`d_B_C=\sqrt{(-1-6)^(2)+(3+1)^(2)}`

`d_B_C=\sqrt{(-7)^(2)+(4)^(2)}`

`d_B_C=√(65)\ units`

step 3

Find the distance AC

`A(-4,0),C(3,-1)`

substitute in the formula

`d_A_C=\sqrt{(-1-0)^(2)+(3+4)^(2)}`

`d_A_C=\sqrt{(-1)^(2)+(7)^(2)}`

`d_A_C=√(50)\ units`

step 4

Find the perimeter

`P=AB+BC+AC`

substitute the values

`P=√(45)+√(65)+√(50)=21.8\ units`