Question

What is the perimeter of a triangle with vertices located at (1.4) (2,7) and (0,5)

Answer 1

The perimeter of triangle is
`P=(√(10)+3√(2))\ units`

Explanation:

Let

`A(1.4),B(2,7),C(0,5)`

we know that

The perimeter of the triangle 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(1.4),B(2,7)`

substitute the values

`AB=\sqrt{(7-4)^(2)+(2-1)^(2)}`

`AB=\sqrt{(3)^(2)+(1)^(2)}`

`AB=√(10)\ units`

step 2

Find the distance BC

`B(2,7),C(0,5)`

substitute the values

`BC=\sqrt{(5-7)^(2)+(0-2)^(2)}`

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

`BC=√(8)\ units`

`BC=2√(2)\ units`

step 3

Find the distance AC

`A(1.4),C(0,5)`

substitute the values

`AC=\sqrt{(5-4)^(2)+(0-1)^(2)}`

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

`AC=√(2)\ units`

step 4

Find the perimeter

`P=AB+BC+AC`

`P=√(10)+2√(2)+√(2)`

`P=(√(10)+3√(2))\ units`