Question

What is the perimeter of a triangle with corners at `(7 ,6 )`, `(4 ,5 )`, and `(3 ,1 )`?

Answer 1

`Perimeter=sqrt10+sqrt17+sqrt41`

Explanation:

Equation for finding distance between `2` coordinates `=` `sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Let,
`(7,6)=`point `A`

`(4,5)=`point `B`

`(3,1)=`point `C`

Distance between `AB`

`x_1=7,x_2=4,y_1=6,y_2=5`

`AB=sqrt((4-7)^2+(5-6)^2)`

`=>sqrt((-3)^2+(-1)^2`

`=>sqrt(9+1)`

`=>sqrt10`

`AB=sqrt10`

Distance between `BC`

`x_1=4,x_2=3,y_1=5,y_2=1`

`BC=sqrt((3-4)^2+(1-5)^2)`

`=>sqrt((-1)^2+(-4)^2`

`=>sqrt(1+16)`

`sqrt17`

`BC=sqrt17`

Distance between `AC`

`x_1=7,x_2=3,y_1=6,y_2=1`

`AC=sqrt((3-7)^2+(1-6)^2)`

`=>sqrt((-4)^2+(-5)^2`

`=>sqrt(16+25)`

`=>sqrt41`

`AC=sqrt41`

Perimeter of a triangle is the sum of its `3` sides, that `AB+BC+AC.`

`Perimeter=sqrt10+sqrt17+sqrt41`