Question

If vertices of a triangle are $(0,4),(4,1)$ and $(7,5)$, find its perimeter.

Answer 1

Let A(0,4) B(4,1) and C(7,5)

Using distance formula

$AB=\sqrt{4-0{)}^2+(1-4{)}^2}=\sqrt{16+9}=5$

$BC=\sqrt{7-4{)}^2+(5-1{)}^2}=\sqrt{9+16}=5$

$CA=\sqrt{0-7{)}^2+(4-5{)}^2}=\sqrt{49+1}=5\sqrt{2}$

Now,

Perimeter of $△=5+5+5\sqrt{2}$

$=10+7.07=17.07units$