Question

Find the perimeter of a triangle whose vertices have the coordinates $(3,10),(5,2)$ and $(14,12)$.

Answer 1

Solution:

Let $A(3,10),B(5,2)$ and $C(14,12)$ be the coordinates of the vertices of a triangle.

$AB=\sqrt{(3-5{)}^2+(10-2{)}^2}$

$=\sqrt{4+64}$

$=\sqrt{68}$

$=2\sqrt{17}$ units $=8.25$ units

$BC=\sqrt{(14-5{)}^2+(12-2{)}^2}$

$=\sqrt{81+100}$

$=\sqrt{181}$

$=13.45$ units

$AC=\sqrt{(3-14{)}^2+(10-12{)}^2}$

$=\sqrt{121+4}$

$=\sqrt{125}$

$=5\sqrt{5}$ units

$=11.180$ units

So, perimeter of the required triangle $=AB+BC+AC$

$=8.25+13.45+11.180$ units

$=32.880$ units