# What is the perimeter of a triangle with corners at (5 ,2 ), (2 ,7 ), and (1 ,4 )?

Jul 30, 2017

Perimeter of the triangle is $13.46 \left(2 \mathrm{dp}\right)$ unit.

#### Explanation:

Let $A \left(5 , 2\right) , B \left(2 , 7\right) , C \left(1 , 4\right)$ are the three corners of the triangle.

Length $A B = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}} = \sqrt{{\left(5 - 2\right)}^{2} + {\left(2 - 7\right)}^{2}}$

$= \sqrt{9 + 25} = \sqrt{34} \approx 5.83 \left(2 \mathrm{dp}\right)$

Length $B C = \sqrt{{\left({x}_{2} - {x}_{3}\right)}^{2} + {\left({y}_{2} - {y}_{3}\right)}^{2}} = \sqrt{{\left(2 - 1\right)}^{2} + {\left(7 - 4\right)}^{2}}$

$= \sqrt{1 + 9} = \sqrt{10} \approx 3.16 \left(2 \mathrm{dp}\right)$

Length $C A = \sqrt{{\left({x}_{3} - {x}_{1}\right)}^{2} + {\left({y}_{3} - {y}_{1}\right)}^{2}} = \sqrt{{\left(1 - 5\right)}^{2} + {\left(4 - 2\right)}^{2}}$

$= \sqrt{16 + 4} = \sqrt{20} \approx 4.47 \left(2 \mathrm{dp}\right)$

Perimeter of the triangle is $P = A B + B C + C A \approx \left(5.83 + 3.16 + 4.47\right)$ or

$P \approx 13.46 \left(2 \mathrm{dp}\right)$ unit. [Ans]