# What is the perimeter of a triangle with corners at (3 ,9 ), (5 ,7 ), and (1 ,4 )?

Apr 14, 2018

Perimeter of the triangle is $13.22$ unit

#### Explanation:

$A \left(3 , 9\right) , B \left(5 , 7\right) , C \left(1 , 4\right)$ are the three vertices of triangle.

Distance between two points $\left({x}_{1} , {y}_{1}\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right)$ is

$D = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$

$A {B}^{2} = {\left(3 - 5\right)}^{2} + {\left(9 - 7\right)}^{2} = 8 \therefore A B = \sqrt{8} \approx 2.83$ unit

$B {C}^{2} = {\left(5 - 1\right)}^{2} + {\left(7 - 4\right)}^{2} = 25 \therefore B C = 5$ unit

$C {A}^{2} = {\left(1 - 3\right)}^{2} + {\left(4 - 9\right)}^{2} = 29 \therefore C A = \sqrt{29} \approx 5.39$ unit

Perimeter of the triangle is $P = A B + B C + C A$ or

$P = 2.83 + 5.0 + 5.39 \approx 13.22$ unit [Ans]