# What is the perimeter of a triangle with corners at (1 ,4 ), (6 ,7 ), and (4 ,2 )?

Perimeter $= \sqrt{34} + \sqrt{29} + \sqrt{13} = 3.60555$

#### Explanation:

$A \left(1 , 4\right)$ and $B \left(6 , 7\right)$ and $C \left(4 , 2\right)$ are the vertices of the triangle.

Compute for the length of the sides first.

Distance AB

${d}_{A B} = \sqrt{{\left({x}_{A} - {x}_{B}\right)}^{2} + {\left({y}_{A} - {y}_{B}\right)}^{2}}$

${d}_{A B} = \sqrt{{\left(1 - 6\right)}^{2} + {\left(4 - 7\right)}^{2}}$

${d}_{A B} = \sqrt{{\left(- 5\right)}^{2} + {\left(- 3\right)}^{2}}$

${d}_{A B} = \sqrt{25 + 9}$

${d}_{A B} = \sqrt{34}$

Distance BC

${d}_{B C} = \sqrt{{\left({x}_{B} - {x}_{C}\right)}^{2} + {\left({y}_{B} - {y}_{C}\right)}^{2}}$

${d}_{B C} = \sqrt{{\left(6 - 4\right)}^{2} + {\left(7 - 2\right)}^{2}}$

${d}_{B C} = \sqrt{{\left(2\right)}^{2} + {\left(5\right)}^{2}}$

${d}_{B C} = \sqrt{4 + 25}$

${d}_{B C} = \sqrt{29}$

Distance BC

${d}_{A C} = \sqrt{{\left({x}_{A} - {x}_{C}\right)}^{2} + {\left({y}_{A} - {y}_{C}\right)}^{2}}$

${d}_{A C} = \sqrt{{\left(1 - 4\right)}^{2} + {\left(4 - 2\right)}^{2}}$

${d}_{A C} = \sqrt{{\left(- 3\right)}^{2} + {\left(2\right)}^{2}}$

${d}_{A C} = \sqrt{9 + 4}$

${d}_{A C} = \sqrt{13}$

Perimeter $= \sqrt{34} + \sqrt{29} + \sqrt{13} = 3.60555$

God bless....I hope the explanation is useful.