# What is the perimeter of a triangle with corners at (1 ,9 ), (8, 2 ), and (9 ,4 )?

Apr 8, 2016

Perimeter$= 7 \sqrt{2} + \sqrt{5} + \sqrt{89}$, units

#### Explanation:

Distance between any two points $\left({x}_{1} , {y}_{1}\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right)$, is given by the formula
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

Using this formula distance ${d}_{1}$ between two corners of triangle $\left(1 , 9\right) \mathmr{and} \left(8 , 2\right)$ is

${d}_{1} = \sqrt{{\left(8 - 1\right)}^{2} + {\left(2 - 9\right)}^{2}}$
or ${d}_{1} = \sqrt{{\left(7\right)}^{2} + {\left(- 7\right)}^{2}}$
or ${d}_{1} = \sqrt{49 + 49}$
or ${d}_{1} = \sqrt{2 \times 49}$
or ${d}_{1} = 7 \sqrt{2}$

Similarly distance ${d}_{2}$ between points $\left(8 , 2\right) \mathmr{and} \left(9 , 4\right)$ is
${d}_{2} = \sqrt{{\left(9 - 8\right)}^{2} + {\left(4 - 2\right)}^{2}}$
or ${d}_{2} = \sqrt{{\left(1\right)}^{2} + {\left(2\right)}^{2}}$
or ${d}_{2} = \sqrt{1 + 4}$
or ${d}_{2} = \sqrt{5}$

And distance ${d}_{3}$ between points $\left(9 , 4\right) \mathmr{and} \left(1 , 9\right)$ is
${d}_{3} = \sqrt{{\left(1 - 9\right)}^{2} + {\left(9 - 4\right)}^{2}}$
or ${d}_{3} = \sqrt{{\left(- 8\right)}^{2} + {\left(5\right)}^{2}}$
or ${d}_{3} = \sqrt{64 + 25}$
or ${d}_{3} = \sqrt{89}$

Now perimeter of the triangle is sum of all three sides.
Perimeter$= {d}_{1} + {d}_{2} + {d}_{3}$
Perimeter$= 7 \sqrt{2} + \sqrt{5} + \sqrt{89}$
or Perimeter$= 7 \sqrt{2} + \sqrt{5} + \sqrt{89}$