What is the perimeter of a triangle with corners at (6 ,0 ), (5 ,2 ), and (5 ,4 )?

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Feb 1, 2018

Perimeter of the triangle $P = \textcolor{b r o w n}{8. 5953}$

Explanation:

Given : $A \left(6 , 0\right) , B \left(5 , 2\right) , C \left(5 , 4\right)$

Distance between two points, given the two end points' coordinates is given by the formula,

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

$a = \sqrt{{\left(5 - 6\right)}^{2} + {\left(2 - 0\right)}^{2}} = \sqrt{{1}^{2} + {2}^{2}} \approx \textcolor{g r e e n}{2.2361}$

$b = \sqrt{{\left(6 - 5\right)}^{2} + {\left(0 - 4\right)}^{2}} = \sqrt{{1}^{2} + {4}^{2}} \approx \textcolor{g r e e n}{4.1231}$

$c = \sqrt{{\left(6 - 5\right)}^{2} + {\left(0 - 2\right)}^{2}} = \sqrt{{1}^{2} + {2}^{2}} = \textcolor{g r e e n}{2.2361}$

It is an isosceles triangle with sides a & c = 2.2361.

Perimeter of the triangle $P = \frac{a + b + c}{2} = \frac{2.2361 + 4.1231 + 2.2361}{2} = \textcolor{b r o w n}{8. 5953}$

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