How do you find the perimeter of a triangle whose coordinates are (0,0) (6,8) ( 12,0)?

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Answer 1 · 2018-05-12T19:14:14

$P = 32$ $P=32$

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$A (0, 0), B (6, 8), C (12, 0)$ $A(0,0),B (6,8),C (12,0)$

We use the distance formula to find the length of each side. Then we add them together. The distance formula is:

$d = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$ $d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$A B = \sqrt{{(6 - 0)}^{2} + {(8 - 0)}^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10$ $AB=sqrt((6-0)^2+(8-0)^2)=sqrt(36+64)=sqrt100=10$

$B C = \sqrt{{(12 - 6)}^{2} + {(0 - 8)}^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10$ $BC=sqrt((12-6)^2+(0-8)^2)=sqrt(36+64)=sqrt100=10$

$A C = \sqrt{{(12 - 0)}^{2} + {(0 - 0)}^{2}} = \sqrt{144 + 0} = \sqrt{144} = 12$ $AC=sqrt((12-0)^2+(0-0)^2)=sqrt(144+0)=sqrt144=12$

$P = A B + B C + A C = 10 + 10 + 12 = 32$ $P=AB+BC+AC=10+10+12=32$