Question 596007
How do I find the perimeter of the triangle whose vertices are (4,10), (-6,2), and (-6, 8) by simplifying without using decimals?

Hi, there--
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The perimeter of a triangle is the sum of the lengths of the three sides. To solve this problem, we can use the distance formula to find the distance between each pair of vertices. Then we find the sum of the three distances.
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I'm going to label the three vertices to make it easier to explain what's happening.

A (4,10)
B (-6,2)
C (-6,8)
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[I] Distance from A to B, (AB)
The distance formula is
{{{AB=sqrt((x[B]-x[A])^2+(y[B]-y[A])^2)}}}
{{{AB=sqrt((-6-4)^2+(2-10)^2)}}}
{{{AB=sqrt((-10)^2+(-8)^2)}}}
{{{AB=sqrt(100+64)}}}
{{{AB=sqrt(164)}}}
{{{AB=2*sqrt(41)}}}
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[II] Distance from B to C, (BC)
{{{BC=sqrt((x[C]-x[B])^2+(y[C]-y[B])^2)}}}
{{{BC=sqrt((-6-(-6))^2+(8-2)^2)}}}
{{{BC=sqrt((0)^2+(6)^2)}}}
{{{BC=sqrt(36)}}}
{{{BC=6}}}
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[III] Distance from A to C, (AC)
{{{AC=sqrt((x[C]-x[A])^2+(y[C]-y[A])^2)}}}
{{{AC=sqrt((-6-4)^2+(8-10)^2)}}}
{{{AC=sqrt((-2)^2+(-2)^2)}}}
{{{AC=sqrt(4+4)}}}
{{{AC=sqrt(8)}}}
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[IV] Find the perimeter.
{{{P=AB+BC+AC}}}
{{{P=2*sqrt(41)+6+sqrt(8)}}}
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That's it. This expression is the exact perimeter of this triangle. Hope this helps! Feel free to email if you have any questions about the explanation.
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Ms.Figgy
math.in.the.vortex@gmail.com