Two corners of an isosceles triangle are at $(5 ,6 )$ and $(4 ,8 )$ . If the triangle's area is $36$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-07-09T11:07:29

The lengths of the sides are $=2.24, 32.21,32.21$

The length of the base is

$b=sqrt((4-5)^2+(8-6)^2)=sqrt(1+4)=sqrt5$

The area of the triangle is

$A=1/2*b*h=36$

So,

The altiude is $h=36*2/b=72/sqrt5$

We apply Pythagoras' theorem

The length of the side is

$l=sqrt((b/2)^2+(h)^2)$

$=sqrt((5/4+72^2/5))$

$=sqrt(1038.05)$

$=32.21$

Two corners of an isosceles triangle are at (5 ,6 )(5 ,6 ) and (4 ,8 )(4 ,8 ). If the triangle's area is 36 36 , what are the lengths of the triangle's sides?