Two corners of an isosceles triangle are at $(2 ,4 )$ and $(1 ,8 )$ . If the triangle's area is $64$ , what are the lengths of the triangle's sides?

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Answer 1 · 2018-05-20T15:56:51

$color(blue)((5sqrt(44761))/34 , (5sqrt(44761))/34, sqrt(17)$

Let $A=(2,4), and B=(1,8)$

Then side $c=AB$

Length of $AB=sqrt((1-2)^2+(8-4)^2)=sqrt(17)$

Let this be the base of the triangle:

Area is:

$1/2ch=64$

$1/2sqrt(17)(h)=64$

$h=128/sqrt(17)$

For isosceles triangle:

$a=b$

Since the height bisects the base in this triangle:

$a=b=sqrt((c/2)^2+(h^2))$

$a=b=sqrt((sqrt(17)/2)^2+(128/sqrt(17))^2)=(5sqrt(44761))/34~~31.11$

Sides are:

$color(blue)((5sqrt(44761))/34 , (5sqrt(44761))/34, sqrt(17)$

Two corners of an isosceles triangle are at (2 ,4 )(2 ,4 ) and (1 ,8 )(1 ,8 ). If the triangle's area is 64 64 , what are the lengths of the triangle's sides?