Question

Two corners of an isosceles triangle are at `(1 ,6 )` and `(2 ,9 )`. If the triangle's area is `36`, what are the lengths of the triangle's sides?

Answer 1

`sqrt(10),sqrt(520.9), sqrt(520.9)~=3.162,22.823,22.823`

Explanation:

The length of the given side is
`s=sqrt((2-1)^2+(9-6)^2)=sqrt(1+9)=sqrt(10)~=3.162`

From the formula of the triangle's area:
`S=(b*h)/2` => `36=(sqrt(10)*h)/2` => `h=72/sqrt(10)~=22.768`

Since the figure is an isosceles triangle we could have Case 1 , where the base is the singular side, ilustrated by Fig. (a) below

Or we could have Case 2 , where the base is one of the equal sides, ilustrated by Figs. (b) and (c) below

For this problem Case 1 always applies, because:

`tan(alpha/2)=(a/2)/h` => `h=(1/2)a/tan(alpha/2)`

But there's a condition so that Case 2 apllies:

`sin(beta)=h/b` => `h=bsin beta`
Or `h=bsin gamma`
Since the highest value of `sin beta` or `sin gamma` is `1`, the highest value of `h`, in Case 2, must be `b`.

In the present problem h is longer than the side to which it is perpendicular, so for this problem only the Case 1 applies.

Solution considering Case 1 (Fig. (a))

`b^2=h^2+(a/2)^2`
`b^2=(72/sqrt(10))^2+(sqrt(10)/2)^2`
`b^2=5184/10+10/4=(5184+25)/10=5209/10` => `b=sqrt(520.9)~=22.823`