Question

Two corners of an isosceles triangle are at `(9 ,4 )` and `(3 ,2 )`. If the triangle's area is `48`, what are the lengths of the triangle's sides?

Answer 1

`2sqrt(10),sqrt(1202/5),sqrt(1202/5)~=6.325,15.505,15.505`

Explanation:

The length of the given side is
`s=sqrt((9-3)^2+(4-2)^2)=sqrt(40)=2sqrt(10)~=6.325`

From the formula of the triangle's area:
`S=(b*h)/2` => `48=(cancel(2)sqrt(10)*h)/cancel(2)` => `h=48/sqrt(10)~=15.179`

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 applies:

`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=(48/sqrt(10))^2+(cancel(2)sqrt(10)/cancel(2))^2`
`b^2=2304/10+10=2404/10=1202/5` => `b=sqrt(1202/5)~=15.505`