An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of #8 # and the triangle has an area of #40 #. What are the lengths of sides A and B?

Let side A=side B= #x#

Then drawing a right-angled triangle, with #x# as your hypotenuse and #4# as the length of your bottom side and the final side be #h#

Using pythagoras theorem,

#x^2# = #h^2+4^2#

#x^2=h^2+16#

#h^2=x^2-16#

#h=+-sqrt(x^2-16)#

But since h is a length, it can only be positive

#h=sqrt(x^2-16)#

Area of triangle =

#1/2times8timessqrt(x^2-16) = 40#

#4sqrt(x^2-16)=40#

#sqrt(x^2-16)=10#

#x^2-16=100#

#x^2=116#

#x=+-10.77#

But since x is a side, then it can only be positive

#x=10.77#

Hey, small letters for triangle sides.

Isosceles triangle #a=b#, #c=8#, Area #mathcal{A}=40.#

Let's see if we can do this different ways. First, the way the teacher probably expects:

The foot #F# of the altitude #h# from vertex #C# to side #c# is the midpoint of #AB#. So

#mathcal{A} = 1/2 ch#

#40 = 1/2 (8) h#

#h = 10#

#AF=c/2=4# and we have a right triangle with the altitude:

#(c/2)^2 + h^2 = a^2#

#4^2 + 10^2 = a^2#

#a = b = sqrt{116} = 2 sqrt{29}#

Let's just do it directly from Archimedes' Theorem:

#16 mathcal{A} ^2 = 4a^2 c^2 - (b^2 - c^2-a^2)^2#

The sides can be arbitrarily interchanged; I picked the form that gets some cancelling when #a=b#:

#16 mathcal{A} ^2 = 4a^2 c^2 - c^4#

# a^2 = {c^4 + 16 mathcal{A}^2}/{4c^2} #

# a = \sqrt{ {8^4 + 16(40)^2}/{4(8^2)}} =2 sqrt{29} quad sqrt#

That was easier.