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 #16 # and the triangle has an area of #40 #. What are the lengths of sides A and B?

Small letters for triangle sides please, dearies. #c=16, a=b#.

My favorite formula for the area of a triangle #A# with sides #a,b,c# is

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

Let #a=b# for an isosceles triangle.

# 16 A^2 =4a^2 c^2 - c^4 #

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

# a^2 = {16(40)^2 + 16^4}/{4(16^2)} = 89 #

#a = b = sqrt{89} #

Check:

The altitude #h# splits an isosceles triangle with common side #a# and base #c# is two right triangles, #(c/2)^2 + h^2 = a^2# or #h = sqrt{a^2 - c^2/4}# and an area of

#text{area} = 1/2 ch = c/2 \ sqrt{a^2 - c^2/4}#

# = 1/2 (16) sqrt{ 89 - 64} = 8 sqrt{ 25} = 40 quad sqrt#

Isosceles Triangle means two sides are equal.
The area of triangle is
Area = #(bxxh)/2#
The given are base and the area.
first determine the height before finding A and B.
Substitute Area = 40 and Base C = 16
to the formula
Area = #(bxxh)/2#
#40=(16xxh)/2#

or #(40) (2) = 16 h#
or #80 = 16 h#
divide both side by #16#
to get the #h# (height)
#80/16# = #(16 h)/16#
#h = 5#
since you have the height
cut into two congruent lengths by the height, since we are dealing with an Isosceles triangle, the entire base measures 16, half of the base measures 8.
Now we have two Right Triangles
using Pythagorean Theorem
#a^2 + b^2 = c^2#
#8^2 + 5^2 = c^2#
64 + 25 = #c^2#
89 = #c^2#
#sqrt(89)# = C
89 is a prime number
#sqrt(89)= 9.4339#
#A = 9.4339#
#B = 9.4339#