99% of my Geometry class can’t solve this problem. I’m in the 99%

Question

I need help with this geometry problem, number 22.

I’m in Chapter 3, so we should not be using Pythogerean or Trig just yet. I’m not allowed to use anything past congruent triangles and closely related theorems, as this is a culmulative review in a popular textbook.

Answer 1

The problem is wrong. Looking at AC, x must be greater than 40, or else the segment has non-positive length. AD is shown as x-10 and must be shorter than AB, which is shown as sqrt(x). The problem is, x-10 > sqrt(x) for all x > 40. There is no x that satisfies this problem.

I wonder how 1% of your class managed to solve it?

Answer 2

Using the Pythagorean theorem: (I know that you say you shouldn't need it, but)

$(\frac {x}{10}-4)^2 + (0.08x)^2 = x\\ 0.0164 x^2 - 1.8x + 16 = 0\\ x = 100$

$AC = \frac {100}{10}-4 = 6\\ BC = 0.08\cdot 100 = 8\\ AB = \sqrt {100} = 10$

But, this is inconsistent with the statement $AD = x-10 = 90$