Hello, ExpoSport!

The problem is a bit more complicated than that.

I'll pick the dimensions for my convenience.

Suppose the right triangle has sides: 12, 16, 20 Code:

A o
| *
| *
| *
| *
| * 20
12 | *
| *
| *
| *
| *
| *
C o - - - - - - - - - - - - - - - - - o B
16

Suppose we cut a 1-inch strip from the hypotenuse.

The new (smaller) triangle is *not* 11 by 15 . . .

Code:

A o E
| o
|θ/ *
|/ 1 *
D o *
| * *
| * *
| * *
| * * G
| * o
| * 1/ *
| * / θ *
C o - - - - - - - - - - - o - - - - - o B
F

Let $\displaystyle \theta = \angle B,\:\text{ then: }\angle ADE = \theta$

Then: .$\displaystyle \begin{array}{c}\sin\theta \:=\:\dfrac{12}{20} \:=\:\dfrac{3}{5} \\ \\[-3mm] \cos\theta \:=\:\dfrac{16}{20} \:=\:\dfrac{4}{5} \end{array}$

$\displaystyle DE = FG = 1$ is the width of the strip.

In right triangle $\displaystyle FGB\!:\;\;\sin\theta \:=\:\frac{1}{FB} \quad\Rightarrow\quad \frac{3}{5} \:=\:\frac{1}{FB} \quad\Rightarrow\quad FB \:=\:\frac{5}{3}$

In right triangle $\displaystyle AED\!:\;\;\cos\theta \:=\:\frac{1}{AD} \quad\Rightarrow\quad \frac{4}{5} \:=\:\frac{1}{AD} \quad\Rightarrow\quad AD \:=\:\frac{5}{4}$

Hence: .$\displaystyle \begin{array}{ccccccc}CF \:=\:16 - \frac{5}{3} \:=\:14\frac{1}{3} \\ \\[-3mm] CD \:=\:12 - \frac{5}{4} \:=\:10\frac{3}{4} \end{array}$

Get the idea?