\section*{Exercise 2 - 161022}

For a right triangle \(A B C\), where \(C D\) is the height on the hypotenuse, the length of the cathetus \(b=A C=4 \mathrm{~cm}\) and the length \(p=B D=1.8 \mathrm{~cm}\) are given.

Calculate the lengths of the remaining sides of the triangle, the height \(C D=h\), and the length \(q=A D\).
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For the sought lengths \(a=BC, c=AB, h, q\) it holds: (1) \({ }^{2}+b^{2}=c^{2}\) (according to the Pythagorean theorem)

(2) \(c q=b^{2}\) (according to the leg theorem)

(3) \(p q=h^{2}\) (according to the altitude theorem)

(4) \(p+q=c\) (since \(D\) lies on \(AB\)).

From (2), (4) it follows that \(q^{2}+p q-b^{2}=0\). For the numerical value \(x\) of the length \(q\) measured in \(\mathrm{cm}\), it therefore holds that \(x^{2}+\) \(1.8 x-16=0\) and \(x>0\). With the solutions

\[
x_{1,2}=-0.9 \pm \sqrt{0.81+16}=-0.9 \pm 4.1
\]

it follows that \(q=3.2 \mathrm{~cm}\) and thus for the remaining quantities \(c=5 \mathrm{~cm}, a=3 \mathrm{~cm}\) and \(h=2.4 \mathrm{~cm}\).