\section*{Exercise 3 - 201013}

Given the side lengths \(a=\overline{B C}=15 \mathrm{~cm}, b=\overline{A C}=14 \mathrm{~cm}, c=\overline{A B}=13 \mathrm{~cm}\) of a triangle \(A B C\).

Calculate the length \(h_{b}\) of the height through \(B\) and the area \(F\) of this triangle!
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![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1429.jpg?height=342&width=332&top_left_y=523&top_left_x=268)

Let the foot of the given height be \(D\); let \(u = AD, v = CD\). Then, by the Pythagorean theorem,

\[
h_{b}^{2} + v^{2} = a^{2} \quad(1) \quad ; \quad h_{b}^{2} + u^{2} = c^{2}
\]

From this, it follows that

\[
v^{2} - u^{2} = a^{2} - c^{2} \quad ; \quad (v - u)(v + u) = 56 \text{ cm}
\]

If \(D\) were not between \(A\) and \(C\), then \(v - u = b = 14 \text{ cm}\); from this and (3), it would follow that \(v + u = 4 \text{ cm}\), leading to the contradiction \(v + u < v \mid u\). Therefore, \(D\) lies between \(A\) and \(C\), and it holds that

\[
v + u = b = 14 \text{ cm}
\]

From this and (3), it follows that \(v - u = 4 \text{ cm}\) (5). From (4) and (5), we get \(2v = 18 \text{ cm}, v = 9 \text{ cm}\); from this and (1), it follows that

\[
h_{b} = \sqrt{a^{2} - v^{2}} = 12 \text{ cm}
\]

Thus, we also obtain \(F = \frac{b \cdot h_{b}}{2} = 84 \text{ cm}^{2}\).