Question

question 19 (5 points)
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what is the length of the altitude cd in the figure?

Explanation:

Step1: Use the geometric mean theorem (altitude-on-hypotenuse theorem)

In a right - triangle \(ABC\) with right angle at \(C\) and altitude \(CD\) to the hypotenuse \(AB\), we know that \(AC^{2}=AD\times AB\). Let \(AD = 5\), \(AC = 6\). First, find \(AB\). Let \(DB=x\), then \(AB=5 + x\). By the Pythagorean theorem in \(\triangle ACD\), \(CD^{2}=AC^{2}-AD^{2}\), and also by the geometric mean theorem \(CD^{2}=AD\times DB\).

Another way: Using the area formula. The area of right - triangle \(ABC\) can be expressed in two ways. \(S=\frac{1}{2}\times AC\times BC=\frac{1}{2}\times AB\times CD\). First, find \(BC\) using the Pythagorean theorem in \(\triangle ABC\). Let \(AB = c\), \(AC = b=6\), \(BC=a\), \(AD = 5\). By the Pythagorean theorem in \(\triangle ACD\), \(CD=\sqrt{AC^{2}-AD^{2}}=\sqrt{36 - 25}=\sqrt{11}\). But we can also use the geometric mean formula \(CD=\sqrt{AD\times DB}\).

Let's use the formula \(CD=\sqrt{AD\times(AB - AD)}\). First, find \(AB\) using the fact that in right - triangle \(ABC\), \(\cos A=\frac{AD}{AC}=\frac{AC}{AB}\) (since \(\triangle ACD\sim\triangle ABC\)). \(\frac{5}{6}=\frac{6}{AB}\), then \(AB=\frac{36}{5}\). Then \(DB=AB - AD=\frac{36}{5}-5=\frac{36 - 25}{5}=\frac{11}{5}\).

Step2: Calculate \(CD\)

By the formula \(CD=\sqrt{AD\times DB}\), substituting \(AD = 5\) and \(DB=\frac{11}{5}\), we have \(CD=\sqrt{5\times\frac{11}{5}}=\sqrt{11}\)

Answer:

\(\sqrt{11}\)