Question

question 16 of 28 what is the value of m in the figure below? if necessary, round your answer to the nearest tenth of a unit.

Explanation:

Step1: Use the geometric - mean theorem in right - triangles

In right - triangle $ABC$ with right - angle at $B$ and altitude $BD$ from the right - angle to the hypotenuse $AC$, we know that $BD^{2}=AD\times DC$. Also, we can use the similarity of triangles. Triangles $ABD$ and $ABC$ are similar. And by the property of similar triangles, $\frac{BD}{DC}=\frac{AD}{BD}$. Another way is to use the Pythagorean theorem in right - triangle $BDC$. First, in right - triangle $ABD$, using the Pythagorean theorem, $BD=\sqrt{AB^{2}-AD^{2}}$. In right - triangle $ABC$, let $AB = h$, $BC=m$, $AC=AD + DC=13 + 4=17$.
In right - triangle $BDC$, we know that $\triangle BDC\sim\triangle ABC$. Then $\frac{BC}{AC}=\frac{DC}{BC}$, which means $BC^{2}=DC\times AC$.

Step2: Substitute the given values

We are given $DC = 4$ and $AC=17$. Substituting into the formula $m^{2}=4\times17$.
$m^{2}=68$.

Step3: Solve for $m$

Take the square - root of both sides: $m=\sqrt{68}\approx 8.246$. Rounding to the nearest tenth, $m\approx8.2$.

Answer:

A. 8.2