Question

solve the following problems. given: △abc, (overline{cm}perpoverline{ab}), (bc = 5), (ab = 7), (ca=4sqrt{2}). find: cm

Explanation:

Step1: Use Heron's formula to find area of $\triangle ABC$.

Let $a = 5$, $b = 4\sqrt{2}$, $c = 7$. Semi - perimeter $s=\frac{a + b + c}{2}=\frac{5+4\sqrt{2}+7}{2}=6 + 2\sqrt{2}$.
Area $S=\sqrt{s(s - a)(s - b)(s - c)}=\sqrt{(6 + 2\sqrt{2})(1 + 2\sqrt{2})(6 - 2\sqrt{2})(2\sqrt{2}- 3)}$.
Also, area $S=\frac{1}{2}\times AB\times CM$.

Step2: Calculate area and solve for $CM$.

$S=\frac{1}{2}\times7\times CM$.
First, $(6 + 2\sqrt{2})(6 - 2\sqrt{2})=36-8 = 28$ and $(1 + 2\sqrt{2})(2\sqrt{2}-3)=4 - 3 - 6\sqrt{2}+2\sqrt{2}=1 - 4\sqrt{2}$.
$S=\sqrt{28(1 - 4\sqrt{2})}$ is wrong. Let's use another way.
Let $BM=x$, then $AM = 7 - x$.
By Pythagorean theorem: $BC^{2}-BM^{2}=CA^{2}-AM^{2}$.
$25 - x^{2}=32-(7 - x)^{2}$.
$25 - x^{2}=32-(49 - 14x+x^{2})$.
$25 - x^{2}=32 - 49 + 14x - x^{2}$.
$14x=42$, $x = 3$.
In right - triangle $BCM$, by Pythagorean theorem, $CM=\sqrt{BC^{2}-BM^{2}}=\sqrt{25 - 9}=4$.

Answer:

$4$