## Median to 16

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side.  You are given a triangle with side lengths of 11 cm, 13 cm, and 16 cm.  Determine the length of the median that joins the side of length 16 cm and the opposite vertex.
Source: mathcontest.olemiss.edu 11/2/2009

SOLUTION

Let $AB=16 \textup { cm}, AC=11 \textup{ cm and } BC=13 \textup{ cm}$. Let $\overline{CD}$ be the median joining vertex $C$ to the midpoint $D \textup{ of } \overline{AB}$. From vertex $C$ draw altitude $\overline{CE}$ to side $\overline{AB}$.

Length of $\overline{DE}$
In right triangle $ECB$, by the Pythagorean theorem we have
$EB^2+EC^2=BC^2=13^2=169$

In right triangle $EAC$, we have by the same theorem
$EA^2+EC^2=11^2=121$

Multiplying the second equation by $-1$ then adding the two equations yields
$EB^2+EC^2=169$
$-EA^2-EC^2=-121$
—————————–
$EB^2-EA^2=48$

$\left (EB+EA\right )\left (EB-EA\right )=48$
$AB\left (EB-EA\right )=48$
$16\left (EB-EA\right )=48$
$EB-EA=3$

Since $EB=BD+DE=DA+DE \textup{ and } EA=DA-DE$, rewrite the above equation as follows:
$EB-EA=3$
$DA+DE-\left (DA-DE\right )=3$
$DA+DE-DA+DE=3$
$2DE=3$
$DE=1.5$

Calculate $EC^2$
$EA=DA-DE$
$=8-1.5$
$=6.5$

In right triangle $EAC$:
$EA^2+EC^2=11^2=121$
$6.5^2+EC^2=121$
$42.25+EC^2=121$
Thus, $EC^2=121-42.25=78.75$

Length of median $\overline {CD}$
In right triangle $DEC$:
$DE^2+EC^2=CD^2$
$1.5^2+78.75=CD^2$
$2.25+78.75=CD^2$
$81=CD^2$

Thus, $CD=9$.