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.
 
Answer: 9.
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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , . Bookmark the permalink.

2 Responses to Median to 16

  1. You might enjoy a very different solution based on what I call an “almost pythagorean” relationship that exists in all triangles.

    • mvtrinh says:

      Thank you very much for the alternative solution. Your suggestion has certainly enlarged my knowledge of geometry. The almost pythagorean relationship is quite unexpected and interesting. I have not encountered it in any book during my highschool years. Thanks again.

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