In a trapezoid, the diagonals have lengths $10$ and $24$, and are perpendicular to each other. Find the length of the median?
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Take $a$ and $b$ to be the lengths of the segments of the length-10 and length-24 diagonals respectively that are on one side of the intersection point, as shown. By similarity of triangles we have $$\frac ab=\frac{10-a}{24-b}$$ $$(24-b)a=b(10-a)$$ $$24a=10b\qquad12a=5b$$ Without loss of generality we may take $a=5$ and $b=12$. Then the trapezium is a rhombus and its median length is the same as its side length – $\sqrt{5^2+12^2}=13$. The sum of the parallel sides' lengths remains the same (26) regardless of how the two diagonals are positioned, so the median length stays constant as well.
$\endgroup$$\begingroup$Say $ABCD$ is the trapezoid $AB||CD$. Draw a parallel to $BD$ through $C$ which cuts line $AB$ at $E$. Then $$AE = \sqrt{10^2+24^2} =26$$ But it is easy to see that middle line of the triangle $AEC$ (parallel to $AE$) has the same length as the middle line od trapezoid $ABCD$ and thus it is $13$.
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