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A median of a triangle is a line segment from a vertex of a triangle to the midpoint of the opposite side of the triangle. The medians to the legs of a certain right triangle have lengths 13 and 19. What is the length of the hypotenuse of the triangle?

I don't know how to find the length of the hypotenuse with such little information given. Any ideas?

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Draw a picture. Let our triangle be $ABC$, with right angle at $C$. Let the legs be $a$ and $b$. Let the median to side $a$ have length $13$. By the Pythagorean Theorem, we have $\frac{a^2}{4}+b^2=13^2$. Similarly, $a^2+\frac{b^2}{4}=19^2$. Solve.

Or better, don't solve. Add. We get $\frac{5}{4}(a^2+b^2)=13^2+19^2$.

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diagram

In the above diagram, The relevant information is that $ADC$ and $ABE$ are right triangles too, and their hypotenuses are known and their legs are each one of the main triangle legs and half the other main triangle leg. Therefore, you get two equations using the Pythagorean Theorem, where $a=AB, b=AC, c=BC, d=DC, e=BE$ $$d^2=\frac{a^2}{4}+b^2$$ $$e^2=a^2+\frac{b^2}{4}$$ Solving this yields $$a=2\sqrt{\frac{(4e^2-d^2)}{15}}$$ $$b=2\sqrt{\frac{(4d^2-e^2)}{15}}$$ Using the Pythagorean Theorem again, this means the hypotenuse is: $$c=2\sqrt{\frac{(d^2+e^2)}{5}}$$

EDIT: Fixed some silly algebra mistakes I made. My apologies.

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