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The altitude drawn to the hypotenuse of a right triangle divides the hypotenuse into segments such as their lengths are in ratio of 1:4. If the length of the altitude is 8, find the length of: (a) Each segment of the hypotenuse (b) The longer leg of the triangle.

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I happen to know that when the foot F of the altitude divides the hypotenuse into two lengths
d and
e, the altitude
h is the geometric mean of
d and
e, that is
h^2= de.


But you probably didn't know that. We'll take our triangle labeled the usual way, right angle C, hypotenuse
c = d + e. There are three right triangles, which I'll write without the benefit of the figure. Let's say sides
a=BC and
d=BF so



d^2 + h^2 = a^2



e^2 + h^2 = b^2



a^2 + b^2 = c^2 = (d + e)^2 = d^2 + 2de + e^2


Adding all three equations lots of stuff cancels,



2h^2 = 2de



h^2 = de


That proves the geometric mean thing. In our problem we have
h=8, d:e=1:4 which just means
e=4d.



8^2 = d(4d) = 4d^2



d=4



e=16



b^2=e^2+h^2= 16^2+8^2=5(8^2)



b=8√(5)


Part a)
d=4,e=16


Part b)
b=8\sqrt 5


Check:



a^2=d^2+h^2=16+64=80=16\cdot 5


The triangle is
a=4√(5), b=8√(5) so
c=√(16\cdot 5 + 64\cdot 5)=√(400)=20 \quad\checkmark



User Andrey Bienkowski
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