If one of the legs of a certain right triangle is three times the other, in what ratio does the altitude on the hypotenuse divide the hypotenuse?

Guest Nov 16, 2014

#1**+10 **

Geometric solution:

Triangle ADB is similar to triangle ABC so BD/AD = BC/AB ...(1)

Triangle BDC is similar to triangle ABC so CD/BD = BC/AB ...(2)

Multiply (1) and (2) together: (BD/AD)*(CD/BD) = (BC/AB)^{2}

or CD/AD = (3/1)^{2}

so CD/AD = 9/1

.

Alan
Nov 16, 2014

#2

#3**+5 **

Yes that is really 'neat' Alan.

I had a much more long winded approach

I just let the sides be 1 and 3 so the hypotenuse is sqrt(10)

I dropped that altitude from the hypotenuse and let it be h units long.

This cut the hypotenuse in the ratio x to sqrt(10)-x

so

$$\\1^2+h^2=x^x\\

and\\

3^2+h^2=(\sqrt{10}-x)^2$$

solving this simultaneously you get $$x=\frac{1}{\sqrt{10}}$$

so you have the ratio

$$\\\frac{1}{\sqrt{10}}:\sqrt{10}-\frac{1}{\sqrt{10}}\\\\

\frac{1}{\sqrt{10}}:\frac{10-1}{\sqrt{10}}\\\\

1:9$$

Alan's method is neater - This is just an alternative. I specialise in 'LONG' methods.

Although this was not particularly long.

Melody
Nov 17, 2014