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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

Best Answer 

 #4
avatar+26741 
+10

Actually, I also solved it algebraically at first, Melody, then realised there was a nice geometric treatment of it.

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Alan  Nov 17, 2014
 #1
avatar+26741 
+10

Geometric solution:

triangle

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

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Alan  Nov 16, 2014
 #2
avatar+87293 
0

Very nice, Alan....I always like these "clean" geometric proofs.........!!!!!

 

CPhill  Nov 16, 2014
 #3
avatar+92751 
+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
 #4
avatar+26741 
+10
Best Answer

Actually, I also solved it algebraically at first, Melody, then realised there was a nice geometric treatment of it.

.

Alan  Nov 17, 2014
 #5
avatar+87293 
+5

Both answers are very good.....!!!!

 

 

CPhill  Nov 17, 2014

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