The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Trisected Hypotenuse of a Triangle

Date: 12/20/98 at 14:30:43
From: Will Nuse
Subject: Trisected hypotenuse of a triangle

Here's the problem:

In right triangle ABC, with C as the right angle, it is known that 
AD = DE = EB (the three segments of the hypotenuse) and that D and E 
are points on hypotenuse AB. It is also known that CD = 7 and that 
CE = 9.

What is the length of AB (the hypotenuse)?       

I tried to work this problem using the law of cosines because at first 
I thought that line segments CD and CE trisected the 90 degrees into 
30 degrees. However, my trig teacher showed me that there is more than 
one way to draw this problem, and therefore it would be almost 
impossible to get all three angles ACD, DCE, and ECB equal to 30 

I'm stumped, along with my entire trig class.  Can you help me out?

Thank you very much.  I appreciate your help.

Will Nuse

Date: 12/22/98 at 17:05:13
From: Doctor Floor
Subject: Re: Trisected hypotenuse of a triangle

Hi Will,

Thanks for your question!

For naming, let a = BC, b = BC, and c = AB. Let's take a look at a 
sketch of the situation:

   F E
   |  \
   G   D        I added the points F and G that trisect BC,
   |    \
   C-H-I-A      and H and I that trisect AC.

It is not difficult to see that, for instance, FE = CH. But that means 
that CE^2 = EF^2 + FC^2, and thus
          81 = ((1/3)a)^2 + ((2/3)b)^2

          81 = (1/9)a^2   + (4/9)b^2     [multiply by 9]

          a^2 + 4b^2 = 729  [1]

In the same way DC^2 = CI^2 + ID^2, and thus

          49 = ((2/3)a)^2 + ((1/3)b)^2

          49 = (4/9)a^2   + (1/9)b^2     [multiply by 9]

          4a^2 + b^2 = 441 [2]

Adding [1] and [2] gives that 

          5a^2 + 5b^2 = 1170, so then a^2 + b^2 = 234. 

So the hypotenuse must be sqrt(234).

Best regards,

- Doctor Floor, The Math Forum   

Date: 06/03/02 at 10:26:14
From: Doctor Ian
Subject: Re: Trisected hypotenuse of a triangle

Grant Woodward suggests another approach.  Here is a drawing of the triangle:


Start with a few simple observations:

  cosA = b/3x      
  cosB = a/3x

  a^2 + b^2 = 9x^2

Apply the Law of Cosines to triangles BCE and ACD, respectively:

  9^2 = x^2 + a^2 - 2ax cosB

  7^2 = x^2 + b^2 - 2bx cosA

Adding these two equations gives

  9^2 + 7^2 = 2x^2 + a^2 + b^2 - 2ax cosB - 2bx cosA

            = 2x^2 + a^2 + b^2 - 2x(a cosB + b cosA)

We noted earlier that a^2 + b^2 = 9x^2, so substituting gives

            = 2x^2 + 9x^2 - 2x(a cosB + b cosA)

            = 11x^2 - 2x(a cosB + b cosA)

Substituting the values we found earlier for the cosines gives

            = 11x^2 - 2x(a * a/(3x) + b * b/(3x))

            = 11x^2 - 2x((a^2 + b^2)/(3x))

Substituting again for a^2 + b^2 gives

            = 11x^2 - 2x(9x^2/(3x))

            = 11x^2 - 6x^2

            = 5x^2

So we end up with

  9^2 + 7^2 = 5x^2

                    9^2 + 7^2 
          x = sqrt( --------- )

            = sqrt(26)

So the length of the hypotenuse must be 3 sqrt(26). 

- Doctor Ian, The Math Forum   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
High School Trigonometry

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.