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
degrees.

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,

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

B
|\
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
http://mathforum.org/dr.math/
```

```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:

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

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( --------- )
5

= sqrt(26)

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

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
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
Math Forum Home || Math Library || Quick Reference || Math Forum Search