My Math Forum  

Go Back   My Math Forum > High School Math Forum > Trigonometry

Trigonometry Trigonometry Math Forum


Thanks Tree1Thanks
  • 1 Post By skeeter
Reply
 
LinkBack Thread Tools Display Modes
December 17th, 2014, 08:48 PM   #1
Newbie
 
Joined: Dec 2014
From: Ohio

Posts: 4
Thanks: 0

Help on Pythagorean Proof

Hello,

So I'm a little stuck here. If a=3 and b=4 then squaring provides a large square of 7x7, giving an area of 49. However adding all the areas of the squares provides an area of 50 (3^2=9, 4^2=16, 5^2=25; 9+16+25=50). It seems to me that this would indicate the Pythagorean Theorem is not correct, however I'm quite sure I am just not conceptualizing it correctly... where am I going wrong?
modus1985 is offline  
 
December 17th, 2014, 09:39 PM   #2
Math Team
 
Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Hello, modus1985!

You really don't understand Pythagoras, do you?


Quote:
So I'm a little stuck here.
If a=3 and b=4, then squaring provides a large square of 7x7,
giving an area of 49.$\;\;\color{red}{?}\;\color{blue}{[1]}$

However adding all the areas of the squares provides an area of 50
(3^2 = 9, 4^2 = 16, 5^2 = 25; 9 + 16 + 25 = 50).$\;\;\color{red}{?}\;\color{blue}{[2]}$

In a right triangle: $\:a^2+b^2 \:=\:c^2$, where $c$ is the hypotenuse.

In $\color{blue}{[1]}$, we have: $\,a = 3,\;b = 4.$
Then you added $a$ and $b\!:\;3+4 \,=\,7$
But the theorem says nothing about $a+b$, does it?

In $\color{blue}{[2]}$, we have: $\,a = 3,\;b =4,\:c=5.$
Then you added all three squares: $\:a^2+b^2+c^2$ . . . Why?
The theorem says nothing about the sum of all three squares.

Pythagoras says the sum of the two smaller squares
$\quad$ equals the largest square.
So: $\:3^2+4^2 \,=\,5^2 \quad\Rightarrow\quad 9 + 16 \,=\,25$

Last edited by skipjack; July 31st, 2015 at 10:14 PM.
soroban is offline  
December 18th, 2014, 01:09 AM   #3
Senior Member
 
Joined: Apr 2014
From: Glasgow

Posts: 2,157
Thanks: 731

Math Focus: Physics, mathematical modelling, numerical and computational solutions
Don't worry, it's okay not to understand something first time and by posting your query here you did the right thing

However, Soroban is right. You are interpreting Pythagoras' theorem incorrectly.

In short, you should the square the lengths of the shorter sides, then add them together. Read more below if you have the time.

------------------------------------------------------------------------

First, let's remind ourselves of the order of operations:

BIDMAS:
1. Brackets
2. Indices (powers)
3. Division/Multiplication
4. Addition/Subtraction

This is true also in algebra. We must stick to this, otherwise we would get different answers for the same thing and mathematics becomes a horrible mess.

Okay... let's now look at Pythagoras' formula. The formula is

$\displaystyle a^2 + b^2 = c^2$

where $\displaystyle a$ and $\displaystyle b$ are the lengths of the shorter sides and $\displaystyle c$ is the length of the hypoteneuse, which is the longest side. Try to remember that the hypoteneuse is always on its own on whilst the squares of the shorter sides are always added together.

The formula tells us that there is a relationship between the sides of a right-angled triangle. We can say that the lengths of the sides of a right-angled triangle are related. Some people even call equations like this one relations just to make the point

Pythagoras' theorem is so incredibly useful... ancient engineers used it to make 3-4-5 wedges of various sizes, safe in the knowledge that they were right-angled, so they could build walls that were exactly vertical with the ground

Anyway... by using the order of operations, we can see that the first thing we need to do is square things, because step 2 is Indices/powers (we don't have any brackets), then we add the results together, which is step 4.

Here's an example.

1. What is the length of the longest side of a right-angled triangle if the lengths of the shorter sides are 3cm and 4cm?

We start with

$\displaystyle a = 3$cm
$\displaystyle b = 4$cm
$\displaystyle c = ?$cm

Pythagoras' theorem states that

$\displaystyle a^2 + b^2 = c^2$,

so, we swap the $\displaystyle a$ for a 3 and the $\displaystyle b$ for a 4 (this swapping is called substitution). Then we leave $\displaystyle c$ alone and try and find what it is.

$\displaystyle 3^2 + 4^2 = c^2$

Because $\displaystyle 3^2 = 3 \times 3 = 9$ and $\displaystyle 4^2 = 4 \times 4 = 16$, we get

$\displaystyle 9 + 16 = c^2$
$\displaystyle 25 = c^2$

This means that

$\displaystyle c = \sqrt{25} = 5$cm

Have a go at some exercises. Practise will help you figure it out. Start with exercises that ask you to find the hypoteneuse, then do some exercises that ask you to find the length of a shorter side if you know the hypoteneuse and the other short side. It will test your algebra skills

Last edited by Benit13; December 18th, 2014 at 01:25 AM.
Benit13 is offline  
December 18th, 2014, 04:15 PM   #4
Newbie
 
Joined: Dec 2014
From: Ohio

Posts: 4
Thanks: 0

I appreciate the responses but I don't think I was clear in my initial post. My question was more on the common proof given for the Pythagorean Theorem as seen in the link below.


Pythagorean Theorem Proof

So, say if we take a^2+b^2 as being 3^2+4^2, The proof would show a perfect square with an area of 49 (length=7, width=7) yet the area of the triangles within the perfect square would be 50 (a^2+b^2+c^2=3^2+4^2+5^2=9+16+25=50.... Note: This is just for showing the area of the objects that prove the theorem....the Pythagorean theorem is this example would obviously as stated above provide a hypotenuse of 25). Please help me figure this out, it's been driving me nuts!!!
modus1985 is offline  
December 18th, 2014, 05:07 PM   #5
Math Team
 
skeeter's Avatar
 
Joined: Jul 2011
From: Texas

Posts: 3,002
Thanks: 1587



Quote:
the area of the triangles within the perfect square would be 50 (a^2+b^2+c^2=3^2+4^2+5^2=9+16+25=50
no, the entire figure has an area of only 49.


each blue triangle has area $\displaystyle \frac{1}{2}ab$ ... there are four triangles, so the area of all four is $\displaystyle 4 \cdot \frac{1}{2}ab = 2ab$

the yellow square has area $\displaystyle c^2$, so 4 blue triangles + 1 yellow square has total area $\displaystyle 2ab + c^2$

the large square also has area $\displaystyle (a+b)^2$ ...

$\displaystyle (a+b)(a+b) = 2ab + b^2$

$\displaystyle a^2+2ab+b^2 = 2ab+b^2$

$\displaystyle a^2 + b^2 = c^2$

now, using 3, 4, and 5 for a, b, and c respectively ...

$\displaystyle 2\cdot3\cdot4 + 5^2 = 49$

$\displaystyle (a+b)(a+b) = (3+4)(3+4) = (7)(7) = 49$
Thanks from perfect_world
skeeter is offline  
December 18th, 2014, 07:14 PM   #6
Newbie
 
Joined: Dec 2014
From: Ohio

Posts: 4
Thanks: 0

Thanks Skeeter!! Boy do I feel like an idiot lol.... much simpler than I thought.
modus1985 is offline  
December 18th, 2014, 07:17 PM   #7
Newbie
 
Joined: Dec 2014
From: Ohio

Posts: 4
Thanks: 0

Quote:
Originally Posted by modus1985 View Post
I appreciate the responses but I don't think I was clear in my initial post. My question was more on the common proof given for the Pythagorean Theorem as seen in the link below.


Pythagorean Theorem Proof

So, say if we take a^2+b^2 as being 3^2+4^2, The proof would show a perfect square with an area of 49 (length=7, width=7) yet the area of the triangles within the perfect square would be 50 (a^2+b^2+c^2=3^2+4^2+5^2=9+16+25=50.... Note: This is just for showing the area of the objects that prove the theorem....the Pythagorean theorem is this example would obviously as stated above provide a hypotenuse of 25). Please help me figure this out, it's been driving me nuts!!!
Correction... the hypotenuse would of course be 5!!
modus1985 is offline  
December 19th, 2014, 09:44 PM   #8
Global Moderator
 
Joined: Dec 2006

Posts: 20,916
Thanks: 2199

Enjoy this article.
skipjack is offline  
Reply

  My Math Forum > High School Math Forum > Trigonometry

Tags
proof, pythagorean



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Pythagorean theorem Perlita Geometry 1 May 19th, 2014 08:46 AM
Pythagorean once more! Denis New Users 2 April 21st, 2013 09:02 PM
Primitive Pythagorean Proof question.! eChung00 Number Theory 2 September 10th, 2012 05:13 PM
pythagorean theorem moonrains Geometry 2 January 7th, 2009 03:46 PM
new Pythagorean theorem mohanned karkosh Geometry 1 October 22nd, 2007 05:13 AM





Copyright © 2019 My Math Forum. All rights reserved.