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? 
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:
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.  
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 rightangled triangle. We can say that the lengths of the sides of a rightangled 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 345 wedges of various sizes, safe in the knowledge that they were rightangled, 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 rightangled 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. 
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!!! 
December 18th, 2014, 05:07 PM  #5  
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587  Quote:
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 ... $\displaystyle 2\cdot3\cdot4 + 5^2 = 49$ $\displaystyle (a+b)(a+b) = (3+4)(3+4) = (7)(7) = 49$  
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.

December 18th, 2014, 07:17 PM  #7  
Newbie Joined: Dec 2014 From: Ohio Posts: 4 Thanks: 0  Quote:
 
December 19th, 2014, 09:44 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,916 Thanks: 2199 
Enjoy this article.


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