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- anonymous

Ok can anyone please help me understand the pythagorean theorom please ?

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- anonymous

Ok can anyone please help me understand the pythagorean theorom please ?

- jamiebookeater

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- anonymous

In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two legs.

- pokemon23

|dw:1328570206293:dw|

- anonymous

|dw:1328570208471:dw|
\[h=\sqrt{a^2+b^2}\]
\[a=\sqrt{h^2-b^2}\]
\[b=\sqrt{h^2-a^2}\]

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- anonymous

\[h^2=a^2+b^2\]

- TuringTest

here is a nice visual proof
http://www.google.com/imgres?q=pythagorean+theorem&hl=en&sa=X&biw=1024&bih=655&tbm=isch&prmd=imvns&tbnid=0a8M5qnfEI1lGM:&imgrefurl=http://www.math-aids.com/Pythagorean_Theorem/&docid=mrZX0eSmrJiEWM&imgurl=http://www.math-aids.com/images/pythagorean-definition-04.png&w=612&h=792&ei=GGAwT4XWHqLb0QGnhLHUCg&zoom=1&iact=hc&vpx=560&vpy=273&dur=0&hovh=255&hovw=197&tx=110&ty=274&sig=107113811444028023167&page=1&tbnh=130&tbnw=88&start=0&ndsp=16&ved=1t:429,r:13,s:0

- anonymous

All right i understand all that but where do the square roots come in like if i give you a problem would you be able to help ?

- pokemon23

where c represents the length of the hypotenuse, and a and b represent the lengths of the other two sides.

- pokemon23

In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).

- anonymous

The square roots come in because the pythagorean theorem tells you the sum of the squares of the legs equal the square of the hypotenuse. You have to use the square root to find the number.
\[(\sqrt{a^2})=a\]

- pokemon23

|dw:1328570575319:dw|

- anonymous

|dw:1328570529625:dw| how would i find the answers to that ?

- anonymous

Here, you wouldn't use the pythagorean theorem until you use some trig to find one side.

- pokemon23

for example in this question we need to find c^2
by finding c^2 we plug in the formula a^2+b^2=c^2
6^2+7^2=c^2
36+48=c^2
\[\sqrt{36}\]+\[\sqrt{48}\]

- pokemon23

I mean 49*

- anonymous

\[\tan(60º)=\frac{9\sqrt{3}}{x}\]
\[x=\frac{9\sqrt{3}}{\tan(60º)}\]

- anonymous

When you find x, now you can use the pythagorean theorem.
\[y=\sqrt{x^2+(9\sqrt{3})^2}\]

- anonymous

oh man my teacher never told me we had to use tan to solve this answer

- anonymous

Well, this is a special right triangle, so I suppose you don't have to use trig. It is a 30-60-90 triangle, did your teacher tell you how to solve these?

- anonymous

http://www.cliffsnotes.com/study_guide/Special-Right-Triangles.topicArticleId-18851,articleId-18821.html
Look at this link, it tells you the relationship between sides in special right triangles so you don't have to use trig.

- anonymous

ok i understand that now do you think you could give me a problem so i could try to answer it

- anonymous

|dw:1328571722764:dw|

- anonymous

Can you do that one?

- anonymous

for this one what do i have to find ?

- anonymous

so the short leg would be 5 and the hyp would 25 ?

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