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#1 2023-09-18 17:13:04

sologuitar
Member
Registered: 2022-09-19
Posts: 467

Sides of A Right Triangle

College Algebra
Section R.3

Suppose that m and n are positive integers with m > n.
If a = (m^2 - n^2), b = 2mn, and c = (m^2 + n^2), show that a, b, and c are the lengths of the sides of a right triangle.

Note: This formula can be used to find the sides of a right triangle that are integers such as 3, 4, 5; 5, 12, 13; and so on. Such triplets of integers are called Pythagorean triplets.

NOTE: Looking for the set up only.

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#2 2023-09-19 04:30:10

Bob
Registered: 2010-06-20
Posts: 10,514

Re: Sides of A Right Triangle

You need to show that two of these when squared and added make the same result as the third when squared.

But which to choose? I can see that c is bigger than a.  So I suggest working out c^2 and write that down. Then get an expression for a^2 + b^2. If this comes to the same as c^2 then you are done.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

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#3 2023-09-19 08:37:48

e_jane_aran
Member
Registered: 2023-09-15
Posts: 10

Re: Sides of A Right Triangle

harpazo1965 wrote:

Suppose that m and n are positive integers with m > n.
If a = (m^2 - n^2), b = 2mn, and c = (m^2 + n^2), show that a, b, and c are the lengths of the sides of a right triangle.

Note: This formula can be used to find the sides of a right triangle that are integers such as 3, 4, 5; 5, 12, 13; and so on. Such triplets of integers are called Pythagorean triplets.

NOTE: Looking for the set up only.

I suspect that they're expecting you to notice how the given side-length expressions hint of connections to perfect-square trinomiais.

If you add two of the above expressions, you get the third of the above expressions. Which pair works?

"They are fast. Faster than you can believe. Don't turn your back. Don't look away. And most of all, don't blink." -the 10th Doctor

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#4 2023-09-19 08:39:57

e_jane_aran
Member
Registered: 2023-09-15
Posts: 10

Re: Sides of A Right Triangle

I apologize for my failure at formatting in the above. I though [ imath ] tags would work. Kindly please ignore the "\qquad" part at the beginning of each line. D'oh!

"They are fast. Faster than you can believe. Don't turn your back. Don't look away. And most of all, don't blink." -the 10th Doctor

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#5 2023-09-19 10:36:38

sologuitar
Member
Registered: 2022-09-19
Posts: 467

Re: Sides of A Right Triangle

Bob wrote:

You need to show that two of these when squared and added make the same result as the third when squared.

But which to choose? I can see that c is bigger than a.  So I suggest working out c^2 and write that down. Then get an expression for a^2 + b^2. If this comes to the same as c^2 then you are done.

Bob

Thank you, Bob.

Last edited by sologuitar (2023-09-23 01:00:12)

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#6 2023-09-19 19:08:02

Bob
Registered: 2010-06-20
Posts: 10,514

Re: Sides of A Right Triangle

to both of you:

I have edited the above post so that it displays properly.  The correct command is square brackets math

When a member makes a post, it is visible to anyone who logs in to the site.  And anyone can respond to any post.  If someone disobeys our rules then I will take action. This may involve any of the following: Issuing a warning; Editing the offending part of a post; Deleting the post; Banning for a limited period; Removing the person completely from the membership. This last means that all that person's posts get deleted too.

At the moment I can see no reason to do any of these things.  Let's keep it that way.

Best wishes,

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

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#7 2023-09-23 01:00:57

sologuitar
Member
Registered: 2022-09-19
Posts: 467

Re: Sides of A Right Triangle

Ok. No problem. Let's get back to math.

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#8 2023-09-23 01:11:44

sologuitar
Member
Registered: 2022-09-19
Posts: 467

Re: Sides of A Right Triangle

e_jane_aran wrote:
harpazo1965 wrote:

Suppose that m and n are positive integers with m > n.
If a = (m^2 - n^2), b = 2mn, and c = (m^2 + n^2), show that a, b, and c are the lengths of the sides of a right triangle.

Note: This formula can be used to find the sides of a right triangle that are integers such as 3, 4, 5; 5, 12, 13; and so on. Such triplets of integers are called Pythagorean triplets.

NOTE: Looking for the set up only.

I suspect that they're expecting you to notice how the given side-length expressions hint of connections to perfect-square trinomiais.

If you add two of the above expressions, you get the third of the above expressions. Which pair works?

If I add a^2 + c^2, the middle term cancels out. This does not leave me with much to play with.

If I do that, I am left with m^4 + n^4 twice. In other words, I get
2(m^4 + n^4) = 2m^4 + 2n^4, which does not represent b^2.

Last edited by sologuitar (2023-09-23 01:12:18)

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#9 2023-09-23 16:37:56

amnkb
Member
Registered: 2023-09-19
Posts: 253

Re: Sides of A Right Triangle

harpazo1965 wrote:

Suppose that m and n are positive integers with m > n.
If a = (m^2 - n^2), b = 2mn, and c = (m^2 + n^2), show that a, b, and c are the lengths of the sides of a right triangle.

Note: This formula can be used to find the sides of a right triangle that are integers such as 3, 4, 5; 5, 12, 13; and so on. Such triplets of integers are called Pythagorean triplets.

NOTE: Looking for the set up only.

e_jane_aran wrote:

I suspect that they're expecting you to notice how the given side-length expressions hint of connections to perfect-square trinomiais.

If you add two of the above expressions, you get the third of the above expressions. Which pair works?

harpazo1965 wrote:

If I add a^2 [and] c^2, the middle term cancels out. This does not leave me with much to play with.

If I do that, I am left with m^4 + n^4 twice. In other words, I get
2(m^4 + n^4) = 2m^4 + 2n^4, which does not represent b^2.

so maybe try a different pair of terms? (there r 2 other pairs; 1 of them works)

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#10 2023-09-23 17:25:44

sologuitar
Member
Registered: 2022-09-19
Posts: 467

Re: Sides of A Right Triangle

amnkb wrote:
harpazo1965 wrote:

Suppose that m and n are positive integers with m > n.
If a = (m^2 - n^2), b = 2mn, and c = (m^2 + n^2), show that a, b, and c are the lengths of the sides of a right triangle.

Note: This formula can be used to find the sides of a right triangle that are integers such as 3, 4, 5; 5, 12, 13; and so on. Such triplets of integers are called Pythagorean triplets.

NOTE: Looking for the set up only.

e_jane_aran wrote:

I suspect that they're expecting you to notice how the given side-length expressions hint of connections to perfect-square trinomiais.

If you add two of the above expressions, you get the third of the above expressions. Which pair works?

harpazo1965 wrote:

If I add a^2 [and] c^2, the middle term cancels out. This does not leave me with much to play with.

If I do that, I am left with m^4 + n^4 twice. In other words, I get
2(m^4 + n^4) = 2m^4 + 2n^4, which does not represent b^2.

so maybe try a different pair of terms? (there r 2 other pairs; 1 of them works)

If I add a^2 + b^2 I get c^2. So, the addition of the two legs of this triangle
yields the value of the hypotenuse. You say?

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