Someone help me with my maths homework please? Watch

3mma_gal96
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Hi, does anyone have the slightest clue how to work this out please?!
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TenOfThem
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(Original post by 3mma_gal96)
Hi, does anyone have the slightest clue how to work this out please?!
Pythagoras
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steve2005
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(Original post by 3mma_gal96)
Hi, does anyone have the slightest clue how to work this out please?!
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This might be easier to read.

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Boro123
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Not certain, but I think the answer might be 7+ root7. Using Pythagoras, you can find the hypotenuse of each triangle Pattern 1: (1^2)+(1^2)=2, therefore hypotenuse=root2. Follow this pattern, and the hypotenuse of each triangle is the root of the pattern number. To cut a long story short, the hypotenuse of the 6th triangle in pattern 6 will be root7. Add the other sides (1x7), you get 7. I'm not sure if that is clear enough for you, but if you use Pythagoras each time (a^2)+(b^2)=c^2, you can't go wrong.
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LShirley95
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One of the sides of the next triangle is the hypotenuse of the last. Find the final length using pythagoras 6 times and then add 7 for the extra 1 cm sides in the perimeter. For example, with pattern 4:

Spoiler:
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P = \sqrt{\sqrt{\sqrt{\sqrt{1^2 + 1^2}^2 + 1^2}^2 + 1^2}^2 +1^2} +5



P = \sqrt{5} + 5
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TenOfThem
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(Original post by Boro123)
Not certain, but I think the answer might be 7+ root7. Using Pythagoras, you can find the hypotenuse of each triangle Pattern 1: (1^2)+(1^2)=2, therefore hypotenuse=root2. Follow this pattern, and the hypotenuse of each triangle is the root of the pattern number. To cut a long story short, the hypotenuse of the 6th triangle in pattern 6 will be root7. Add the other sides (1x7), you get 7. I'm not sure if that is clear enough for you, but if you use Pythagoras each time (a^2)+(b^2)=c^2, you can't go wrong.

(Original post by LShirley95)
One of the sides of the next triangle is the hypotenuse of the last. Find the final length using pythagoras 6 times and then add 7 for the extra 1 cm sides in the perimeter. For example, with pattern 4:

Spoiler:
Show
P = \sqrt{\sqrt{\sqrt{\sqrt{1^2 + 1^2}^2 + 1^2}^2 + 1^2}^2 +1^2} +5



P = \sqrt{5} + 5
Perhaps you are not aware of the forum guidelines

We are asked not to give full solutions

And it is frowned upon to do another poster's homework for them
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LShirley95
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(Original post by TenOfThem)
Perhaps you are not aware of the forum guidelines

We are asked not to give full solutions

And it is frowned upon to do another poster's homework for them
I showed her how to do pattern 4, it is up to her to pattern 6? I understand I may have pushed it a bit far but I didn't go all out, sorry.
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3mma_gal96
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(Original post by TenOfThem)
Perhaps you are not aware of the forum guidelines

We are asked not to give full solutions

And it is frowned upon to do another poster's homework for them
Yeah, I'm not saying "someone tell me the answers", I'm asking for "some help."

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TenOfThem
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(Original post by 3mma_gal96)
Yeah, I'm not saying "someone tell me the answers", I'm asking for "some help."
I did not suggest that you were at fault
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Boro123
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(Original post by TenOfThem)
Perhaps you are not aware of the forum guidelines

We are asked not to give full solutions

And it is frowned upon to do another poster's homework for them
Oops. Sorry about that, although I'm not certain if I am right or not. Anyways, no more solutions in future.
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