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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,711

Yes for 7, that is what I did.

I am not following 8, where did you put C?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,121

AMC = 90

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

I put C on the circle so from m to b that is 1

One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

ooooh but the radius of the circle is half the diameter So from M to c that is 2

One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,711

Okay, I think Bob has something for you.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,121

Q8

AMC = 90 You had that in the original diagram.

AM = MC = 1 = radius

So use Pythagoras

Bob

ps. 6F, 7A, 9D all seem to be done and correct.

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

So that makes it a right triangle So that means I can use the Pythagorean theorum.

1^2 + b^2 = 2^2

1+b^2= 4

b^2= 4-1

b^2=3

b= √3

One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

lol sorry I posted that before reading post #31 Thank you !

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,121

That's OK No apology needed. Great that you spotted this yourself.

But look again at your calculation.

Which side (AM,MC or AC) is the hypotenuse?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

AC is going to be the hypotenuse

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

So 2^2 +2^2= c^2

4+4= c^2

c^2=8

sqrt 8

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

#10 I like my answer (D) because I know that 9 is the length of DJ and it is equal to 3√3 *√3

So the other side is 3√3 and the hypotenuse = 3√3 *2 which gives me 6√3 which is 6*sqrt(3)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,711

Hi;

Isn't 8) 1^2 + 1^2 = c^2?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

but AM is 2 and Mi is 2 so isn't AI going to be 2^2+2^2

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,711

Hi;

8. If I drew a line segment from A to C, and the radius of circle M was 1, what would line segment AC be?

A sqrt 7

B sqrt 2

C sqrt 8

D sqrt 13

E sqrt 4

F sqrt 3

Aren't they 1 not 2

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

if it 1^2 + 1^2 = c^2 that makes my first answer (b) correct

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,711

From your original drawing AM is 1 and MC is 1 because they are radii. AC is the hypotenuse because it is opposite the right angle. B should be correct. Why did you change?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

1 is the radius but the diameter is 2

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,711

Hi;

You do not need the diameter.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

So I add the radius not the diameter to get the hypotenuse ?

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,121

hi zee-f

I thought you were still trying to do 8.

I've looked back, and yes you did say (b) hours ago.

Oh dear .... got in a muddle there.

As for 10, that is strange. I cannot find the answer I've got for this.

What's your working?

Bob

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,121

No wait. My mistake

D is good!

Bob

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

9 is the length of DJ and it is equal to 3√3 *√3

So the other side is 3√3 and the hypotenuse = 3√3 *2 which gives me 6√3 which is 6*sqrt(3)

I used the Special Right Triangles formula

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

When I do to much math problems I sometimes get mixed up But thanks for the help guys !

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

I have 10 more left but I am still working on then lol

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