Circles: Chords, Radii, and Arcs

bobbym · 2012-10-30 03:44:31

Yes for 7, that is what I did.

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

Bob · 2012-10-30 03:49:22

AMC = 90

Bob

zee-f · 2012-10-30 03:50:20

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

zee-f · 2012-10-30 03:51:40

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

bobbym · 2012-10-30 03:52:42

Okay, I think Bob has something for you.

Bob · 2012-10-30 03:55:19

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.

zee-f · 2012-10-30 04:01:33

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

zee-f · 2012-10-30 04:02:59

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

Bob · 2012-10-30 04:07:51

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

zee-f · 2012-10-30 04:28:07

AC is going to be the hypotenuse

zee-f · 2012-10-30 04:33:03

So 2^2 +2^2= c^2
4+4= c^2
c^2=8
sqrt 8

zee-f · 2012-10-30 04:34:09

#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)

bobbym · 2012-10-30 04:36:49

Hi;

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

zee-f · 2012-10-30 04:48:51

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

bobbym · 2012-10-30 04:51:01

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

zee-f · 2012-10-30 04:51:02

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

bobbym · 2012-10-30 04:53:46

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?

zee-f · 2012-10-30 04:55:14

1 is the radius but the diameter is 2

bobbym · 2012-10-30 04:58:28

Hi;

You do not need the diameter.

zee-f · 2012-10-30 05:00:35

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

Bob · 2012-10-30 05:01:14

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

Bob · 2012-10-30 05:03:02

No wait. My mistake

D is good!

Bob

zee-f · 2012-10-30 05:03:06

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

zee-f · 2012-10-30 05:04:19

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

zee-f · 2012-10-30 05:05:35

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

Math Is Fun Forum

#26 2012-10-30 03:44:31

Re: Circles: Chords, Radii, and Arcs

#27 2012-10-30 03:49:22

Re: Circles: Chords, Radii, and Arcs

#28 2012-10-30 03:50:20

Re: Circles: Chords, Radii, and Arcs

#29 2012-10-30 03:51:40

Re: Circles: Chords, Radii, and Arcs

#30 2012-10-30 03:52:42

Re: Circles: Chords, Radii, and Arcs

#31 2012-10-30 03:55:19

Re: Circles: Chords, Radii, and Arcs

#32 2012-10-30 04:01:33

Re: Circles: Chords, Radii, and Arcs

#33 2012-10-30 04:02:59

Re: Circles: Chords, Radii, and Arcs

#34 2012-10-30 04:07:51

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#35 2012-10-30 04:28:07

Re: Circles: Chords, Radii, and Arcs

#36 2012-10-30 04:33:03

Re: Circles: Chords, Radii, and Arcs

#37 2012-10-30 04:34:09

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#38 2012-10-30 04:36:49

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#39 2012-10-30 04:48:51

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#40 2012-10-30 04:51:01

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#41 2012-10-30 04:51:02

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#42 2012-10-30 04:53:46

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#43 2012-10-30 04:55:14

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#44 2012-10-30 04:58:28

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#45 2012-10-30 05:00:35

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