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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

I have a question here asking for the Perimeter of a shaded section of a circle:

The full circle is shown with center O. One sector is labled with a radius of 20 and angle Pi/2 (90 degrees). The rest of the circle is shaded and that is the part I need to work out the perimeter of.

I know that a full circle is 2Pi radians (360 degrees) and since the unshaded sector is Pi/2 radians I can deduce

2Pi - Pi/2 = 3Pi/2 (270 degrees)

Now I just use the regular formula for working out the arc length which is

s = rθ

(let θ = the angle in radians)

s = 20 * 3Pi/2 = 60Pi/2 = 94.2 (3.s.f)

Is this correct? The answers at the back of the book states134.2!

Aloha Nui means Goodbye.

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

Do I need to set my calculator to radian mode? How do I do this? It's a casio fx-83wa.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Notice something. Your answer is off by 134.2 - 94.2 = 40. Coincidence? I think not. That number should sound familiar. What piece are you missing?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

40? Hum... I can see that the first sector (with the angle Pi/2) has an arc length of 10Pi and the other sector has an arc of 30Pi, that adds up to 40Pi but I do not see how it is related?

P.S. This is the second question I've ever done of this module so I do not know much about it

*Last edited by rickyoswaldiow (2006-06-29 02:42:34)*

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

I could look at it this way:

The first sector has angle Pi/2 and radius 20. s=rθ=20*Pi/2=10Pi

Since Pi/2 = 90 degrees which is 1/4 of the full circle, the shaded sector is the other 3/4 of the circle. 1/4 of the circle = 10Pi and thus 3/4 of it would = 30Pi, 94.2.

Aloha Nui means Goodbye.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

40 = 2*20 = 2*r. You have to not only measure the arc length but also the...

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

Aha! Of course

The perimeter stretches along the two straight edges of the unshaded sector and not just round the arc of the shaded area.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Right!

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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