• May 1st 2006, 05:39 AM
kenj
Please help, I have my exam on Friday and really need to know how to solve the 2003 Standard Grade Paper, Question 10.

question: A sheep shelter is part of a cylinder thats is 6 meters
wide and 2 meters high. The cross section of the shelter is a segment
of a circle with centre, O. Calculate the radius of the circle?
• May 1st 2006, 05:53 AM
kenj
Circle Attachment
Maybe this pic will help?
• May 1st 2006, 06:00 AM
kenj
And the next step (I Think!)
I think this is the right direction ?
• May 1st 2006, 07:24 AM
earboth
Quote:

Originally Posted by kenj
Maybe this pic will help?

Hello,

You see a right triangle with the hypotenuse of 2r. Then the half chord is the height in the right triangle.

In a right triangle the product of the two parts of the hypotenuse equals the square of the height:
$2\cdot (2r-2)=3^2 \Longleftrightarrow 4r-4=9 \Longleftrightarrow r = \frac{13}{4}$

Greetings

EB
• May 1st 2006, 07:30 AM
kenj
Many thanks.
• May 1st 2006, 08:50 AM
CaptainBlack
Quote:

Originally Posted by kenj
Please help, I have my exam on Friday and really need to know how to solve the 2003 Standard Grade Paper, Question 10.

question: A sheep shelter is part of a cylinder thats is 6 meters
wide and 2 meters high. The cross section of the shelter is a segment
of a circle with centre, O. Calculate the radius of the circle?

Use the intersecting chord theorem.

In this case it tell us that:

$
3\times 3=2\times(2r-2)
$

where $r$ is the radius of the circle.

RonL
• May 1st 2006, 10:04 AM
kenj
Many thanks for your help. Thanks to your posts I also completed it just by using Pythagoros. Such that:

3sq + (r-2)sq = rsq

So 9 + rsq -2r – 2r + 4 = rsq

13 = 4r

r = 13 / 4 = 3.25