story problem using quadratic formula

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Jodene222

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When mineral depostis formed a coating 1 mm thick on the inside of a pipe, the area through which fluid can flow was reduced by 20%. Find thte original inside diamerer of the pipe.
Remember: Area of a circle = pie r ^2 and diameter = 2r


Please help I do not know how to set up this problem.
 
Edit: made a mistake...deleted what I said sense Mrspi gave the correct answer.
 
Let r = original radius

Original area = pi * r[sup:2cgco1h4]2[/sup:2cgco1h4]

The mineral buildup reduces the radius by 1 mm. So, r - 1 = new radius

New area = pi * (r - 1)[sup:2cgco1h4]2[/sup:2cgco1h4]

We know that the area is reduced by 20%. The new area then is 80% of the original area (100% - 20% = 80%).

New area = 80% of original area

pi * (r - 1)[sup:2cgco1h4]2[/sup:2cgco1h4] = 0.8 * pi * r[sup:2cgco1h4]2[/sup:2cgco1h4]

Now, you can solve that for r, the original radius. But remember that you're asked for the original diameter, so you'll need to multiply the value of r by 2.
 
I did not get the correct answer using your suggestion. The answer is 18.94.
Please help...i am determined and have been working on this problem for 4 hours.
 
Jodene222 said:
I did not get the correct answer using your suggestion. The answer is 18.94.
Please help...i am determined and have been working on this problem for 4 hours.
Click to expand...

If solved correctly (you'll want to use the quadratic formula), the equation I gave earlier WILL produce an answer of 18.94 mm.

Please show us your work....after spending 4 hours on the problem, you should surely have something to show for it. We can't tell where you might be making a mistake unless we see what you actually did.
 
here is what I have:

3.14(r^2-2r+1) = .8(3.14r^2)

3.14r^2 -6.28r + 3.14 = 2.51r^2

.63r^2 - 6.28r + 3.14 = 0

a + .63 b = 6.28 c = 3.14

when i use the quadratic equation I do not get 18.94. what am I doing wrong?
 
pi * (r - 1)^2 = 0.8 * pi * r^2

Jodene, you are COMPLICATING something simple; stary by dividing each side by pi:
(r - 1)^2 = .8r^2 ; now you'll get:
r^2 - 2r + 1 = .8r^2
.2r^2 - 2r + 1 = 0 ; divide by .2:
r^2 - 10r + 5 = 0

OK???
 
Jodene222 said:
here is what I have:

3.14(r^2-2r+1) = .8(3.14r^2)

3.14r^2 -6.28r + 3.14 = 2.51r^2

.63r^2 - 6.28r + 3.14 = 0

a + .63 b = 6.28 c = 3.14 error here....b = -6.28

when i use the quadratic equation I do not get 18.94. what am I doing wrong?
Click to expand...

Using the quadratic formula, with your numbers, I get r = 9.44 and 2r (the diameter) is 18.88....that's pretty close to the book answer. If you START solving the equation by dividing both sides by pi, like so:

[pi*(r - 1)[sup:28bd1vn0]2[/sup:28bd1vn0] ] / pi = ( 0.8*pi*r[sup:28bd1vn0]2[/sup:28bd1vn0] ) / pi

you will get this.....

(r - 1)[sup:28bd1vn0]2[/sup:28bd1vn0] = 0.8 r[sup:28bd1vn0]2[/sup:28bd1vn0]

This will eliminate any round-off errors introduced by using 3.14 for pi, and will also get rid of some of those nasty decimals.

Double-check your arithmetic in the quadratic formula....your error may lie there.

You're too fast for me, Denis.
 
THANK YOU!!!!!!!!!!!!!!
I was able to solve the problem both ways. It is much easier dividing each side by pi. Thank you for all your help.
 
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