masters level geomentry help needed!

H

hsmith82

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Jul 10, 2007
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Hi all,

Here is the problem:
"Farmer John stores grain in a large silo located at the edge of his farm. The cylinder-shaped silo has one flat, rectangular face that rests against the side of his barn. The height of the silo is 30 feet and the face resting against the barn is 10 feet wide. If the barn is approximately 5 feet from the center of the silo, determine the capacity of Farmer John’s silo in cubic feet of grain?"(Walden University).

This is what I did:
1-diameter=10
2-radius=5
3-area=78.5-based on the formula pi(5)^2
4-then I calculated for volume by multipling the area by the height and came up with 2,355 cubic feet.

This seems too high to me? Also, I can't figure out how to calculate for the capacity in cubic feet? I also think this is wrong because I never used the information about the silo being 5 feet from the barn.

Any help is appreciated!!
Thanks :oops:
 
I'm not seeing where the exercise gives you that the diameter is ten feet, especially since this would make the one flat side actually cut the silo into a half-cylinder...?

Please reply with clarification. Thank you.

Eliz.
 
hsmith82 said:
"Farmer John stores grain in a large silo located at the edge of his farm. The cylinder-shaped silo has one flat, rectangular face that rests against the side of his barn. The height of the silo is 30 feet and the face resting against the barn is 10 feet wide. If the barn is approximately 5 feet from the center of the silo, determine the capacity of Farmer John’s silo in cubic feet of grain?"
This is what I did:
1-diameter=10
2-radius=5
3-area=78.5-based on the formula pi(5)^2
4-then I calculated for volume by multipling the area by the height and came up with 2,355 cubic feet.
This seems too high to me? Also, I can't figure out how to calculate for the capacity in cubic feet? I also think this is wrong because I never used the information about the silo being 5 feet from the barn.
Click to expand...
The 10 feet face against the barn is NOT the diameter, but is a chord

The 5 feet distance of barn from silo's center is the perpendicular height from
silo's center to the chord, splitting the chord in half; this creates right triangle
with both legs = 5 feet: so radius = sqrt(5^2 + 5^2) = sqrt(50)

So if silo was a full silo, you'd have a cylinder of height 30 and circular base
of radius sqrt(50): you can easily find volume of that, right?

Then you need to reduce that volume by the volume of the portion of the silo
that's missing (behind the 10 feet chord); so by the volume related to the
segment created by the 10 feet chord; if you're not familiar with "segment of
circle", do a google search.
 
Hello, hsmith82!

Farmer John stores grain in a large silo located at the edge of his farm.
The cylinder-shaped silo has one flat, rectangular face that rests against the side of his barn.
The height of the silo is 30 feet and the face resting against the barn is 10 feet wide.
If the barn is approximately 5 feet from the center of the silo,
determine the capacity of Farmer John’s silo in cubic feet of grain?
Click to expand...
Code: 
              * * *
        A *     C  5  * B
     ---*-------*-------*---
       *  \     :     /  *
            \  5:45°/R
      *       \ : /       *
      *         *         *
      *         O         *

       *                 *
        *               *
          *           *
              * * *

The center of the circular silo is \(\displaystyle O\).
The flat wall is \(\displaystyle AB:\;AB\,=\,10,\;AC\,=\,CB\,=\,5\)
We are told that \(\displaystyle OC\,=\,5\).
\(\displaystyle OA\) and \(\displaystyle OB\) are radii, \(\displaystyle R\).
. . Note that \(\displaystyle \angle BOC\,=\,45^o\).

Pythagorus tell us that: \(\displaystyle \:5^2\,+\,5^2\:=\:R^2\;\;\Rightarrow\;\;R\,=\,\sqrt{50}\)


The area of the entire circular floor is: \(\displaystyle \:A \:=\:\pi R^2\:=\:50\pi \:\approx\:157\) ft².

The area of the quarter-circle is: \(\displaystyle \:\frac{1}{4}\,\times\,50\pi \:=\:\frac{25\pi}{2}\)
The area of right triangle \(\displaystyle AOB\) is: \(\displaystyle \:\frac{1}{2}\cdot\sqrt{50}\cdot\sqrt{50} \:=\:25\)
Hence, the area of the segment is: \(\displaystyle \:\frac{25\pi}{2}\,-\,25 \:=\:14.26990817 \:\approx\:14.27\) ft².

Then the area of the floor is: \(\displaystyle \:157\,-\,14.27 \:=\:152.73\) ft².

Therefore, the volume of the silo is: \(\displaystyle \:30\,\times\,152.73 \:=\:4581.9\) ft³.

 
thanks

I really do appreciate the help but it is sill confusing! Can you explain it to me as if I were five years old? I am so very confused. Geomentry has always been so tough for me. Thanks again!
 
It really cannot be done without finger-pointing.

Follow Soroban's example, one piece at a time.

Look at each phase. Contemplate each calculation.
 
i am okay until i get tothe last part. i can find the area of the entire floor an even 1/4 of the floor. any further clarification after that part???
 
You must stare at the drawing until you see it. What is the central angle pointing toward the building? 45º on one side and 45º on the other. That's 90º. How many degrees in the WHOLE circle? 360º. 90º/360º = 1/4

The finding of an unusual area often requires one to chop it up into comprehensible pieces.

1) There's the whole circle. Whoops, that's too much.
2) Throw out the quarter circle. Whoops, we threw away too much.
3) Put back the Triangle. There, we're done.
 
Someone told me all I had to do was find the volume of a cylinder using the formula piR^2(H). That gives me 2,355 cu ft.
Why is this wrong? What does this give me? The volume of the entire cyliner?
 
> Someone told me all I had to do was find the volume of a cylinder using the
> formula piR^2(H). That gives me 2,355 cu ft.
> Why is this wrong? What does this give me? The volume of the entire cyliner?

R^2 = sqrt(50)^2 = 50; so pi * 50 * 30 = 4712.38898....
How d'heck did you get 2355 ?
Ahhh...you're using radius = 5: IT IS NOT :shock:

> why is the answer ^3 not ^2?

answer shown as ft^3 simply means cubic feet;
like your "2,355 cu ft" is same as 2,355 ft^3
 
Thanks for the clarification!!!

BIGGER PROBLEM NOW!! My instructor said the answer is:


"140.41 sq feet times 30ft tall silo = 4,212.3 cubic feet in the silo!"

My answer was 4581.9ft^3
 
I do not agree with what your instructor got, but it's close ...

Area of silo floor = (3/4 of a circle with radius sqrt(50)) + area of the right triangle

A = (3/4)(50pi) + 25 = approx 142.81 ft<sup>2</sup>

volume of the silo = (floor area)(silo height) = (142.8 ft<sup>2</sup>)(30 ft) = approx 4284 ft<sup>3</sup>
 
Well, that's kind of what you get for just copying, rather than understanding and checking. If you look carefully at soroban's work, you will see an error in the subtraction right before the end. This will get you closer. You'll have to prove it one way or the other.
 
the professor caluclated like this:
153.86 (area of a circle)-13.45 (unecessary portion)= 140.41
140.41 x 30 (height)=4.212.3 cubic feet

I calculated
157 (area of a circle)-14.27(unnecessary)=152.73
152.73 x 30 =4,581.9 cubic feet
 
even if i corect that mistake I get-4,281.9
That is still not correct.

Please understand that I have spent at least six hours doing this problem. I am in no way copying! Looking at lines, angles and space is very hard for me. I just do not think that way. Even with the numbers in place I am having a hard time making sense of it all.
 
hsmith82 said:
even if i corect that mistake I get-4,281.9
That is still not correct.
Click to expand...
~4,281.9 IS CORRECT, Heather!!

Your prof is WRONG: area of circle is NOT 153.86; pi * 50 = ~157.08
To get 153.86, he used a radius of ~48.97, which means "his barn"
is a bit lesser than 5 feet away!
 
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