Calling all math geniuses (Foil :P)

[flip]

This problem is stumping my wife. Any help is appreciated. Cannot use the Pythagorean theorem.

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.

[Foil]

Ack, just now saw this... and I'm just about to head home from work. I'll be back online later this evening.

In the meantime... is she allowed to use the formulas for the area of a circle and/or area of a section of a circle?

[flip]

I can use anything other than the Pythagorean theorem. I must include an explanation and use of radius squares as well. I appreciate your help. Figuring out the radius of the circle is my main obstacle.

[Lothar]

The problem reduces fairly simply to \"area of base * 30\" (provided the silo doesn't grow out above the roof of the barn, which it doesn't seem to be doing.)

So the question is, what's the area of the base?

The face against the barn is 10 feet wide, or 5 feet to each side of the centerline, and it's 5 feet from the center. That means the angle between the center of the silo and the barn-silo intersection is 45 degrees to each side, or 90 degrees total, and the radius of the silo is 5*sqrt(2).

One way to think of the silo shape is as 3/4 of a circle (everything but the 90 degrees that faces the barn, in dark gray in the image I'll upload as soon as they're done doing maintenance on the server) plus a triangle (the 90 degrees that faces the barn, in light gray on the same image.)

The area of a whole circle would be pi*r^2, or pi*(5*sqrt(2))^2, or pi*50. So 3/4 of the circle has an area of 150*pi/4, or 75*pi/2. The area of the triangle is base*height/2, or 10*5/2 = 25. So the sum total area is 25+75*pi/2, and the overall volume is 30*(25+75*pi/2) or approximately 4284.29 cubit feet, provided I've done all my calculations right.

[ccb056]

Nothing a little trig can't solve.

We can't use pythagorean's theorem, but lets make a triangle anyway.

Draw a line from the center of the circle straight up to the barn, then draw to the left, then draw to the circle. You now have a 5,5,h triangle. To find the hypotenuse of the triangle (radius of the circle) use the following trig formulas:

tan(theta)=5/5
theta=45 degrees
sin(45)=5/h

h is solved to be 5*sqrt(2)

The volume of the silo is equal to:

(Area of large circle - .5*(area of elipsoid))*height
...

(pi*(5*sqrt(2))^2-.5*(pi*5*(5*sqrt(2)-5)))*30

[ccb056]

damn, lothar beat me

[ccb056]

One way to think of the silo shape is as 3/4 of a circle (everything but the 90 degrees that faces the barn, in dark gray in the image I'll upload as soon as they're doing doing maintenance on the server) plus a triangle (the 90 degrees that faces the barn, in light gray on the same image.)

?

[Lothar]

One way to think of the silo shape is as 3/4 of a circle (everything but the 90 degrees that faces the barn, in dark gray in the image I'll upload as soon as they're doing doing maintenance on the server) plus a triangle (the 90 degrees that faces the barn, in light gray on the same image.)

?

Apologies for the typo.

I have an image ready to upload, it's just that the server I was going to put it on is currently being worked on.

[ccb056]

I was questioning the logic behind the 3/4 value.

You and I get slightly different results.

[SuperSheep]

In an isosceles right triangle the sides are in the ratio 1:1: sqrt(2), therefore 5*sqrt(2) is your radius.

[Grendel]

Uhm, isn't using 5*sqrt(2) as r based on the pythagorean theorem ? 5*sqrt(sq(1)+sq(1)) ...

[ccb056]

Not if you use trig.

[Lothar]

the dark gray area in my above image is 3/4 of the circle, total area 75*pi/2. The light gray area is a triangle with area of 25. This gives you a net area of 75*pi/2 + 25.

You can get the same value by computing the value of the whole circle and then subtracting the partial segment. That segment is not an ellipsoid, but a circle segment; the formula for area is R*R*(t-sin t)/2, or 50*(pi/2 - 1)/2.

This should give you a final area of pi*50 - 25*(pi/2-1) = 75*pi/2 + 25.

EDIT: to find the radius, you can either cheat and use the pythagorean theorem, use the knowledge of a 1:1:sqrt(2) triangle, or use trig. Knowing that you have a triangle with 2 sides of length 1, find the angle whose tangent (opposite/adjacent) = 1; you'll find this angle is pi/4. Since sin(pi/4) = cos(pi/4) = sqrt(2)/2 (opp/hyp or adj/hyp), you can find the hypotenuse or radius is 5*sqrt(2).

[Grendel]

I was thinking of this and this

[Foil]

Thinking of the base as 3/4 area of a circle, plus the little triangular area... bingo!

I figured someone would post the solution before I got back online.

[SuperSheep]

Well, the sin(45) = 0.707... which is equal to 1/sqrt(2) so I suppose either way you go, the same ratio would be returned. The area is obvious after determining the radius.

The only other way would be to calculate the area using reduction, but I don't think that's what teacher wants.

[fliptw]

http://mathworld.wolfram.com/Chord.html

[snoopy]

I have another question/problem for you guys, that I did a long time ago, and have always felt like I was missing something:

You have a cylindrical tank, laying on its side. (So if there was a hill, it'd start rolling.) You want to know the volume of liquid inside, and the only way you have to measure is by dipping a ruler in it, and measuring the depth from the bottom of the cylinder to the level of the liquid.

I solved it using calculus.

Without using calculus, can you come up with a formula, Volume(Radius, Length, Depth)=?

[Insurrectionist]

I keep hearing planes flying over my head.

[Foil]

You have a cylindrical tank, laying on its side.
...
Without using calculus, can you come up with a formula, Volume(Radius, Length, Depth)=?

The only idea that comes quickly to mind for me is essentially using an approximation method that converges toward the solution... but that's essentially just doing calculus the hard way.

[Lothar]

You want to know the volume of liquid inside, and the only way you have to measure is by dipping a ruler in it, and measuring the depth from the bottom of the cylinder to the level of the liquid.

This again reduces to "find the area of the segment of the circle, and then multiply by the length of the tank". You can find the area of the segment of the circle using the "circle segment" formulas, where x is the depth of the liquid:

t = 2 arccos((R-x)/R)

A=(R^2)*(t-sin t)/2

V=A*L

Essentially, you use trig to find out how wide of an angle the liquid is making across the bottom of the tank. Once you have that, you can compute the area of the circle segment, which is equal to the area of a circular wedge minus the area of the triangular part. Then just multiply by the length of the tank.

[flip]

Thank you for the posts. Here's how I solved using the radius square:
To find the radius of the circle without using the Pythagorean Theorem, I solved by using the formula for the area of the square using A=2^r. The area of the square is 100 sqft so r=7.07. Plugging that value into the formula for the area of the circle results in 157 sqft. I then subtracted the area of the square from the area of the circle. That meant the area of the circle not covered by the square was 57 sqft. Dividing that by 4 came to 14 sqft, the area of the missing segment. So, 157 sqft-14 sqft=143 sqft. This is the area of the base. Using the formula for area of a cylinder (v=ah)told me that the silo would hold 4290 cubic ft of grain.

[Foil]

Ah, flip used a method no one mentioned yet! Using the 100 sq.ft. square inscribed in the circle works pretty well, as the missing section of the circle is exactly 1/4 of the area of the circle outside the square.

Just goes to show how many mathematical methods can work for a single problem.

-----------

P.S. Was that answer accepted, flip? Rounding the numbers off like you did made your answer somewhat different than the exact one.

[flip]

I'm sorry I need to clarify this. After the first post I let my wife continue the discussion using my account. I may just get her a separate account of her own to avoid any confusion in the future. She asked me this question and the first person I thought about was Foil and told her she should ask. So, aside from this post and the very first one she has been the one participating. Her name is Dori by the way.

[ccb056]

I like the inscribed square method

[Foil]

Ah, hi Dori! Well done, by the way.

[Lobber]

verified:

using trigonometry

h2 = x2+y2
h2 = ( 5)2+( 5)2
h = [ ( 25.00)+( 25.00) ] ½
h = 7.0710678118654755

hypotenuse = radius

circle area = 157.07963267948967329653735064999
75% of circle area = 117.80972450961725497240301298749
area of triangle (10 * 5)/2 = 25
25 + 117.80972450961725497240301298749 = 142.80972450961725497240301298749

142.80972450961725497240301298749 * 30 = 4284.2917352885176491720903896248 cubic feet

[Neo]

Here's how to do it using calculus, for anyone who's interested (edit: made a slight error but it's now fixed, turned it into a link):

Let A be the total area
Let V be the volume of the silo
Let Δz be the height of the silo

EPIC WIN