Results 1 to 2 of 2

Math Help - Farmer John Grain Silo

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    1

    Farmer John Grain Silo

    Hello,

    I hope that all is well. I am having difficulty solving two Geometry problems from your course 6572. I am trying to get the correct calculations in order to complete my portfolio. If possible, will you look over my calculations and tell me if I am correct and if I am incorrect could you point out where the mistakes occur? Thanking you in advance for your help.

    Jacob


    PROBLEM 1
    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.


    The volume of any prism is the area of the face multiplied by the height. The height is given, so all you need to find is the area of the base. I started off by drawing a circle. One side has a chord of length 10 feet cut across it. Its given that the chord is 5 feet from the center of the circle. Therefore, you draw a line from the center bisecting the chord, as well as a line from the center to each end of the chord (the radius), and you have two equal isosceles triangles with two sides of length 5. Pythagorean's relation tells us that the third side has a length of 5*sqroot(2) ft. Now that you have the radius, you can find the area of the circle. However, its not a true circle you're concerned with. You want to exclude the area on the outside of the chord. Using trig, you can find that the total angle subtended by the chord is 90°. Therefore, the area that you're concerned with is the circular area for the outside 270° plus the area of the triangles. 270° is (3/4) of the circle, so its area is (3/4) the area of the entire circle. A = (3/4)*π*r² = (3/4)*π*(5*sqroot(2)ft)² = 118ft². The area of each of the smaller triangles is given by (1/2)*b*h. Since there's two triangles, the total area of the two is 2*(1/2)*b*h = b*h = 5ft*5ft = 25ft². The total area of the base is 118ft² + 25ft² = 143ft². Multiplied by the height will give the total volume of the prism, 143ft²*30ft = 4290ft³.

    PROBLEM 2

    Regular pentagon ABCDE is enclosed in a circle. If the radius of the circle is 5 and each side AB = BC = CD = DE = EA = 5.88, find the area of the pentagon.
    (In a 90 degree triangle, using pythagorean theorem c = square root of a2 + b2)

    AO = BO = CO = DO = EO = 5

    Therefore, we have 5 isosceles triangles. The altitude of an
    isosceles triangle intersects the side with which it makes
    the 90 degree angle.




    H (Height) of each triangle = square root of 5^2-3^2
    = 4.05

    Area = 5 (1/2)(4.05)(5.88)
    = 59.54 cm

    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by jasmith56 View Post
    Hello,

    I hope that all is well. I am having difficulty solving two Geometry problems from your course 6572. I am trying to get the correct calculations in order to complete my portfolio. If possible, will you look over my calculations and tell me if I am correct and if I am incorrect could you point out where the mistakes occur? Thanking you in advance for your help.

    Jacob


    PROBLEM 1
    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.


    The volume of any prism is the area of the face multiplied by the height. The height is given, so all you need to find is the area of the base. I started off by drawing a circle. One side has a chord of length 10 feet cut across it. Its given that the chord is 5 feet from the center of the circle. Therefore, you draw a line from the center bisecting the chord, as well as a line from the center to each end of the chord (the radius), and you have two equal isosceles triangles with two sides of length 5. Pythagorean's relation tells us that the third side has a length of 5*sqroot(2) ft. Now that you have the radius, you can find the area of the circle. However, its not a true circle you're concerned with. You want to exclude the area on the outside of the chord. Using trig, you can find that the total angle subtended by the chord is 90°. Therefore, the area that you're concerned with is the circular area for the outside 270° plus the area of the triangles. 270° is (3/4) of the circle, so its area is (3/4) the area of the entire circle. A = (3/4)*π*r² = (3/4)*π*(5*sqroot(2)ft)² = 118ft². The area of each of the smaller triangles is given by (1/2)*b*h. Since there's two triangles, the total area of the two is 2*(1/2)*b*h = b*h = 5ft*5ft = 25ft². The total area of the base is 118ft² + 25ft² = 143ft². Multiplied by the height will give the total volume of the prism, 143ft²*30ft = 4290ft³.

    PROBLEM 2

    Regular pentagon ABCDE is enclosed in a circle. If the radius of the circle is 5 and each side AB = BC = CD = DE = EA = 5.88, find the area of the pentagon.
    (In a 90 degree triangle, using pythagorean theorem c = square root of a2 + b2)

    AO = BO = CO = DO = EO = 5

    Therefore, we have 5 isosceles triangles. The altitude of an
    isosceles triangle intersects the side with which it makes
    the 90 degree angle.




    H (Height) of each triangle = square root of 5^2-3^2
    = 4.05

    Area = 5 (1/2)(4.05)(5.88)
    = 59.54 cm

    you have a little bit of round-off error in your final solutions, but your methods are correct.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cow tied outside a circular silo
    Posted in the Calculus Forum
    Replies: 6
    Last Post: May 6th 2012, 12:44 PM
  2. Reduction of grain by 80%.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 4th 2011, 05:06 AM
  3. farmer land
    Posted in the Algebra Forum
    Replies: 9
    Last Post: September 8th 2010, 02:08 AM
  4. A farmer and a house
    Posted in the Pre-Calculus Forum
    Replies: 10
    Last Post: August 27th 2010, 06:14 PM
  5. Farmer's Story Problem
    Posted in the Algebra Forum
    Replies: 7
    Last Post: May 3rd 2008, 12:23 PM

Search Tags


/mathhelpforum @mathhelpforum