Question 213809: 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.
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! A rough sketch of the cross section of the silo might help.
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It is basically a circle with a chord of 10 ft. Perpendicular to this chord and thru the center of the circle is a line ,5 ft long. Connect another line from the center to each of the ends of the chord.
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This makes a circle with two triangles inside.
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The problem is to find the area of the cross section of the silo, that holds grain. This is the circle, less the sector of the circle, plus the area of the two triangles.
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To find the radius of the silo, note that the triangle base is 10 ft total, but 5 ft for each of the two triangles. With the other leg also 5 ft, this makes an isosocles triangle with 2 legs both 5Ft. Pythagorous Theorem , c^2 = a^2 + b^2, finds the radius to be
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R = sqrt{ 5^2 +5^2} = sqrt 50 = 7.071 ft.
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Note also that the central angle of the large triangle is 90 degrees,
(two 45 's).
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Computing the cross sectional area of the usable silo,
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A= pi*r^2 - (90/360)*pi*r^2 + (1/2) bh
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A= pi*(7.071)^2 - (1/4)*pi*(7.071)^2 + (1/2)*10*5 = 142.81 sq ft
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With the height of the silo, 30 ft,
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Volume = B*h = 142.81 * 30 = 4284.29 cu ft
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