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Math Help - Geometry Problems

  1. #1
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    Geometry Problems

    I have two questions that I've been stuck on for a very long time and need any help I can get!

    1. A certain tower has a mass of 7.2 million kilograms and a height of 324 meters. Its base is square with a side length of 125 meters. The steel used to make the tower occupies a volume of 930 cubic meters. Air has a density of 1.225 kg per cubic meter. Suppose the tower was contained in a cylinder. Find the mass of the air in the cylinder. (Assume the dimensions of the cylinder minimize the volume. Round your answer to three decimal places.)

    I know the answer should be __ million kg.

    2. Aleko's Pizza has delivered a beautiful 18 inch diameter pie to Lee's dorm room. The pie is sliced into 8 equal sized pieces, but Lee is such a non-conformist he cuts off an edge as pictured. John then takes one of the remaining triangular slices.

    Geometry Problems-eeaf8b930633dcdf1d3a3857807435.gif

    Who has more pizza and by how much? (Round your answer to two decimal places.)

    I know Lee should have more I just need help finding how much.


    Again, any help with how to solve these is appreciated! Thank you very much!
    Last edited by Phresh; September 30th 2012 at 12:27 AM.
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  2. #2
    MHF Contributor
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    Re: Geometry Problems

    Hey Phresh.

    First off, do you know the formula for the area of a triangle (with ones that include right angles) and also the formula for a portion of a circle with a given radius and a known arc?
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  3. #3
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    Re: Geometry Problems

    Quote Originally Posted by chiro View Post
    Hey Phresh.

    First off, do you know the formula for the area of a triangle (with ones that include right angles) and also the formula for a portion of a circle with a given radius and a known arc?
    I've been applying the formula for right triangles to both problems but I just can't get the right answer to any of them. And no I don't for the circle with a given radius and known arc.
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  4. #4
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    Re: Geometry Problems

    The area for a portion of a circle is (1/2)*r^2*theta where theta is the angular interval measure. So for example a quarter of a circle has theta = pi/2 and a full revolution is 2*pi which if you plug in gives the area of a circle to be 1/2*2*pi*r^2 = pi*r^2.

    So if you have triangles pi/4 radians each then the total arc corresponds to a pi/2 angular measure.
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