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Topscore Test 1 Math question # 12

Discussion in 'DAT Discussions' started by Awuah29, Apr 17, 2007.

  1. Awuah29

    Awuah29 Christian predent 7+ Year Member

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    Oct 13, 2003
    New Jersey
    Heh guys,

    Shouldn't the solution to # 12 Math Test 1 Topscore be
    16pi-64?

    Ok, Area of square is X^2 radius=4 diameter=8

    so area of rectangle= 8^2 =64

    area of circle= Pi r^2 so pi 4^2 = 16

    I am missing here a something? Anyone , please check it out and let me know:thumbup:
     
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  3. Streetwolf

    Streetwolf Ultra Senior Member Dentist 7+ Year Member

    1,801
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    Oct 25, 2006
    NJ
    I'm assuming the square is outside of the circle. Then it should be 64 - 16pi.
     
  4. Awuah29

    Awuah29 Christian predent 7+ Year Member

    294
    0
    Oct 13, 2003
    New Jersey
    Heh streetwolf,
    what do you mean if square is outside? the square is inside and the solution is 16 pi -32, but why? don't get it:(
     
  5. Benoir

    Benoir 2+ Year Member

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    0
    Apr 7, 2007
    I'm not sure if this is right, and it would help if the entire problem was given.

    Assuming....
    The square is inscribed inside of the circle.

    So...
    The radius line should touch a point on the square and the circle.
    The radius lines going to two adjacent corners form a 90 degree angle.
    a^2 + b^2 = c^2 and c is one of the sides.

    One of the sides of the square is equal to (r^2 + r^2)^0.5, but since you are going to multiply it by itself anyway....
    (r^2 + r^2)^0.5 * (r^2 + r^2)^0.5 = Area of the square = (r^2 + r^2)

    (4^2 + 4^2) = 32
     
  6. Streetwolf

    Streetwolf Ultra Senior Member Dentist 7+ Year Member

    1,801
    5
    Oct 25, 2006
    NJ
    Or since the radius is 4 then the diagonal of the square is 8, so the side is 8/root(2). When you square that to get the area of the square you get 64/2 = 32.
     

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