Monday, June 15, 2009

Problem 50 - GMAT Geometry


6 comments:

  1. Two things will solve this guy faster: 1) know that a regular hexagon can be divided into six equal equilateral triangles, 2) know the formula used to calculate the area of an equilateral triangle using the side as a variable.

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  2. I am getting 32Pi..
    area of equilateral triangle = 4root3
    =0.5 X x/2 X (root3)/2 X x
    x=4root2
    area of circle = pi x X x
    = 32pi

    what gives?

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  3. area of an equil triangle is [(side^2)*(sq root 3)]/4. so 6*[(side^2)*(sq root 3)]/4 = 24*(sq root 3). Solve for side = 4, which also = radius of circle. pi r^2 = 16 pi.

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  4. Area of 6 EQ triangle = 24 sqrt 3
    Area of 1 Eq Triangle = 4 sqrt 3
    0.5 * (2 x)* sqrt (3) x = 4 sqrt 3
    x = 2
    Area of circle pi (2 x)^2 = 16 pi
    Assumed The length of equ triangle = 2x which equals the radii of the circle

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