Area Problem #17

quad_area

The red circle has diameter FC.
The regular pentagon has side length 2cm.
Find the area of the quadrilateral FGCH.

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2 thoughts on “Area Problem #17

  1. Nice problem! Thanks for posting.
    Since FC is the diameter of the circle we know that both CGF and CHF are right angles. By symmetry triangle CGF is congruent to triangle CHF. So the area of the quadrilateral is twice the area of triangle CGF = 2 * 1/2 * length of CG * length of GF or just length CG * length of GF.
    So if we can find those two lengths we can calculate the area.
    The third side of triangle CGF, namely FC the diameter of the circle, is equal to the height of the pentagon minus the height of the equilateral triangle. We know that the ratio of the length of a diagonal of a pentagon to one of its side is the golden ratio (1+sqrt(5))/2. So the length CE is twice the golden ratio, or 1+sqrt(5). By Pythagoras we can get the height of this pentagon = sqrt(5+2*sqrt(5)). Splitting the equilateral triangle we can get its height using a 30/60/90 triangle as sqrt(3). So the length of FC = height of pentagon – height of equilateral triangle = sqrt(5+2*sqrt(5)) – sqrt(3).
    The external angle of the pentagon is 2pi/5, so its internal angle is 3pi/5 so angle GCF is 3*pi/10. Consider the isosceles triangle CBD which has two sides of length 2 and one side of length 1+sqrt(5) (twice the golden ratio). Cut the isosceles triangle into two right angled triangles. It’s easy to see that Sin(3pi/10) is (1+sqrt(5))/4. By Pythagoras Cos(3pi/10) is sqrt(1^2-((1+sqrt(5))/4)^2) = sqrt(16/16 – (1+2*sqrt(5)+5)/16) = sqrt(16-6-2sqrt(5))/4 = sqrt(10-2*sqrt(5))/4.
    So we can calculate length of CG = length of FC * cos(GCF)
    = (sqrt(5+2*sqrt(5)) – sqrt(3)) * sqrt(10-2*sqrt(5))/4
    And the length of GF = length of FC * sin(GCF)
    = (sqrt(5+2*sqrt(5)) – sqrt(3)) * (1+sqrt(5))/4
    Area of FGCH = length of CG * length of GF = 0.861052 cm^2 which looks about right since GF is a fraction longer than 1cm and FC is a bit shorter than 1cm.

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