Two parallel chords of a circle with lengths of 8 and 10 serve as bases of a trapezoid inscribed in the circle. If the length of a radius of the circle is 12, what is the largest possible area of such a described inscribed trapezoid?

1 Answer
Jan 23, 2016

#72*sqrt(2)+9*sqrt(119)~=200.002#

Explanation:

Consider Figs. 1 and 2

I created this figure using MS Excel

I created this figure using MS Excel

Schematically, we could insert a parallelogram ABCD in a circle, and on condition that sides AB and CD are chords of the circles, in the way of either figure 1 or figure 2.

The condition that the sides AB and CD must be chords of the circle implies that the inscribed trapezoid must be an isosceles one because

  • the trapezoid's diagonals (#AC# and #CD#) are equal because
  • #A hat B D=B hat A C=B hatD C= A hat C D#
    and the line perpendicular to #AB# and #CD# passing through the center E bisects these chords (this means that #AF=BF# and #CG=DG# and the triangles formed by the intersection of the diagonals with bases in #AB# and #CD# are isosceles).

But since the area of the trapezoid is
#S=(b_1+b_2)/2*h#, where #b_1# stands for base-1, #b_2# for base-2 and #h# for height, and #b_1# is parallel to #b_2#

And since the factor #(b_1+b_2)/2# is equal in the hypotheses of the Figures 1 and 2, what matters is in which hypothesis the trapezoid has a longer height (#h#). In the present case, with chords smaller than the circle's radius, there's no doubt that in the hypothesis of the figure 2 the trapezoid has a longer height and therefore it has a higher area.

According to Figure 2, with #AB=8#, #CD=10# and #r=12#
#triangle_(BEF) -> cos alpha =((AB)/2)/r=(8/2)/12=4/3=1/3#
#->sin alpha = sqrt(1-1/9)=sqrt(8)/3=2sqrt(2)/3#
#->tan alpha=(sin alpha)/cos alpha=(2sqrt(2)/cancel(3))/(1/cancel(3))=2sqrt(2)#
#tan alpha = x/((AB)/2)# => #x=8/cancel(2)*cancel(2)sqrt(2)# => #x=8sqrt(2)#

#triangle_(ECG) -> cos beta=((CD)/2)/r=(10/2)/12=5/12#
#-> sin beta =sqrt (1-25/144)=sqrt(119)/12#
#-> tan beta=(sin beta)/cos beta=(sqrt(119))/cancel(12))/(5/cancel(12))=sqrt(119)/5#
#tan beta=y/((CD)/2)# => #y=10/2*sqrt(119)/5# => #y=sqrt(119)#

Then
#h=x+y#
#h=8sqrt(2)+sqrt(119)#

#S=(b_1+b_2)/2*h=(8+10)/2(8sqrt(2)+sqrt(119))=72sqrt(2)+9sqrt(119)~=200.002#