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?

Question

Allison Allen · Accepted Answer

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

Consider Figs. 1 and 2

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#

Autumn Armstrong · Answer

undefined To find the largest possible area of the inscribed trapezoid, you need to maximize the height of the trapezoid. The height of the trapezoid is the perpendicular distance between the two parallel chords.

Using the properties of circles and chords, you can find the height of the trapezoid using the Pythagorean theorem, since the radius and the perpendicular distance from the center of the circle to the chords form a right triangle.

Let $ r $ be the radius of the circle, $ h $ be the height of the trapezoid, and $ d $ be the perpendicular distance between the chords.

Using the Pythagorean theorem:

$$
\begin{align*}
r^2 &= \left(\frac{{d}}{2}\right)^2 + h^2 \
12^2 &= \left(\frac{{d}}{2}\right)^2 + h^2 \
144 &= \frac{{d^2}}{4} + h^2 \
144 &= \frac{{d^2 + 4h^2}}{4} \
576 &= d^2 + 4h^2 \
4h^2 &= 576 - d^2 \
h^2 &= \frac{{576 - d^2}}{4}
\end{align*}
$$

Given that the lengths of the chords are 8 and 10 units, the distance between them (the length of the trapezoid's top base) is 10 - 8 = 2 units.

The area $ A $ of a trapezoid is given by:

$$
A = \frac{{\text{{sum of bases}}}}{2} \times \text{{height}}
$$

So, the area $ A $ of the trapezoid can be expressed as:

$$
A = \frac{{8 + 10}}{2} \times \sqrt{\frac{{576 - 2^2}}{4}}
$$

$$
A = 9 \times \sqrt{\frac{{576 - 4}}{4}}
$$

$$
A = 9 \times \sqrt{\frac{{572}}{4}}
$$

$$
A = 9 \times \sqrt{143}
$$

$$
A \approx 9 \times 11.958
$$

$$
A \approx 107.62
$$

So, the largest possible area of the inscribed trapezoid is approximately $ 107.62 $ square units.