Problem:
Let ABCD be a trapezoid with AB∥CD,AB=11,BC=5,CD=19, and DA=7. Bisectors of ∠A and ∠D meet at P, and bisectors of ∠B and ∠C meet at Q. What is the area of hexagon ABQCDP ?
Answer Choices:
A. 283
B. 303
C. 323
D. 353
E. 363 Solution:
Let M and N be the midpoints of sides AD and BC. Set ∠BAD=2y and ∠ADC=2x. We have x+y=90∘, from which it follows that ∠APD=90∘. Hence in triangle APD,MP is the median to the hypotenuse AD, so AM=MD=MP and ∠MPA=∠MAP=∠PAB. Thus, MP∥AB. Likewise, QN∥AB. It follows that M,P,Q, and N are collinear, and
PQ=MN−MP−QN=2AB+CD−AD−BC=9
The area of ABQCDP is equal to the sum of the areas of two trapezoids ABQP and CDPQ. Let F be the foot of the perpendicular from A to CD. Then the area of ABQCDP is equal to
2AB+PQ⋅2AF+2CD+PQ⋅2AF=12AF
Let E lie on DC so that AE∥BC. Then AE=BC=5 and DE=CD−CE=CD−AB=8. We have AD2−DF2=AF2=AE2−EF2=AE2−(DE−DF)2, or 49−DF2=25−(8−DF)2. Solving the last equation gives DF=211. Thus AF=253 and the area of ABQCDP is 12AF=303.
OR
As in the first solution, conclude that AE=5 and DE=8. Apply the Law of Cosines to △ADE to obtain
cos(∠AED)=2⋅8⋅582+52−72=21
Therefore ∠AED=60∘, so AF=53/2, and the area of ABQCDP is 303.
