Problem:
In △PQR,PR=15,QR=20, and PQ=25. Points A and B lie on PQ, points C and D lie on QR, and points E and F lie on PR, with PA=QB=QC=RD=RE=PF=5. Find the area of hexagon ABCDEF.
Solution:
Triangle PQR is a right triangle with area 21⋅15⋅20=150. Each of △PAF, △QCB, and △RED shares an angle with △PQR. Because the area of a triangle with sides a,b, and included angle γ is 21a⋅b⋅sinγ, it follows that the areas of △PAF,△QCB, and △RED are each 21⋅5⋅5⋅ab150, where a and b are the lengths of the sides of △PQR adjacent to the shared angle. Thus the sum of the areas of △PAF,△QCB, and △RED is
5⋅5⋅15⋅25150+5⋅5⋅25⋅20150+5⋅5⋅20⋅15150=25(52+103+21)=30
Therefore ABCDEF has area 150−30=120.
The problems on this page are the property of the MAA's American Mathematics Competitions