Problem:
In β³PQR,PR=15,QR=20, and PQ=25. Points A and B lie on PQβ, points C and D lie on QRβ, 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 21ββ
15β
20=150. Each of β³PAF, β³QCB, and β³RED shares an angle with β³PQR. Because the area of a triangle with sides a,b, and included angle Ξ³ is 21βaβ
bβ
sinΞ³, it follows that the areas of β³PAF,β³QCB, and β³RED are each 21ββ
5β
5β
ab150β, where a and b are the lengths of the sides of β³PQR adjacent to the shared angle. Thus the sum of the areas of β³PAF,β³QCB, and β³RED is
5β
5β
15β
25150β+5β
5β
25β
20150β+5β
5β
20β
15150β=25(52β+103β+21β)=30
Therefore ABCDEF has area 150β30=120.
The problems on this page are the property of the MAA's American Mathematics Competitions