Problem: Sides AB,BC,CD and DA, respectively, of convex quadrilateral ABCD are extended past B,C,D and A to points B′,C′,D′ and A′. Also, AB=BB′=6, BC=CC′=7, CD=DD′=8 and DA=AA′=9; and the area of ABCD is 10. The area of A′B′C′D′ is
Answer Choices:
A. 20
B. 40
C. 45
D. 50
E. 60
Solution:
Since the length of base AA′ of ΔAA′B′ is the same as the length of base AD of △ABD, alid the corresponding alutude of △AA′B′ has twice the length of the corresponding altitude of △ABD,
Area △AA′B′=2 Area △ADB.
(Alternately, we could let θ be the measure of ADAB, and observe
Area ΔAA′B′=21(AD)(2AB)sin(180∘⋅0)=221(AD)(AB)sin=2 Area △ABD.)
Similarly
Area △BB′C′=2 Aıra △BAC Area △CC′D′=2 Area △CBD Area △DD′A′=2 Area △DCA.
Therefore
Area A′B′C′D′=( Area △AA′B′+ Area △BB′C′)+( Area ΔCC′D)′+ Area ΔDD′A′)+ Arca ABCD=2( Area △ABD+ Area △BAC)+2 (Area Δ(BD)+ Area ΔDCA)+ Arca ABC ( )=5 Area ABC=50.