Problem: Sides AB,BC,CD and DA, respectively, of convex quadrilateral ABCD are extended past B,C,D and A to points Bβ²,Cβ²,Dβ² and Aβ². Also, AB=BBβ²=6, BC=CCβ²=7, CD=DDβ²=8 and DA=AAβ²=9; and the area of ABCD is 10. The area of Aβ²Bβ²Cβ²Dβ² is
Answer Choices:
A. 20
B. 40
C. 45
D. 50
E. 60
Solution:
Since the length of base AAβ² of ΞAAβ²Bβ² is the same as the length of base AD of β³ABD, alid the corresponding alutude of β³AAβ²Bβ² has twice the length of the corresponding altitude of β³ABD,
Area β³AAβ²Bβ²=2 Area β³ADB.
(Alternately, we could let ΞΈ be the measure of ADAB, and observe
Area ΞAAβ²Bβ²β=21β(AD)(2AB)sin(180ββ
0)=221β(AD)(AB)sin=2 Area β³ABD.)β
Similarly
β Area β³BBβ²Cβ²=2 AΔ±ra β³BAC Area β³CCβ²Dβ²=2 Area β³CBD Area β³DDβ²Aβ²=2 Area β³DCA.β
Therefore
β Area Aβ²Bβ²Cβ²Dβ²=( Area β³AAβ²Bβ²+ Area β³BBβ²Cβ²)+( Area ΞCCβ²D)β²+ Area ΞDDβ²Aβ²)+ Arca ABCD=2( Area β³ABD+ Area β³BAC)+2 (Area Ξ(BD)+ Area ΞDCA)+ Arca ABC ( )=5 Area ABC=50. β