Problem: Located inside equilateral triangle ABC is point P such that PA=6,PB=8, and PC=10. To the nearest integer the area of triangle ABC is:
Answer Choices:
A. 159
B. 131
C. 95
D. 79
E. 50
Solution:
Rotate CP through 60∘ to position CP′. Draw BP′. This is equivalent to rotating △CAP into position CBP′. In a similar manner rotate △ABP into position ACP′′ and △BCP into position BAP"'.
On the one hand hexagon AP′′BP′CP′′ consists of △ABC and △CBP′,ACP ", BAP′′′. Letting K represent area, we have, from the congruence relations, K(ABC)=K(CBP′)+K(ACP′′)+K(BAP′′′)∴K(ABC)=21K (hexagon).
On the other hand the hexagon consists of three quadrilaterals PCP′B, PBP′′′A, and PAP′′C, each of which consists of the 6−8−10 right triangle and an equilateral triangle. ∴K (hexagon) =3(21⋅6⋅8)+41⋅1023+41⋅823+41⋅623=72+503∴K(ABC)=36+253≈79.
OR
Applying the Law of Cosines to △APB wherein ∠APB=150∘, we have s2=62+82−2⋅6⋅8cos150∘=100+483. Therefore, K(ABC)=4s23=253+36≈79.
.jpg)