Let △ABC have side lengths AB=13, BC=14, and CA=15. Triangle A′B′C′ is obtained by rotating △ABC about its circumcenter so that A′C′ is perpendicular to BC, with A′ and B not on the same side of line B′C′. Find the integer closest to the area of hexagon AA′CC′BB′.
By Shoelace Theorem[AA′CC′BB′]=215(893)+14(−827)+0(−4043)−12(881)−(893)14−(40201)5=218465−4189−10129−2243−4651−8201=218264−4840−10129−2243=2133−210+(10−129−1215)=21−177−5672=215−885−672=101557=155.7⟹Ans=156
