过点A作AD⊥BC交BC为点D
设BD为x,则DC为9-x.
AB²-BD²=AC²-DC²,
即8²-x²=7²-(9-x)²
解得x=16/3.
AD=√8²-(16/3)²
=(8√5)/3
∴S=1/2×9×(8√5)/3
=12√5≈27