In △ ABC, BC = 14, ab = 15. It is known that the area of △ ABC is 84, and the length of AC is obtained What's inside? I won't draw any pictures,

Crossing point a as the high crossing point of BC is better than that of D
The area of △ ABC is 84, so adxbc / 2 = 84 ∵ BC = 14 ∵ ad = 12
In △ abd, BD & # 178; = AB & # 178; - Ad & # 178; = 225-144 = 81, BD = 9, CD = bc-bd = 14-9 = 5
In △ ADC, AC & # 178; = ad & # 178; + DC & # 178; = 144 = 25 = 169, AC = 13

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