题目在Rt△ABCRt\triangle ABCRt△ABC中,∠ACB=90∘\angle ACB=90^{\circ}∠ACB=90∘,AC=9AC=9AC=9,BC=12BC=12BC=12,CCC点到ABABAB的距离是( )A.365\frac{36}{5}536B.1225\frac{12}{25}2512C.94\frac{9}{4}49D.34\frac{3}{4}43知识点:点到直线的距离章节:未标注