题目如图,若△ABC\triangle ABC△ABC和△DEF\triangle DEF△DEF的面积分别为S1S_{1}S1、S2S_{2}S2,则( )A.S1=12S2S_{1}=\frac{1}{2}S_{2}S1=21S2B.S1=72S2S_{1}=\frac{7}{2}S_{2}S1=27S2C.S1=S2S_{1}=S_{2}S1=S2D.S1=85S2S_{1}=\frac{8}{5}S_{2}S1=58S2知识点:解直角三角形章节:未标注