点A.B.C.D是直线上顺次四点,且AB;BC;CD=2;3;4,点P.Q分别是线段AB.CD的终点,若PQ=1

问题描述:

点A.B.C.D是直线上顺次四点,且AB;BC;CD=2;3;4,点P.Q分别是线段AB.CD的终点,若PQ=1

点A、B、C、D是直线/上顺次四点,切AB:BC:CD=2:3:4,点P、Q分别是线段AB、CD的中点,若PQ=12cm:
①求线段AB的长; 解:依题意,得
l__l__l______l____l____l2+3+29 AD - 29 × 12 AD = 23 AD
AP BCQD则 AD = 18 ( cm )BC = 39 AB = 6 ( cm )
答:BC的长为6 cm.