如图,人的身高CG=EH=1.7m,路灯高为AB,第一次影子长为CD=1.3m,第二次影子长为EF=1.8m,2s内人前进的距离CE=1m/s×2s=2m,根据题意得:
(1)△GCD∽△ABD,∴
CG |
AB |
CD |
BC+CD |
1.7m |
AB |
1.3m |
BC+1.3m |
(2)△HEF∽△ABF,∴
HE |
AB |
EF |
BC+CE+EF |
1.7m |
AB |
1.8m |
BC+2m+1.8m |
(1)(2)两式联立解得:BC=5.2m,AB=8.5m.
故答案为:8.5.
CG |
AB |
CD |
BC+CD |
1.7m |
AB |
1.3m |
BC+1.3m |
HE |
AB |
EF |
BC+CE+EF |
1.7m |
AB |
1.8m |
BC+2m+1.8m |