The height of the two buildings is 34 m and 29 m respectively. The distance between their top is 13 m, then the distance between the two buildings is
The vertical buildings AB and CD are 34 m and 29 m respectively.
Draw DE ⊥ AB.
Then AE=AB–EB but EB=DC
Therefore AE=34−29=5 m
Now, AED is a right angled triangle and right angled at E.
Therefore,
AD2=AE2+ED2
⇒132=52+ED2
⇒169=25+ED2
⇒ED2=169−25
⇒ED=√144
⇒ED=12
Therefore, the distance between the two buildings = 12 m