A15m long ladder is placed against the wall to reach a window 12m high.Find the distance of the foot of the ladder from the wall
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let AC be the ladder =15m
let AB be the height = 12 m
let the distance between foot of ladder from the wal be BC
by Pythagoras
AC^2= AB^2 + BC^2
15^2= 12^2 +BC^2
Therefore
225-144= BC^2
so, BC^2 = 81
therefor, BC=9 m
let AB be the height = 12 m
let the distance between foot of ladder from the wal be BC
by Pythagoras
AC^2= AB^2 + BC^2
15^2= 12^2 +BC^2
Therefore
225-144= BC^2
so, BC^2 = 81
therefor, BC=9 m
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