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प्रश्न
A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
उत्तर
Let AC be the ladder of length 5 m and BC = 4 m be the height of the wall, which ladder is placed.
If the foot of the ladder is moved 1.6 m towards the wall i.e, AD = 1.6 m,
Then the ladder is slide upward i.e., CE = x m.
In right angled ∆ABC,
AC2 = AB2 + BC2 ...[By pythagoras theorem]
⇒ (5)2 = (AB)2 + (4)2
⇒ AB2 = 25 – 16 = 9
⇒ AB = 3 m
Now, DB = AB – AD
= 3 – 1.6
= 1.4 m
In right angled ∆EBD,
ED2 = EB2 + BD2 ...[By pythagoras theorem]
⇒ (5)2 = (EB)2 + (14)2 ...[BD = 1.4]
⇒ 25 = (EB)2 + 1.96
⇒ (EB)2 = 25 – 1.96 = 23.04
⇒ EB = `sqrt(23.04)` = 4.8
Now, EC = EB – BC
= 4.8 – 4
= 0.8
Hence, the top of the ladder would slide upwards on the wall at distance is 0.8 m.
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