Latest Miscellaneous Company Exam Aptitude Question SOLUTION: A ladder 25m long reaches a window of building 20m above the ground. Then the distance of the foot of the ladder from the building is? A) 35m b)45m c)30m d)15m

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(#M40019704) MISCELLANEOUS COMPAN Numerical Ability question Distance Keep an EYE

A ladder 25m long reaches a window of building 20m above the ground. Then the distance of the foot of the ladder from the building is?
A) 35m b)45m c)30m d)15m

Asked In Miscellaneous Company Exam Chithra (10 years ago)

Unsolved Read Solution (6)

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Read Solution (6) :

veneetha  (10 years ago)

let it forms a right angled triangle and follow c^2=a^2+b^2
then 25^2=20^2+b^2
b^2=625-400
b=15

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mohan lal garg  (10 years ago)

this question is solve using Pythagoras theorem of right angle tringle
so let y is distance from building
25^2=20^2+y^2
y^2=625-400=225
y=15
d.15 is right ans

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pramod dusad  (10 years ago)

the distance of window from the ground is 20 meter.
and the length of ladder is 25 meter.

so the laddermake the right hand triangle from the wall so from the pythagoras theorem


(the distance of the foot of ladder from the building)^2 =
(length of ladder)^2 -(distance of window from ground)^2


so (the distance of the foot of ladder from the building)^2 = (25)^2 - (20)^2
= 625 - 400
= 225 so the distance of the foot of ladder from the building = 15 meter

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Prashanth  (10 years ago)

((heights)^2)+((base)^2))=((hypotenuse)^2)
((base)^2))=((hypotenuse)^2)-((heights)^2)
((base)^2))=(25^2)-(20^2)
((base)^2))=625-400=225
base =sqrt(225)
base =15m is the distance of the foot of the ladder from the building.

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vijayaragavan  (10 years ago)

25^2=20^2+x^2 625-400=x^2 x=15

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ravikumar  (10 years ago)

it is pithagirus theorem format s0 15m is the ans

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