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A foot of a ladder is 6m away from a wall and its top reaches a window 8m above the ground. If the ladder is shifted in such a way that its foot is 8m away

from the wall, to what height does the top reach ?


Let the ladder in initial position be represented by a straight line AB with point A on the wall and B on the ground. Let O be the point where perpendiculars from point A on the wall and from point B on the ground meet the common edge of the wall and ground.
Hence AO = 8, BO = 6, therefore triangle AOB being a right angle , AB, the hypotenuse = √(8² + 6²) = 10 (Please calculate and verify yourself from Pythagoras theorem).

Now let ladder's position be CD in second position with C on the wall and D on the ground., such that OD = 8.  Now triangle COD is a right angle triangle, with hypotenuse CD = 10, one side OD = 8, you can use Pythagoras theorem and find 
CO = √(10² - 8²) = 6.
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  initial distance between wall and ladder is 6m
  ht. of wall at which the ladder touches is 8m
 so by pythagoros theorem,length of ladder=√ 8²+6²
length of ladder =10m
distance of ladder from ground =8m
and we know that lt. of ladder =10m
by pythagoros theorem,
ht. of wall at which the ladder touches is √10²-8²

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