Answer :
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.
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.
given
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
=10
length of ladder =10m
now
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
=6m
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
=10
length of ladder =10m
now
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
=6m