1st case:

Length of the ladder = 10 foot

Distance from the ground to point the ladder touches = 8 feet

So the distance of the base of the ladder from the wall =\(\sqrt{10^2 - 8^2} = 6\)

2nd case:

Now the distance of the base slips 1 foot away from the wall i.e = 6 + 1 =7

and the length of the ladder = 10 foot

SO the distance from the base to the point the ladder touches can be calculated by Pythagoras theorem = \(\sqrt{10^2 - 7^2} = \sqrt{51}\)

Now \(\sqrt{49} < \sqrt{51} < \sqrt {64}\)

or \(7 < \sqrt{51} < 8\)

so the new value of the base lies between 7 and 8

so the distance the ladder slides down = 8 - (number greater than 7 but less than 8) = less than 1

SO \(QTYA > QTY B\)
_________________

By Pythagoras theorem

Ans B

Please fix the typo

considering the original condition, a 10 ft. ladder resting on a wall with its top 8 ft. above the ground. A right-angle triangle with a hypotenuse of 10 ft. (length of ladder) and height of 8 ft. can be drawn, by applying Pythagoras theorem we can derive the base of the triangle as 6 ft.
After the ladder has shifted by 1 ft. away from the wall (as shown), we can conclude that the top of the ladder has also slid down the wall (lets assume that distance as "d"). In the new triangle, the hypotenuse remains 10 ft., the base becomes (6+1) = 7 ft. and the height will be now (8-d) ft. Again on applying the Pythagoras theorem, we can calculate d=~0.86 ft., which is less than 1.

Hence, column (A) is greater than (B). Right answer is (A)

Thank you

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