A 25 ft long ladder is placed against the wall with
its base 7 ft from the wall. The base of the ladder is drawn out so that the
top comes down by half the distance that the base is drawn out. This distance
is in the range:

(1)(2,7) (2)(5,8) (3)(9,10) (4)(3,7) (5)None of these

Solution follows here:

__Solution:__

Ladder in Original position is designated as A

_{1}B_{1}=> A_{1}B_{1}=25
Ladder in new position is designated as A

_{2}B_{2}=> A_{2}B_{2}=25
Let the base of the ladder is drawn out by ‘x’ => A

_{1}A_{2}=x
Originally the ladder is placed against the wall with
its base 7 ft from the wall => OA

_{1}= 7
OA

_{2}= 7+x
Applying Pythagoras to the right-triangle OA

_{1}B_{1}=> OB_{1}^{2}= A_{1}B_{1}^{2}– OA_{1}^{2}= 25^{2}-7^{2}= 625-49 = 576
=> OB

_{1}= 24
OB

_{2}= 24 - x/2
Applying Pythagoras to the right-triangle OA

_{2}B_{2}=> OA_{2}^{2}+ OB_{2}^{2}= A_{2}B_{2}^{2}
=> (7+x)

^{2}+(24 - x/2)^{2}= 25^{2}=> 4(7+x)^{2}+(48-x)^{2}= 4*25^{2}=> 4*7^{2}+4x^{2}+56x+48^{2}+x^{2}-96x = 50^{2}
=> 5x

^{2}-40x = 50^{2}-48^{2}-196 = (50+48)(50-48)-196 = (98)(2)-196 = 0
=> x

^{2}= 8x => x = 8 (As x could not be zero)**Answer (5)**
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