## Friday, 25 November 2011

### Geometry -18 (XAT-2011) (2 Marks)

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 A1B1 => A1B1 =25
Ladder in new position is designated as A2B2       => A2B2 =25
Let the base of the ladder is drawn out by ‘x’       => A1A2 =x
Originally the ladder is placed against the wall with its base 7 ft from the wall => OA1 = 7
OA2 = 7+x
Applying Pythagoras to the right-triangle OA1B1 => OB12 = A1B12 – OA12 = 252-72 = 625-49 = 576
=> OB1 = 24
OB2 = 24 - x/2
Applying Pythagoras to the right-triangle OA2B2 => OA22 + OB22 = A2B22
=> (7+x)2+(24 - x/2)2 = 252 => 4(7+x)2+(48-x)2 = 4*252 => 4*72+4x2+56x+482+x2-96x = 502
=> 5x2-40x = 502-482-196 = (50+48)(50-48)-196 = (98)(2)-196 = 0
=> x2 = 8x => x = 8 (As x could not be zero)