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)

Answer (5)