Problem Solving

18. A ladder 25 feet long is leaning against a wall that is perpendicular to level ground. The bottom of the ladder is 7 feet from the base of the wall. If the top of the ladder slips down 4 feet, how many feet will the bottom of the ladder slip?
(A) 4 (B) 5 (C) 8
(D) 9 (E) 15

Please help! OA is C

Thanks.

right triangle with hipotenuse=25 and short leg = 7,
hence the the distance to the point of contact of ladder to wall X is
sqrt(25^2-7^2) because x^2+7^2=25^2
x= sqrt(625-49) = 24

if the ladder slipps 4 feet x=20 (distance of ground to the point of contact with wall)
so y^2+20^2=25^2 y=sqrt(25^2-20^2)=sqrt(625-400) = sqrt(225) = 15.

but wait, what is asked is by how much did the ladder slip, so 15-7=8.
C is the answer