A 25 feet long ladder is placed against the wall with its base 7 feet

Given base of triangle is \(7\) and hypotenuse \(25\).

We can find the third side of the triangle. Let the third side be \(a\).

\(25^2 = a^2 + 7 ^2\)

\(625 = a^2 + 49\)

\(a^2 = 625 - 49 = 576\)

\(a = 24\)

Given the base of the ladder is drawn out so that the top comes down by half the distance that the base is drawn out.

Let the top distance drawn out be \(x\) ; then the base distance drawn out is \(2x\).

Hypotenuse is the same \(= 25\)

New base \(= 7 + 2x\)

New value of other side \(= 24 - x\)

\((7 + 2x)^2 + (24-x)^2 = 25^2\)

\(4x^2 + 28x + 49 + x^2 - 48x + 576 = 625\)

\(5x^2 - 20x + 625 = 625\)

\(5x^2 - 20x = 625 - 625 => 5x^2- 20x = 0\)

\(x^2 - 4x = 0\)

\(x(x-4) = 0\)

\(x = 0\) or \(x = 4\)

\(x\) cannot be \(0\) because its given the ladder is drawn out hence has moved some distance. Therefore \(x = 4\)

\(x = 4 ; 2x = 8\)

Sides of new triangle will be;

\(7 + 2x = 7 + 8 = 15\)

\(24 - x = 24 - 4 = 20\)

Answer E