A 25-foot ladder is placed against a vertical wall of a building, with the bottom of the ladder standing on concrete 7 feet from the base of the building. If the top of the ladder slips down 4 feet, then the bottom of the ladder will slide out how many feet?

A.) 4 feet

B.) 5 feet

C.) 6 feet

D.) 7 feet

E.) 8 feet

A.) 4 feet

B.) 5 feet

C.) 6 feet

D.) 7 feet

E.) 8 feet

Pythagorean Theorem

a² + b² = c² {the sum of the squares of the legs equals the square of the hypotenuse}

**Before the ladder slips:**

A right triangle is formed with the hypotenuse being the ladder, itself, and the base of the building being one of the legs.

a² + 7² = 25² {substituted into Pythagorean Theorem}

a² + 49 = 625 {evaluated exponents}

a² = 576 {subtracted 49 from both sides}

a = 24 {took square root of both sides}

24 is the height of the ladder against building

So, before it slips, the right triangle consists of:

the ladder = 25ft {hypotenuse}

the height of ladder against building = 24 ft {one leg}

the distance of ladder from building = 7 ft {other leg}

After it slips, the right triangle consists of:

the ladder = 25 ft

the height of ladder against building = 20 ft {one leg} {it slipped down 4 ft}

the distance of ladder from building = unknown {other leg}

a² + 20² = 25² {substituted information "after slip" into Pythagorean Theorem}

a² + 400 = 625 {evaluated exponents}

a² = 225 {subtracted 400 from both sides}

a = 15 {took square root of both sides}

After the slip, the ladder is 15 ft away from the building {and someone is probably on the ground!}

Therefore, in answering the original question how many feet will the ladder slide out:

It was originally 7 feet from the base and now it is 15 feet from the base, so it will slide out 8 feet.

**E.) 8 feet**

*- Algebra House*