Let's set up the problem with two different right triangle configurations created by the ladder as it shifts. Let:

• $x$ be the length of the ladder (hypotenuse in both cases).
• In the first case, when the ladder is shifted so that the bottom is 14 units closer, let the height of the ladder on the wall be $y$.
• In the second case, when the ladder is shifted 21 units further away, let the height of the ladder be $y - 27$.

We have two equations based on the Pythagorean theorem from each configuration:

Equation 1

When the bottom of the ladder is 14 units closer: $x^2 = (14)^2 + (y + 8)^2$ Simplifying: $x^2 = 196 + (y + 8)^2$

Equation 2

When the bottom of the ladder is 21 units further away: $x^2 = (21)^2 + (y - 27)^2$ Simplifying: $x^2 = 441 + (y - 27)^2$

Now we have two equations:

1. $x^2 = 196 + (y + 8)^2$
2. $x^2 = 441 + (y - 27)^2$

Since both expressions are equal to $x^2$, we can set them equal to each other:

$196 + (y + 8)^2 = 441 + (y - 27)^2$

Expanding and Simplifying

Expanding each squared term: $196 + y^2 + 16y + 64 = 441 + y^2 - 54y + 729$

Combine like terms: $260 + y^2 + 16y = 1170 + y^2 - 54y$

Cancel $y^2$ from both sides: $260 + 16y = 1170 - 54y$

Bring terms involving $y$ to one side: $70y = 910$

Solve for $y$: $y = 13$

Substitute $y$ to Find $x$

Now, substitute $y = 13$ into one of the original equations to find $x$. Using Equation 1:

$x^2 = 196 + (13 + 8)^2$ $x^2 = 196 + 21^2$ $x^2 = 196 + 441$ $x^2 = 637$ $x = \sqrt{637} \approx 25.24$

Conclusion

The length of the ladder is approximately 25.24 units.

Would you like further clarification on any part of this solution?

Here are five related questions that expand on this information:

1. What would happen if the ladder was moved an additional distance from the wall?
2. How would the Pythagorean theorem apply if the wall or ground was not perpendicular?
3. Could we solve this using trigonometric identities for angles in each configuration?
4. What if we knew the initial distance of the ladder from the wall before it was shifted?
5. How would the solution change if the wall and ground formed an acute angle?

Tip: When solving problems with shifting distances, set up equations based on each scenario to find common variables using substitution or elimination.