Answer:
16 ft.
Explanation:
Let's denote the height of the wall as H and the distance between the foot of the ladder and the base of the wall as D.
According to the problem, the ladder is leaning against the wall and reaches the top of the wall, forming a right triangle. The ladder, the wall, and the ground form a right triangle.
Using the Pythagorean theorem, we can relate the length of the ladder (20 feet), the height of the wall (H), and the distance between the foot of the ladder and the base of the wall (D):
20^2 = D^2 + H^2
We also know that the height of the wall is 4 feet more than the distance (H = D + 4).
Substituting H = D + 4 into the equation above, we have:
20^2 = D^2 + (D + 4)^2
400 = D^2 + D^2 + 8D + 16
Combining like terms, we get:
2D^2 + 8D + 16 - 400 = 0
2D^2 + 8D - 384 = 0
Dividing the equation by 2 to simplify it, we have:
D^2 + 4D - 192 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
D = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 4, and c = -192. Plugging in these values, we have:
D = (-4 ± √(4^2 - 4(1)(-192))) / (2(1))
D = (-4 ± √(16 + 768)) / 2
D = (-4 ± √784) / 2
D = (-4 ± 28) / 2
Simplifying further, we get two possible solutions for D:
D1 = (-4 + 28) / 2 = 24 / 2 = 12 feet
D2 = (-4 - 28) / 2 = -32 / 2 = -16 feet (not a valid solution since distance can't be negative)
Therefore, the distance between the foot of the ladder and the base of the wall is 12 feet.
Since the height of the wall (H) is 4 feet more than the distance, we have:
H = D + 4 = 12 + 4 = 16 feet
Therefore, the height of the wall is 16 feet.