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A ladder is leaning against a building. The distance from the bottom of the ladder to the building is 9 ft less than the length...Asked by Ous
A ladder is leaning against a building. The distance from the bottom of the ladder to the building is 10 ft shorter than the length of the ladder. How high up the side of the building is the top of the ladder if that distance is 5 ft less than the length of the ladder?
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Answered by
GPT-4o mini
Let's denote the length of the ladder as \( L \) ft. According to the problem, we have two conditions based on the description:
1. The distance from the bottom of the ladder to the building is 10 ft shorter than the length of the ladder:
\[
d = L - 10
\]
where \( d \) is the distance from the bottom of the ladder to the building.
2. The height the ladder reaches up the building is 5 ft less than the length of the ladder:
\[
h = L - 5
\]
where \( h \) is the height the ladder reaches.
Since we have a right triangle formed by the ladder, the distance from the bottom of the ladder to the building, and the height up the building, we can apply the Pythagorean theorem:
\[
L^2 = d^2 + h^2
\]
Substituting the expressions for \( d \) and \( h \):
\[
L^2 = (L - 10)^2 + (L - 5)^2
\]
Now, let's expand the expressions on the right side:
- Expanding \( (L - 10)^2 \):
\[
(L - 10)^2 = L^2 - 20L + 100
\]
- Expanding \( (L - 5)^2 \):
\[
(L - 5)^2 = L^2 - 10L + 25
\]
Now, we can combine these results:
\[
L^2 = (L^2 - 20L + 100) + (L^2 - 10L + 25)
\]
\[
L^2 = 2L^2 - 30L + 125
\]
Next, let's rearrange the equation:
\[
L^2 - 2L^2 + 30L - 125 = 0
\]
\[
-L^2 + 30L - 125 = 0
\]
Multiplying through by -1 gives:
\[
L^2 - 30L + 125 = 0
\]
Now, let's solve this quadratic equation using the quadratic formula:
\[
L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1, b = -30, c = 125 \):
\[
L = \frac{30 \pm \sqrt{(-30)^2 - 4 \cdot 1 \cdot 125}}{2 \cdot 1}
\]
\[
L = \frac{30 \pm \sqrt{900 - 500}}{2}
\]
\[
L = \frac{30 \pm \sqrt{400}}{2}
\]
\[
L = \frac{30 \pm 20}{2}
\]
We have two potential solutions:
1. \( L = \frac{50}{2} = 25 \)
2. \( L = \frac{10}{2} = 5 \) (not valid since the distance and height can't be negative)
Thus, the length of the ladder is \( L = 25 \) ft.
Now, we can find the height \( h \):
\[
h = L - 5 = 25 - 5 = 20 \text{ ft}
\]
Thus, the height up the side of the building is \( \boxed{20} \) feet.
1. The distance from the bottom of the ladder to the building is 10 ft shorter than the length of the ladder:
\[
d = L - 10
\]
where \( d \) is the distance from the bottom of the ladder to the building.
2. The height the ladder reaches up the building is 5 ft less than the length of the ladder:
\[
h = L - 5
\]
where \( h \) is the height the ladder reaches.
Since we have a right triangle formed by the ladder, the distance from the bottom of the ladder to the building, and the height up the building, we can apply the Pythagorean theorem:
\[
L^2 = d^2 + h^2
\]
Substituting the expressions for \( d \) and \( h \):
\[
L^2 = (L - 10)^2 + (L - 5)^2
\]
Now, let's expand the expressions on the right side:
- Expanding \( (L - 10)^2 \):
\[
(L - 10)^2 = L^2 - 20L + 100
\]
- Expanding \( (L - 5)^2 \):
\[
(L - 5)^2 = L^2 - 10L + 25
\]
Now, we can combine these results:
\[
L^2 = (L^2 - 20L + 100) + (L^2 - 10L + 25)
\]
\[
L^2 = 2L^2 - 30L + 125
\]
Next, let's rearrange the equation:
\[
L^2 - 2L^2 + 30L - 125 = 0
\]
\[
-L^2 + 30L - 125 = 0
\]
Multiplying through by -1 gives:
\[
L^2 - 30L + 125 = 0
\]
Now, let's solve this quadratic equation using the quadratic formula:
\[
L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1, b = -30, c = 125 \):
\[
L = \frac{30 \pm \sqrt{(-30)^2 - 4 \cdot 1 \cdot 125}}{2 \cdot 1}
\]
\[
L = \frac{30 \pm \sqrt{900 - 500}}{2}
\]
\[
L = \frac{30 \pm \sqrt{400}}{2}
\]
\[
L = \frac{30 \pm 20}{2}
\]
We have two potential solutions:
1. \( L = \frac{50}{2} = 25 \)
2. \( L = \frac{10}{2} = 5 \) (not valid since the distance and height can't be negative)
Thus, the length of the ladder is \( L = 25 \) ft.
Now, we can find the height \( h \):
\[
h = L - 5 = 25 - 5 = 20 \text{ ft}
\]
Thus, the height up the side of the building is \( \boxed{20} \) feet.
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