A pole 23 m long reaches a window which is 3 \(\sqrt5\) m above the ground on one side of a street. Keeping its foot at the same point, the pole is turned to the other side of the street to reach a window 4 \(\sqrt15\) m high. What is the width (in m) of the street?
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Understanding the Geometry Problem
This problem involves a pole used to reach windows on opposite sides of a street. The key insight is that the pole, the side of the building (height to the window), and the ground form a right-angled triangle on each side of the street. The base of each triangle is the horizontal distance from the point where the pole's foot is placed to the base of the building on that side. The width of the street is the sum of these two horizontal distances.
We can use the Pythagorean theorem to solve this problem. The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. Mathematically, this is expressed as:
\(a^2 + b^2 = c^2\)
Where:
- \(a\) is the length of one leg (e.g., height to the window)
- \(b\) is the length of the other leg (e.g., distance from the pole's foot to the building)
- \(c\) is the length of the hypotenuse (the pole length)
Step-by-Step Solution for Street Width Calculation
Let's break down the problem into two parts, one for each side of the street.
Side 1: Reaching the First Window
On the first side, the pole reaches a window that is \(3 \sqrt{5}\) m above the ground. The pole is 23 m long. Let the distance from the foot of the pole to the building on this side be \(d_1\).
- Height to window (\(a\)): \(3 \sqrt{5}\) m
- Pole length (\(c\)): 23 m
- Distance from pole to building (\(d_1\)): Unknown (\(b\))
Using the Pythagorean theorem:
\((3 \sqrt{5})^2 + d_1^2 = 23^2\)
Calculate the squares:
\((3 \sqrt{5})^2 = 3^2 \times (\sqrt{5})^2 = 9 \times 5 = 45\)
\(23^2 = 529\)
Substitute these values back into the equation:
\(45 + d_1^2 = 529\)
Solve for \(d_1^2\):
\(d_1^2 = 529 - 45\)
\(d_1^2 = 484\)
Now, find \(d_1\) by taking the square root of both sides:
\(d_1 = \sqrt{484}\)
\(d_1 = 22\) m (Since distance must be positive)
Side 2: Reaching the Second Window
On the other side, the pole reaches a window that is \(4 \sqrt{15}\) m high. The pole length remains 23 m. Let the distance from the foot of the pole to the building on this side be \(d_2\).
- Height to window (\(a\)): \(4 \sqrt{15}\) m
- Pole length (\(c\)): 23 m
- Distance from pole to building (\(d_2\)): Unknown (\(b\))
Using the Pythagorean theorem:
\((4 \sqrt{15})^2 + d_2^2 = 23^2\)
Calculate the squares:
\((4 \sqrt{15})^2 = 4^2 \times (\sqrt{15})^2 = 16 \times 15 = 240\)
\(23^2 = 529\)
Substitute these values back into the equation:
\(240 + d_2^2 = 529\)
Solve for \(d_2^2\):
\(d_2^2 = 529 - 240\)
\(d_2^2 = 289\)
Now, find \(d_2\) by taking the square root of both sides:
\(d_2 = \sqrt{289}\)
\(d_2 = 17\) m (Since distance must be positive)
Calculating the Width of the Street
The width of the street is the sum of the distances from the foot of the pole to the buildings on each side (\(d_1 + d_2\)).
Street width = \(d_1 + d_2 = 22 + 17 = 39\) m
Summary of Calculations
| Parameter | Side 1 | Side 2 |
|---|---|---|
| Pole Length (Hypotenuse) | 23 m | 23 m |
| Window Height (Leg 1) | \(3 \sqrt{5}\) m | \(4 \sqrt{15}\) m |
| Height Squared | \((3 \sqrt{5})^2 = 45\) | \((4 \sqrt{15})^2 = 240\) |
| Pole Length Squared | \(23^2 = 529\) | \(23^2 = 529\) |
| Distance Squared (Leg 2) | \(529 - 45 = 484\) | \(529 - 240 = 289\) |
| Distance from Pole (Leg 2) | \(\sqrt{484} = 22\) m | \(\sqrt{289} = 17\) m |
The width of the street is the sum of the distances: \(22 + 17 = 39\) m.
Revision Table: Pole, Heights, and Street Width
| Concept | Description | Application in Problem |
|---|---|---|
| Right-angled Triangle | A triangle with one angle equal to 90 degrees. | Formed by the pole, the building side (height), and the ground. |
| Pythagorean Theorem | \(a^2 + b^2 = c^2\) for sides a, b, and hypotenuse c. | Used to find the horizontal distance (base of the triangle) on each side of the street. |
| Hypotenuse | The longest side of a right-angled triangle, opposite the right angle. | The length of the pole (23 m) serves as the hypotenuse in both triangles. |
| Legs (of a right triangle) | The two sides that form the right angle. | The window height and the horizontal distance from the pole's foot to the building are the legs. |
Additional Information: Real-World Pythagoras
The Pythagorean theorem is a fundamental concept in geometry and has many real-world applications beyond just finding distances in simple scenarios like this pole problem. It's used in construction, navigation, surveying, architecture, and even in computer graphics. Any time you need to calculate distances or check for right angles in a two-dimensional plane, Pythagoras is likely involved. For example, builders use it to ensure corners are square (90 degrees), and navigators use it to calculate the shortest distance between two points.
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