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?
39
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.
Similar Questions
A person standing at a distance looks at a building having a height of 1000 metres. The angle between the top of the building and the ground is 30°. At what approximate distance (in metres) is the person standing away from the building.
Two ships are on the opposite of a light house such that all three of them are collinear. The angles of depression of the two ships from the top of the light house are 30° and 60°. If the ships are 230√3 m apart, then find the height of the light house (in m).
The angle of elevation of the top of a building at a distance of 70 m from its foot on a horizontal plane is found to be 60°. Find the height of the building.
From the top of an upright pole 17.75 m high, the angle of elevation of the top of an upright tower was 60⁰. If the tower was 57.75 m tall, how far away (in m) from the foot of the pole was the foot of the tower?
From the top of an upright pole 24√3 feet high, the angle of elevation of the top of an upright tower was 60°. If the foot of the pole was 60 feet away from the foot of the tower, what tall (in feet) was the tower?
The angle of elevation of the top of a tower from the top of a building whose height is 680 m is 45° and the angle of elevation of the top of same tower from the foot of the same building is 60°. What is the height (in m) of the tower?
- The angle of elevation of the top of an unfinished tower at a point distant 78 m from its base is 30°. How much higher must the tower be raised (in m) so that the angle of elevation of the top of the finished tower at the same point will be 60º?
A kite is attached to a string. Find the length of the string (in m) when the height of the kite is 90 m and the string makes an angle of 30° with the ground.
A vertical pole and a vertical tower are on the same level ground in such a way that, from the top of the pole, the angle of elevation of the top of the tower is 60° and the angle of depression of the bottom of the tower is 30°. If the height of the pole is 24 m, then find the height of the tower (in m).
The angle of elevation of the top of a tall building from the points M and N at the distances of 72 m and 128 m, respectilvely, from the base of the building and in the same straight line with it, are complementary. The height of the building (in m) is:
Important Questions from Heights and Distances
Two ships are sailing in the sea on the two sides of a lighthouse. The angles of elevation of the top of the lighthouse as observed from the ships are 45 ° and 60° respectively. If the lighthouse is 81 m high, then the distance between two ships is:
The horizontal distance between two towers is 40√3 m. The angle of depression of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 130 m, find the height of the first tower.
The angle of elevation of a ladder leaning against a house is 60° and the foot of the ladder is 6.5 metres from the house. The length of the ladder is
A kite is flying at a height of 50 m. If the length of the string is 100 m then the inclination of the string to the horizontal ground in degree measures is:
A. 90
B. 45
C. 60
D. 30
Two poles of the height 15 m and 20 m stand vertically upright on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
A. 11 m
B. 12 m
C. 13 m
D. 14 m
