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A building is shaped like a right square pyramid with base side length \(160\,\text{m}\) and height \(60\,\text{m}\). A spherical ventilation chamber is centered on the pyramid's vertical axis and is tangent to the floor and all four sloping roof faces.
Find the chamber's radius and the height of its center above the floor.
Hints
- Use a vertical cross-section through the apex and the midpoints of opposite base sides.
- In that cross-section, the sphere becomes a circle tangent to all three sides of a triangle.
- Find the equal side lengths of the triangle with the Pythagorean theorem.
- Use the relationship \(A=rs\) for a triangle's area, inradius, and semiperimeter.
Solution
1. Take a vertical cross-section through the apex and the midpoints of two opposite sides of the square base. The cross-section is an isosceles triangle with base \(160\,\text{m}\) and height \(60\,\text{m}\). The sphere appears as the triangle's incircle.
2. Each equal side of the triangle has length \(\sqrt{80^2+60^2}=100\,\text{m}\).
3. The triangle's area is \(\frac12\cdot160\cdot60=4800\,\text{m}^2\), and its semiperimeter is \(\frac{160+100+100}{2}=180\,\text{m}\).
4. For a triangle with inradius \(r\), \(A=rs\). Therefore, \(r=\frac{4800}{180}=\frac{80}{3}\,\text{m}\approx26.67\,\text{m}\).
5. Because the chamber is tangent to the floor, its center is one radius above the floor. Its center is therefore \(\frac{80}{3}\,\text{m}\approx26.67\,\text{m}\) high.
Answer
The radius is \(\frac{80}{3}\,\text{m}\approx26.67\,\text{m}\), and the center is the same distance above the floor.
