Question

answer the following questions.
write your answers in terms of π.
give exact answers (not decimal approximations).
(a) the square pyramid and the cone below both have a height of 11 mm.
the pyramid has a base length of ( 2sqrt{pi} ) mm, and the base of the cone has
a radius of 2 mm. a plane parallel to the bases crosses both solids at
4 mm from the top. the resulting cross sections (shaded) have the same
area.
for each solid, the top portion (which has the highlighted cross section as
its base) is similar to the entire solid.
use this fact to find the areas of the cross sections.

(b) the height of the pyramid is 11 mm. find the volume of the pyramid.

(c) the pyramid and the cone in part (a) have the same height. their cross
sections at every level have the same area. following from cavalieris
principle, the pyramid and the cone must have the same volume.
using these facts, choose the equation that gives the volume in terms of r,
the radius of the base of the cone, and h, the height of the cone.
( \bigcirc ) volume ( =pi r^{2} h )
( \bigcirc ) volume ( =\frac{pi r^{2} h}{3} )
( \bigcirc ) volume ( =2 pi r h )
( \bigcirc ) volume ( =\frac{pi r h}{3} )

Explanation:

Part (a)

For the square pyramid:

• The ratio of the height of the top - similar pyramid to the height of the original pyramid is \(k=\frac{4}{11}\).
• The area of the base of the original square pyramid \(B=(2\sqrt{\pi})^2 = 4\pi\).
• If two similar solids have a scale factor \(k\), the ratio of their areas is \(k^{2}\). Let \(A_{1}\) be the area of the cross - section (area of the base of the top - similar pyramid). Then \(A_{1}=B\times k^{2}\).
• Substitute \(B = 4\pi\) and \(k=\frac{4}{11}\) into the formula: \(A_{1}=4\pi\times(\frac{4}{11})^{2}=\frac{16\pi}{121}\).

For the cone:

• The ratio of the height of the top - similar cone to the height of the original cone is \(k = \frac{4}{11}\).
• The area of the base of the original cone \(B=\pi r^{2}\), where \(r = 2\), so \(B = 4\pi\).
• Using the formula for the area of similar solids (\(A=B\times k^{2}\)), substitute \(B = 4\pi\) and \(k=\frac{4}{11}\) into the formula. Then \(A_{2}=4\pi\times(\frac{4}{11})^{2}=\frac{16\pi}{121}\).

Part (b)

• The volume formula for a pyramid is \(V=\frac{1}{3}Bh\).
• The base area \(B=(2\sqrt{\pi})^2=4\pi\) and the height \(h = 11\).
• Substitute \(B = 4\pi\) and \(h = 11\) into the formula: \(V=\frac{1}{3}\times4\pi\times11=\frac{44\pi}{3}\).

Part (c)

• The volume of a pyramid is \(V=\frac{1}{3}Bh\). For a cone, since the cross - sectional areas at every level are the same as the cross - sectional areas of a pyramid with the same height (by Cavalieri's Principle) and the base area of a cone \(B=\pi r^{2}\).
• The volume of a cone \(V=\frac{1}{3}Bh=\frac{1}{3}\pi r^{2}h\).

Answer:

• (a) The area of the cross - section of the square pyramid is \(\frac{16\pi}{121}\text{ mm}^2\), and the area of the cross - section of the cone is \(\frac{16\pi}{121}\text{ mm}^2\).
• (b) The volume of the pyramid is \(\frac{44\pi}{3}\text{ mm}^3\).
• (c) The correct equation is \(Volume=\frac{\pi r^{2}h}{3}\).