



Solutions for Session 8, Part B
See solutions for Problems: B1  B2  B3  B4  B5  B6  B7  B8  B9  B10

Problem B1  
For the solids in parts (a) and (b), it does not matter which face is the base. In the third solid, it does matter, because the crosssection method only works with solids that have two parallel congruent bases and where the cross section is congruent to the base. The only face that can be a base here is a pentagon (made from the rectangle and triangle).
<< back to Problem B1




Problem B2  
a.  A horizontal cross section has area 9 • 5 = 45 square units, and the height is 2. So the volume is 90 cubic units. 
b.  A horizontal cross section has area 8 • 11 = 88 square units, and the height is 15. So the volume is 1,320 cubic units. 
c.  It is 1,980 cubic units. Find the area of the base (pentagon) by first finding the area of the rectangle (8 • 12 = 96) plus the area of the triangle (6 • 12 • 1/2 = 36) so the area of the base is 96 + 36 = 132 square units. Multiply the crosssection area (same as the base area; i.e., 132) by height (15) and get 1,980 cubic units. 
<< back to Problem B2





Problem B3  
We can approximate the volume by selecting several cross sections at random, determining their areas, averaging the measures, and multiplying the average area by the height. Using the described method, find the area of each cross section and average them. The areas are 100, 196, 324, 625, and 400. The average is 329, so a good guess at the volume is 329 • 100, or, using an approximate value for , we get V 329 • 3.14 • 100 103,306 cubic units.
<< back to Problem B3







Problem B6  
The volume of a cylinder is three times the volume of a cone with equal height and radius. The volume of a sphere is two times the volume of a cone with equal height and radius.
So the ratio of volumes is 3:1:2. In other words, it takes three times as much rice to fill the cylinder as it does to fill the cone, and twice as much to fill the sphere as it does to fill the cone.
<< back to Problem B6
<< back to Problem B6 noninteractive version





Problem B7  
a.  The area of the base is r^{2}. The formula for the volume of a cylinder is V = r^{2}h. Since, in our case, h is equal to 2r, we have V = 2r^{3}. 
b.  Based on the ratio observations in an earlier problem, the formula for the volume of a cone is V = (1/3) r^{2} h. Since in this particular case h = 2r, we have V = (2/3)r^{3}. 
c.  Based on the ratios observed earlier, the volume of a sphere is twothirds of the volume of a cylinder, so the formula is V = (2/3) r^{2}h. Since the height of our sphere is twice the radius, we have V = (2/3) r^{2} (2r), or V = (4/3) r^{3}. 
<< back to Problem B7
<< back to Problem B7 noninteractive version





Problem B8  


Pair 

Solids 

What's the same? 

How are the volumes related? 


1 

A, B 



Volume B is onethird of A. 

2 
C, D 

Volume D is onethird of C. 

3 
E, F 

Volume F is onethird of E. 

4 
G, H 
Same base, height of H is twice the height of G 

Volume H is twothirds of G. 

5 
I, J 
Same height, base J is half the area of base I 

Volume J is onesixth of I. 

6 
I, K 

Volume K is onethird of I. 



The relationship between a prism and a pyramid parallels the relationship you discovered between a cone and cylinder: The volume of a pyramid is onethird the volume of a prism with the samesized base and the same height. If, however, the heights of the two solids are not the same, as with Solids G and H, or if the bases are not identical, as with Solids I and J, then the relationships differ.
<< back to Problem B8
<< back to Problem B8 noninteractive version





Problem B9  
If a solid comes to a point (a cone or a pyramid), its volume is onethird of the equivalent complete solid (cylinder or prism). Notice that the base and the height of the two corresponding solids must be the same in order for the relationship to hold.
<< back to Problem B9
<< back to Problem B9 noninteractive version





Problem B10  
The volume of a pyramid is V = (1/3)Bh, where B is the area of the base. This is related to a cone's volume, since the cone's base is a circle with area • r^{2}. For a square pyramid, the area of the base is s^{2} (where s is the length of a side of the square base). So the volume is (1/3)s^{2}h. For a triangular pyramid, the area of the base is (1/2)ab (base and height of the triangle  here we use a for altitude so we don't confuse the height of the triangle and the height of the pyramid). We get a volume formula of (1/6)abh.
<< back to Problem B10
<< back to Problem B10 noninteractive version


