CK-12 Geometry Concepts

(Elliott) #1

11.9. Area and Volume of Similar Solids http://www.ck12.org


Concept Problem Revisited


The coffee mugs are similar because the heights and radii are in a ratio of 2:3, which is also their scale factor. The
volume of Dad’s mug is 54πin^3 and Mom’s mug is 16πin^3. The ratio of the volumes is 54π: 16π, which reduces to
8:27.


Vocabulary


Two solids aresimilarif they are the same type of solid and their corresponding radii, heights, base lengths, widths,
etc. are proportional.


Guided Practice



  1. Determine if the two triangular pyramids similar.

  2. If the ratio of the volumes of two similar prisms is 125:8, what is their scale factor?

  3. Two similar right triangle prisms are below. If the ratio of the volumes is 343:125, find the missing sides in both
    figures.

  4. The ratio of the surface areas of two similar cylinders is 16:81. If the volume of the smaller cylinder is 96πin^3 ,
    what is the volume of the larger cylinder?


Answers:



  1. Let’s match up the corresponding parts.
    6
    8 =


12
16 =

3
4 however,

8
12 =

2
3.
Because one of the base lengths is not in the same proportion as the other two lengths, these right triangle pyramids
are not similar.



  1. We need to take the cubed root of 125 and 8 to find the scale factor.


√ 3


125 :


√ 3


8 =5 : 2



  1. If the ratio of the volumes is 343:125, then the scale factor is 7:5, the cubed root of each. With the scale factor,
    we can now set up several proportions.


7


5


=


7


y

7


5


=


x
10

7


5


=


35


w
72 +x^2 =z^2

7


5


=


z
v
y= 5 x= 14 w= 25 72 + 142 =z^2

z=


245 = 7



5


7


5


=


7



5


v
→v= 5


5



  1. First we need to find the scale factor from the ratio of the surface areas. If we take the square root of both numbers,
    we have that the ratio is 4:9. Now, we need cube this to find the ratio of the volumes, 4^3 : 9^3 =64 : 729. At this point
    we can set up a proportion to solve for the volume of the larger cylinder.


64


729


=


96 π
V
64 V= 69984 π
V= 1093. 5 πin^3
Free download pdf