c (3 l + 5r):(4 l + 4r) and (3 l + 5r):(l + r), where r is the radius and l is the slant height of the small cone.
D
d 7:8 and 7:1
Practice makes perfect
a A cone is cut by a plane parallel to its base. The small cone on top is similar to the large cone. The ratio of the slant heights of the cones is 1:2.
The scale factor is 2 and 1 from large cone to small cone. We want to find the ratio of their surface areas and to do so let's recall the Areas and Volumes of Similar Solids Theorem.
Areas and Volumes of Similar Solids Theorem
If the scale factor of two similar solids is ab or a:b then the ratio of their areas is a^2b^2 or a^2:b^2.
To find the ratio of their surface areas we will substitute in the scale factor values.
The ratio of their surface areas is 41, and can also be written as 4:1.
b A cone is cut by a plane parallel to its base. The small cone on top is similar to the large cone. The ratio of the slant heights of the cones is 1:2.
To find the ratio of their volumes we will substitute in the scale factor values and follow the same steps as before. The scale factor is 2 and 1, from large cone to small cone. Let's use the Areas and Volumes of Similar Solids Theorem again to recall the relationship between the scale factor and volume.
Areas and Volumes of Similar Solids Theorem
If the scale factor of two similar solids is ab or a:b then the ratio of their volumes is a^3b^3 or a^3:b^3.
To find the ratio of their volumes we will substitute in the scale factor values.
The ratio of their volumes is 81 and can also be written as 8:1.
c Remember that a frustum is the left-over lower part of the pyramid after the upper part has been cut off by a plane parallel to its base.
We want to find the ratio of the surface area of the frustum to that of the large cone and to that of the small cone. The surface area of the frustum is the sum of the areas of the two circles plus the lateral surface area.
Let's look and determine what is part of the surface area of the small cone.
The surface area of the small cone is the sum of the area of the smaller circle and the smaller lateral surface area.
S.A._(small cone) = π r^2 + π r l
The ratio of the surface area of the frustum to the surface area of the small cone is the following.
Therefore, the ratio of the surface area of the frustum to that of the large cone is 5r + 3 l : 4r + 4l, and the ratio of the surface area of the frustum to that of the small cone is 5r + 3 l : r + l.
d We want to find the ratio of the volume of the frustum to that of the large cone and to that of the small cone. The surface area of the frustum is the sum of the areas of the two circles plus the lateral surface area.
