MATH SOLVE

4 months ago

Q:
# The height of one solid limestone square pyramid is 12 m. A similar solid limestone square pyramid has a height of 15 m. The volume of the larger pyramid is 16,000 m^3. Determine each of the following, showing all your work and reasoning. (a) The scale factor of the smaller pyramid to the larger pyramid in simplest form.(b) The ratio of the area of the base of the smaller pyramid to the larger pyramid.(c) Ratio of the volume of the smaller pyramid to the larger.(d) The volume of the smaller pyramid.

Accepted Solution

A:

Answers:
In this case, we have two pyramids:
-A small pyramid, which height is [tex]h_{small}=12m[/tex]
-A larger pyramid, which height is [tex]h_{large}=15m[/tex]. We also have the information about its volume [tex]V_{large}=16000{m}^{3}[/tex]
Now, before solving these questions, it is important to have clear that ratio is a quantified relationship between two magnitudes that reflects their proportion. In other words, ratio shows the relative sizes of two values.
In this context, we are asked to scale factor of the smaller pyramid to the larger pyramid in simplest form, this means we need to write the ratio between the heights of both pyramids and then simplify it:
(a) [tex]\frac{h_{small}}{h_{large}}=\frac{12m}{15m}[/tex]
Dividing numerator and denominator by 3 (simplifying):
[tex]\frac{4}{5}[/tex]>>>>This is the ratio of the length of both pyramids (b) If we want to know the ratio of the area, we only have to square the ratio we calculated in part (a):
[tex]{(\frac{4}{5})}^{2}=\frac{16}{25}[/tex]
(c) Now we need to know the of the volume of the smaller pyramid to the larger, in this case we have to cubic the ratio we calculated in part (a):
[tex]{(\frac{4}{5})}^{3}=\frac{64}{125}[/tex]
(d) In order to caculate the volume of the smaller pyramid, we will use the volume ratio from part (c) to write the following relation:
[tex]\frac{V_{small}}{V_{large}}=\frac{64}{125}[/tex]
We know the value of [tex]V_{large}[/tex] which is [tex]16000{m}^{3}[/tex], we have to find [tex]V_{small}[/tex]. So, rewriting the equation:
[tex]\frac{V_{small}}{16000{m}^{3}}=\frac{64}{125}[/tex]
Finally:
[tex]V_{small}=8192{m}^{3}[/tex] which is the volume of the smaller pyramid.