Answer:
the ratio of the surface area of Pyramid A to Pyramid B is: [tex]\frac{36}{49}[/tex]
Step-by-step explanation:
Given the information:
- Pyramid A : 648 [tex]m^{3}[/tex]
- Pyramid B : 1,029 [tex]m^{3}[/tex]
- Pyramid A and Pyramid B are similar
As we know that:
If two solids are similar, then the ratio of their volumes is equal to the cube
of the ratio of their corresponding linear measures.
<=> [tex]\frac{Volume of A }{Volume of B}[/tex] =[tex](\frac{a}{b}) ^{3}[/tex] = [tex]\frac{684}{1029}[/tex] = [tex]\frac{216}{343}[/tex]
<=> [tex]\frac{a}{b} =[/tex][tex]\frac{6}{7}[/tex]
Howver, If two solids are similar, then the
n ratio of their surface areas is equal to the square of the ratio of their corresponding linear measures
<=> [tex]\frac{surface area of A}{surface area of B} =( \frac{a}{b}) ^{2}[/tex]
= [tex]\frac{36}{49}[/tex]
So the ratio of the surface area of Pyramid A to Pyramid B is: [tex]\frac{36}{49}[/tex]
