Answer:
Mike is not right
Step-by-step explanation:
we know that
If two figures are similar, then the ratio of its surface areas is equal to the scale factor squared
Let
z----> the scale factor
x----> surface area of the enlarged rectangular prism
y-----> surface area of the original rectangular prism
[tex]z^{2}=\frac{x}{y}[/tex]
so
In this problem we have
[tex]z=2[/tex]
substitute
[tex]2^{2}=\frac{x}{y}[/tex]
[tex]4=\frac{x}{y}[/tex]
[tex]x=4y[/tex]
so
The surface area of the enlarged rectangular prism is 4 times the surface area of the original rectangular prism
therefore
Mike is not right
Verify with an example
we have a rectangular prism
[tex]L=5\ m[/tex]
[tex]W=2\ m[/tex]
[tex]H=3\ m[/tex]
The surface area of the prism is equal to
[tex]SA=2(LW)+(2L+2W)H[/tex]
substitute the values
[tex]SA=2*(5*2)+(2*5+2*2)*3=62\ m^{2}[/tex]
If he doubles each dimension of any rectangular prism
then
the new dimensions will be
[tex]L=5*2=10\ m[/tex]
[tex]W=2*2=4\ m[/tex]
[tex]H=3*2=6\ m[/tex]
The new surface area will be
[tex]SA=2*(10*4)+(2*10+2*4)*6=248\ m^{2}[/tex]
[tex]248/62=4[/tex]
therefore
The surface area of the enlarged rectangular prism is 4 times the surface area of the original rectangular prism
