To solve for [tex]\( a \)[/tex] from the given formula for the volume [tex]\( V \)[/tex] of a right square pyramid, follow these steps:
The formula for the volume of a right square pyramid is:
[tex]\[ V = \frac{1}{3} a^2 h \][/tex]
1. Isolate [tex]\( a^2 \)[/tex]:
[tex]\[ V = \frac{1}{3} a^2 h \][/tex]
Multiply both sides of the equation by 3 to get rid of the fraction:
[tex]\[ 3V = a^2 h \][/tex]
2. Solve for [tex]\( a^2 \)[/tex]:
Divide both sides by [tex]\( h \)[/tex] to isolate [tex]\( a^2 \)[/tex]:
[tex]\[ \frac{3V}{h} = a^2 \][/tex]
3. Solve for [tex]\( a \)[/tex]:
Take the square root of both sides to solve for [tex]\( a \)[/tex]:
[tex]\[ a = \sqrt{\frac{3V}{h}} \][/tex]
Now, let's examine the given options:
A. [tex]\( a = 3 \sqrt{\frac{h}{V}} \)[/tex]
This does not match our derived formula because the fraction inside the square root is inverted and multiplied by 3.
B. [tex]\( a = \sqrt{\frac{s}{4}} \)[/tex]
This has no relation to our variables [tex]\( V \)[/tex] and [tex]\( h \)[/tex] and does not match our derived formula.
C. [tex]\( a = \sqrt{\frac{8Y}{6}} \)[/tex]
This also does not match our variables and is incorrect.
D. [tex]\( a = 3 \sqrt{\frac{V}{h}} \)[/tex]
This is correct! It matches our derived formula when considering [tex]\( 3 \sqrt{\frac{V}{h}} \)[/tex] is equivalent to [tex]\( \sqrt{\frac{3V}{h}} \)[/tex].
Therefore, the correct answer is:
D. [tex]\( a = 3 \sqrt{\frac{V}{h}} \)[/tex]
Hence, the correct choice is option D.
