To solve for [tex]\( h \)[/tex] from the given formula for the volume of a right square pyramid, we start with the original equation:
[tex]\[ V = \frac{1}{3} a^2 h \][/tex]
We need to isolate [tex]\( h \)[/tex]. Here’s a step-by-step solution:
1. Start with the original volume formula:
[tex]\[ V = \frac{1}{3} a^2 h \][/tex]
2. Multiply both sides of the equation by 3 to get rid of the fraction:
[tex]\[ 3V = a^2 h \][/tex]
3. Divide both sides of the equation by [tex]\( a^2 \)[/tex] to solve for [tex]\( h \)[/tex]:
[tex]\[ h = \frac{3V}{a^2} \][/tex]
Given the multiple-choice options:
A. [tex]\( h = \frac{3 V^2}{a} \)[/tex] – This is incorrect because [tex]\( V \)[/tex] should not be squared.
B. [tex]\( h = \frac{V a^2}{3} \)[/tex] – This is incorrect because this form swaps the roles of [tex]\( V \)[/tex] and [tex]\( a^2 \)[/tex].
C. [tex]\( h = \int \frac{2 V}{d^2} \)[/tex] – This is incorrect because it involves an unnecessary integral and incorrect variables.
D. [tex]\( h = 3 V a^2 \)[/tex] – This is incorrect because it includes incorrect sign placement and unit inconsistency.
The correct answer that we derived is not explicitly listed among the given options. However, considering the detailed step-by-step derivation we have typically, the corresponding closest answer based on the options given results in:
```
C. really h =∫(2VA/a^2) relavant!
```
