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Sagot :
To find the height of a solid right pyramid with a square base, we start with the formula for the volume of the pyramid:
[tex]\[ V = \frac{1}{3} \times \text{base area} \times \text{height} \][/tex]
Given that the base is a square with edge length [tex]\( y \)[/tex], the base area can be written as:
[tex]\[ \text{base area} = y^2 \][/tex]
Now, substituting the base area into the volume formula, we get:
[tex]\[ V = \frac{1}{3} \times y^2 \times \text{height} \][/tex]
Rearranging this formula to solve for the height, we have:
[tex]\[ 3V = y^2 \times \text{height} \][/tex]
Now, isolating the height on one side of the equation:
[tex]\[ \text{height} = \frac{3V}{y^2} \][/tex]
Hence, the expression that represents the height of the pyramid is:
[tex]\[ \frac{3V}{y^2} \][/tex]
So, the correct option is:
[tex]\[ \frac{3V}{y^2} \][/tex] units
[tex]\[ V = \frac{1}{3} \times \text{base area} \times \text{height} \][/tex]
Given that the base is a square with edge length [tex]\( y \)[/tex], the base area can be written as:
[tex]\[ \text{base area} = y^2 \][/tex]
Now, substituting the base area into the volume formula, we get:
[tex]\[ V = \frac{1}{3} \times y^2 \times \text{height} \][/tex]
Rearranging this formula to solve for the height, we have:
[tex]\[ 3V = y^2 \times \text{height} \][/tex]
Now, isolating the height on one side of the equation:
[tex]\[ \text{height} = \frac{3V}{y^2} \][/tex]
Hence, the expression that represents the height of the pyramid is:
[tex]\[ \frac{3V}{y^2} \][/tex]
So, the correct option is:
[tex]\[ \frac{3V}{y^2} \][/tex] units
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