Respuesta :
Square base pyramid
We know that the volume of a square base pyramid is 144m³. And we know we can find it using the following equation:
[tex]\text{vol}=\frac{1}{3}base\text{ area}\cdot h[/tex]where h: height.
Since the base is a square, the area of the base is given by its width
area = w · w = w²
then, replacing in the first equation:
[tex]\text{vol}=\frac{1}{3}w^2h=144m[/tex]where w: width.
On the other hand, we have that its height, h, is twice its width, w:
2w = h
↓
w = h/2
Then, replacing w by h in our equation:
[tex]\begin{gathered} \text{vol}=\frac{1}{3}(\frac{h}{2})^2\cdot h=144m^3 \\ \\ \\ \\ \\ \\ \end{gathered}[/tex]since the exponent is distributed to every part of the fraction we have:
[tex]\begin{gathered} \\ \\ (\frac{h}{2})^2=\frac{h^2}{2^2}=\frac{h^2}{4} \\ \downarrow \\ \frac{1}{3}(\frac{h}{2})^2\cdot h=\frac{1}{3}\cdot\frac{h^2}{4}^{}\cdot h \\ \\ \end{gathered}[/tex]in order to operate them we can conver h into a fraction:
[tex]\begin{gathered} \\ \\ h=\frac{h}{1} \\ \downarrow \\ \frac{1}{3}\cdot\frac{h^2}{4}^{}\cdot h=\frac{1}{3}\cdot\frac{h^2}{4}^{}\cdot\frac{h}{1}=\frac{1\cdot h^2\cdot h}{3\cdot4\cdot1}^{}=\frac{h^3}{12} \\ \downarrow \\ \frac{h^3}{12}=144m^3 \end{gathered}[/tex]Now, we can solve the equation for h:
[tex]\begin{gathered} \frac{h^3}{12^{}}=144m^3 \\ \downarrow \\ h^3=144m^3\cdot12=1728m^3 \\ \downarrow \\ h=\sqrt[3]{1728m^3} \end{gathered}[/tex]We have that
[tex]\sqrt[3]{1728m^3}=12m[/tex]Then, h = 12 m (and w = 6m)
Answer: its height is 12m
