Answer:
[tex]h \approx 12.7\,cm[/tex]
Step-by-step explanation:
The side of the square base is:
[tex]l = \sqrt{64\,cm^{2}}[/tex]
[tex]l = 8\,cm[/tex]
The formula for the surface area of the pyramid is:
[tex]A_{s} = \frac{1}{2}\cdot (3\cdot l)\cdot \sqrt{0.25\cdot l^{2}+h^{2}} + l^{2}[/tex]
The height is cleared in the previous expression:
[tex]A_{s} - l^{2} = \frac{3}{2}\cdot l\cdot \sqrt{0.25\cdot l^{2}+h^{2}}[/tex]
[tex]\frac{2}{3}\cdot \frac{A_{s}-l^{2}}{l} = \sqrt{0.25\cdot l^{2} + h^{2}}[/tex]
[tex]\frac{4}{9}\cdot \left(\frac{A_{s}-l^{2}}{l} \right)^{2} = \frac{1}{4} \cdot l^{2} + h^{2}[/tex]
[tex]h = \sqrt{\frac{4}{9}\cdot \left(\frac{A_{s}-l^{2}}{l} \right)^{2} - \frac{1}{4}\cdot l^{2}}[/tex]
[tex]h = \sqrt{\frac{4}{9}\cdot \left[\frac{224\,cm^{2}-(8\,cm)^{2}}{8\,cm}\right]^{2}-\frac{1}{4}\cdot (8\,cm)^{2} }[/tex]
[tex]h \approx 12.7\,cm[/tex]