Surface of pyramid

In a regular quadrilateral pyramid, the height of the sidewall is equal to the length of the edge of the base. The content of the sidewall is 32 cm2. What is the surface of the pyramid?

Correct result:

S =  160 cm2

Solution:

S2=32 cm2 S2=a⋅ h2 a=h S2=a2/2 a=2⋅ S2=2⋅ 32=8 cm S1=a2=82=64 cm2  S=4⋅ S2+S1 S=5⋅ S2=5⋅ 32=160 cm2S_{2}=32 \ \text{cm}^2 \ \\ S_{2}=\dfrac{ a \cdot \ h }{ 2 } \ \\ a=h \ \\ S_{2}=a^2/2 \ \\ a=\sqrt{ 2 \cdot \ S_{2} }=\sqrt{ 2 \cdot \ 32 }=8 \ \text{cm} \ \\ S_{1}=a^2=8^2=64 \ \text{cm}^2 \ \\ \ \\ S=4 \cdot \ S_{2} + S_{1} \ \\ S=5 \cdot \ S_{2}=5 \cdot \ 32=160 \ \text{cm}^2

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