Regular quadrilateral pyramid

Find the surface area of a regular quadrilateral pyramid with a volume of 24 dm3 and a height of 45 cm.

Final Answer:

S =  5539.5431 cm2

Step-by-step explanation:

V=24 dm3→ cm3=24⋅ 1000  cm3=24000 cm3 h=45 cm  V = 31​ ⋅  S1​ ⋅  h  S1​=3⋅ V/h=3⋅ 24000/45=1600 cm2  S1​ = a2  a=S1​​=1600​=40 cm  (a/2)2 + h2 = h22​  h2​=(a/2)2+h2​=(40/2)2+452​=5 97​ cm≐49.2443 cm  S2​=2a⋅ h2​​=240⋅ 49.2443​=100 97​ cm2≐984.8858 cm2  S=4⋅ S2​+S1​=4⋅ 984.8858+1600=5539.5431 cm2

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