Find the

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

Correct result:

S =  5539.5431 cm2

Solution:

V=24 dm3→ cm3=24⋅ 1000  cm3=24000 cm3 h=45 cm  V=13⋅ 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=a⋅ h22=40⋅ 49.24432=100 97 cm2≐984.8858 cm2  S=4⋅ S2+S1=4⋅ 984.8858+1600=5539.5431 cm2

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