The perimeter

The perimeter of the base of a regular quadrilateral pyramid is the same as its height. The pyramid has a volume of 288 dm³. Calculate its surface area round the result to the whole dm².

Correct answer:

S =  326 dm2

Step-by-step explanation:

V=288 dm3  4a = h V = 31 a2 h V = 31 a2 4 a  a=343 V=343 288=6 dm h=4 a=4 6=24 dm  S1=a2=62=36 dm2  h2=h2+(a/2)2=242+(6/2)2=3 65 dm24.1868 dm S2=2a h2=26 24.1868=9 65 dm272.5603 dm2  S=S1+4 S2=36+4 72.5603326.2413 dm2   Verifying Solution:   V2=31 S1 h=31 36 24=288 dm3

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