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The Moscow Papyrus gives a formula for the volume of a truncated square pyramid as (a² + ab + b³) V = where h denotes the height of the truncated pyramid, a the length of the lower base and b the length of the upper base, respectively. Assuming that the volume of a square pyramid is given by V ha? /3, prove that the above formula is correct.

The Moscow Papyrus gives a formula for the volume of a truncated square pyramid as (a² + ab + b³) V = where h denotes the height of the truncated pyramid, a the length of the lower base and b the length of the upper base, respectively. Assuming that the volume of a square pyramid is given by V ha? /3, prove that the above formula is correct.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The Moscow Papyrus gives a formula for the volume of a truncated square pyramid as h V (a² + ab + b²) 3 where h denotes the height of the truncated pyramid, a the length of the lower base and b the length of the upper base, respectively. Assuming that the volume of a square pyramid is given by V = ha?/3, prove that the above formula is correct.
Transcribed Image Text:The Moscow Papyrus gives a formula for the volume of a truncated square pyramid as h V (a² + ab + b²) 3 where h denotes the height of the truncated pyramid, a the length of the lower base and b the length of the upper base, respectively. Assuming that the volume of a square pyramid is given by V = ha?/3, prove that the above formula is correct.
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