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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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![**The Moscow Papyrus and the Volume of a Truncated Square Pyramid**
The Moscow Papyrus gives a formula for the volume of a truncated square pyramid as:
\[ V = \frac{h}{3} (a^2 + ab + b^2) \]
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 \( V = \frac{ha^2}{3} \), prove that the above formula is correct.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdb708fa5-116d-42c3-bb62-31dd00678e29%2F6539524a-48b0-46b4-b38f-5bd00b9a2570%2Fjy2kilzb_processed.png&w=3840&q=75)
Transcribed Image Text:**The Moscow Papyrus and the Volume of a Truncated Square Pyramid**
The Moscow Papyrus gives a formula for the volume of a truncated square pyramid as:
\[ V = \frac{h}{3} (a^2 + ab + b^2) \]
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 \( V = \frac{ha^2}{3} \), prove that the above formula is correct.
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