# [Solid Mensuration] please help meeee!! :)

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• March 23rd 2014, 02:54 AM
mventurina1
[Solid Mensuration] please help meeee!! :)
(Rofl) Please help me on how to solve diz XD

"A block of Granite is in the form of the frustum of a regular square pyramid whose upper and base lower base edges are 3ft and 7ft.., Respectively, if each of the lateral faces is inclined at an angle of 62o 30' to the base , Find the volume of the granite in block"

as seen on the book "Solid Mensuration By Kern And Bland 2nd Edition" #9 page 71

Thanks in advance !!(Hi)
• March 23rd 2014, 06:31 AM
earboth
Re: [SolidMensuration] please help meeee!! :)
Quote:

Originally Posted by mventurina1
(Rofl) Please help me on how to solve diz XD

"A block of Granite is in the form of the frustum of a regular square pyramid whose upper and base lower base edges are 3ft and 7ft.., Respectively, if each of the lateral faces is inclined at an angle of 62o 30' to the base , Find the volume of the granite in block"
as seen on the book "Solid Mensuration By Kern And Bland 2nd Edition" #9 page 71
Thanks in advance !!(Hi)

Hello,

draw a cross-section of the frustrum (see attachment)

With the right triangle(in grey) you'll get: $\frac h2 = \tan(62^\circ30')~\implies~h \approx 3.842'$

Then use the formula to calculate the volume of a frustrum: If $B_l$ denotes the base area and $B_u$ the upper area then the volume is:

$V=\frac h3 \cdot \left(B_l + \sqrt{B_l \cdot B_u}+B_u \right)$

I've got $V \approx 101.173\ cuft$
• March 24th 2014, 04:05 AM
mventurina1
Re: [Solid Mensuration] please help meeee!! :)
thank you sir for the accurate answer and for the explanation. !!!