Frustum of Regular Pyramid..FlowerPot

**cutiemike1** · September 6th 2010, 04:07 AM

The flowerpot shown in the sketch has a uniform thickness of 3in. and is 1ft high. Find the amount of material necessary to construct 1000 such pots. (Neglect the drain in the bottom)

What I did is get the volume of the bigger frustum which results to:

V = [h(18^2 + 12^2 + sqrt (18 * 12))]/3
V = 1930.79 cu. in

I don't know now the next step...

the answer should be 1098 cu. ft.

Help please THANKS!

**sa-ri-ga-ma** · September 6th 2010, 04:45 AM

Inner dimensions will be 12 in, 6 in and height 9 in. Find the volume V' of the inneer space.

V - V' will give the volume of the material.

**cutiemike1** · September 6th 2010, 01:39 PM

Originally Posted by sa-ri-ga-ma

Inner dimensions will be 12 in, 6 in and height 9 in. Find the volume V' of the inneer space.

V - V' will give the volume of the material.

It will not come up with the right answer. By the way thanks for help.

**Soroban** · September 6th 2010, 03:42 PM

Hello, cutiemike1!

Two problems: You are misreading the volume formula.
. . . . . . . . . . .I don't agree with the given answer.

The flowerpot shown in the sketch has a uniform thickness of 3in. and is 1ft high.
Find the amount of material necessary to construct 1000 such pots.

The answer should be 1098 cubic feet. . ??

The volume of a frustum of a pyramid is: . $V \;=\;\dfrac{h}{3}\left(B_1 + B_2 + \sqrt{B_1B_2}\right)$
where: $h$ is the height of the frustum
. . . . . and $B_1,B_2$ are the areas of the two bases.

The outer frustum has: . $\begin{Bmatrix} h &=& 12 \\ B_1 &=& 18^2 &=& 324 \\ B_2 &=& 12^2 &=& 144 \end{Bmatrix}\;\text{ inches}$

The outer volume is: . $V_o \;=\;\dfrac{12}{3}\left(324 + 144 + \sqrt{324\cdot144}\right)$

. . . . . . . . . . . . . . . . . . $=\;4(468 + \sqrt{46,\!656}) \;=\;4(468 + 216)$

. . . . . . . . . . . . . . . . . . $=\;4(684) \;=\;2736\text{ in}^3$

The inner frustum has: . $\begin{Bmatrix}h &=& 9 \\ B_1 &=& 12^2 &=& 144 \\ B_2 &=& 6^2 &=& 36 \end{Bmatrix}\;\text{ inches}$

The inner volume is: . $V_i \;=\;\frac{9}{3}\left(144 + 36 + \sqrt{144\cdot36}\right)$

. . . . . . . . . . . . . . . . . . $=\; 3(180 + \sqrt{5184}) \;=\;3(180 + 72)$

. . . . . . . . . . . . . . . . . . $=\;3(252) \;=\;756\text{ in}^3$

One pot requires: . $2736 - 756 \:=\:1980\text{ in}^3$ of material.

A thousand pots will require: . $1,\!980,\!000\text{ in}^3$ of material.

. . which equals: . $\dfrac{1,\!980,\!000}{1728} \;=\;1145\frac{5}{6}\text{ ft}^3$ of material.

**cutiemike1** · September 8th 2010, 01:00 PM

Thank You Sir Soroban...Sorry for the late reply to this thread.

**chemeng** · March 8th 2011, 07:13 AM

HI! i think the B2 of the inner frustum is 7.5 which results to a final answer of 1079.42 cubic feet.

**compe2014** · May 9th 2011, 05:14 AM

i am really confused with this problem. my prof really argue that the answer on the book solid mensuration which happens to be bland and kern he said though the book ages the answer dont .. lol he really stands that the answer is 1098.0 cu.ft. which is really annoyingly unsolvable

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