A cylindrical pillar, with a regular nonagonal cross-section, has to be carved out of a right circular cylindrical sandal wood. If the height is 10' and diameter of the base is 1', how do you prove that the minimum possible scrap is 0.623 cft, nearly?

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A cylindrical pillar, with a regular nonagonal cross-section, has to be carved out of a right circular cylindrical sandal wood. If the height is 10' and diameter of the base is 1', how do you prove that the minimum possible scrap is 0.623 cft, nearly?

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Answer 1 · 2017-01-07T06:49:38

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The nonagonal cross section of the pillar has been shown in the above figure. Here $r$ represents the radius of the original cylindrical pillar. So $r= 1/2ft ("given")$

So area of the nonagonal cross section is equal to the total area of 9 identical isosceles triangles each having area equal to the area of $DeltaABO$

So the nonagonal cross sectional area

$A=9xxDeltaABO=9xx1/2r^2xxsin/_AOB$

$=9/2xx(1/2)^2xxsin40^@=9/8xxsin40^@sqft$

Height of the pillar $h=10ft$

So volume of the pillar of nonagonal cross section

$V_2=Ah=90/8xxsin40^@cuft$

The original volume of the cylindrical pillar
$V_1=pir^2h=pi(1/2)^2xx10cuft$

So the volume of the minimum possible scrap is

$V_"scrap"=V_1-V_2$

$=pi(1/2)^2xx10cuft-90/8xxsin40^@cuft~~0.623cft$

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A cylindrical pillar, with a regular nonagonal cross-section, has to be carved out of a right circular cylindrical sandal wood. If the height is 10' and diameter of the base is 1', how do you prove that the minimum possible scrap is 0.623 cft, nearly?

1 Answer

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