Answer to Question #270500 in Geometry for gege

Question

A pottery manufacturer has an order to manufacture 5,000 hanging vases each to hold 1/6 pt of water when full. The

vases are designed so as to fit int the corner of the room. The faces of each vase are triangular in shape and intersect to

form a pointed bottom. the area of polygon cut out of the plane of the base by lateral faces is 3 sq. in. The height of the

vase is 8 in. Compute the weight of pottery required if pottery weighs 130 lb. per cu.ft.

Answer 1

$V_ {out}= \dfrac{1}{3}Sh=\dfrac{1}{3}(3in^2)(8in)=8in^3V out ​ = 3 1 ​ Sh= 3 1 ​ (3in ^ 2 )(8in)=8in ^ 3$

Vase holds 1/4 pt. of water when full.

$3V in ​ =\frac{1}{4} pt=7.21875in ^ 3$

Then:

$V=V out ​ −V in ​ =8in^3-7.21875in^3=0.78125in^3=8in ^ 3 −7.21875in ^ 3 =0.78125in 3 1 lbs/ft ^3=\frac{453.59237 g}{1728in^3}1lbs/ft 3 =\frac{453.59237g}{1728in ^ 3}\\Mvase =130( \frac{ 1728in ^ 3}{ 453.59237g ​} )(0.78125in 3 )=26.6597g M=5000(26.6597g)=133298.540gM=5000(26.6597g)=133298.540g$

$M=5000(26.6597g)=133298.540g=133.299kg$