Question 1208118
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1 ft = 12 in
7 ft = 7*12 = 84 inches is the diameter, half that (42) is the radius
1 & 1/3 = 4/3
Volume of cylinder = {{{pi*r^2*h}}}
Volume of cylinder = {{{pi*(42)^2*(4/3)}}}
Volume of cylinder = {{{2352pi}}} cubic inches


"A plate 1 ft. square and 1 in. thick weighs 37.5 lbs"
means we have a 1 ft by 1 ft by 1 inch rectangular slab of iron
i.e. we have a 12 inch by 12 inch by 1 inch slab of iron
The volume of this would be length*width*height = 12*12*1 = 144 cubic inches.
144 cubic inches of iron weighs 37.5 pounds which will help set up a useful conversion factor. 


{{{(matrix(1,2,2352pi,"cubic_inches"))(matrix(1,2,37.5,"pounds")/matrix(1,2,144,"cubic_inches"))}}}


={{{(matrix(1,2,2352pi,cross("cubic_inches")))(matrix(1,2,37.5,"pounds")/matrix(1,2,144,cross("cubic_inches")))}}}


={{{matrix(1,2,2352pi*(37.5/144),"pounds")}}}


={{{matrix(1,2,612.5pi,"pounds")}}}


This is the exact weight of the circular disc in terms of pi.


Here are some select approximations of this weight when we try various approximations of pi
<table border = "1" cellpadding = "5"><tr><td>rounding precision</td><td>pi</td><td>weight</td></tr><tr><td>2</td><td>3.14</td><td>1923.25</td></tr><tr><td>3</td><td>3.142</td><td>1924.475</td></tr><tr><td>4</td><td>3.1416</td><td>1924.23</td></tr><tr><td>5</td><td>3.14159</td><td>1924.223875</td></tr><tr><td>6</td><td>3.141593</td><td>1924.2257125</td></tr><tr><td>7</td><td>3.1415927</td><td>1924.22552875</td></tr><tr><td>8</td><td>3.14159265</td><td>1924.225498125</td></tr><tr><td>9</td><td>3.141592654</td><td>1924.225500575</td></tr><tr><td>10</td><td>3.1415926536</td><td>1924.22550033</td></tr></table>
As we use more decimal digits of pi, the weight seems to approach roughly 1924.2255 pounds.
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