Question 1208118
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The methods shown by the other tutors are fine.<br>
Here is another....<br>
Start with the 37.5 pounds that the plate, 1 foot square and 1 inch thick, weighs.<br>
The cross sectional area of the plate is 1 square foot.
The cross sectional area of the disc is {{{(pi)(r^2)=(pi)(7/2)^2=(49/4)pi}}}.
The cross sectional area of the disc is greater than the cross sectional area of the plate by a factor of {{{(49/4)pi}}}<br>
The thickness of the plate is 1 inch; the thickness of the disc is 1 1/3 inches.
The disc is thicker than the plate by a factor of 1 1/3 = 4/3.<br>
Multiply the weight of the plate by the factors by which the cross sectional area and thickness of the disc exceed those of the plate:<br>
ANSWER: {{{(37.5)((49/4)pi)(4/3)}}} = 612.5pi = 1924 to the nearest whole number.<br>