Hat Day - The Solutions

January 14, 2008

The brim of the hat is essentially made up of a circle with a hole in the middle.  That hole will be the same size as the base of the cylinder that makes up the column portion of the hat. Knowing this, we can solve this problem by subtracting the area covered by the base of the cylinder (or the area of the hole) from the area of the “circle.”

Since circumference equals 2πr, we know the radius of the cylinder is 12 ÷ π, or 3.819719… inches. To find the radius of the “circle” that includes the brim, we add 3 inches. This gives us a radius of 6.819719… inches.

The area of the “circle” is 146.111 square inches.

A = π(6.819719)2

A = 146.111

The area of the base of the cylinder (the opening for Fredrick’s head) is 45.837 square inches.

A = π(3.819719)2

A = 45.837

This means the area of the bottom face of the brim is 100.3 square inches.

146.111 – 45.837 = 100.3, to the nearest tenth.

NOTE:  It would be best – in this problem and the ones that follow – to use the radii

and (+3) and areas π(+3)2 and π ()instead of rounding.

V = πr2h

V = π(3.819719)2(18)

V = 825.1 cubic inches, to the nearest tenth.

Since we know the volume of dirt left and the radius of the space it was meant to fill, we can use the volume formula to figure out how high it would have reached (or the depth of the space it was meant to fill).

229.2 = π(3.819719)2h

229.2 = 45.83663h

h = 5.0 inches, to the nearest tenth.

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