Kirsten used a log to make an umbrella stand. The log was 28 in. tall and had a radius of 7.5 in. Kirsten cut a cylindrical shape, with a radius of 5.5 in., out of the center of the log. After Kirsten cuts out the cylinder, what is the volume of the remaining portion of the log? Use 3.14 to approximate pi, and express your answer to two decimal places. in3 A diagram showing the outline of a cylinder with a horizontal line stretching from the middle of the cylinder's base to its edge. Next to the line is the label seven and one-half inches. Next to the vertical edge of the cylinder is the label twenty-eight inches. Inside, at the bottom of, the cylinder is the dotted outline of a smaller cylinder with a dotted horizontal line stretching from the middle of the smaller cylinder's base to its edge. Next to the dotted line is the label five and one-half inches.

Question

Virginia Salinas

Mathematics

28 July, 11:41

Kirsten used a log to make an umbrella stand. The log was 28 in. tall and had a radius of 7.5 in. Kirsten cut a cylindrical shape, with a radius of 5.5 in., out of the center of the log. After Kirsten cuts out the cylinder, what is the volume of the remaining portion of the log? Use 3.14 to approximate pi, and express your answer to two decimal places. in3 A diagram showing the outline of a cylinder with a horizontal line stretching from the middle of the cylinder's base to its edge. Next to the line is the label seven and one-half inches. Next to the vertical edge of the cylinder is the label twenty-eight inches. Inside, at the bottom of, the cylinder is the dotted outline of a smaller cylinder with a dotted horizontal line stretching from the middle of the smaller cylinder's base to its edge. Next to the dotted line is the label five and one-half inches.

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Answers (1)

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Leah Wells · Answer 1

Volume of a cylinder = π r² h

π = 3.14

Log: r = 7.5 inches; h = 28 inches

Volume of the log = 3.14 * (7.5in) ² * 28in

V = 3.14 * 56.25 in² * 28 in

V = 4,945.5 in³

cylindrical shape: r = 5.5 inches; h = 28 inches

V = 3.14 * (5.5in) ² * 28in

V = 3.14 * 30.25 in² * 28in

V = 2,659.58 in³

Volume of the remaining portion of the log:

4,945.50 in³ - 2,659.58 in³ = 2,285.92 in³