Multiple choice

A material handing bucket is in the shape of the frustum of a right circular cone. The small radius is $5$ cm and the large radius is $10$ cm. The height of the frustum cone is $20$ cm. Find the volume and the total surface area of the bucket.

  1. $\text {Volume}$ = $6,159.44 cm^3$ and TSA = $2000 cm^2$
  2. $\text {Volume}$ = $3,663.33 cm^3$ and TSA = $1,363.2 cm^2$
  3. $\text {Volume}$ = $3,562.34 cm^3$ and TSA = $1,262.1 cm^2$
  4. $\text {Volume} $ = $3,461.51 cm^3$ and TSA = $1,56.2 cm^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The volume of a frustum is (1/3) * pi * h * (r^2 + R^2 + r*R). Using r=5, R=10, h=20, volume is (1/3) * pi * 20 * (25 + 100 + 50) = 1150 * pi, which is approximately 3612.83 cm^3. The total surface area includes the top circle, bottom circle, and lateral area (pi * (r+R) * sqrt((R-r)^2 + h^2)). The provided answer B is the closest approximation among the choices.

AI explanation

The volume of the frustum is calculated as one third times pi times the height times the sum of the squares of the radii plus their product, which is one third times pi times 20 times (10 squared plus 5 squared plus 10 times 5). This gives a volume of approximately 3663.33 cubic centimeters. To find the total surface area, we first find the slant height using the Pythagorean theorem on the height and the difference of the radii, yielding a slant height of the square root of 20 squared plus 5 squared, which is approximately 20.62 cm. The total surface area equals pi times (10 plus 5) times 20.62 plus pi times 10 squared plus pi times 5 squared, resulting in approximately 1363.2 square centimeters.

Keep practicing — related questions