2. The seniors at our high school decided to play a prank on the principal by completely filling his office with

basketballs. To determine the number of basketballs needed the students measured the room after moving out

the furniture. If the room measured 15 ft by 20 ft by 10 ft, approximately how many basketballs did the students

put in the principal's office? The basketballs had a circumference of 29 inches.

A. 56,000

B. 36,600

C. 12,600

D. 9,600

C. 12,600

Step-by-step explanation:

You can try a rough approximation by finding the volume of the room, finding the volume of one basketball, an dividing the first volume by the second one. This method will give a large estimate since there will be empty space between the balls.

volume of room:

V = L * W * H = 15 * 20 * 10 ft^3 = 3,000 ft^3

volume of one ball:

circumference = 29 in.

From the circumference we find the radius.

circumference = 2 (pi) r

2 (pi) r = 29 in.

r = (29 in.) / (2pi) = 4.61549 in.

Now we convert inches to feet.

4.61549 in. = 4.61549/12 ft = 0.384624 ft

Now we use the radius of a ball in ft to find the volume of a ball in cubic feet.

V = (4/3) (pi) r^3 = (4/3) (3.14159) (0.384624 ft) ^3 = 0.23834 ft^3

Now we divide the volume of the room by the volume of a ball.

number of balls = (3,000 ft^3) / (0.23834 ft^3) = 12,587

The closest answer is C. 12,600