Activity 3 (Small-Group Activity)
Volume and Surface Area

Students compute the volume and surface area of objects. Use with the Student Copy and the Reference Copy.

Introduction

• To estimate, make, and use measurements to describe and compare phenomena.
• To extend understanding of the concepts of area and volume.

The Wooden Box

1. Surface area measurement is in square units (e.g., cm2 or ft2).

2. The surface area to paint is 3754 cm2. See individual calculations below:

• 2 side pieces (57 cm x 9 cm each) = 1026 cm2
• 2 end pieces (34 cm x 9 cm each) = 612 cm2
• 1 bottom piece (57 cm x 34 cm)= 1938 cm2
• 4 top rims (2 at 57 cm2 and 2 at 32 cm2) = 178 cm2

The area above assumes that the box edges are angled so that there is no overlapping. Student answers may vary depending on how the construction of the box is defined.
3. The total surface area of the box is 6906 cm2. The outside surface area was calculated in #2a. The individual inside calculations follow. The inside measurements for length and width are 2 cm less than the outside (55 cm and 32 cm respectively); the inside bottom measures 55 cm x 32 cm. The height is 1 cm less than the outside (8 cm).

• 2 side pieces with inside area (55 cm x 8 cm each) =880 cm2
• 2 end pieces with inside area (32 cm x 8 cm each) = 512 cm2
• 1 bottom piece with inside area (55 cm x 32 cm) = 1760 cm2

4. Answers may vary since there will be some overlap needed when wrapping the paper around the box. A piece of paper that will go completely around the box with an extra 2 cm to tape it shut and enough to cover the sides is 88 cm x 68 cm.

5. The inside dimensions of the side and end of the box are 55 cm and 32 cm, so 1760 one-centimeter cubes fill the bottom of the box.

6. There are 8 layers needed to fill the box, so the volume of the box is 8 cm x 55 cm x 32 cm or 14,080 cubic cm.

7. The biggest ball that will fit in the box without rising above the rim is 8 cm in diameter since the height of the box is 8 cm.

In all, 27 such balls can fit in the box; 6 rows with 4 balls to a row and one row with 3 balls. Explanation: It is easy to see that 4 balls will fit along the 32 cm width of the box and that 6 such rows will fill 48 cm in length. The remaining 7 cm can accommodate 3 balls that fit between adjacent balls in the last row of 4 balls. See diagram below.

The distance from the center of the last row of 4 balls to the end of the box, x + 4 cm, must be less than 11 cm for the balls to fit. To find x, use the Pythagorean Theorem, . Since, , the last 3 balls just fit.

Closing Discussion