Math  /  Geometry

Question5. For Figures 1-3, answer the questions below.
Figure 1
Figure 2
Figure 3 a. Which of these nets could be folded along the lines to form a closed rectangular box? b. For the figures that form a closed rectangular box, use the unit square shown to help you find the dimensions of the box. c. For the figures that form a closed rectangular box, find the total area, in square units, of all of the faces of the box. d. For the figures that form a closed rectangular box, find the number of unit cubes it would take to fill the box. a. What are the dimensions of the box at the right? b. On centimeter grid paper, sketch two nets for the box. c. Find the area, in square centimeters, of each net. d. Find the total area of all the faces of the box. How does your answer compare with the areas you found in part (c)? Investigation 4 Measuring Surface Area and Volur

Studdy Solution

STEP 1

What is this asking? We need to figure out which of these flat shapes can fold into a box, what the box's dimensions are, its surface area, and how many unit cubes fit inside!
Plus, we'll explore dimensions, nets, and surface area of a separate given box. Watch out! Don't mix up surface area (the outside area) with volume (the inside space)!
Also, make sure you're counting the right squares when figuring out dimensions.

STEP 2

1. Analyze the Nets
2. Dimensions and Surface Area of Figure 1
3. Dimensions and Surface Area of Figure 3
4. Exploring the Given Box

STEP 3

Let's look at Figure 1.
Imagine folding it.
It totally makes a box!

STEP 4

Now, Figure 2.
Hmm, if we fold this one, those two flaps on the top would overlap.
No box here!

STEP 5

Finally, Figure 3.
This one folds into a box too!

STEP 6

The **length** of the box from Figure 1 is \( 4 \text{ units} \), the **width** is \( 1 \text{ unit} \), and the **height** is \( 1 \text{ unit} \).

STEP 7

To get the **surface area**, we need to add up the areas of all six sides.
We have two sides that are 41=4 4 \cdot 1 = 4 square units, two sides that are 11=1 1 \cdot 1 = 1 square unit, and two more sides that are 41=4 4 \cdot 1 = 4 square units.
So, the **total surface area** is 24+21+24=8+2+8=18 2 \cdot 4 + 2 \cdot 1 + 2 \cdot 4 = 8 + 2 + 8 = 18 square units.

STEP 8

The **volume** is just length times width times height: 411=4 4 \cdot 1 \cdot 1 = 4 cubic units.
So, we can fit **4 unit cubes** inside!

STEP 9

The box from Figure 3 has a **length** of \( 3 \text{ units} \), a **width** of \( 2 \text{ units} \), and a **height** of \( 1 \text{ unit} \).

STEP 10

For the **surface area**, we have two sides that are 32=6 3 \cdot 2 = 6 square units, two sides that are 31=3 3 \cdot 1 = 3 square units, and two sides that are 21=2 2 \cdot 1 = 2 square units.
Adding those up, we get 26+23+22=12+6+4=22 2 \cdot 6 + 2 \cdot 3 + 2 \cdot 2 = 12 + 6 + 4 = 22 square units.

STEP 11

The **volume** is 321=6 3 \cdot 2 \cdot 1 = 6 cubic units.
We can fit **6 unit cubes**!

STEP 12

The given box has dimensions 2 cm×1 cm×4 cm 2 \text{ cm} \times 1 \text{ cm} \times 4 \text{ cm} .

STEP 13

I'll leave sketching the nets to you, but remember a net is just a flattened version of the box!

STEP 14

Each net will have the same area, which is the surface area of the box.
We have two sides that are 21=2 2 \cdot 1 = 2 square cm, two sides that are 24=8 2 \cdot 4 = 8 square cm, and two sides that are 14=4 1 \cdot 4 = 4 square cm.
The **total surface area** is 22+28+24=4+16+8=28 2 \cdot 2 + 2 \cdot 8 + 2 \cdot 4 = 4 + 16 + 8 = 28 square cm.

STEP 15

Figure 1 and Figure 3 form boxes.
Figure 1: dimensions 4x1x1 units, surface area 18 square units, volume 4 cubic units.
Figure 3: dimensions 3x2x1 units, surface area 22 square units, volume 6 cubic units.
The given box: dimensions 2x1x4 cm, surface area 28 square cm.

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