Math worksheet with geometry problems on area, volume, and surface area.

Problem Analysis and Solution

The worksheet involves calculating areas, surface areas, and volumes of geometric shapes. Let's solve each problem step by step.

---

#### Problem 1:
A triangle that is very special to you has an area of 12 square feet. The height of this figure is six feet. What is the length of the base?

- Formula for the area of a triangle:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]

- Given:
\[
\text{Area} = 12 \, \text{square feet}, \quad \text{height} = 6 \, \text{feet}
\]

- Substitute the values into the formula:
\[
12 = \frac{1}{2} \times \text{base} \times 6
\]

- Simplify:
\[
12 = 3 \times \text{base}
\]

- Solve for the base:
\[
\text{base} = \frac{12}{3} = 4 \, \text{feet}
\]

- Answer:
\[
\boxed{4}
\]

---

#### Problem 2:
You find a triangle that has an area of 10 square feet. The height of the triangle is ten feet. What is the length of the base of the triangle?

- Formula for the area of a triangle:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]

- Given:
\[
\text{Area} = 10 \, \text{square feet}, \quad \text{height} = 10 \, \text{feet}
\]

- Substitute the values into the formula:
\[
10 = \frac{1}{2} \times \text{base} \times 10
\]

- Simplify:
\[
10 = 5 \times \text{base}
\]

- Solve for the base:
\[
\text{base} = \frac{10}{5} = 2 \, \text{feet}
\]

- Answer:
\[
\boxed{2}
\]

---

#### Problem 3:
You have a cooking triangle at your camp. It makes great hot dogs on a campfire. The triangle has an area of 16 square feet. The height is four feet. What is the length of the base?

- Formula for the area of a triangle:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]

- Given:
\[
\text{Area} = 16 \, \text{square feet}, \quad \text{height} = 4 \, \text{feet}
\]

- Substitute the values into the formula:
\[
16 = \frac{1}{2} \times \text{base} \times 4
\]

- Simplify:
\[
16 = 2 \times \text{base}
\]

- Solve for the base:
\[
\text{base} = \frac{16}{2} = 8 \, \text{feet}
\]

- Answer:
\[
\boxed{8}
\]

---

#### Problem 4:
The surface area of a cube is 72 in². What is the volume of the cube?

- Surface area of a cube:
A cube has 6 faces, and each face is a square. If the side length of the cube is \( s \), then:
\[
\text{Surface Area} = 6s^2
\]

- Given:
\[
\text{Surface Area} = 72 \, \text{in}^2
\]

- Substitute the value into the formula:
\[
72 = 6s^2
\]

- Solve for \( s^2 \):
\[
s^2 = \frac{72}{6} = 12
\]

- Solve for \( s \):
\[
s = \sqrt{12} = 2\sqrt{3} \, \text{inches}
\]

- Volume of a cube:
\[
\text{Volume} = s^3
\]

- Substitute \( s = 2\sqrt{3} \):
\[
\text{Volume} = (2\sqrt{3})^3 = 2^3 \times (\sqrt{3})^3 = 8 \times 3\sqrt{3} = 24\sqrt{3} \, \text{in}^3
\]

- Answer:
\[
\boxed{24\sqrt{3}}
\]

---

#### Problem 5:
The surface area of a cube is 42 in². What is the volume of the cube?

- Surface area of a cube:
\[
\text{Surface Area} = 6s^2
\]

- Given:
\[
\text{Surface Area} = 42 \, \text{in}^2
\]

- Substitute the value into the formula:
\[
42 = 6s^2
\]

- Solve for \( s^2 \):
\[
s^2 = \frac{42}{6} = 7
\]

- Solve for \( s \):
\[
s = \sqrt{7} \, \text{inches}
\]

- Volume of a cube:
\[
\text{Volume} = s^3
\]

- Substitute \( s = \sqrt{7} \):
\[
\text{Volume} = (\sqrt{7})^3 = 7\sqrt{7} \, \text{in}^3
\]

- Answer:
\[
\boxed{7\sqrt{7}}
\]

---

#### Problem 6:
An oversized colorful puzzle cube has the surface area of 36 in². What is the volume of the puzzle cube?

- Surface area of a cube:
\[
\text{Surface Area} = 6s^2
\]

- Given:
\[
\text{Surface Area} = 36 \, \text{in}^2
\]

- Substitute the value into the formula:
\[
36 = 6s^2
\]

- Solve for \( s^2 \):
\[
s^2 = \frac{36}{6} = 6
\]

- Solve for \( s \):
\[
s = \sqrt{6} \, \text{inches}
\]

- Volume of a cube:
\[
\text{Volume} = s^3
\]

- Substitute \( s = \sqrt{6} \):
\[
\text{Volume} = (\sqrt{6})^3 = 6\sqrt{6} \, \text{in}^3
\]

- Answer:
\[
\boxed{6\sqrt{6}}
\]

---

Final Answers:

1. \(\boxed{4}\)
2. \(\boxed{2}\)
3. \(\boxed{8}\)
4. \(\boxed{24\sqrt{3}}\)
5. \(\boxed{7\sqrt{7}}\)
6. \(\boxed{6\sqrt{6}}\)