### Determining the Volume of a Storage Shed **Problem 15: A Volume Problem** The front and back of a storage shed are shaped like isosceles right triangles with two sides of length 10 ft, as shown on the left of **Figure 13.37**. The storage shed is 40 ft long. Determine the volume of the storage shed. --- **Solution Process:** 1. **Using the Pythagorean Theorem:** - Qing first uses the Pythagorean theorem to determine that the length of the unknown side of the triangle is \(\sqrt{200}\) ft long. - **Figure 13.37 Front View:** - A right isosceles triangle is shown with each of the equal sides measuring 10 ft. 2. **Calculating the Height of the Triangle:** - Qing uses the Pythagorean theorem again to calculate the height of the triangle \(h\) if the base is the side of length \(\sqrt{200}\). Qing carries this out as follows: \[ \left(\frac{\sqrt{200}}{2}\right)^2 + h^2 = 10^2 \] \[ h^2 = 100 - \frac{200}{4} = 100 - 50 = 50 \] \[ h = \sqrt{50} \] 3. **Determining the Floor Area:** - Next, Qing determines that the floor area of the storage shed is: \[ 40 \cdot \sqrt{200} \text{ square feet} \] 4. **Calculating the Volume:** - Finally, Qing calculates that the volume of the storage shed is given by the floor area of the shed times the height of the triangle: \[ 40 \cdot \sqrt{200} \cdot \sqrt{50} \] --- **Discussion Questions:** a. **Is Qing’s method of calculation correct? Discuss Qing’s work. Which parts (if any) are correct; which parts (if any) are incorrect?** b. **Solve the volume problem in a different way than Qing did, explaining your reasoning.** --- Figure 13.37: This figure shows a front view of the storage shed's triangular face. The triangle is an isosceles right triangle

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### Determining the Volume of a Storage Shed

**Problem 15: A Volume Problem**

The front and back of a storage shed are shaped like isosceles right triangles with two sides of length 10 ft, as shown on the left of **Figure 13.37**. The storage shed is 40 ft long. Determine the volume of the storage shed.

---

**Solution Process:**

1. **Using the Pythagorean Theorem:**
    - Qing first uses the Pythagorean theorem to determine that the length of the unknown side of the triangle is \(\sqrt{200}\) ft long.
    - **Figure 13.37 Front View:**
        - A right isosceles triangle is shown with each of the equal sides measuring 10 ft.

2. **Calculating the Height of the Triangle:**
    - Qing uses the Pythagorean theorem again to calculate the height of the triangle \(h\) if the base is the side of length \(\sqrt{200}\). Qing carries this out as follows:
        \[
        \left(\frac{\sqrt{200}}{2}\right)^2 + h^2 = 10^2
        \]
        \[
        h^2 = 100 - \frac{200}{4} = 100 - 50 = 50
        \]
        \[
        h = \sqrt{50}
        \]

3. **Determining the Floor Area:**
    - Next, Qing determines that the floor area of the storage shed is:
        \[
        40 \cdot \sqrt{200} \text{ square feet}
        \]

4. **Calculating the Volume:**
    - Finally, Qing calculates that the volume of the storage shed is given by the floor area of the shed times the height of the triangle:
        \[
        40 \cdot \sqrt{200} \cdot \sqrt{50}
        \]

---

**Discussion Questions:**

a. **Is Qing’s method of calculation correct? Discuss Qing’s work. Which parts (if any) are correct; which parts (if any) are incorrect?**

b. **Solve the volume problem in a different way than Qing did, explaining your reasoning.**

---

Figure 13.37: This figure shows a front view of the storage shed's triangular face. The triangle is an isosceles right triangle
Transcribed Image Text:### Determining the Volume of a Storage Shed **Problem 15: A Volume Problem** The front and back of a storage shed are shaped like isosceles right triangles with two sides of length 10 ft, as shown on the left of **Figure 13.37**. The storage shed is 40 ft long. Determine the volume of the storage shed. --- **Solution Process:** 1. **Using the Pythagorean Theorem:** - Qing first uses the Pythagorean theorem to determine that the length of the unknown side of the triangle is \(\sqrt{200}\) ft long. - **Figure 13.37 Front View:** - A right isosceles triangle is shown with each of the equal sides measuring 10 ft. 2. **Calculating the Height of the Triangle:** - Qing uses the Pythagorean theorem again to calculate the height of the triangle \(h\) if the base is the side of length \(\sqrt{200}\). Qing carries this out as follows: \[ \left(\frac{\sqrt{200}}{2}\right)^2 + h^2 = 10^2 \] \[ h^2 = 100 - \frac{200}{4} = 100 - 50 = 50 \] \[ h = \sqrt{50} \] 3. **Determining the Floor Area:** - Next, Qing determines that the floor area of the storage shed is: \[ 40 \cdot \sqrt{200} \text{ square feet} \] 4. **Calculating the Volume:** - Finally, Qing calculates that the volume of the storage shed is given by the floor area of the shed times the height of the triangle: \[ 40 \cdot \sqrt{200} \cdot \sqrt{50} \] --- **Discussion Questions:** a. **Is Qing’s method of calculation correct? Discuss Qing’s work. Which parts (if any) are correct; which parts (if any) are incorrect?** b. **Solve the volume problem in a different way than Qing did, explaining your reasoning.** --- Figure 13.37: This figure shows a front view of the storage shed's triangular face. The triangle is an isosceles right triangle
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