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Let's work through each problem step-by-step.

### Problem 10
We need to find a third number so that the three numbers form a right triangle. For any right triangle, the numbers should satisfy the Pythagorean theorem: \( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse.

1. **(i) Numbers 9 and 41**
   - Let's assume 41 is the hypotenuse, so we have \( 9^2 + x^2 = 41^2 \).
   - Solving:
     \[
     9^2 + x^2 = 41^2
     \]
     \[
     81 + x^2 = 1681
     \]
     \[
     x^2 = 1600
     \]
     \[
     x = \sqrt{1600} = 40
     \]
   - Therefore, the third number is **40**.

2. **(ii) Numbers 13 and 85**
   - Assuming 85 is the hypotenuse:
     \[
     13^2 + x^2 = 85^2
     \]
     \[
     169 + x^2 = 7225
     \]
     \[
     x^2 = 7056
     \]
     \[
     x = \sqrt{7056} = 84
     \]
   - The third number is **84**.

### Problem 11
A right triangle has a hypotenuse of 13 and one leg of 5. We need to find the other leg.

- Using the Pythagorean theorem:
  \[
  5^2 + x^2 = 13^2
  \]
  \[
  25 + x^2 = 169
  \]
  \[
  x^2 = 144
  \]
  \[
  x = \sqrt{144} = 12
  \]
- The other leg is **12**.

### Problem 12
Two joggers run 8 miles north and then 5 miles west. To find the shortest distance back, we use the Pythagorean theorem, where the legs are 8 and 5 miles.

- Calculate the hypotenuse:
  \[
  8^2 + 5^2 = c^2
  \]
  \[
  64 + 25 = c^2
  \]
  \[
  c^2 = 89
  \]
  \[
  c = \sqrt{89} \approx 9.4 \text{ miles}
  \]
- The shortest distance is **9.4 miles**.

### Problem 13
Oscar's dog house has slanted sides of 5 feet, and the base is 6 feet across. This forms an isosceles triangle with a height we need to find.

1. The base is 6 feet, so each half of the base is 3 feet.
2. Using the Pythagorean theorem with one of the right triangles formed:
   \[
   3^2 + h^2 = 5^2
   \]
   \[
   9 + h^2 = 25
   \]
   \[
   h^2 = 16
   \]
   \[
   h = \sqrt{16} = 4
   \]
- The height is **4 feet**.

### Problem 14
To avoid a pond, the path involves walking 34 meters south and 41 meters east. We need to find the direct distance.

- Using the Pythagorean theorem:
  \[
  34^2 + 41^2 = c^2
  \]
  \[
  1156 + 1681 = c^2
  \]
  \[
  c^2 = 2837
  \]
  \[
  c = \sqrt{2837} \approx 53.3 \text{ meters}
  \]
- Distance saved by walking through the pond:
  \[
  (34 + 41) - 53.3 = 75 - 53.3 = 21.7 \text{ meters}
  \]
- **21.7 meters** would be saved.

### Problem 15
The suitcase measures 24 inches long with a diagonal of 30 inches. We need to find the width.

- Using the Pythagorean theorem:
  \[
  24^2 + w^2 = 30^2
  \]
  \[
  576 + w^2 = 900
  \]
  \[
  w^2 = 324
  \]
  \[
  w = \sqrt{324} = 18 \text{ inches}
  \]
- The width is **18 inches**.

Would you like further details on any of these solutions? Here are related questions to explore:

1. How can you determine if three numbers form a right triangle?
2. What is the significance of the Pythagorean theorem in geometry?
3. How does one calculate the distance saved in problem-solving involving direct and indirect paths?
4. What is the process of finding heights in isosceles triangles with given base and side lengths?
5. How can understanding triangles help in real-life applications?

**Tip:** When working with right triangles, always check if the side lengths satisfy \( a^2 + b^2 = c^2 \) for correct triangle configuration.

Find solutions to geometry and trigonometry problems involving right triangles, distance calculations, and area coverage, including questions like finding a third side to form a right triangle, calculating distances and height using the Pythagorean theorem, and finding the width of a suitcase with a known diagonal.

Explore detailed solutions to geometry problems involving the Pythagorean theorem. Learn how to find missing sides in right triangles, calculate direct and indirect distances, determine the height of an isosceles triangle, and calculate dimensions in practical applications. Suitable for students in Grades 8-10.

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