Question
Ashlyn wants to build a bridge that will include a triangle made of metal pieces. She has two wooden pieces, one of which measures 8 inches and the other of which measures 12 inches. Which could be the length of the third piece of week in order for her to form a triangle?
• 1 inch
• 4 inches
• 5 inches
• 20 inches
• 1 inch
• 4 inches
• 5 inches
• 20 inches
Answers
Answered by
GPT-4o mini
To determine the possible lengths of the third side of a triangle given two sides, we can use the triangle inequality theorem. According to this theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
In Ashlyn's case, we have two sides measuring 8 inches and 12 inches. Let's denote the length of the third side as \( x \). The triangle inequality gives us the following three conditions:
1. \( 8 + 12 > x \)
2. \( 8 + x > 12 \)
3. \( 12 + x > 8 \)
Now, let's solve each inequality:
1. From \( 8 + 12 > x \):
\[
20 > x \quad \text{or} \quad x < 20
\]
2. From \( 8 + x > 12 \):
\[
x > 12 - 8 \quad \text{or} \quad x > 4
\]
3. From \( 12 + x > 8 \):
\[
x > 8 - 12 \quad \text{or} \quad x > -4 \quad \text{(always true since } x \text{ will be positive)}
\]
Combining the relevant conditions from inequalities 1 and 2, we find that \( x \) must satisfy:
\[
4 < x < 20
\]
Now we can evaluate the possible lengths for the third piece:
- 1 inch: **Not valid** (since \( 1 < 4 \))
- 4 inches: **Not valid** (since \( 4 \) is not greater than \( 4 \))
- 5 inches: **Valid** (since \( 4 < 5 < 20 \))
- 20 inches: **Not valid** (since \( 20 \) is not less than \( 20 \))
Therefore, the only possible length for the third piece that would allow Ashlyn to form a triangle is **5 inches**.
In Ashlyn's case, we have two sides measuring 8 inches and 12 inches. Let's denote the length of the third side as \( x \). The triangle inequality gives us the following three conditions:
1. \( 8 + 12 > x \)
2. \( 8 + x > 12 \)
3. \( 12 + x > 8 \)
Now, let's solve each inequality:
1. From \( 8 + 12 > x \):
\[
20 > x \quad \text{or} \quad x < 20
\]
2. From \( 8 + x > 12 \):
\[
x > 12 - 8 \quad \text{or} \quad x > 4
\]
3. From \( 12 + x > 8 \):
\[
x > 8 - 12 \quad \text{or} \quad x > -4 \quad \text{(always true since } x \text{ will be positive)}
\]
Combining the relevant conditions from inequalities 1 and 2, we find that \( x \) must satisfy:
\[
4 < x < 20
\]
Now we can evaluate the possible lengths for the third piece:
- 1 inch: **Not valid** (since \( 1 < 4 \))
- 4 inches: **Not valid** (since \( 4 \) is not greater than \( 4 \))
- 5 inches: **Valid** (since \( 4 < 5 < 20 \))
- 20 inches: **Not valid** (since \( 20 \) is not less than \( 20 \))
Therefore, the only possible length for the third piece that would allow Ashlyn to form a triangle is **5 inches**.