Question

Classify each triangle as acute, right, or obtuse
6, 2sqrt55, 17

Answer 1

To classify the triangle with sides 6, \(2\sqrt{55}\), and 17, we can use the triangle inequality theorem and the relationship between the lengths of the sides to determine whether the triangle is acute, right, or obtuse.

1. **Check if it's a valid triangle**:
- The sum of the lengths of any two sides should be greater than the length of the third side:
- \(6 + 2\sqrt{55} > 17\)
- \(6 + 17 > 2\sqrt{55}\)
- \(2\sqrt{55} + 17 > 6\)

- Calculate \(2\sqrt{55} \approx 2 \times 7.416 = 14.832\) (since \(\sqrt{55} \approx 7.416\))
- \(6 + 14.832 \approx 20.832 > 17\) (holds true)
- \(6 + 17 = 23 > 14.832\) (holds true)
- \(14.832 + 17 \approx 31.832 > 6\) (holds true)

Since all inequalities hold, it is a valid triangle.

2. **Classify the triangle**:
- We can use the relationship \(a^2 + b^2\) compared to \(c^2\) (where \(c\) is the longest side):
- Let \(a = 6\), \(b = 2\sqrt{55}\), \(c = 17\).
- Calculate \(a^2 + b^2\) and \(c^2\):
- \(a^2 = 6^2 = 36\)
- \(b^2 = (2\sqrt{55})^2 = 4 \times 55 = 220\)
- \(c^2 = 17^2 = 289\)

- Now, compute \(a^2 + b^2\):
- \(a^2 + b^2 = 36 + 220 = 256\)

- Compare \(a^2 + b^2\) with \(c^2\):
- We find \(256 < 289\).

3. **Conclusion**:
- Since \(a^2 + b^2 < c^2\), the triangle is classified as **obtuse**.

Thus, the triangle with sides 6, \(2\sqrt{55}\), and 17 is an **obtuse triangle**.