Question

directions: classify the triangle with the given side lengths as acute, right, obtuse, or not a triangle.

1. 11, 13, 25
2. 9, 15, 19
3. 21, 28, 35
4. 8, 12, 14

not a △
acute
right
obtuse
not a △
acute
right
obtuse
not a △
acute
right
obtuse
not a △
acute
right
obtuse

Explanation:

Step1: Check triangle inequality

For a triangle with side lengths \(a,b,c\) (\(c\) is the longest side), the triangle inequality \(a + b>c\) must hold.

For \(7:11,13,25\)

\(11+13 = 24<25\), so it is not a triangle.

For \(8:9,15,19\)

\(9 + 15=24>19\). Now, use the Pythagorean - like formula \(a^{2}+b^{2}\) and \(c^{2}\) (\(c = 19\), \(a = 9\), \(b = 15\)).
\(a^{2}+b^{2}=9^{2}+15^{2}=81 + 225=306\), \(c^{2}=19^{2}=361\). Since \(a^{2}+b^{2}

For \(9:21,28,35\)

\(21+28 = 49>35\). \(a = 21\), \(b = 28\), \(c = 35\).
\(a^{2}+b^{2}=21^{2}+28^{2}=441+784 = 1225\), \(c^{2}=35^{2}=1225\). Since \(a^{2}+b^{2}=c^{2}\), the triangle is right - angled.

For \(10:8,12,14\)

\(8 + 12=20>14\). \(a = 8\), \(b = 12\), \(c = 14\).
\(a^{2}+b^{2}=8^{2}+12^{2}=64 + 144=208\), \(c^{2}=14^{2}=196\). Since \(a^{2}+b^{2}>c^{2}\), the triangle is acute.

Answer:

1. Not a \(\triangle\); 8. Obtuse; 9. Right; 10. Acute