Question

the sides of a triangle have lengths 41, 39, and 28. what kind of triangle is it?
acute
right
obtuse

Explanation:

Step1: Identify the longest side

The sides are 41, 39, 28. The longest side is 41.

Step2: Apply the Pythagorean theorem converse

For a triangle with sides \(a\), \(b\), \(c\) (where \(c\) is the longest side), we check \(a^{2}+b^{2}\) vs \(c^{2}\).
Calculate \(39^{2}+28^{2}\):
\(39^{2}=1521\), \(28^{2} = 784\), so \(39^{2}+28^{2}=1521 + 784=2305\)
Calculate \(41^{2}=1681\)
Wait, no, wait. Wait, I made a mistake. Wait, 39 squared is 1521, 28 squared is 784, sum is 2305. 41 squared is 1681? No, 40 squared is 1600, 41 squared is \(40^{2}+2\times40\times1 + 1^{2}=1600 + 80+1 = 1681\)? Wait, no, that can't be. Wait, 39, 28, 41. Wait, maybe I mixed up. Wait, let's recalculate. 39 squared: 3939. 4040=1600, minus 40 - 39 = 1, so (40 - 1)^2=40^2 - 2401+1^2=1600 - 80 + 1=1521. 28 squared: 2828 = 784. Sum: 1521+784=2305. 41 squared: 4141. 4041=1640, 141=41, so 1640 + 41=1681? Wait, that's less than 2305? No, that can't be. Wait, no, 41 is longer than 39 and 28, but 39 and 28 sum squared is more than 41 squared? Wait, no, the formula is: if \(a^{2}+b^{2}>c^{2}\), then acute; \(a^{2}+b^{2}=c^{2}\), right; \(a^{2}+b^{2}41: 4040=1600, 401=40, 140=40, 11=1, so (40 + 1)^2=40^2+2401 + 1^2=1600+80 + 1=1681. Wait, but 39 and 28: 39 is 39, 28 is 28. Wait, no, maybe I got the longest side wrong. Wait, 41 is the longest side? Wait, 39 is 39, 28 is 28, 41 is 41. So 41 is the longest. But 39^2 +28^2=2305, 41^2=1681. Wait, that would mean \(a^{2}+b^{2}>c^{2}\), so the triangle is acute? But that seems odd. Wait, no, wait, maybe I mixed up the sides. Wait, maybe the longest side is 39? No, 41 is longer than 39. Wait, 41>39>28. So c=41, a=39, b=28. Then \(a^{2}+b^{2}=39^{2}+28^{2}=1521 + 784=2305\), \(c^{2}=41^{2}=1681\). Since \(2305>1681\), so \(a^{2}+b^{2}>c^{2}\), so the triangle is acute? But that contradicts. Wait, no, wait, I think I made a mistake in 41 squared. Wait, 4141: 4041=1640, 141=41, 1640+41=1681. Yes. 3939=1521, 2828=784. 1521+784=2305. 2305>1681. So according to the Pythagorean theorem converse, if \(a^{2}+b^{2}>c^{2}\), the triangle is acute. So the triangle is acute.

Wait, but let's check again. Maybe I had the sides wrong. The problem says 41, 39, 28. So 41 is the longest. So \(39^{2}+28^{2}=1521 + 784 = 2305\), \(41^{2}=1681\). Since \(2305>1681\), so the triangle is acute.

Answer:

acute