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the lengths of two sides of a right triangle are 5 inches and 8 inches.…

Question

the lengths of two sides of a right triangle are 5 inches and 8 inches. what is the difference between the two possible lengths of the third side of the triangle? round your answer to the nearest tenth. 3.2 inches 10.0 inches 15.7 inches 2.4 inches

Explanation:

Step1: Case 1 - 8 is hypotenuse

Let the third side be \(x\). By Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), here \(c = 8\), \(a = 5\), so \(x^{2}+5^{2}=8^{2}\).
\(x^{2}=8^{2}-5^{2}=64 - 25=39\), then \(x=\sqrt{39}\approx6.2\) inches.

Step2: Case 2 - 8 is a leg

Let the third side (hypotenuse) be \(y\). Then \(y^{2}=5^{2}+8^{2}=25 + 64 = 89\), so \(y=\sqrt{89}\approx9.4\) inches.

Step3: Find the difference

The two possible lengths are approximately \(6.2\) and \(9.4\) (wait, no, earlier miscalculation. Wait, when 8 is hypotenuse, third side is \(\sqrt{8^{2}-5^{2}}=\sqrt{64 - 25}=\sqrt{39}\approx6.2\). When 8 is a leg, hypotenuse is \(\sqrt{5^{2}+8^{2}}=\sqrt{25 + 64}=\sqrt{89}\approx9.4\)? No, wait, no, 5 and 8 are two sides, so two cases: either 8 is hypotenuse (so third side is shorter leg: \(\sqrt{8^{2}-5^{2}}=\sqrt{39}\approx6.2\)) or 8 is a leg (so hypotenuse is \(\sqrt{5^{2}+8^{2}}=\sqrt{89}\approx9.4\))? Wait, no, that can't be. Wait, no, 5 and 8: if 8 is hypotenuse, then third side is \(\sqrt{8^{2}-5^{2}}=\sqrt{64 - 25}=\sqrt{39}\approx6.2\). If 8 is a leg, then hypotenuse is \(\sqrt{5^{2}+8^{2}}=\sqrt{25 + 64}=\sqrt{89}\approx9.4\)? Wait, but the options are 3.2, 10.0, 15.7, 2.4. Wait, I must have made a mistake. Wait, maybe 8 is a leg, and the other leg is 5, so hypotenuse is \(\sqrt{5^{2}+8^{2}}=\sqrt{25 + 64}=\sqrt{89}\approx9.4\)? No, wait, 5 and 8: maybe I messed up the cases. Wait, another way: two cases for right triangle: third side \(c\), either \(c\) is hypotenuse (\(c^{2}=5^{2}+8^{2}\)) or \(8\) is hypotenuse (\(8^{2}=5^{2}+c^{2}\)). So first case: \(c=\sqrt{25 + 64}=\sqrt{89}\approx9.4\)? No, \(\sqrt{89}\) is about 9.4? Wait, 9 squared is 81, 10 squared is 100, so \(\sqrt{89}\approx9.4\). Second case: \(c=\sqrt{8^{2}-5^{2}}=\sqrt{39}\approx6.2\). Then the difference is \(9.4 - 6.2 = 3.2\)? Wait, no, 9.4 - 6.2 is 3.2? Wait, 9.4 - 6.2 = 3.2. So the difference is approximately 3.2 inches. Wait, but let's recalculate:

First case: 8 is hypotenuse, third side \(a\): \(a=\sqrt{8^{2}-5^{2}}=\sqrt{64 - 25}=\sqrt{39}\approx6.245\)

Second case: 8 is a leg, third side (hypotenuse) \(b\): \(b=\sqrt{5^{2}+8^{2}}=\sqrt{25 + 64}=\sqrt{89}\approx9.434\)

Difference: \(9.434 - 6.245\approx3.2\) inches. So the answer is 3.2 inches.

Answer:

3.2 inches