Question

find the length of the third side. if necessary, round to the nearest tenth. 6 8

Explanation:

Step1: Identify triangle type

It's a right triangle, so use Pythagorean theorem: \(a^2 + b^2 = c^2\), where \(c\) is hypotenuse, \(a = 6\), \(b =?\), \(c = 8\)? Wait, no—wait, the right angle is between 6 and the unknown side, so 6 and unknown are legs, 8 is hypotenuse? Wait, no, let's check. Wait, the triangle has legs 6 and \(x\), hypotenuse 8? Wait, no, maybe I mixed up. Wait, in a right triangle, hypotenuse is the longest side. Wait, 8 is longer than 6, so 8 is hypotenuse, 6 is one leg, let the other leg be \(x\). Then by Pythagoras: \(6^2 + x^2 = 8^2\)? Wait, no, wait—wait, maybe the legs are 6 and the unknown, and hypotenuse is 8? Wait, no, 6 and 8: wait, no, the right angle is between 6 and the unknown side, so 6 and unknown are legs, hypotenuse is 8? Wait, no, 6² + x² = 8²? Wait, 6² is 36, 8² is 64. Then x² = 64 - 36 = 28? No, that can't be. Wait, maybe I got the legs wrong. Wait, maybe 6 and 8 are legs, hypotenuse is \(x\). Oh! That makes more sense. Because 6 and 8 are the two legs, right angle between them, so hypotenuse is \(x\). So Pythagorean theorem: \(a^2 + b^2 = c^2\), where \(a = 6\), \(b = 8\), \(c = x\).

Step2: Apply Pythagorean theorem

\(x^2 = 6^2 + 8^2\)
\(x^2 = 36 + 64\)
\(x^2 = 100\)
\(x = \sqrt{100}\)
\(x = 10\)

Wait, but wait, the diagram: the right angle is between 6 and the unknown side, and the other side is 8. Wait, maybe I misread the diagram. Wait, the triangle has a right angle, one leg 6, hypotenuse 8? No, that would mean the other leg is \(\sqrt{8^2 - 6^2} = \sqrt{64 - 36} = \sqrt{28} \approx 5.3\), but that contradicts. Wait, maybe the 8 is a leg, 6 is a leg, hypotenuse is \(x\). Then \(x = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\). That makes sense, because 10 is longer than 8 and 6, so hypotenuse. So the third side (hypotenuse) is 10.

Answer:

10