Question

question
find the length of the third side. if necessary, write in simplest radical form.
(image of a right triangle with legs 7 and \\(\sqrt{51}\\))

Explanation:

Step1: Identify the triangle type

The triangle is a right - triangle, so we can use the Pythagorean theorem. The Pythagorean theorem states that for a right - triangle with legs of lengths \(a\) and \(b\) and hypotenuse of length \(c\), \(c^{2}=a^{2}+b^{2}\) (if we are finding the hypotenuse) or \(a^{2}=c^{2}-b^{2}\) (if we are finding a leg), where \(c\) is the hypotenuse (the side opposite the right angle) and \(a\) and \(b\) are the legs.

In the given triangle, we have one leg \(a = \sqrt{51}\), the other leg \(b\) (let's say) and the hypotenuse \(c = 7\)? Wait, no. Wait, the right angle is between the side of length \(7\) and \(\sqrt{51}\)? Wait, no, the hypotenuse is the side opposite the right angle. Wait, maybe I got the legs and hypotenuse wrong. Let's assume that the two legs are \(x\) (the unknown side) and \(\sqrt{51}\), and the hypotenuse is \(7\)? No, that can't be, because \((\sqrt{51})^{2}=51\) and \(7^{2} = 49\), and \(51>49\), so the hypotenuse must be the side with length \(7\)? No, that's impossible. Wait, no, I must have misidentified. Wait, the side of length \(7\) and the side of length \(\sqrt{51}\) are the legs, and we need to find the hypotenuse? Wait, no, let's check:

Let the two legs be \(a=\sqrt{51}\) and \(b = x\) (the unknown side), and the hypotenuse \(c = 7\). But then by Pythagorean theorem, \(a^{2}+b^{2}=c^{2}\), so \((\sqrt{51})^{2}+x^{2}=7^{2}\), \(51 + x^{2}=49\), \(x^{2}=49 - 51=- 2\), which is impossible. So the hypotenuse must be the side of length \(7\)? No, that can't be. Wait, maybe the leg is \(7\) and the other leg is \(\sqrt{51}\), and we need to find the hypotenuse. Let's try that.

Let \(a = 7\), \(b=\sqrt{51}\), and \(c\) be the hypotenuse. Then by Pythagorean theorem \(c^{2}=a^{2}+b^{2}\)

Step2: Apply the Pythagorean theorem

Substitute \(a = 7\) and \(b=\sqrt{51}\) into the formula \(c^{2}=a^{2}+b^{2}\)

\(a^{2}=7^{2}=49\), \(b^{2}=(\sqrt{51})^{2}=51\)

So \(c^{2}=49 + 51=100\)

Then \(c=\sqrt{100}=10\)

Wait, no, wait. Wait, maybe the hypotenuse is \(7\) and one leg is \(\sqrt{51}\), and we need to find the other leg. But as we saw earlier, that would give a negative value under the square root, which is impossible. So the correct approach is that the two legs are \(7\) and \(\sqrt{51}\), and we are finding the hypotenuse.

Wait, let's re - examine the triangle. The right angle is between the side of length \(7\) and the side of length \(\sqrt{51}\), so the hypotenuse \(c\) is given by \(c^{2}=7^{2}+(\sqrt{51})^{2}\)

\(7^{2}=49\), \((\sqrt{51})^{2}=51\)

\(c^{2}=49 + 51=100\)

\(c=\sqrt{100}=10\)

Answer:

\(10\)