Question

find the length of the third side. if necessary, write in simplest radical form.
6
2√13

Explanation:

Step1: Identify triangle type

It's a right triangle, so use Pythagorean theorem: \(a^2 + b^2 = c^2\), where \(c\) is hypotenuse, \(a,b\) legs. Here, hypotenuse \(c = 2\sqrt{13}\)? Wait, no—wait, one leg is 6, hypotenuse? Wait, no, maybe I mixed. Wait, the side \(2\sqrt{13}\) and leg 6? Wait, no, let's check: right triangle, so let's denote the unknown leg as \(x\), hypotenuse as \(2\sqrt{13}\)? Wait, no, \(2\sqrt{13}\) squared is \(4\times13 = 52\), 6 squared is 36. Wait, if \(c\) is hypotenuse, then \(a^2 + b^2 = c^2\). Wait, maybe the hypotenuse is \(2\sqrt{13}\)? No, 6 is a leg, another leg \(x\), hypotenuse \(2\sqrt{13}\)? Wait, no, \(6^2 + x^2 = (2\sqrt{13})^2\)? Wait, no, that would be \(36 + x^2 = 52\), so \(x^2 = 16\), \(x=4\). Wait, but maybe the hypotenuse is 6? No, \(2\sqrt{13}\) is about 7.21, which is longer than 6? Wait, no, \(2\sqrt{13} \approx 7.21\), 6 is shorter. Wait, so hypotenuse is \(2\sqrt{13}\)? No, 6 is a leg, another leg \(x\), hypotenuse \(c\). Wait, the diagram: right angle, so two legs and hypotenuse. Let's assume the hypotenuse is \(2\sqrt{13}\)? No, 6 is a leg, so let's let the unknown leg be \(x\), hypotenuse \(c = 2\sqrt{13}\)? Wait, no, \(6^2 + x^2 = (2\sqrt{13})^2\)? Wait, \(6^2 = 36\), \((2\sqrt{13})^2 = 4\times13 = 52\). So \(36 + x^2 = 52\) → \(x^2 = 16\) → \(x = 4\). Wait, but maybe the hypotenuse is 6? No, \(2\sqrt{13}\) is longer than 6, so hypotenuse must be \(2\sqrt{13}\), and one leg is 6, so the other leg is 4. Wait, let's confirm:

Step1: Apply Pythagorean theorem

Let the unknown side be \(x\), right triangle, so \(a^2 + b^2 = c^2\). Let’s say \(a = 6\), \(c = 2\sqrt{13}\) (hypotenuse), then \(6^2 + x^2 = (2\sqrt{13})^2\).

Step2: Calculate squares

\(6^2 = 36\), \((2\sqrt{13})^2 = 4 \times 13 = 52\).

Step3: Solve for \(x^2\)

\(36 + x^2 = 52\) → \(x^2 = 52 - 36 = 16\).

Step4: Find \(x\)

\(x = \sqrt{16} = 4\).

Answer:

4