QUESTION IMAGE
Question
find the length of side x in simplest radical form with a rational denominator.
Step1: Identify the triangle type
The triangle is an isosceles right triangle (two legs equal, right angle), so we use the Pythagorean theorem: \(a^2 + b^2 = c^2\). Here, \(a = b=\sqrt{11}\), \(c = x\).
Step2: Apply Pythagorean theorem
Substitute \(a = b=\sqrt{11}\) into the formula: \(x^2 = (\sqrt{11})^2 + (\sqrt{11})^2\).
Step3: Simplify the equation
Calculate \((\sqrt{11})^2 = 11\), so \(x^2 = 11 + 11 = 22\). Then \(x=\sqrt{22}\), but simplify \(\sqrt{22}\) as \(\sqrt{2\times11}\), which can't be simplified further, but wait, no—wait, in an isosceles right triangle, the hypotenuse is leg \(\times\sqrt{2}\). Wait, maybe I messed up. Wait, the legs are equal, so if legs are \(l\), hypotenuse \(h = l\sqrt{2}\). Here, legs are \(\sqrt{11}\), so \(x=\sqrt{11}\times\sqrt{2}=\sqrt{22}\)? Wait, no, wait: Pythagorean theorem: \(x^2 = (\sqrt{11})^2 + (\sqrt{11})^2 = 11 + 11 = 22\), so \(x = \sqrt{22}\)? But \(\sqrt{22}\) is already simplified? Wait, no, wait, maybe the triangle is isosceles right, so legs are equal, right angle, so hypotenuse is leg \(\times\sqrt{2}\). So leg is \(\sqrt{11}\), so hypotenuse \(x=\sqrt{11}\times\sqrt{2}=\sqrt{22}\). Wait, but let's check again.
Wait, the two legs are both \(\sqrt{11}\) (since the marks indicate they are equal), and it's a right triangle. So Pythagorean theorem: \(x^2 = (\sqrt{11})^2 + (\sqrt{11})^2 = 11 + 11 = 22\), so \(x = \sqrt{22}\). But \(\sqrt{22}\) can be written as \(\sqrt{2\times11}\), which can't be simplified, but wait, maybe I made a mistake. Wait, no, the problem says "simplest radical form with a rational denominator", but here there's no denominator. Wait, maybe the triangle is a right isosceles, so legs are equal, so hypotenuse is leg sqrt(2). So leg is sqrt(11), so hypotenuse is sqrt(11)sqrt(2) = sqrt(22). Yes, that's correct.
Wait, but let's re-express:
\(x^2 = (\sqrt{11})^2 + (\sqrt{11})^2\)
\(x^2 = 11 + 11\)
\(x^2 = 22\)
\(x = \sqrt{22}\) (since length is positive)
But \(\sqrt{22}\) is already in simplest radical form, and there's no denominator, so that's it.
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\(\sqrt{22}\)