Question 10 asks to find the length of side $x$ in simplest radical form with a rational denominator, given that the triangle is isosceles and the hypotenuse has length $\sqrt{11}$.

1. Problem Description

Question 10 asks to find the length of side $x$ in simplest radical form with a rational denominator, given that the triangle is isosceles and the hypotenuse has length $\sqrt{11}$.

2. Solution Steps

Since the triangle is isosceles and a right triangle, the two legs must be equal in length, both having length $x$.

Using the Pythagorean theorem, we have:

$x^2 + x^2 = (\sqrt{11})^2$

$2x^2 = 11$

$x^2 = \frac{11}{2}$

$x = \sqrt{\frac{11}{2}}$

To rationalize the denominator, multiply the numerator and denominator inside the square root by 2:

$x = \sqrt{\frac{11 \times 2}{2 \times 2}} = \sqrt{\frac{22}{4}} = \frac{\sqrt{22}}{\sqrt{4}} = \frac{\sqrt{22}}{2}$

3. Final Answer

$x = \frac{\sqrt{22}}{2}$