Question

the triangle below is isosceles. find the length of side x in simplest radical form with a rational denominator.
image of a right isosceles triangle with one leg labeled 3 and the hypotenuse labeled x
answer attempt 1 out of 2
x =
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Explanation:

Step1: Identify triangle type

It's a right isosceles triangle, so legs are equal (both 3).

Step2: Apply Pythagoras theorem

$x^2 + 3^2 = (\text{hypotenuse})^2$? No, wait—wait, if legs are equal, hypotenuse is $3\sqrt{2}$, but wait no—wait, no: if it's right isosceles, legs are equal, so if one leg is 3, the other leg is also 3, hypotenuse is $3\sqrt{2}$. But wait, the question says find side x—assuming x is the hypotenuse? Wait no, wait: wait, no, maybe I mixed up. Wait, no—wait, if it's a right isosceles triangle, the two legs are equal, so if one leg is 3, the other leg is 3, hypotenuse is $3\sqrt{2}$. But wait, the problem says "simplest radical form with rational denominator"—but $3\sqrt{2}$ has no denominator. Wait, maybe x is a leg and hypotenuse is 3? Oh, right! Maybe I misread: if hypotenuse is 3, then legs are equal: let legs be x, so $x^2 + x^2 = 3^2$ → $2x^2 =9$ → $x^2=9/2$ → $x=3/\sqrt{2}$ → rationalize denominator: $x=3\sqrt{2}/2$. That makes sense (since it needs rational denominator).

Step1: Set up Pythagoras equation

$x^2 + x^2 = 3^2$

Step2: Simplify the equation

$2x^2 = 9$ → $x^2 = \frac{9}{2}$

Step3: Solve for x and rationalize

$x = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$

Answer:

$\frac{3\sqrt{2}}{2}$