Question

the triangle below is isosceles. find the length of side x in simplest radical form with a rational denominator.
triangle image with legs 10 and x? wait, no, the triangle has a right angle, one leg 10, hypotenuse x? wait, the image shows a right isosceles triangle? wait, the ocr text: the triangle below is isosceles. find the length of side x in simplest radical form with a rational denominator. the image is a right triangle (right angle at the bottom), one leg is 10, the other leg (wait, no, in a right isosceles triangle, legs are equal. wait, the triangle is isosceles and right-angled? so its a right isosceles triangle, so legs are equal, hypotenuse is leg sqrt(2). wait, the side labeled 10 is a leg, so the other leg is also 10? no, wait, the side labeled x is the hypotenuse? wait, the image: the triangle has a right angle, one leg is 10, the other leg (wait, no, the side labeled x is a side, and the leg is 10. wait, maybe its a right isosceles triangle, so legs are equal, so if one leg is 10, the other leg is 10, then hypotenuse x = 10sqrt(2). but the ocr text is as above, with the image of the triangle (right-angled, one leg 10, side x).

Explanation:

Step1: Identify triangle type

It's a right - isosceles triangle, so the two legs are equal. One leg is 10, so the other leg is also 10.

Step2: Apply Pythagorean theorem

For a right triangle, $a^{2}+b^{2}=c^{2}$, where $a = 10$, $b = 10$, and $c=x$.
So $x^{2}=10^{2}+10^{2}=100 + 100=200$.

Step3: Simplify the radical

$x=\sqrt{200}=\sqrt{100\times2}=10\sqrt{2}$ (rational denominator as is, since $\sqrt{2}$ in denominator would need rationalizing, but here we have $10\sqrt{2}$ which has a rational denominator).

Answer:

$10\sqrt{2}$