Question

question
find the length of side x in simplest radical form with a rational denominator.
(an image of a right triangle with two 45° angles, one leg labeled 1 and the other leg labeled x)
answer attempt 1 out of 2
x = input box submit answer

Explanation:

Step1: Identify triangle type

The triangle is a right - isosceles triangle (two angles are $45^{\circ}$, one is $90^{\circ}$), so the two legs are equal, and the hypotenuse $h$ and leg $l$ have the relation $h = l\sqrt{2}$.

Step2: Apply trigonometric ratio or Pythagorean theorem

We can use the sine function. $\sin(45^{\circ})=\frac{x}{1}$ (since the hypotenuse is 1). We know that $\sin(45^{\circ})=\frac{\sqrt{2}}{2}$, so $x = \sin(45^{\circ})\times1=\frac{\sqrt{2}}{2}$. Or using Pythagorean theorem: Let the legs be $x$ and 1 (wait, no, in a 45 - 45 - 90 triangle, legs are equal. Wait, the side with length 1 is a leg? Wait, no, looking at the triangle, the two angles of $45^{\circ}$ are at the base and the other vertex. Wait, actually, in a right - isosceles triangle, if the hypotenuse is $c$ and legs are $a$ and $b$ ($a = b$), then $c=a\sqrt{2}$. But here, if one leg is 1, and we want to find the other leg $x$? Wait, no, maybe I mixed up. Wait, the angle of $45^{\circ}$, so $\sin(45^{\circ})=\frac{opposite}{hypotenuse}$. Wait, the right angle is between $x$ and the side of length 1? Wait, no, the triangle has a right angle, and two $45^{\circ}$ angles. So the two legs are $x$ and 1, and the hypotenuse is... Wait, no, in a triangle with angles $45^{\circ},45^{\circ},90^{\circ}$, the legs are equal. Wait, maybe the side of length 1 is the hypotenuse? Wait, no, the right angle is between $x$ and the side adjacent to the $45^{\circ}$ angle. Wait, let's re - examine. The triangle has a right angle, and two angles of $45^{\circ}$. So it's an isosceles right triangle, so the two legs are equal. Let's assume that the hypotenuse is 1? No, the side labeled 1 is a leg? Wait, no, the angle of $45^{\circ}$: $\sin(45^{\circ})=\frac{x}{hypotenuse}$. Wait, maybe the side with length 1 is the hypotenuse. Then, if the hypotenuse $c = 1$, and the legs are $x$ (equal legs), then from $c=x\sqrt{2}$, we get $x=\frac{c}{\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$ (rationalizing the denominator). Yes, that makes sense. So if the hypotenuse is 1, then each leg $x=\frac{\sqrt{2}}{2}$.

Answer:

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