Question

find the length of side x in simplest radical form with a rational denominator. answer attempt 1 out of 2 x =

Explanation:

Step1: Identify Triangle Type

The triangle is a right - isosceles triangle (two angles are \(45^{\circ}\), so the two legs are equal? Wait, no, wait. Wait, in a right - angled triangle with angles \(45^{\circ},45^{\circ},90^{\circ}\), the legs are equal, and the hypotenuse \(h\) and leg \(l\) are related by \(h = l\sqrt{2}\), or \(l=\frac{h}{\sqrt{2}}\). Wait, let's check the sides. The right angle is between the two legs, and the angles opposite the legs: one leg is \(x\), the other leg? Wait, no, the side labeled 1: wait, the triangle has a right angle, one angle \(45^{\circ}\), another \(45^{\circ}\), so it's an isosceles right triangle. So the two legs are equal? Wait, no, wait the side labeled 1: is that a leg or the hypotenuse? Wait, the angles: the angle of \(45^{\circ}\) is opposite to side \(x\)? Wait, no, let's label the triangle. Let the right angle be at the top, one angle at the bottom left is \(45^{\circ}\), bottom right is \(45^{\circ}\). So the sides: the side between the right angle and the bottom left \(45^{\circ}\) angle is \(x\) (a leg), the side between the right angle and the bottom right \(45^{\circ}\) angle: wait, no, the side opposite the bottom left \(45^{\circ}\) angle: wait, the side labeled 1 is the side between the two \(45^{\circ}\) angles? No, the side labeled 1 is the base, between the two non - right angles. Wait, in a right - angled isosceles triangle, the legs are equal, and the hypotenuse is leg\(\times\sqrt{2}\). Wait, maybe I got the sides wrong. Let's use trigonometry. Let's take the angle of \(45^{\circ}\), the adjacent side to the \(45^{\circ}\) angle (bottom right) is \(x\), and the opposite side? Wait, no, the right angle is at the top. So the legs are the two sides forming the right angle, and the hypotenuse is the side opposite the right angle. Wait, the angle at the bottom right is \(45^{\circ}\), so for that angle, the adjacent leg is \(x\), and the opposite leg: wait, no, the side labeled 1: is that the hypotenuse? Wait, no, in a right - angled triangle, the hypotenuse is the longest side. Wait, maybe the side labeled 1 is a leg, and \(x\) is the other leg? But no, the angles are \(45^{\circ}\), so it's isosceles, so legs are equal. Wait, no, maybe I made a mistake. Wait, let's use sine or cosine. Let's take the angle of \(45^{\circ}\), \(\sin(45^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos(45^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). Wait, if we take the angle at the bottom right (\(45^{\circ}\)), the adjacent side to this angle is \(x\), and the opposite side: wait, the side opposite to the \(45^{\circ}\) angle (bottom right) is \(x\)? No, wait, the right angle is at the top. So the sides: the leg from top to bottom left is \(x\) (length \(x\)), the leg from top to bottom right: let's call it \(y\), and the hypotenuse is from bottom left to bottom right (length 1). In an isosceles right triangle, \(x = y\), and by Pythagoras, \(x^{2}+y^{2}=1^{2}\). Since \(x = y\), we have \(2x^{2}=1\), so \(x^{2}=\frac{1}{2}\), \(x=\frac{\sqrt{2}}{2}\). Wait, that makes sense. Because in an isosceles right triangle, legs are equal, hypotenuse \(h = l\sqrt{2}\), so \(l=\frac{h}{\sqrt{2}}\). If \(h = 1\), then \(l=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\) (after rationalizing the denominator).

Step2: Rationalize the Denominator (if needed)

We had \(x=\frac{1}{\sqrt{2}}\). To rationalize the denominator, we multiply numerator and denominator by \(\sqrt{2}\): \(x=\frac{1\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}=\frac{\sqrt{2}}{2}\).

Answer:

\(\frac{\sqrt{2}}{2}\)