Question

find the length of side x to the nearest tenth.

Explanation:

Step1: Identify Triangle Type

The triangle has a right angle and two \(45^\circ\) angles, so it's a 45 - 45 - 90 triangle. In such triangles, the legs are equal, and the hypotenuse \(c = a\sqrt{2}\) (where \(a\) is a leg). Wait, no—wait, here, the side labeled 1: wait, no, let's check angles. The right angle, and two \(45^\circ\) angles, so the two legs are equal? Wait, no, wait the side labeled 1: wait, maybe I misread. Wait, the triangle: right - angled, two \(45^\circ\) angles, so it's an isosceles right triangle. So the legs are equal? Wait, no, wait the side with length 1: is that a leg or the hypotenuse? Wait, in a 45 - 45 - 90 triangle, the hypotenuse is \(leg\times\sqrt{2}\). Wait, let's see: the angles are \(45^\circ\), \(45^\circ\), \(90^\circ\). So the two legs are equal, and the hypotenuse is \(leg\sqrt{2}\). Wait, but in the diagram, the side labeled 1: is that a leg or the hypotenuse? Wait, the right angle is at the bottom - right, so the two legs are the one with \(x\) and the vertical leg? Wait, no, the angles: one \(45^\circ\) at the top, one \(45^\circ\) at the bottom - left, right angle at bottom - right. So the two legs are the side with \(x\) (bottom leg) and the right - hand leg (vertical). The side with length 1 is the hypotenuse? Wait, no, wait: in a triangle, the side opposite the \(90^\circ\) angle is the hypotenuse. So the hypotenuse is the side opposite the right angle, so the side with length 1 is the hypotenuse? Wait, no, that can't be. Wait, no, the angles: \(45^\circ\), \(45^\circ\), \(90^\circ\). So the sides: legs (opposite \(45^\circ\)) are equal, hypotenuse (opposite \(90^\circ\)) is \(leg\sqrt{2}\). Wait, maybe I got it reversed. Let's use trigonometry. Let's take the angle at the bottom - left, \(45^\circ\). The side adjacent to it is \(x\), the hypotenuse is 1? No, wait, no—wait, the side labeled 1: let's see the triangle again. The triangle has vertices: top ( \(45^\circ\) ), bottom - left ( \(45^\circ\) ), bottom - right (right angle). So the sides: from bottom - left to bottom - right is \(x\) (leg), bottom - right to top is a leg (let's say \(y\)), and bottom - left to top is 1 (hypotenuse). Wait, no, in a right - triangle, the hypotenuse is the longest side. So if it's a 45 - 45 - 90 triangle, the legs are equal, and hypotenuse is \(leg\sqrt{2}\). So if we have a leg \(x\), then hypotenuse \(= x\sqrt{2}\). But in the diagram, the side with length 1: is that a leg or the hypotenuse? Wait, maybe I made a mistake. Wait, let's use sine. \(\sin(45^\circ)=\frac{opposite}{hypotenuse}\). The angle at the bottom - left is \(45^\circ\), the opposite side is the right - hand leg (vertical), and the hypotenuse is 1? No, that would mean \(opposite = \sin(45^\circ)\times1=\frac{\sqrt{2}}{2}\approx0.7\), but that doesn't make sense. Wait, no, maybe the side with length 1 is a leg. So if it's a leg, then the other leg \(x\) is equal to it (since it's a 45 - 45 - 90 triangle, legs are equal). Wait, that must be it. Because in a 45 - 45 - 90 triangle, the two legs are congruent. So if one leg is 1, the other leg \(x\) is also 1? But that seems too simple. Wait, no, maybe I misread the diagram. Wait, the side labeled 1: is it a leg or the hypotenuse? Wait, the right angle is at the bottom - right, so the two legs are the horizontal ( \(x\) ) and vertical (let's say \(y\)) sides, and the hypotenuse is the side connecting the top and bottom - left. So the angles: top angle \(45^\circ\), bottom - left angle \(45^\circ\), so the triangle is isosceles with legs \(x\) and \(y\) equal, and hypotenuse \(1\). Wait, that would mean \(x = y\), and by Pythagoras, \(x^{2}+x^{2}=1^{2}\), so \(2x^{2}=1\), \(x^{2}=\frac{1}{2}\), \(x=\frac{\sqrt{2}}{2}\approx0.7\). Wait, that's different. So which is it? Let's clarify: in a 45 - 45 - 90 triangle, the ratio of legs to hypotenuse is \(1:1:\sqrt{2}\). So if the hypotenuse is \(c\), then each leg \(a = \frac{c}{\sqrt{2}}\). If a leg is \(a\), then hypotenuse \(c = a\sqrt{2}\). So in the diagram, the side with length 1: is it a leg or the hypotenuse? The right angle is at the bottom - right, so the hypotenuse is the side opposite the right angle, which is the side connecting the top vertex and the bottom - left vertex (length 1). So the two legs are \(x\) (bottom - right to bottom - left) and the vertical leg (bottom - right to top). So since the triangle has two \(45^\circ\) angles, the legs are equal, so \(x=\) vertical leg. Then by Pythagoras, \(x^{2}+x^{2}=1^{2}\), so \(2x^{2}=1\), \(x^{2}=\frac{1}{2}\), \(x = \frac{\sqrt{2}}{2}\approx0.7\) (to the nearest tenth, \(0.7\) or \(0.71\approx0.7\)? Wait, \(\sqrt{2}\approx1.414\), so \(\frac{\sqrt{2}}{2}\approx0.707\approx0.7\) when rounded to the nearest tenth. Wait, but maybe the side with length 1 is a leg. Let's check that case. If the side with length 1 is a leg, then the other leg \(x\) is also 1 (since it's a 45 - 45 - 90 triangle, legs are equal), and the hypotenuse is \(1\times\sqrt{2}\approx1.4\). But which is more likely? Let's look at the diagram again. The side with length 1 is the side connecting the bottom - left and top vertices, which is opposite the right angle, so it's the hypotenuse. So the legs are \(x\) and the vertical leg, equal in length. So \(x=\frac{1}{\sqrt{2}}\approx0.7\).

Step2: Calculate \(x\)

Using the property of 45 - 45 - 90 triangles: if hypotenuse \(c = 1\), then leg \(a=\frac{c}{\sqrt{2}}\). So \(x=\frac{1}{\sqrt{2}}\). Rationalizing the denominator, \(x = \frac{\sqrt{2}}{2}\approx0.707\approx0.7\) (to the nearest tenth). Alternatively, using trigonometry: \(\sin(45^\circ)=\frac{x}{1}\)? No, wait, \(\sin(45^\circ)=\frac{opposite}{hypotenuse}\). The opposite side to the \(45^\circ\) angle (bottom - left) is the vertical leg, and \(x\) is the adjacent side. Wait, \(\cos(45^\circ)=\frac{adjacent}{hypotenuse}=\frac{x}{1}\), so \(x = \cos(45^\circ)=\frac{\sqrt{2}}{2}\approx0.7\).

Answer:

\(0.7\) (or more precisely, \(\frac{\sqrt{2}}{2}\approx0.7\) when rounded to the nearest tenth)