Question

find the length of side x to the nearest tenth.

Explanation:

Step1: Identify Triangle Type

The triangle is isosceles right - angled (two equal sides, right angle). In an isosceles right - angled triangle, if the hypotenuse is \( c \) and the legs are \( a = b=x \), the Pythagorean theorem is \( a^{2}+b^{2}=c^{2} \). Here, \( c = \sqrt{8} \), and \( a = b=x \), so \( x^{2}+x^{2}=(\sqrt{8})^{2} \).

Step2: Simplify the Equation

Simplify the left - hand side and the right - hand side. The left - hand side is \( 2x^{2} \), and the right - hand side is \( 8 \) (since \( (\sqrt{8})^{2}=8 \)). So we have the equation \( 2x^{2}=8 \).

Step3: Solve for \( x^{2} \)

Divide both sides of the equation \( 2x^{2}=8 \) by 2. We get \( x^{2}=\frac{8}{2}=4 \)? Wait, no, wait. Wait, maybe I misidentified the triangle. Wait, the hypotenuse is \( \sqrt{8} \)? Wait, no, in the diagram, the two legs are equal (marked with the same tick), and the hypotenuse is \( \sqrt{8} \)? Wait, no, maybe the legs are \( x \) and the other leg is equal to \( x \), and the hypotenuse is \( \sqrt{8} \). Wait, Pythagorean theorem: \( x^{2}+x^{2}=(\sqrt{8})^{2} \), so \( 2x^{2}=8 \), then \( x^{2}=4 \), \( x = 2 \)? But that seems too simple. Wait, maybe I got the hypotenuse wrong. Wait, maybe the side with length \( \sqrt{8} \) is a leg? Wait, the right angle is at the bottom left, so the two legs are \( x \) and the other leg (equal to \( x \)), and the hypotenuse is the side opposite the right angle, which is \( \sqrt{8} \)? Wait, no, if it's an isosceles right - angled triangle, the hypotenuse \( c=\sqrt{2}x \), where \( x \) is the length of the leg. So if \( c = \sqrt{8} \), then \( \sqrt{2}x=\sqrt{8} \).

Step4: Correct the Approach

Let's use the correct formula for isosceles right - angled triangle: hypotenuse \( c=\sqrt{2}x \), where \( x \) is the leg length. Given \( c = \sqrt{8} \), we can solve for \( x \) by the formula \( x=\frac{c}{\sqrt{2}} \). Substitute \( c = \sqrt{8} \) into the formula: \( x=\frac{\sqrt{8}}{\sqrt{2}} \).

Step5: Simplify the Fraction

Simplify \( \frac{\sqrt{8}}{\sqrt{2}} \). We know that \( \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}} \) (for \( a\geq0,b > 0 \)), so \( \frac{\sqrt{8}}{\sqrt{2}}=\sqrt{\frac{8}{2}}=\sqrt{4}=2 \). Wait, but that's 2.0 when rounded to the nearest tenth.

Answer:

\( 2.0 \)