Question

the figure below is a square. find the length of side ( x ) to the nearest tenth.

Explanation:

Step1: Identify the triangle type

The square's diagonal splits it into two right - isosceles triangles. In a square, the diagonal \(d\) and side \(x\) are related by the Pythagorean theorem. For a square, if the side is \(x\) and diagonal is \(d\), then \(d^{2}=x^{2}+x^{2}=2x^{2}\) (since in a right - triangle formed by two sides of the square and the diagonal, the two legs are of length \(x\) and the hypotenuse is the diagonal \(d\)). Here, the diagonal \(d = 1\).

Step2: Solve for \(x\)

We have the equation \(d^{2}=2x^{2}\). Substitute \(d = 1\) into the equation:
\[1^{2}=2x^{2}\]
\[1 = 2x^{2}\]
Then, solve for \(x^{2}\):
\[x^{2}=\frac{1}{2}\]
Take the square root of both sides:
\[x=\sqrt{\frac{1}{2}}\]
Simplify \(\sqrt{\frac{1}{2}}=\frac{\sqrt{1}}{\sqrt{2}}=\frac{1}{\sqrt{2}}\approx\frac{1}{1.4142}\approx0.7071\)
To the nearest tenth, we look at the hundredth place. The number \(0.7071\) has a hundredth digit of \(0\). Since \(0\lt5\), we round down. But wait, actually \(\sqrt{\frac{1}{2}}\approx0.707\), and to the nearest tenth, we consider the first decimal place. The value \(0.707\) is closer to \(0.7\)? Wait, no, wait. Wait, \(\sqrt{\frac{1}{2}}\approx0.707\), when rounding to the nearest tenth, we look at the digit in the hundredth place. The digit in the hundredth place is \(0\), but wait, no, \(\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\approx\frac{1.414}{2} = 0.707\). Wait, maybe I made a mistake. Wait, let's re - calculate.

Wait, the Pythagorean theorem: in a square, diagonal \(d\), side \(s\). Then \(d^{2}=s^{2}+s^{2}=2s^{2}\). So \(s=\frac{d}{\sqrt{2}}\). Given \(d = 1\), so \(s=\frac{1}{\sqrt{2}}\approx0.707\). Rounding to the nearest tenth: the tenths place is \(7\), the hundredths place is \(0\). Wait, no, \(0.707\): the first decimal is \(7\) (tenths place), the second is \(0\) (hundredths), the third is \(7\) (thousandths). When rounding to the nearest tenth, we look at the hundredths digit. Since the hundredths digit is \(0\), which is less than \(5\), we keep the tenths digit as it is? Wait, no, \(0.707\) is approximately \(0.7\) when rounded to the nearest tenth? Wait, no, wait, \(\sqrt{\frac{1}{2}}\approx0.707\), and \(0.707\) is closer to \(0.7\) than to \(0.8\)? Wait, no, \(0.707\) is \(0.7 + 0.007\), so it's closer to \(0.7\). But wait, maybe I messed up the initial step. Wait, the diagonal of the square is \(1\), so the two sides of the right triangle are \(x\) (the side of the square) and the hypotenuse is \(1\). So by Pythagoras: \(x^{2}+x^{2}=1^{2}\), \(2x^{2}=1\), \(x^{2}=\frac{1}{2}\), \(x = \sqrt{\frac{1}{2}}\approx0.7\) (to the nearest tenth). Wait, but let's check with a calculator: \(\sqrt{0.5}\approx0.7071\), which rounds to \(0.7\) when rounded to the nearest tenth.

Answer:

\(0.7\)