Question

the figure below is a square. find the length of side x in simplest radical form with a rational denominator.
figure of a square (diamond shape) with a diagonal of length 10 and side length x

Explanation:

Step1: Analyze the square's diagonals

A square's diagonals are equal and bisect each other at 90 degrees. Here, the diagonal is split into two equal parts, each of length \( \frac{10}{2}=5 \). The triangle formed by two sides of the square and half the diagonal is a right - isosceles triangle (since in a square, the sides are equal and the diagonal bisects the angles, so the two legs of the right triangle are equal).

Step2: Apply the Pythagorean theorem

For a right - triangle with legs of length \( a \) and \( b \) and hypotenuse of length \( c \), the Pythagorean theorem is \( a^{2}+b^{2}=c^{2} \). In our right - isosceles triangle, \( a = b=x \) (the side of the square) and the two legs formed by half - diagonals are both 5? Wait, no. Wait, actually, when we draw a diagonal of a square, it divides the square into two right - triangles. The diagonal of the square is 10. Let the side of the square be \( x \). In a square, using the Pythagorean theorem, if the side is \( x \) and the diagonal is \( d \), then \( x^{2}+x^{2}=d^{2} \) (since in a square, the two legs of the right - triangle are the sides of the square). Given \( d = 10 \), so \( 2x^{2}=10^{2}=100 \).

Step3: Solve for \( x \)

First, divide both sides of the equation \( 2x^{2}=100 \) by 2: \( x^{2}=\frac{100}{2} = 50 \). Then, take the square root of both sides: \( x=\sqrt{50} \). Simplify \( \sqrt{50}=\sqrt{25\times2}=5\sqrt{2} \).

Answer:

\( 5\sqrt{2} \)