To find the length of the diagonal [tex]\( x \)[/tex] of the rectangular napkin, we can use the relationship between the length [tex]\( L \)[/tex], width [tex]\( w \)[/tex], and the properties of a right triangle. Here's the step-by-step solution:
1. Let the width of the napkin be [tex]\( w \)[/tex].
2. The length [tex]\( L \)[/tex] is given to be twice the width, so:
[tex]\[
L = 2w
\][/tex]
3. The diagonal of the rectangle forms a right triangle with the length and width as the two legs. According to the Pythagorean Theorem:
[tex]\[
x^2 = L^2 + w^2
\][/tex]
4. Substitute the expressions for [tex]\( L \)[/tex]:
[tex]\[
x^2 = (2w)^2 + w^2
\][/tex]
5. Simplify the equation:
[tex]\[
x^2 = 4w^2 + w^2
\][/tex]
[tex]\[
x^2 = 5w^2
\][/tex]
6. To solve for [tex]\( x \)[/tex], take the square root of both sides:
[tex]\[
x = \sqrt{5w^2}
\][/tex]
7. Simplify the square root expression:
[tex]\[
x = w\sqrt{5}
\][/tex]
Given that we need to express [tex]\( x \)[/tex] in the form [tex]\(\frac{\sqrt{a}}{b}\)[/tex], where [tex]\( w \)[/tex] is a part of the geometric dimensions:
[tex]\[
x = \frac{\sqrt{5} \cdot w}{1}
\][/tex]
However, for the question at hand, we are specifically asked for the form:
[tex]\[
x = \frac{\sqrt{a}}{b}
\][/tex]
where [tex]\( w \)[/tex] is implicitly included. In this case, the answer for [tex]\( a \)[/tex] and [tex]\( b \)[/tex] is:
[tex]\[
\boxed{5}
\][/tex]
[tex]\( \boxed{1} \)[/tex]
Thus, the correct answer in the box is:
[tex]\[
x = \frac{\sqrt{5}}{1}
\][/tex]
