Answer:
[tex]x=\dfrac{\sqrt{30}}{2}[/tex]
Step-by-step explanation:
The line segment x represents the altitude of an equilateral triangle with side lengths measuring [tex]\sqrt{10}[/tex] units.
The altitude of an equilateral triangle divides it into two congruent right triangles. In each right triangle, the hypotenuse is equal to the side length of the equilateral triangle, and the base of the right triangle is half the side length of the equilateral triangle. Consequently, to determine the length of x, we can apply the Pythagorean Theorem.
[tex]\boxed{\begin{array}{l}\underline{\sf Pythagorean \;Theorem} \\\\\Large\text{$a^2+b^2=c^2$}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$a$ and $b$ are the legs of the right triangle.}\\\phantom{ww}\bullet\;\textsf{$c$ is the hypotenuse (longest side) of the right triangle.}\\\end{array}}[/tex]
In this case:
[tex]a = \dfrac{\sqrt{10}}{2}[/tex]
[tex]b = x[/tex]
[tex]c = \sqrt{10}[/tex]
Substitute these values into the formula and solve for x:
[tex]\begin{aligned}\left(\dfrac{\sqrt{10}}{2}\right)^2+x^2&=\left(\sqrt{10}\right)^2\\\\\dfrac{10}{4}+x^2&=10\\\\\dfrac{5}{2}+x^2&=10\\\\x^2&=10-\dfrac{5}{2}\\\\x^2&=\dfrac{20}{2}-\dfrac{5}{2}\\\\x^2&=\dfrac{15}{2}\\\\x&=\sqrt{\dfrac{15}{2}}\\\\x&=\dfrac{\sqrt{15}}{\sqrt{2}}\\\\x&=\dfrac{\sqrt{15}\cdot \sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}\\\\x&=\dfrac{\sqrt{30}}{2}\end{aligned}[/tex]
Therefore, the length of x in simplest radical form with a rational denominator is:
[tex]\Large\boxed{\boxed{x=\dfrac{\sqrt{30}}{2}}}[/tex]
Additional Notes
Note that we only consider the positive square root of 15/2 when determining the value of x since length is always positive.
