the triangle below is equilateral. find the length of side $x$ to the nearest tenth.

Explanation:

Step1: Recall properties of equilateral triangle

In an equilateral triangle, if we consider the right - triangle formed by the altitude. Let the side length of the equilateral triangle be (a). The altitude divides the equilateral triangle into two right - triangles, and the hypotenuse of the right - triangle is the side of the equilateral triangle, and the other two sides are half of the side of the equilateral triangle and the altitude. Here, assume the side length of the equilateral triangle is (a), and the right - triangle has hypotenuse (a), one side (\frac{a}{2}) and the other side (x). Using the Pythagorean theorem (a^{2}=x^{2}+\left(\frac{a}{2}\right)^{2}). Also, from the right - triangle with hypotenuse (\sqrt{10}), we know that if the side of the equilateral triangle is (a), then (\sqrt{10}) is related to the side of the equilateral triangle. In the right - triangle formed by the altitude of the equilateral triangle, if the side of the equilateral triangle is (a), and the right - triangle with hypotenuse (\sqrt{10}) has sides (x) and (\frac{a}{2}).

Step2: Apply Pythagorean theorem

Let the side of the equilateral triangle be (a). In the right - triangle with hypotenuse (\sqrt{10}), by the Pythagorean theorem (a^{2}=x^{2}+\left(\frac{a}{2}\right)^{2}), which can be rewritten as (a^{2}-\frac{a^{2}}{4}=x^{2}), so (\frac{3a^{2}}{4}=x^{2}). Also, in the right - triangle with hypotenuse (\sqrt{10}), we have (10 = x^{2}+\left(\frac{a}{2}\right)^{2}). Since (a^{2}=x^{2}+\left(\frac{a}{2}\right)^{2}), we know that (10=x^{2}+\frac{a^{2}}{4}). Substitute (a^{2}=\frac{4x^{2}}{3}) into (10=x^{2}+\frac{a^{2}}{4}), we get (10=x^{2}+\frac{1}{4}\times\frac{4x^{2}}{3}).

Step3: Solve the equation for (x)

[ \begin{align*} 10&=x^{2}+\frac{x^{2}}{3}\ 10&=\frac{3x^{2}+x^{2}}{3}\ 10&=\frac{4x^{2}}{3}\ x^{2}&=\frac{30}{4} = 7.5\ x&=\sqrt{7.5}\approx2.7 \end{align*} ]

Answer:

(2.7)