Question

finding the length of an altitude
the sides of an equilateral triangle are 8 units long. what is the length of the altitude of the triangle?
○ ( 5sqrt{2} ) units
○ ( 4sqrt{3} ) units
○ ( 10sqrt{2} ) units
○ ( 16sqrt{8} ) units

Explanation:

Step1: Recall properties of equilateral triangles

In an equilateral triangle, the altitude splits the triangle into two 30 - 60 - 90 right triangles. The side of the equilateral triangle is the hypotenuse of the right triangle, and half of the side of the equilateral triangle is one of the legs. If the side length of the equilateral triangle is \(a = 8\) units, then the length of the leg adjacent to the 30° angle (half of the base) is \(\frac{a}{2}=\frac{8}{2} = 4\) units.

Step2: Apply the Pythagorean theorem

Let \(h\) be the altitude (the other leg of the right triangle). By the Pythagorean theorem, for a right triangle with hypotenuse \(c\) and legs \(a\) and \(b\), \(c^{2}=a^{2}+b^{2}\). Here, \(c = 8\) (hypotenuse, side of equilateral triangle), one leg \(a = 4\) (half of the base), and the other leg \(b=h\) (altitude). So we have \(8^{2}=4^{2}+h^{2}\).

Step3: Solve for \(h\)

First, calculate \(8^{2}=64\) and \(4^{2} = 16\). Then the equation becomes \(64=16 + h^{2}\). Subtract 16 from both sides: \(h^{2}=64 - 16=48\). Take the square root of both sides: \(h=\sqrt{48}\). Simplify \(\sqrt{48}=\sqrt{16\times3}=\sqrt{16}\times\sqrt{3}=4\sqrt{3}\) units.

Answer:

\(4\sqrt{3}\) units (corresponding to the option " \(4\sqrt{3}\) units")