Question

what is the length of the altitude of the equilateral triangle below?
a. $\sqrt{48}$
b. 6
c. $2\sqrt{3}$
d. 2
e. $6\sqrt{3}$
f. 36

Explanation:

Step1: Identify the triangle type

We have a right - angled triangle (one angle is \(90^{\circ}\)) with hypotenuse \(4\sqrt{3}\) and one leg \(2\sqrt{3}\). We can use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(c\) is the hypotenuse and \(a,b\) are the legs of the right - triangle. Let the altitude be \(a\), one leg \(b = 2\sqrt{3}\) and hypotenuse \(c=4\sqrt{3}\).

Step2: Apply the Pythagorean theorem

According to the Pythagorean theorem \(a=\sqrt{c^{2}-b^{2}}\). Substitute \(c = 4\sqrt{3}\) and \(b=2\sqrt{3}\) into the formula:

$$
LATEXBLOCK0
$$

We can also use the properties of a \(30 - 60-90\) triangle. In a \(30 - 60 - 90\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite \(30^{\circ}\) is the shortest side (let's call it \(x\)), the side opposite \(60^{\circ}\) is \(x\sqrt{3}\) and the hypotenuse is \(2x\). Here, the side opposite \(30^{\circ}\) is \(2\sqrt{3}\), so the side opposite \(60^{\circ}\) (which is the altitude \(a\)) should be \(x\sqrt{3}\), where \(x = 2\sqrt{3}\)? Wait, no. Wait, in the right - triangle, the angle of \(30^{\circ}\) has the opposite side \(2\sqrt{3}\), the angle of \(60^{\circ}\) has the opposite side \(a\). In a \(30 - 60-90\) triangle, \(\tan(60^{\circ})=\frac{a}{2\sqrt{3}}\), and since \(\tan(60^{\circ})=\sqrt{3}\), we have \(a = 2\sqrt{3}\times\sqrt{3}=6\).

Answer:

B. 6