Question

1. equilateral triangle abc is inscribed in a circle with center o and a radius of 1 as shown below. the height of the triangle is bd. what is the area of triangle abc?

a. \\(\sqrt{3}\\)
b. \\(3\sqrt{3}\\)
c. \\(\frac{3\sqrt{3}}{2}\\)
d. \\(\frac{3\sqrt{3}}{4}\\)

Explanation:

⚡ Using what you learned: properties of triangles · right triangle trigonometry (soh cah toa)

Step 1: Identify the properties of the equilateral triangle and its height

For an equilateral triangle \( \triangle ABC \) with side length \( s \), the height \( BD \) is perpendicular to the base \( AC \) and bisects it.
This splits \( \triangle ABC \) into two congruent \( 30^\circ \)-\( 60^\circ \)-\( 90^\circ \) right triangles, \( \triangle ABD \) and \( \triangle CBD \).

The relationship between the side length \( s \) and the height \( h = BD \) is:

$$ h = s \frac{\sqrt{3}}{2} $$

Step 2: Relate the radius of the circumscribed circle to the height

The center \( O \) of the circumscribed circle (circumcenter) of an equilateral triangle lies on the height \( BD \).
Since the triangle is equilateral, the circumcenter \( O \) is also the centroid. The centroid divides the median (which is also the height \( BD \)) in a \( 2:1 \) ratio.

Thus, the circumradius \( R \) (the distance from the vertex \( B \) to the center \( O \)) is:

$$ R = BO = 1 $$

The remaining segment \( OD \) is:

$$ OD = \frac{1}{2} R = \frac{1}{2} $$

Therefore, the total height \( BD \) is:

$$ BD = BO + OD = 1 + \frac{1}{2} = \frac{3}{2} $$

Step 3: Calculate the side length of the equilateral triangle

Using the height \( BD = \frac{3}{2} \):

$$ s \frac{\sqrt{3}}{2} = \frac{3}{2} $$

$$ s = \frac{3}{\sqrt{3}} = \sqrt{3} $$

Step 4: Calculate the area of the equilateral triangle

The area \( A \) of an equilateral triangle with side length \( s \) is:

$$ A = \frac{\sqrt{3}}{4} s^2 $$

Substitute \( s = \sqrt{3} \):

$$ A = \frac{\sqrt{3}}{4} (\sqrt{3})^2 = \frac{3\sqrt{3}}{4} $$

Answer:

D. \(\frac{3\sqrt{3}}{4}\)