Harshgmat wrote:

An equilateral triangle has a height of 4\(\sqrt{3}\). What is the area of this triangle?

A) 16

B) \(16√3\)

C) \(24√3\)

D) 48

E) \(32√3\)

Attachment:

equitriHfindA 09.16.18.jpg [ 37.97 KiB | Viewed 210 times ]
Good question +1 to refresh a handy tool: the altitude of an equilateral triangle divides it into two congruent 30-60-90 triangles

I. 30-60-90 triangles[/u]

This method is as quick as the formula below and IMO easier to remember.

In an equilateral triangle, each altitude = height = median

Drop an altitude, BD, from ∠B to side AC

The height, BD, divides ∆ABC into

two congruent 30-60-90 triangles. • Altitude BD is a perpendicular bisector of vertex B and opposite side AC

Resultant angles are 30, 60, and 90 degrees

• Original angles A, B, and C = 60°

• ∠ B is bisected into two 30° angles

• At D, the altitude forms two 90° angles

• Angles A and C = 60°

• Both smaller triangles are 30-60-90

30-60-90 triangles have corresponding sides opposite those angles in the ratio

\(x:x\sqrt{3}:2x\)

Height of \(4\sqrt{3}\), opposite the 60° angle, corresponds with \(x\sqrt{3}\)

\(4\sqrt{3}=x\sqrt{3}\)

\(x=4\)

Base AC = BC = \(2x=8\)

(In ∆BCD, side BC is opposite the 90° angle. Sides are equal.)

Area, \(A=\frac{b*h}{2}=\frac{8*4\sqrt{3}}{2}=16\sqrt{3}\)

Answer B

II. FormulaFor an equilateral triangle with side \(a\),

Height = \(\frac{1}{2}a*\sqrt{3}\)

Given: height = \(4\sqrt{3}\)

Side length:

\(4\sqrt{3}=\frac{1}{2}a*\sqrt{3}\)

\(8\sqrt{3}=a\sqrt{3}\)

\(a=8\) = base (all sides are equal)

Area, \(A=\frac{b*h}{2}=\frac{8*4\sqrt{3}}{2}=16\sqrt{3}\)

Answer B