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 339 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
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