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# An equilateral triangle has a height of

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Director
Joined: 08 Jun 2013
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An equilateral triangle has a height of  [#permalink]

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16 Sep 2018, 07:25
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5% (low)

Question Stats:

88% (00:50) correct 13% (01:20) wrong based on 32 sessions

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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) $$32√3$$

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Re: An equilateral triangle has a height of  [#permalink]

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16 Sep 2018, 08:04
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) $$32√3$$

let each side be a
the height will divide the triangle into two
each of the two will be a right triangle
using Pythagoras , we have
16*3 = $$a^2$$ - ($$a^2$$/4)
upon solving we get a = 8
area = $$\frac{1}{2}$$ 8 * 4$$\sqrt{3}$$ = $$16√3$$
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Joined: 22 May 2016
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Re: An equilateral triangle has a height of  [#permalink]

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16 Sep 2018, 16:52
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, $$A=\frac{b*h}{2}=\frac{8*4\sqrt{3}}{2}=16\sqrt{3}$$

II. Formula
For 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, $$A=\frac{b*h}{2}=\frac{8*4\sqrt{3}}{2}=16\sqrt{3}$$

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Re: An equilateral triangle has a height of   [#permalink] 16 Sep 2018, 16:52
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