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30-60-90 Triangle Property Question

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Intern
Joined: 08 Nov 2012
Posts: 19

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14 Apr 2013, 10:57
The height of an equilateral triangle splits the triangle into two 30-60-90 triangles (Each 30-60-90 triangle has sides in the ratio of 1: square root of 3: 2). Because of this, the area for an equilateral triangle can be expressed in terms of one side. If we call the side of the equilateral triangle, s, the height must be (s multiplied by square root of 3) / 2 (using the 30-60-90 relationships).

Can someone explain why the height which is the square root of 3 is being divided by 2? The explanation then states... "The area of a triangle = 1/2 × base × height, so the area of an equilateral triangle can be expressed as: 1/2 × s × (s multiplied by square root of 3) / 2 .

The question i pulled this from is asking about the area of the triangle. MGMAT's explanation does not make sense to me of why the height is being divided by 2 initially.

Thanks!!!!!!
Director
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Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: 30-60-90 Triangle Property Question  [#permalink]

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14 Apr 2013, 11:23
1
Richard0715 wrote:
The height of an equilateral triangle splits the triangle into two 30-60-90 triangles (Each 30-60-90 triangle has sides in the ratio of 1: square root of 3: 2). Because of this, the area for an equilateral triangle can be expressed in terms of one side. If we call the side of the equilateral triangle, s, the height must be (s multiplied by square root of 3) / 2 (using the 30-60-90 relationships).

Can someone explain why the height which is the square root of 3 is being divided by 2? The explanation then states... "The area of a triangle = 1/2 × base × height, so the area of an equilateral triangle can be expressed as: 1/2 × s × (s multiplied by square root of 3) / 2 .

The question i pulled this from is asking about the area of the triangle. MGMAT's explanation does not make sense to me of why the height is being divided by 2 initially.

Thanks!!!!!!

Lets say that the side is $$s$$. so the side of the base of the 30-60-90 triangle is s/2. Do you agree? Now the side of the 30-60-90 triangle are in the proportion
$$1x$$ : $$x\sqrt{3}$$ : $$2x$$. because $$1x$$ is refered to the base and the base = s/2 we can conclude that x=s/2
So the height will be $$x\sqrt{3}$$ where x=s/2 so H=$$\frac{s}{2}\sqrt{3}$$

Let me know if it's clear now
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4485
Re: 30-60-90 Triangle Property Question  [#permalink]

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17 Apr 2013, 10:54
Richard0715 wrote:
The height of an equilateral triangle splits the triangle into two 30-60-90 triangles (Each 30-60-90 triangle has sides in the ratio of 1: square root of 3: 2). Because of this, the area for an equilateral triangle can be expressed in terms of one side. If we call the side of the equilateral triangle, s, the height must be (s multiplied by square root of 3) / 2 (using the 30-60-90 relationships).

Can someone explain why the height which is the square root of 3 is being divided by 2? The explanation then states... "The area of a triangle = 1/2 × base × height, so the area of an equilateral triangle can be expressed as: 1/2 × s × (s multiplied by square root of 3) / 2 .

The question i pulled this from is asking about the area of the triangle. MGMAT's explanation does not make sense to me of why the height is being divided by 2 initially.

Thanks!!!!!!

Dear Richard,
It appears that Zarrolou already gave a fine explanation to this specific question. Here are a couple links that provide some more background:
http://magoosh.com/gmat/2012/gmat-math- ... emorizing/
http://magoosh.com/gmat/2012/the-gmats- ... triangles/
Let me know if you have any further questions.
Mike
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Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Manager
Joined: 14 Feb 2016
Posts: 58
GMAT 1: 710 Q48 V40

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30 May 2016, 03:48
Richard0715 wrote:
The height of an equilateral triangle splits the triangle into two 30-60-90 triangles (Each 30-60-90 triangle has sides in the ratio of 1: square root of 3: 2). Because of this, the area for an equilateral triangle can be expressed in terms of one side. If we call the side of the equilateral triangle, s, the height must be (s multiplied by square root of 3) / 2 (using the 30-60-90 relationships).

Can someone explain why the height which is the square root of 3 is being divided by 2? The explanation then states... "The area of a triangle = 1/2 × base × height, so the area of an equilateral triangle can be expressed as: 1/2 × s × (s multiplied by square root of 3) / 2 .

The question i pulled this from is asking about the area of the triangle. MGMAT's explanation does not make sense to me of why the height is being divided by 2 initially.

Thanks!!!!!!

This derivation helped me with the question. When we draw a perpendicular from C to AB , the two triangles ( X and Y) automatically become congruent.
Attachments

Equilateral Triangle.PNG [ 622.05 KiB | Viewed 3626 times ]

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GMAT 1: 620 Q49 V26
GMAT 2: 590 Q49 V21
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Re: 30-60-90 Triangle Property Question  [#permalink]

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30 May 2016, 06:43
Richard0715 wrote:
The height of an equilateral triangle splits the triangle into two 30-60-90 triangles (Each 30-60-90 triangle has sides in the ratio of 1: square root of 3: 2). Because of this, the area for an equilateral triangle can be expressed in terms of one side. If we call the side of the equilateral triangle, s, the height must be (s multiplied by square root of 3) / 2 (using the 30-60-90 relationships).

Can someone explain why the height which is the square root of 3 is being divided by 2? The explanation then states... "The area of a triangle = 1/2 × base × height, so the area of an equilateral triangle can be expressed as: 1/2 × s × (s multiplied by square root of 3) / 2 .

The question i pulled this from is asking about the area of the triangle. MGMAT's explanation does not make sense to me of why the height is being divided by 2 initially.

Thanks!!!!!!

In order to solve your problem, we need to learn some basic formulas related to the area of a triangle.
1. Area of a triangle A = (1/2)*base*height
2. Area of a triangle A = sq.rt(s*(s-a)*(s-b)*(s-c)) where s= (a+b+c)/2
In an equilateral triangle all sides are equal. So s=3a/2. Where a is the side of the triangle
Per second point, A = (sq.rt(3)/4)*a^2
Per 1st point , A = a*height/2
If you compare both of them you will get the height.

Re: 30-60-90 Triangle Property Question   [#permalink] 30 May 2016, 06:43