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by MBA.Aspirant » Thu Jun 30, 2011 1:43 am
It says in a math review that "if a polygon has n sides, it can be divided into n-2 triangles". If you consider a hexagon for example it can be divided into 6 equilateral triangles, not 4. so how's this statement true?
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by GMATGuruNY » Thu Jun 30, 2011 1:59 am
MBA.Aspirant wrote:It says in a math review that "if a polygon has n sides, it can be divided into n-2 triangles". If you consider a hexagon for example it can be divided into 6 equilateral triangles, not 4. so how's this statement true?
If you draw line segments from one vertex to all of the other vertices, the polygon will be divided into n-2 triangles.
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by amit2k9 » Thu Jun 30, 2011 2:47 am
all right good point here.That means the question has to be slightly rephrased.
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by MBA.Aspirant » Thu Jun 30, 2011 5:57 am
GMATGuruNY wrote:
MBA.Aspirant wrote:It says in a math review that "if a polygon has n sides, it can be divided into n-2 triangles". If you consider a hexagon for example it can be divided into 6 equilateral triangles, not 4. so how's this statement true?
If you draw line segments from one vertex to all of the other vertices, the polygon will be divided into n-2 triangles.
Thanks GMATGuru.

I have another question: how do we establish similarity between triangles?

Given this figure we're asked to calculate the area of ADG. Given AB=BC=CD and area of CDE is 42.
we know that DC: DA is 1:3, but how does he get the same ratio for the bases as well and establish similarity for the 3 triangles?

Thanks ahead
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by Frankenstein » Thu Jun 30, 2011 6:11 am
Hi,
If two triangles are similar then the ratios of each of the corresponding sides is same
Two triangles are similar if the set of angles of one triangle is same as the set of angles of another triangle.
Consider triangle CDE and ADG
E = G = 90degrees
D = D(common angle)
So, angle C has to be equal to angle A because sum of angles of any triangle is constant(180 degrees).
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by MBA.Aspirant » Sat Jul 02, 2011 1:38 pm
Thanks Frankenstein. so for 2 triangles to be similar they can have congruent/common angle and/or the ratio of their sides is the same. Does similar shape determine similarity?
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by winniethepooh » Sat Jul 02, 2011 7:50 pm
You can also use the parallel lines property to prove the angles to be congruent!
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by carred » Sun Jul 03, 2011 9:33 am
MBA.Aspirant wrote:
GMATGuruNY wrote:
MBA.Aspirant wrote: I have another question: how do we establish similarity between triangles?

Given this figure we're asked to calculate the area of ADG. Given AB=BC=CD and area of CDE is 42.
we know that DC: DA is 1:3, but how does he get the same ratio for the bases as well and establish similarity for the 3 triangles?

Thanks ahead

So how is the actual area of the bigger triangle calculated?

Thanks a lot
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by GMATGuruNY » Sun Jul 03, 2011 10:13 am
Since similar triangles have the same combination of angles, they bear the same shape. One triangle essentially is a bigger version of the other.

The ratio of the two bases, like that of the two heights, is the same as the ratio of each pair of corresponding sides.

Given 2 similar triangles where each side of the larger triangle is x times the corresponding side of the smaller triangle:
If B = base of the larger triangle and b = base of the smaller triangle, B = xb.
If H = height of the larger triangle and h = height of the smaller triangle, H = xh.
Since the area of the larger triangle = (1/2)BH = (1/2)(xb)(xh) = x²(1/2)bh, the area of the larger triangle is x² times the area of the smaller triangle.

In the problem above, since DA:DC = 3:1, B = 3b and H = 3h, implying that the area of ADG is 3²=9 times the area of CDE.
Since CDE = 42, ADG = 9*42 = 378.
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by MBA.Aspirant » Sun Jul 03, 2011 10:18 am
carred wrote: So how is the actual area of the bigger triangle calculated?

Thanks a lot
AB = BC = CD

so AD = 3 CD

Since triangle CDE and ADG are similar, given their equal and common angles (equal 90 degrees angles, parallel angles at ABC, common angle at D) we know the ratio of their legs is also 1/3.

so AG = 3 CE, DG = 3 DE

since area of CDE = 1/2 DE * CE = 42
then area of ADG = 1/2 DG * AG
= 1/2 3*DE * 3*CE
= 9 (1/2 *DE *CE)
= 9 * 42
= 378
Last edited by MBA.Aspirant on Sun Jul 03, 2011 12:48 pm, edited 2 times in total.
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by carred » Sun Jul 03, 2011 10:20 am
Wow that was super fast and really comprehensible!

Now the learning can go on :D
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