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Re: In the figure above, what is the ratio of the area of triangle ABC to [#permalink]

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08 Jan 2012, 11:05

shk18 wrote:

We need the ratio of areas ie., Area(ABD)/Area(ABC)

Area(ABD)/Area(ABC) = (0.5*b1*h1)/(0.5*b2*h2)

since b1 = b2 (both triangles have same base)

Area(ABD)/Area(ABC) = h1/h2;

Since (1) provides value for this ratio, we can find ratio of area of triangles.

Statement 1 states the ratio of two specific heights of the triangles without saying which ones. Though it could be true that the ratio of the heights you mention are in this ratio, it is also possible that the ratio of 2 other heights are in this ratio ( For example from vertex A and B respectively). Therefore Insufficient.

Yes, I agree with Sony. It makes no sense to talk about *the* height of a triangle. Any triangle has 3 heights, depending which side you choose as your base. There's no reason to assume the side at the bottom of the picture is the base. It's a very badly designed question.
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Re: In the figure above, what is the ratio of the area of triangle ABC to [#permalink]

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08 Jan 2012, 13:31

IanStewart wrote:

Yes, I agree with Sony. It makes no sense to talk about *the* height of a triangle. Any triangle has 3 heights, depending which side you choose as your base. There's no reason to assume the side at the bottom of the picture is the base. It's a very badly designed question.

Thanks for your input Ian. +1

Could you help me with the following please?

I know that if 2 triangles are similar we can assume that all their heights are in proportion. If we know that all the heights of 2 triangles are in proportion, can we assume that the triangles are similar?

Re: In the figure above, what is the ratio of the area of triangle ABC to [#permalink]

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08 Jan 2012, 17:28

RULE: If sides of triangle have a proportion defined, the areas are in equal proportion (or multiples in case of squares...)

In this problem, base is same. So, in area ratios, the base cancels out. Only thing thats left is the height, for which ratios are defined. This is enough.

1. Ratio of height defined, so, area1/area2 = (1/2 x 3 x AB) / (1/2 x 4 x AB) = 3:4. Suff. 2. AB=8. No info on height. Insuff.

A.
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Re: In the figure above, what is the ratio of the area of triangle ABC to [#permalink]

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09 Jan 2012, 02:36

shinbhu wrote:

RULE: If sides of triangle have a proportion defined, the areas are in equal proportion (or multiples in case of squares...)

In this problem, base is same. So, in area ratios, the base cancels out. Only thing thats left is the height, for which ratios are defined. This is enough.

1. Ratio of height defined, so, area1/area2 = (1/2 x 3 x AB) / (1/2 x 4 x AB) = 3:4. Suff. 2. AB=8. No info on height. Insuff.

A.

I believe that you abused the rules of similar triangles.

There are 3 criteria to prove similarity. Only then will the sides of 2 triangles be in proportion.

In this question we only know that 2 of the heights of 2 triangles have a specific ratio. This doesn't tell us that all sides are in this ratio. Moreover, every triangle has 3 heights, 1 from each vertex as Ian pointed out in an earlier post.

If you have 2 random triangles and you divide 2 of their heights you will come up with a ratio. Does that mean that the triangles are similar? Does that mean that all the heights are in the same ratio? I don't think so.

I know that if 2 triangles are similar we can assume that all their heights are in proportion. If we know that all the heights of 2 triangles are in proportion, can we assume that the triangles are similar?

Say you know the lengths of the three heights in your triangle are 24, 30 and 40 (as is the case in a 30-40-50 triangle), and say the corresponding bases are a, b and c respectively. Then the area of the triangle is equal to 24a/2, and is also equal to 30b/2 and to 40c/2. These expressions are all equal, so, for example, 30b/2 = 40c/2, from which we can find the ratio of b to c (it is 3 to 4). Similarly we can find the ratio of a to b. So if we know all of our heights, we can find the ratio of the lengths of all of our sides. If, say, we double all of our heights, we'll still find that our sides are in the same ratio. So the answer to your question is 'yes'; two triangles with three heights in the same ratio must be similar (their sides must be in the same ratio).

All of that said, this is certainly not the kind of fact you would ever benefit from memorizing for the GMAT. GMAT geometry questions only require you to know a very small set of facts - you could list them all on half a page of paper. I certainly did not know the answer to your question until I set about trying to prove it, and I've never needed to know the answer for any GMAT question - the GMAT couldn't ask a question that required you to know about this, since almost no test taker will.
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Re: In the figure above, what is the ratio of the area of triangle ABC to [#permalink]

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09 Jan 2012, 14:45

IanStewart wrote:

SonyGmat wrote:

IanStewart wrote:

Could you help me with the following please?

I know that if 2 triangles are similar we can assume that all their heights are in proportion. If we know that all the heights of 2 triangles are in proportion, can we assume that the triangles are similar?

Say you know the lengths of the three heights in your triangle are 24, 30 and 40 (as is the case in a 30-40-50 triangle), and say the corresponding bases are a, b and c respectively. Then the area of the triangle is equal to 24a/2, and is also equal to 30b/2 and to 40c/2. These expressions are all equal, so, for example, 30b/2 = 40c/2, from which we can find the ratio of b to c (it is 3 to 4). Similarly we can find the ratio of a to b. So if we know all of our heights, we can find the ratio of the lengths of all of our sides. If, say, we double all of our heights, we'll still find that our sides are in the same ratio. So the answer to your question is 'yes'; two triangles with three heights in the same ratio must be similar (their sides must be in the same ratio).

All of that said, this is certainly not the kind of fact you would ever benefit from memorizing for the GMAT. GMAT geometry questions only require you to know a very small set of facts - you could list them all on half a page of paper. I certainly did not know the answer to your question until I set about trying to prove it, and I've never needed to know the answer for any GMAT question - the GMAT couldn't ask a question that required you to know about this, since almost no test taker will.

Thanks for your clear explanation.. and for the tip of not learning unnecessary things - formulas!