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Triangle trouble

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Triangle trouble

by mj78ind » Sun Jun 13, 2010 7:20 pm
What is the height of triangle ABC?
1. BC is the base and the sum of all the sides is 15 cms
2. The ratio of the sides BC:CA:AB::2:2:1

Official answer is C .. ....... but I am not sure

I am more interested in the logic.

Thanks
Last edited by mj78ind on Sun Jun 13, 2010 9:29 pm, edited 1 time in total.
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Source: — Data Sufficiency |

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by Rahul@gurome » Sun Jun 13, 2010 8:27 pm
Statement (1) alone is not suffucient because only the sum of two sides is given. With this obviously we cannot calculate the height.
Statement (2) alone is also not sufficient because only the ratio of sides is given and to know the height, we need absoloute values of sides.
Let us combine both and check.
Let the sides BC, CA and AB be 2x, 2x and x.
So from (1), 2x+x = 3x = 15 since CA and AB are sides.
Or x = 5.
So all the sides are 10, 10 and 5.
So semiperimeter "s" is (10+10+5)/2 = 12.5.
So area of triangle is {s(s-10)(s-10)(s-5)}^(1/2) = k say.
So (1/2)*height*10 = k.
So height is (2k)/10 = k/5.
Here since it a DS question, we need not calculate k.
The correct answer is hence (C).
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by mj78ind » Sun Jun 13, 2010 9:29 pm
Rahul@gurome wrote:Statement (1) alone is not suffucient because only the sum of two sides is given. With this obviously we cannot calculate the height.
Statement (2) alone is also not sufficient because only the ratio of sides is given and to know the height, we need absoloute values of sides.
Let us combine both and check.
Let the sides BC, CA and AB be 2x, 2x and x.
So from (1), 2x+x = 3x = 15 since CA and AB are sides.
Or x = 5.
So all the sides are 10, 10 and 5.
So semiperimeter "s" is (10+10+5)/2 = 12.5.
So area of triangle is {s(s-10)(s-10)(s-5)}^(1/2) = k say.
So (1/2)*height*10 = k.
So height is (2k)/10 = k/5.
Here since it a DS question, we need not calculate k.
The correct answer is hence (C).
@Rahul. Great analysis!!
The statement 1 meant sum of all sides is 15 cms, but I guess the logic still holds. I will make the question explicit as well.

Thanks!
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by GMATGuruNY » Mon Jun 14, 2010 10:55 am
Something about this question troubles me.

Any side of a triangle can be called the base, and each base has a corresponding height. (Once you decide which side you want to call the base, to get the corresponding height, you draw a line from the opposite vertex so that it forms a right angle with the side you're calling the base.)

I can't imagine that GMAT would ever ask for THE height or deem one side THE base. To bypass this issue, GMAT likely would ask for the area of the triangle, since no matter which base and corresponding height are used, the resulting area will be the same.
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by kvcpk » Mon Jun 14, 2010 11:37 am
Hi,

A triangle can be formed if :
all 3 sides are given, or
2 sides , sin of angle between them are given or,
1 side and 2 angles with this side as base are given.

list might be more.. I dont rememberat the top of my mind.

In this case, In A we do not know the lengths so INSUFFICIENT.
in B only ratios are there so Insufficient.

If we combine both the statements, we kow the length of each side. So we can draw a triangle with that. Hence height can be found. Hence C is the answer.
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by Rich@VeritasPrep » Mon Jun 14, 2010 4:39 pm
I agree with Mitch.

This question would never appear on the actual GMAT, because it leaves too much room for ambiguity.

There's no way to ask for THE height, because any given height of a triangle depends on the corresponding base.

I suppose the prompt could refer to a particular base, but I also agree that the GMAT would more than likely ask for the area.

What was the source of this question?
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by mj78ind » Mon Jun 14, 2010 7:05 pm
@Raz, Mitch, Rahul and kvcpk

Thanks guys! I think even though the question may not be exactly "GMAT material" I learnt from it. Stuff like I should revise other formulas for area of triangles - the sine formula, s(s-a) etc....

The source was -
https://www.testpreppractice.net/GMAT/Fr ... ncy-1.aspx it is the 14th question
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