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Re: If two sides of a triangle are 12 and 8 [#permalink]

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24 Oct 2012, 12:19

7

This post was BOOKMARKED

Bunuel - I have few other questions related to this. Please correct me if Im wrong with any of the statements below.

1/ Given 2 sides of the triangle max area can be go by forming 90* between them?

2/ Given the hypotenuse of a right angle triangle I can find out the max area by forming isosceles traingle (I find the other 2 sides using the relationship x:x:x root 2)

3/For a given perimeter equilateral triangle has the largest area. Perimeter = 3+4+5 = 12. The area of an equilateral triangle = (s²/4)√3. (s = 12/3)

4/ For a given area equilateral triangle has the smallest perimeter. Given area = 18 then smallest perimeter possible = 1/2 a^2 = 18 so a = 6 then perimeter 6+6+6 =18....

5/ sum of any 2 sides should be > than the 3side and diff should be < than 3rd side...

Are there any other relationship should I be aware of?

If two sides of a triangle are 12 and 8, which of the following could be the area of the triangle?

I. 35 II. 48 III. 56

A. I only B. I and II only C. I and III only D. II and III only E. I, II, and III

Given two sides of a triangle, the greatest area is obtained when there is 90 degrees between them.

Thus the area cannot be more than 12*8*1/2=48.

Look at the diagram below:

Attachment:

Triangles.png [ 2.36 KiB | Viewed 22258 times ]

If we consider 12 as a base, then you can see that the greatest height is when it coincides with the second side. Now, if we rotate the second side we can make the height close to zero, thus we can have any height from 0 to 8, which implies that we can have any area more than 0 but not more than 48 (12*8*1/2=48).

Re: If two sides of a triangle are 12 and 8, which of the [#permalink]

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23 May 2015, 21:26

Can you explain how we could get an area of 35? Base*Height would have to be equal to 70 right? How could we get B*H to be equal to 70 if the two sides are 12-8 which neither go into 70? Can we assume that the third side doesn't have to be an integer?

Can you explain how we could get an area of 35? Base*Height would have to be equal to 70 right? How could we get B*H to be equal to 70 if the two sides are 12-8 which neither go into 70? Can we assume that the third side doesn't have to be an integer?

Given two sides, the greatest area will be obtained when the sides are at right angle to each other i.e. they form a right angled triangle. For any other type of triangle, the two sides will not form the base and the height, thus \(\frac{1}{2}\) * multiplication of the sides will not give you the area of the triangle.

So for example, for a scalene triangle, one of the sides could be the base and then the height to that base from the opposite vertex would need to be calculated which may not be an integer.

Thus \(\frac{1}{2}\) * base ( either 12 or 8) * height can be equal to 35

This question is an off-shoot of a Triangle Inequality Theorem question - in those questions, you're given 2 of the sides of the triangle and you're asked for what the third side COULD be. The lengths of the 2 sides dictate the minimum and maximum possible values for the third side and that same logic can be used here.

To emphasize the logic, you should sketch out some quick pictures....

First, the greatest area will be formed when the two sides form a RIGHT angle. With sides of 8 and 12, you have an area of (1/2)(8)(12) = 48.

Next, draw a picture in which the "8 side" does NOT form a right angle with the "12 side"....put the "8 side" at an angle so that it's almost "on top of" the "12 side".... This area is clearly SMALLER than 48 (in fact, it's just a little greater than 0).

If you draw a picture with the "8 side" almost "in line, but going in the other direction" with the "12 side" (so that you form a really long, thin triangle), THAT area is also really small (just a little greater than 0).

These examples serve as proof that any area from "almost 0" up to 48, inclusive is possible. In this question, that means that 35 and 48 are possible areas.

If two sides of a triangle are 12 and 8, which of the [#permalink]

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20 Jul 2015, 03:33

1

This post received KUDOS

Jp27 wrote:

Bunuel - I have few other questions related to this. Please correct me if Im wrong with any of the statements below.

1/ Given 2 sides of the triangle max area can be go by forming 90* between them?

2/ Given the hypotenuse of a right angle triangle I can find out the max area by forming isosceles traingle (I find the other 2 sides using the relationship x:x:x root 2)

3/For a given perimeter equilateral triangle has the largest area. Perimeter = 3+4+5 = 12. The area of an equilateral triangle = (s²/4)√3. (s = 12/3)

4/ For a given area equilateral triangle has the smallest perimeter. Given area = 18 then smallest perimeter possible = 1/2 a^2 = 18 so a = 6 then perimeter 6+6+6 =18....

5/ sum of any 2 sides should be > than the 3side and diff should be < than 3rd side...

Are there any other relationship should I be aware of?

I know trignometry is not expected in GMAT, but this particular problem becomes very easy if you know just 2 little concepts of trignometry.

I Concept: Area of a triangle = \((1/2)a.b.sin(\alpha\)), where \(a\) and \(b\) are any two sides of the triangle and \(\alpha\) is the angle contained by these two sides.

II Concept: \(sin(\alpha\)) varies between \(0\) and \(1\) (becomes maximum at \(\alpha\) \(=90∘\)) So, in this case area will vary between \(0\) and \(48\).

In other words, all the values between 0 and 48 are possible.

There are few problems of area maximization in a triangle which are difficult to be visualized but can be dealt quickly by this method.

If two sides of a triangle are 12 and 8, which of the following could be the area of the triangle?

I. 35 II. 48 III. 56

A. I only B. I and II only C. I and III only D. II and III only E. I, II, and III

CONCEPT: - Maximum Area of the Triangle for given two sides will be obtained when the two sides are taken as perpendicular to each other - Minimum Area of the Triangle for given two sides can always be as small as zero

i.e. \(Area_{Max} = (1/2)*12*8 = 48\) i.e. \(Area_{Min} ≈ 0\)

Answer: Option B
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Re: If two sides of a triangle are 12 and 8, which of the [#permalink]

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30 Nov 2015, 10:30

What about the triangles properties?

"any side must be smaller than the sum of the other sides and greater than their difference". I thought it was not possible to have a side way smaller. The properties are not applied in this case?

EMPOWERgmatRichC wrote:

Hi healthjunkie,

This question is an off-shoot of a Triangle Inequality Theorem question - in those questions, you're given 2 of the sides of the triangle and you're asked for what the third side COULD be. The lengths of the 2 sides dictate the minimum and maximum possible values for the third side and that same logic can be used here.

To emphasize the logic, you should sketch out some quick pictures....

First, the greatest area will be formed when the two sides form a RIGHT angle. With sides of 8 and 12, you have an area of (1/2)(8)(12) = 48.

Next, draw a picture in which the "8 side" does NOT form a right angle with the "12 side"....put the "8 side" at an angle so that it's almost "on top of" the "12 side".... This area is clearly SMALLER than 48 (in fact, it's just a little greater than 0).

If you draw a picture with the "8 side" almost "in line, but going in the other direction" with the "12 side" (so that you form a really long, thin triangle), THAT area is also really small (just a little greater than 0).

These examples serve as proof that any area from "almost 0" up to 48, inclusive is possible. In this question, that means that 35 and 48 are possible areas.

That rule applies to ALL triangles, regardless of what the question is asking about. Here though, we're NOT asked for the third side, we're asked for the possible AREAS when we have those two specific side lengths.

With a side of 8 and a side of 12, the third side would fall into the falling range:

Re: If two sides of a triangle are 12 and 8, which of the [#permalink]

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02 Dec 2015, 08:31

hello,

Can anyone tell how area of 0 is possible??as if height reaches 0 it wont be triangle at all right??please explain me how we got 0 as minimum area..may be by drawing some picture..thanks in advance

Re: If two sides of a triangle are 12 and 8, which of the [#permalink]

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02 Dec 2015, 08:50

vaidhaichaturvedi wrote:

hello,

Can anyone tell how area of 0 is possible??as if height reaches 0 it wont be triangle at all right??please explain me how we got 0 as minimum area..may be by drawing some picture..thanks in advance

If you read EMPOWERgmatRichC 's reply above, he is not saying that the area can be = 0 (there is a "<" sign and not \(\leq\)sign next to 0). You are correct that if you bring height to 0 , then the triangle will no longer be a triangle.

Re: If two sides of a triangle are 12 and 8, which of the [#permalink]

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28 Sep 2017, 09:10

EMPOWERgmatRichC wrote:

Hi gabriela2015,

That rule applies to ALL triangles, regardless of what the question is asking about. Here though, we're NOT asked for the third side, we're asked for the possible AREAS when we have those two specific side lengths.

With a side of 8 and a side of 12, the third side would fall into the falling range:

4 < third side < 20

The range of the areas will still end up being:

0 < Area <= 48

GMAT assassins aren't born, they're made, Rich

Could you tell me please how you got the possible range of the third side (4<third side < 20) Thanks

Re: If two sides of a triangle are 12 and 8, which of the [#permalink]

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28 Sep 2017, 09:13

mohd_int wrote:

EMPOWERgmatRichC wrote:

Hi gabriela2015,

That rule applies to ALL triangles, regardless of what the question is asking about. Here though, we're NOT asked for the third side, we're asked for the possible AREAS when we have those two specific side lengths.

With a side of 8 and a side of 12, the third side would fall into the falling range:

4 < third side < 20

The range of the areas will still end up being:

0 < Area <= 48

GMAT assassins aren't born, they're made, Rich

Could you tell me please how you got the possible range of the third side (4<third side < 20) Thanks

There's a Geometry rule called the Triangle Inequality Theorem - and the rule is that the sum of ANY 2 sides of a triangle MUST be greater than the third side. Thus, if we call the 3 sides of a triangle A, B and C, then...

A+B > C A+C > B B+C > A

All 3 inequalities MUST exist for a triangle to exist. Here's a simple example to prove the point: Can you have a triangle with sides of 1, 1 and 100? Try drawing it. You would NOT have an actual triangle, you would have a really long line with 2 tiny lines attached at the ends (and those tiny lines would NOT meet).

So if you have one side of a triangle that is 12, then the other two sides MUST sum to a total that is GREATER than 12. If one of those two sides is an 8, then the third side must be GREATER than 4. In that same way, 12+8 = 20, so the third side must be LESS than 20.

Re: If two sides of a triangle are 12 and 8, which of the [#permalink]

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30 Sep 2017, 02:01

EMPOWERgmatRichC wrote:

Hi mohd_int,

There's a Geometry rule called the Triangle Inequality Theorem - and the rule is that the sum of ANY 2 sides of a triangle MUST be greater than the third side. Thus, if we call the 3 sides of a triangle A, B and C, then...

A+B > C A+C > B B+C > A

All 3 inequalities MUST exist for a triangle to exist. Here's a simple example to prove the point: Can you have a triangle with sides of 1, 1 and 100? Try drawing it. You would NOT have an actual triangle, you would have a really long line with 2 tiny lines attached at the ends (and those tiny lines would NOT meet).

So if you have one side of a triangle that is 12, then the other two sides MUST sum to a total that is GREATER than 12. If one of those two sides is an 8, then the third side must be GREATER than 4. In that same way, 12+8 = 20, so the third side must be LESS than 20.