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If two sides of a triangle are 6 and 12, respectively, which of the fo

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If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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If two sides of a triangle are 6 and 12, respectively, which of the following could NOT be the area of this triangle?

A. 1
B. 6
C. 17
D. 29
E. 38

Bunuel could you help!

[Reveal] Spoiler:
Official answer from veritas:

Solution: E.

If you know two sides of a triangle, the maximum area of the triangle can be obtained by setting these two sides as the base and height of a right triangle. For our triangle this gives us a maximum area of (6 * 12) / 2, or 36, so anything higher than this is invalid and the answer is (E). An alternative approach is to consider the minimum possible value of the triangle. Say we violated our third side rule and set the triangle's sides as 6, 12, and 18, respectively. Our "maximum" area here would be the set the base as 18, but since the triangle has no height -- the three sides only "fit together" as two parallel lines -- the height is 0 and the area is 0. But were we to extend one of the other two sides by an infinitesimally small amount, our area would be infinitesimally greater than 0, so presumably our minimum value is merely above zero. This means the greatest answer choice will be the number that violates the rule, and since no answer violates our minimum, (E) must again be the correct choice.
[Reveal] Spoiler: OA

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Re: If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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New post 29 Nov 2014, 21:27
Consider the triangle with length 12 (say side a) laying flat. One can place the side with length 6 (side b) at different angles with the side a. The triangle will have max area when b is perpendicular to a. And for this case, area is 36. Hence, the area cannot be more than 36.

PS: There is a longer mathematical proof to this, but I don't think we need to go into that detail here.

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Re: If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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New post 30 Nov 2014, 04:26
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pradeepss wrote:
If two sides of a triangle are 6 and 12, respectively, which of the following could NOT be the area of this triangle?

A. 1
B. 6
C. 17
D. 29
E. 38

Bunuel could you help!

[Reveal] Spoiler:
Official answer from veritas:

Solution: E.

If you know two sides of a triangle, the maximum area of the triangle can be obtained by setting these two sides as the base and height of a right triangle. For our triangle this gives us a maximum area of (6 * 12) / 2, or 36, so anything higher than this is invalid and the answer is (E). An alternative approach is to consider the minimum possible value of the triangle. Say we violated our third side rule and set the triangle's sides as 6, 12, and 18, respectively. Our "maximum" area here would be the set the base as 18, but since the triangle has no height -- the three sides only "fit together" as two parallel lines -- the height is 0 and the area is 0. But were we to extend one of the other two sides by an infinitesimally small amount, our area would be infinitesimally greater than 0, so presumably our minimum value is merely above zero. This means the greatest answer choice will be the number that violates the rule, and since no answer violates our minimum, (E) must again be the correct choice.


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*6*1/2 = 36.

Look at the diagram below:
Attachment:
Untitled.png
Untitled.png [ 1.6 KiB | Viewed 5469 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 6, which implies that we can have any area more than 0 but not more than 32 (12*6*1/2 = 32).

Answer: E.
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If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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New post 30 Nov 2014, 18:50
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area of a triangle is 1/2 a*b*sin(C)
Sin(any angle) belongs to (0,1)
so, area of traingle given= 1/2 * 6*12* Sin(C)= 36* Sin(C)
so, area must be <36

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Re: If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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pradeepss wrote:
If two sides of a triangle are 6 and 12, respectively, which of the following could NOT be the area of this triangle?

A. 1
B. 6
C. 17
D. 29
E. 38


For this question it would be helpful to know the largest area that this triangle could be, given the two sides of 6 and 12.

We know that the area of a triangle will be maximized when two sides are perpendicular to each other (consult Bunuel's drawing above). Thus we have a max area being equal to one half the base times the height, either (.5)12 * 6 -or- (.5)6 * 12 will result in a maximum area of 36 for the triangle; therefore, the triangle could never have an area of 38.

Hope that was clear and helpful!
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If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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New post 24 Jun 2016, 21:47
Just a query.

We can also take the sides as 6 6√3 and 12 also right.
This leads to 30 60 90.

Why are we only taking 12 12 6 in this case ?

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Re: If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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New post 25 Jun 2016, 01:41
Shrivathsan wrote:
Just a query.

We can also take the sides as 6 6√3 and 12 also right.
This leads to 30 60 90.

Why are we only taking 12 12 6 in this case ?

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This triangle is also possible and its area is \(18\sqrt{3}\approx 31.17\).
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Re: If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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New post 25 Jun 2016, 04:04
the greatest are of triangle is when it is a right angled triangle.

so if u calculate area max will be 1/2*6*12=36

E

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If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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Let's use the side the measures 12 as the base to calculate the area of the triangle. The height of the triangle cannot exceed either of the lengths of the other 2 sides, so the height cannot be greater than 6, which implies that the area cannot exceed 6x12/2 = 36.

Answer choice E.
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Last edited by HanoiGMATtutor on 25 Jun 2016, 23:16, edited 1 time in total.

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Re: If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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New post 25 Jun 2016, 10:29
It's worth noting as well that since any height between zero and the maximum could be matched with the base, the area could be any value between zero and some maximum. So the only way to get an invalid answer is to have an area that's too big.

So if you see this question again with different values, pick the largest answer choice. No calculation necessary.

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Re: If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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New post 25 Jun 2016, 13:14
pradeepss wrote:
If two sides of a triangle are 6 and 12, respectively, which of the following could NOT be the area of this triangle?

A. 1
B. 6
C. 17
D. 29
E. 38

Bunuel could you help!

[Reveal] Spoiler:
Official answer from veritas:

Solution: E.

If you know two sides of a triangle, the maximum area of the triangle can be obtained by setting these two sides as the base and height of a right triangle. For our triangle this gives us a maximum area of (6 * 12) / 2, or 36, so anything higher than this is invalid and the answer is (E). An alternative approach is to consider the minimum possible value of the triangle. Say we violated our third side rule and set the triangle's sides as 6, 12, and 18, respectively. Our "maximum" area here would be the set the base as 18, but since the triangle has no height -- the three sides only "fit together" as two parallel lines -- the height is 0 and the area is 0. But were we to extend one of the other two sides by an infinitesimally small amount, our area would be infinitesimally greater than 0, so presumably our minimum value is merely above zero. This means the greatest answer choice will be the number that violates the rule, and since no answer violates our minimum, (E) must again be the correct choice.


This is a very poor question. What is the source ?

Consider a line of 12 inch. Consider two line of 6 inches each, kept exactly above the first line and covering it completely.
Can you form a triangle by moving the two 6 inches lines away from the base line of 12.
No you cannot. The moment you move the any of the 6 inch line there would be a gap. The end points of one or more lines will not meet. If you keep the end points of the two 6 inches lines fixed with the end point of the 12 inches line then the other free end of the 6 inch lines will never meet.
This is the basis of triangle Triangle inequality Rule :- THE SUM OF ANY TWO SIDES OF A TRIANGLE IS GREATER THAN THE THIRD.
Therefore even if we assume the unknown side to be second smallest, then this unknown side must be must be greater than 6 . What can be the maximum length of the unknown side. Less than 18
6<unknown side<18
In the earlier case when two lines of 6 inches were lying exactly above the 12 inches line, the end points of all lines were collinear. Such a collection of three collinear points is called a degenerate triangle. (although it just looks like a line having three points marked over its length).
The area of an degenerate triangle is zero.
Any other combination of sides of a triangle (which follows the triangle inequality rule) will have an area of greater than zero.

Now the area of a triangle is given by \(\frac{1}{2} *b*h\)
and we know that the height is greatest when the angle is 90 degree

Keeping the smallest side of 6 as the base, the height can be 6.1 or 12 or 17.99 (there can be infinite values for height. I am just taking the two limiting cases at the either ends 6.1/17.99 and 12 is already given as the second side in question stem )

AND THE AREA CAN BE \(\frac{1}{2} *6 *(6.1\) OR \(12\) OR \(17.99\)) = approx 18.3, 36 or approx 72
SO TECHNICALLY SPEAKING ALL OPTIONS IN YOUR QUESTION ARE WRONG.

Bunuel can you reconfirm my line of reasoning ?
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Re: If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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New post 25 Jun 2016, 16:42
LogicGuru1 wrote:
pradeepss wrote:
If two sides of a triangle are 6 and 12, respectively, which of the following could NOT be the area of this triangle?

A. 1
B. 6
C. 17
D. 29
E. 38

Bunuel could you help!

[Reveal] Spoiler:
Official answer from veritas:

Solution: E.

If you know two sides of a triangle, the maximum area of the triangle can be obtained by setting these two sides as the base and height of a right triangle. For our triangle this gives us a maximum area of (6 * 12) / 2, or 36, so anything higher than this is invalid and the answer is (E). An alternative approach is to consider the minimum possible value of the triangle. Say we violated our third side rule and set the triangle's sides as 6, 12, and 18, respectively. Our "maximum" area here would be the set the base as 18, but since the triangle has no height -- the three sides only "fit together" as two parallel lines -- the height is 0 and the area is 0. But were we to extend one of the other two sides by an infinitesimally small amount, our area would be infinitesimally greater than 0, so presumably our minimum value is merely above zero. This means the greatest answer choice will be the number that violates the rule, and since no answer violates our minimum, (E) must again be the correct choice.


This is a very poor question. What is the source ?

Consider a line of 12 inch. Consider two line of 6 inches each, kept exactly above the first line and covering it completely.
Can you form a triangle by moving the two 6 inches lines away from the base line of 12.
No you cannot. The moment you move the any of the 6 inch line there would be a gap. The end points of one or more lines will not meet. If you keep the end points of the two 6 inches lines fixed with the end point of the 12 inches line then the other free end of the 6 inch lines will never meet.
This is the basis of triangle Triangle inequality Rule :- THE SUM OF ANY TWO SIDES OF A TRIANGLE IS GREATER THAN THE THIRD.
Therefore even if we assume the unknown side to be second smallest, then this unknown side must be must be greater than 6 . What can be the maximum length of the unknown side. Less than 18
6<unknown side<18
In the earlier case when two lines of 6 inches were lying exactly above the 12 inches line, the end points of all lines were collinear. Such a collection of three collinear points is called a degenerate triangle. (although it just looks like a line having three points marked over its length).
The area of an degenerate triangle is zero.
Any other combination of sides of a triangle (which follows the triangle inequality rule) will have an area of greater than zero.

Now the area of a triangle is given by \(\frac{1}{2} *b*h\)
and we know that the height is greatest when the angle is 90 degree

Keeping the smallest side of 6 as the base, the height can be 6.1 or 12 or 17.99 (there can be infinite values for height. I am just taking the two limiting cases at the either ends 6.1/17.99 and 12 is already given as the second side in question stem )

AND THE AREA CAN BE \(\frac{1}{2} *6 *(6.1\) OR \(12\) OR \(17.99\)) = approx 18.3, 36 or approx 72
SO TECHNICALLY SPEAKING ALL OPTIONS IN YOUR QUESTION ARE WRONG.

Bunuel can you reconfirm my line of reasoning ?


There is a huge logical gap between what a side length can be and what the height can be given the dimensions.

For example, in the case where the sides are 6, 12, and 17.99 (let's call it 18 for easier calculations), the height when you choose the 6-side as the base is actually very small. Because the triangle is degenerate, the height is near zero.

In fact, when you choose 6 as the base, the maximum height is achieved when the 12-side is perpendicular to the base, in which case the height is 12, and the area is 36.
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If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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New post 25 Jun 2016, 22:50
Yes apologies , I made a bad mistake.
somewhere during the calculation i Forgot that 6 and 12 are values already mentioned in the question and thus cannot be changed.
I should have simply approached the question with "The greatest area is always when the triangle is right" approach and used 6 and 12 as my base and height.


HanoiGMATtutor wrote:
LogicGuru1 wrote:
pradeepss wrote:
If two sides of a triangle are 6 and 12, respectively, which of the following could NOT be the area of this triangle?

A. 1
B. 6
C. 17
D. 29
E. 38

Bunuel could you help!

[Reveal] Spoiler:
Official answer from veritas:

Solution: E.

If you know two sides of a triangle, the maximum area of the triangle can be obtained by setting these two sides as the base and height of a right triangle. For our triangle this gives us a maximum area of (6 * 12) / 2, or 36, so anything higher than this is invalid and the answer is (E). An alternative approach is to consider the minimum possible value of the triangle. Say we violated our third side rule and set the triangle's sides as 6, 12, and 18, respectively. Our "maximum" area here would be the set the base as 18, but since the triangle has no height -- the three sides only "fit together" as two parallel lines -- the height is 0 and the area is 0. But were we to extend one of the other two sides by an infinitesimally small amount, our area would be infinitesimally greater than 0, so presumably our minimum value is merely above zero. This means the greatest answer choice will be the number that violates the rule, and since no answer violates our minimum, (E) must again be the correct choice.


This is a very poor question. What is the source ?

Consider a line of 12 inch. Consider two line of 6 inches each, kept exactly above the first line and covering it completely.
Can you form a triangle by moving the two 6 inches lines away from the base line of 12.
No you cannot. The moment you move the any of the 6 inch line there would be a gap. The end points of one or more lines will not meet. If you keep the end points of the two 6 inches lines fixed with the end point of the 12 inches line then the other free end of the 6 inch lines will never meet.
This is the basis of triangle Triangle inequality Rule :- THE SUM OF ANY TWO SIDES OF A TRIANGLE IS GREATER THAN THE THIRD.
Therefore even if we assume the unknown side to be second smallest, then this unknown side must be must be greater than 6 . What can be the maximum length of the unknown side. Less than 18
6<unknown side<18
In the earlier case when two lines of 6 inches were lying exactly above the 12 inches line, the end points of all lines were collinear. Such a collection of three collinear points is called a degenerate triangle. (although it just looks like a line having three points marked over its length).
The area of an degenerate triangle is zero.
Any other combination of sides of a triangle (which follows the triangle inequality rule) will have an area of greater than zero.

Now the area of a triangle is given by \(\frac{1}{2} *b*h\)
and we know that the height is greatest when the angle is 90 degree

Keeping the smallest side of 6 as the base, the height can be 6.1 or 12 or 17.99 (there can be infinite values for height. I am just taking the two limiting cases at the either ends 6.1/17.99 and 12 is already given as the second side in question stem )

AND THE AREA CAN BE \(\frac{1}{2} *6 *(6.1\) OR \(12\) OR \(17.99\)) = approx 18.3, 36 or approx 72
SO TECHNICALLY SPEAKING ALL OPTIONS IN YOUR QUESTION ARE WRONG.

Bunuel can you reconfirm my line of reasoning ?


There is a huge logical gap between what a side length can be and what the height can be given the dimensions.

For example, in the case where the sides are 6, 12, and 17.99 (let's call it 18 for easier calculations), the height when you choose the 6-side as the base is actually very small. Because the triangle is degenerate, the height is near zero.

In fact, when you choose 6 as the base, the maximum height is achieved when the 12-side is perpendicular to the base, in which case the height is 12, and the area is 36.

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Re: If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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New post 09 Nov 2016, 00:58
How can the area be 1 or 6? We know that sum of 2 sides should be greater than 3rd side. In this case, the 3rd side should have a value greater than 6.
The lowest possible combination for the sides: 6,12 & 6.1. In this case the area can never be equal to 1. Right?

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Re: If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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New post 09 Nov 2016, 01:53
dnalost wrote:
How can the area be 1 or 6? We know that sum of 2 sides should be greater than 3rd side. In this case, the 3rd side should have a value greater than 6.
The lowest possible combination for the sides: 6,12 & 6.1. In this case the area can never be equal to 1. Right?


No.

The area of a triangle with sides 6, 6.1 and 12 is 4.65709. The area of a triangle with sides 6, 6.0047 and 12 is ~1. Please check the discussion above. For example this post: if-two-sides-of-a-triangle-are-6-and-12-respectively-which-of-the-fo-189233.html#p1449171
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Re: If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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New post 31 Mar 2017, 07:05
I actually remember a formula from other field
S = 1/2 * a*b*Sin(alpha) = 36*sin(). Since Sin cannot exceed 1, you have an upper limit for area.

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New post 31 Mar 2017, 07:12
exactly i also know the theoram

area of triangle will be largest when it right triangle.

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Re: If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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New post 27 Jun 2017, 03:27
Bunuel wrote:
Shrivathsan wrote:
Just a query.

We can also take the sides as 6 6√3 and 12 also right.
This leads to 30 60 90.

Why are we only taking 12 12 6 in this case ?

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This triangle is also possible and its area is \(18\sqrt{3}\approx 31.17\).


Why are we not thinking about the third side here which could be anything between 6+12>third side>12-6???

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If two sides of a triangle are 6 and 12, respectively, which of the fo [#permalink]

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New post 03 Sep 2017, 01:47
siddreal wrote:
Bunuel wrote:
Shrivathsan wrote:
Just a query.

We can also take the sides as 6 6√3 and 12 also right.
This leads to 30 60 90.

Why are we only taking 12 12 6 in this case ?

----
Regards
Shrivathsan


This triangle is also possible and its area is \(18\sqrt{3}\approx 31.17\).


Why are we not thinking about the third side here which could be anything between 6+12>third side>12-6???




Not sure if below might help you -

Going by your case, maximum integer (just to avoid messy calculations) value for the third side will be 17.

So the three sides are 6,12 and 17.

Here, you are automatically taking 1/2 * b * h - we need to consider that they are sides and you can't just rotate any value to get the height.

For sake of calculation, lets consider the formula for area of the triangle involving sides ---> Sq.rt (s*(s-a)*(s-b)*(s-c)) ; s= (a+b+c)/2

by that formula, max. area is approx. 24

If you consider two sides 17 and 12 (one as height and other as base) receptively, then the hypotenuse (i.e. third side) should be near 21 and not 6.

Hence, consider 6 and 12. The third side is well within the range.

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If two sides of a triangle are 6 and 12, respectively, which of the fo   [#permalink] 03 Sep 2017, 01:47
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