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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.
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
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 36 (12*6*1/2 = 36).
Re: If two sides of a triangle are 6 and 12, respectively, which of the fo
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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.
Re: If two sides of a triangle are 6 and 12, respectively, which of the fo
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22 May 2015, 06:59
1
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
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Updated on: 25 Jun 2016, 23:16
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1
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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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.
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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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
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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?
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.
_________________
Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly. FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.
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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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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?
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