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In the diagram, JL = 4 and JK = 6.

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Senior Manager
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In the diagram, JL = 4 and JK = 6. [#permalink] New post 12 Nov 2017, 01:43
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In the diagram, JL = 4 and JK = 6.

Image



A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

Kudos for correct solution.


[Reveal] Spoiler:
Attachment:
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gqgpp_img4.png [ 1.45 KiB | Viewed 557 times ]

Attachment:
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gqgpp_img4b.png [ 1012 Bytes | Viewed 558 times ]
[Reveal] Spoiler: OA
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Re: In the diagram, JL = 4 and JK = 6. [#permalink] New post 12 Nov 2017, 09:53
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I am not sure Magoosh explanation is right because they use two different figures in the text and in the answer. If we keep using the figure in the question, the answer should be B.

Indeed, we have one leg equal to 4 and hypotenuse equal to 6. Then a triangle has maximal area when it is right-angled. Thus, when its second leg is equal to 36-16 = 2sqrt(5). In this case, the area of the triangle is equal to 4sqrt(5) that is less than 12.

If this is the maximal area, there is no way to get something greater than 12 so that quantity B should be greater.
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Re: In the diagram, JL = 4 and JK = 6. [#permalink] New post 07 Dec 2017, 20:06
Bunuel can you explain this,
I am more confused after reading the other people answer,
thank you
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Re: In the diagram, JL = 4 and JK = 6. [#permalink] New post 07 Dec 2017, 21:45
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wongpcla wrote:
Bunuel can you explain this,
I am more confused after reading the other people answer,
thank you


Let me explain this,

We know in triangle with sides A,B and c

Difference of any two sides < Third side < Sum of any two sides

Now consider the triangle above,

The maximum value of the third side must be less 10, or else the triangle rule wont be applicable

Now if we consider the third side =10 (assuming maximum value)

then Area = 1/2 * base * Altitude

= 1/2 * 4 * 10 = 20

Now the minimum value of the third side must be greater than 2.

SO let us take the value of third side = 2

Then Area = 1/2 * 4 * 2 = 4.

Now the area of the triangle is either greater than 12 or less than 12

Hence the option is D. (The diagram in geometry section are not drawn to scale unless specified. Here it looks like right angle triangle but nothing is mentioned so we cannot assume.)

Let me know if you need further clarification
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Re: In the diagram, JL = 4 and JK = 6. [#permalink] New post 07 Dec 2017, 22:04
pranab01 wrote:
wongpcla wrote:
Bunuel can you explain this,
I am more confused after reading the other people answer,
thank you


Let me explain this,

We know in triangle with sides A,B and c

Difference of any two sides < Third side < Sum of any two sides

Now consider the triangle above,

The maximum value of the third side must be less 10, or else the triangle rule wont be applicable

Now if we consider the third side =10 (assuming maximum value)

then Area = 1/2 * base * Altitude

= 1/2 * 4 * 10 = 20

Now the minimum value of the third side must be greater than 2.

SO let us take the value of third side = 2

Then Area = 1/2 * 4 * 2 = 4.

Now the area of the triangle is either greater than 12 or less than 12

Hence the option is D. (The diagram in geometry section are not drawn to scale unless specified. Here it looks like right angle triangle but nothing is mentioned so we cannot assume.)

Let me know if you need further clarification


thank you very much.
now it is clear
Re: In the diagram, JL = 4 and JK = 6.   [#permalink] 07 Dec 2017, 22:04
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In the diagram, JL = 4 and JK = 6.

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