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Updated on: 09 Oct 2019, 21:23
Originally posted by GMAT TIGER on 08 May 2008, 21:28.
Last edited by Bunuel on 09 Oct 2019, 21:23, edited 3 times in total.
Last edited by Bunuel on 09 Oct 2019, 21:23, edited 3 times in total.
Edited the question and added the OA.
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
51% (01:09) correct
49%
(01:08)
wrong
based on 1380
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What is the length of segment BC?
(1) Angle ABC is 90 degrees.
(2) The area of the triangle is 30.
Attachment:
Triangle.jpg [ 28.62 KiB | Viewed 56554 times ]
18 Jul 2012, 05:11
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Responding to a pm.
What is the length of segment BC?
(1) Angle ABC is 90 degrees.
From this, we can derive that \(BC=\sqrt{13^2-5^2}\). Sufficient.
(2) The area of the triangle is 30.
Consider the diagram below:
Notice that segment \(AB_2\) is the mirror reflection of segment \(AB_1\) around the vertical line passing through point A. Now, if the height of triangles \(ACB_1\) and \(ACB_2\) is \(\frac{60}{13}\), then the area of both triangles is \(area=\frac{1}{2}baseheight=\frac{1}{2}13\frac{60}{13}=30\). So, as you can see, we can have different lengths of segment CB (\(CB_1\) and \(CB_2\)). Not sufficient.
Answer: A.
Hope it's clear.
ABC.png [ 6.06 KiB | Viewed 39526 times ]
What is the length of segment BC?
(1) Angle ABC is 90 degrees.
From this, we can derive that \(BC=\sqrt{13^2-5^2}\). Sufficient.
(2) The area of the triangle is 30.
Consider the diagram below:
Notice that segment \(AB_2\) is the mirror reflection of segment \(AB_1\) around the vertical line passing through point A. Now, if the height of triangles \(ACB_1\) and \(ACB_2\) is \(\frac{60}{13}\), then the area of both triangles is \(area=\frac{1}{2}baseheight=\frac{1}{2}13\frac{60}{13}=30\). So, as you can see, we can have different lengths of segment CB (\(CB_1\) and \(CB_2\)). Not sufficient.
Answer: A.
Hope it's clear.
Attachment:
ABC.png [ 6.06 KiB | Viewed 39526 times ]
12 May 2008, 00:37
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Hi all!
My 2 cents....
Let imagine AB and AC as sticks bounded in point A by a hinge. So, we can move AB relatively to AC and change angle BAC form 0 to 180. The area of ABC triangle will change form 0 (angle BAC=0) through maximum value 12*5*13=32.5 (angle BAC=90) to 0 (angle BAC=180). If you imagine that, It will be obvious that for 0<area<32.5 we will get two possible value of BC and for area=32.5 we will get only one value of BC. In our case area=30<32.5. Therefore, the second condition is insufficient.
My 2 cents....
Let imagine AB and AC as sticks bounded in point A by a hinge. So, we can move AB relatively to AC and change angle BAC form 0 to 180. The area of ABC triangle will change form 0 (angle BAC=0) through maximum value 12*5*13=32.5 (angle BAC=90) to 0 (angle BAC=180). If you imagine that, It will be obvious that for 0<area<32.5 we will get two possible value of BC and for area=32.5 we will get only one value of BC. In our case area=30<32.5. Therefore, the second condition is insufficient.
