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# What is the length of segment BC?

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Math Expert
Joined: 02 Sep 2009
Posts: 53657
Re: What is the length of segment BC?  [#permalink]

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29 Oct 2017, 08:07
ashisplb wrote:
Bunuel wrote:
Responding to a pm.

What is the length of segment BC?

(1) Angle ABC is 90 degrees. $$BC=\sqrt{13^2-5^2}$$. Sufficient.

(2) The area of the triangle is 30. Consider the diagram below:
Attachment:
ABC.png
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 $$area=\frac{1}{2}*base*height=\frac{1}{2}*13*\frac{60}{13}=30$$. So, as you can see we can have different lengths of segment CB ($$CB_1$$ and $$CB_2$$). Not sufficient.

Hope it's clear.

Hi Bunuel. I think second statement is also sufficient.

Both triangles are based on same base. so area of both triangles should be same. There will be change in sin angle only. But sin(x) =sin(180-x)

Sent from my ONE A2003 using GMAT Club Forum mobile app

Please check under the spoiler in original post, OA is not D, it's A. Please read the thread to see WHY it's not D.
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Re: What is the length of segment BC?  [#permalink]

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24 Apr 2018, 09:08
Bunuel wrote:
Responding to a pm.

What is the length of segment BC?

(1) Angle ABC is 90 degrees. $$BC=\sqrt{13^2-5^2}$$. Sufficient.

(2) The area of the triangle is 30. Consider the diagram below:
Attachment:
ABC.png
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 $$area=\frac{1}{2}*base*height=\frac{1}{2}*13*\frac{60}{13}=30$$. So, as you can see we can have different lengths of segment CB ($$CB_1$$ and $$CB_2$$). Not sufficient.

Hope it's clear.

brunel
what is the theory behind it?can you please explain how you initiate thinking in such questions?
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Joined: 04 Dec 2015
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Re: What is the length of segment BC?  [#permalink]

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15 Jan 2019, 06:05
Bunuel wrote:
Responding to a pm.

What is the length of segment BC?

(1) Angle ABC is 90 degrees. $$BC=\sqrt{13^2-5^2}$$. Sufficient.

(2) The area of the triangle is 30. Consider the diagram below:
Attachment:
ABC.png
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 $$area=\frac{1}{2}*base*height=\frac{1}{2}*13*\frac{60}{13}=30$$. So, as you can see we can have different lengths of segment CB ($$CB_1$$ and $$CB_2$$). Not sufficient.

Hope it's clear.

First of all, why do we even need to consider an explanation that involves reflection of a side. Why would I do it when the information given in (2) itself is not sufficient to calculate bc. ? Still confused

I cannot understand why! Kindly help !

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Joined: 28 Apr 2014
Posts: 2
Re: What is the length of segment BC?  [#permalink]

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16 Jan 2019, 06:13
For second statement, try obtuse angle triangle using 5, 13 & 15.6205, u will get 30 as the area of triangle
Re: What is the length of segment BC?   [#permalink] 16 Jan 2019, 06:13

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# What is the length of segment BC?

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