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What is the length of segment BC?

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What is the length of segment BC? [#permalink]

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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.
[Reveal] Spoiler: OA

Last edited by alexBLR on 14 Feb 2010, 12:22, edited 2 times in total.

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Re: good one [#permalink]

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New post 14 Feb 2010, 19:21
With the picture... its A....

As S1 gives Angle ABC to be 90.... Pythagoras Theorem.. reveals BC as 12....

S2... again is IN SUFF...[/quote]


could you please explain why S2 is not sufficient

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Re: good one [#permalink]

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New post 14 Feb 2010, 22:20
s2 is also sufficient.
you have 2 sides of the triangle and the area
\(area = \frac{1}{2}* bc SinA\)

\(30 = \frac{1}{2} * 5 * 13 SinA\)

=> Sin A = 12/13
=> A must be 90 degree but those who are still not sure..check the cosine rule.

\(cos A = \frac{5}{13}\) using \(cos^2 A = 1- sin^2 A\)

\(cos A =\frac{(b^2 + c^2 - a^2)}{2bc}\)

This will solve for a
=> \(\frac{5}{13} = \frac{(25 + 169 -a^2)}{2*5*13}\)
=> 50 = 25 + 169 - a^2 ==> a^2 = 144 thus a = 12
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New post 14 Feb 2010, 23:24
gurpreetsingh wrote:
s2 is also sufficient.
you have 2 sides of the triangle and the area
\(area = \frac{1}{2}* bc SinA\)

\(30 = \frac{1}{2} * 5 * 13 SinA\)

=> Sin A = 12/13
=> A must be 90 degree but those who are still not sure..check the cosine rule.

\(cos A = \frac{5}{13}\) using \(cos^2 A = 1- sin^2 A\)

\(cos A =\frac{(b^2 + c^2 - a^2)}{2bc}\)

This will solve for a
=> \(\frac{5}{13} = \frac{(25 + 169 -a^2)}{2*5*13}\)
=> 50 = 25 + 169 - a^2 ==> a^2 = 144 thus a = 12


Answer is A.
To apply cosine rule or sine rule we need to know A angle.
cos A is not 5/13 or sin A is not 12/13 as the question statement does not tell that triangle has one right angle, its only given by statement 1.

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New post 15 Feb 2010, 05:33
Code:
http://www.math10.com/en/geometry/sin-cos-rule.html


Check the link above

\(Area = \frac{1}{2} bc SinA\)holds good for all triangles
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First of all I want to say that trigonometry is not tested in GMAT. You should be able to solve all GMAT geometry problems without it.

Now statement (1) is clearly sufficient as shown in previous posts.

As for statement (2): consider BA to be the base. \(Area=30=\frac{1}{2}*base*height\). If angle B is right angle, then \(height=CB=12\) BUT if angle B is acute and CA is reflected symmetricly about vertical line, then CA still will be 13, height still will be the same and area still will be 30, though in this case CB will be much greater than 12.

So statement (2) is not sufficient.

Answer: A.
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Re: good one [#permalink]

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New post 15 Feb 2010, 07:11
Bunuel wrote:
First of all I want to say that trigonometry is not tested in GMAT. You should be able to solve all GMAT geometry problems without it.

Now statement (1) is clearly sufficient as shown in previous posts.

As for statement (2): consider BA to be the base. \(Area=30=\frac{1}{2}*base*height\). If angle B is right angle, then \(height=CB=12\) BUT if angle B is acute and CA is reflected symmetricly about vertical line, then CA still will be 13, height still will be the same and area still will be 30, though in this case CB will be much greater than 12.

So statement (2) is not sufficient.

Answer: A.


Could you please let me know where my explanation is wrong? coz m not able to judge it.
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New post 15 Feb 2010, 07:32
gurpreetsingh wrote:
Code:
http://www.math10.com/en/geometry/sin-cos-rule.html


Check the link above

\(Area = \frac{1}{2} bc SinA\)holds good for all triangles


We do not know value for angle A.
Also it is not a right angle triangle so cosA cannot be found as 5/13.

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Bunuel wrote:
First of all I want to say that trigonometry is not tested in GMAT. You should be able to solve all GMAT geometry problems without it.

Now statement (1) is clearly sufficient as shown in previous posts.

As for statement (2): consider BA to be the base. \(Area=30=\frac{1}{2}*base*height\). If angle B is right angle, then \(height=CB=12\) BUT if angle B is acute and CA is reflected symmetricly about vertical line, then CA still will be 13, height still will be the same and area still will be 30, though in this case CB will be much greater than 12.

So statement (2) is not sufficient.

Answer: A.



Area = 30 units 1/2*base *height
going by your explanation: Area is same, height same => base has to be same.
We cannot form two triangles with two sides the same, same height and area yet the third side different.
If one side is13 units, base is 5 units and area of triangle is 30 units, the third side has to be 12 units, perpendicular to the base.
IMO D

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gurpreetsingh wrote:
Bunuel wrote:
First of all I want to say that trigonometry is not tested in GMAT. You should be able to solve all GMAT geometry problems without it.

Now statement (1) is clearly sufficient as shown in previous posts.

As for statement (2): consider BA to be the base. \(Area=30=\frac{1}{2}*base*height\). If angle B is right angle, then \(height=CB=12\) BUT if angle B is acute and CA is reflected symmetricly about vertical line, then CA still will be 13, height still will be the same and area still will be 30, though in this case CB will be much greater than 12.

So statement (2) is not sufficient.

Answer: A.


Could you please let me know where my explanation is wrong? coz m not able to judge it.


The formula you posted is correct: \(Area=\frac{1}{2}BA*CA*sinA\) --> \(sinA=\frac{12}{13}\), but from this we cannot calculate angle \(A\), as \(sin A=Sin (180-A)\). Meaning that this won't give us ONLY one value for angle A, thus we won't have ONLY one value for CB --> A can be acute angle, making B right angle AND making CB equal to 12 OR A can be obtuse angle, making B acute angle AND making CB greater then 12.

Hope it's clear.
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bml wrote:
Bunuel wrote:
First of all I want to say that trigonometry is not tested in GMAT. You should be able to solve all GMAT geometry problems without it.

Now statement (1) is clearly sufficient as shown in previous posts.

As for statement (2): consider BA to be the base. \(Area=30=\frac{1}{2}*base*height\). If angle B is right angle, then \(height=CB=12\) BUT if angle B is acute and CA is reflected symmetricly about vertical line, then CA still will be 13, height still will be the same and area still will be 30, though in this case CB will be much greater than 12.

So statement (2) is not sufficient.

Answer: A.



Area = 30 units 1/2*base *height
going by your explanation: Area is same, height same => base has to be same.
We cannot form two triangles with two sides the same, same height and area yet the third side different.
If one side is13 units, base is 5 units and area of triangle is 30 units, the third side has to be 12 units, perpendicular to the base.
IMO D


We can form two different triangles with same base, same height, same second side and same area.

I found this drawing in the web:
Attachment:
Triangles.jpg
Triangles.jpg [ 30.8 KiB | Viewed 5644 times ]

Hope it helps.
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Re: good one [#permalink]

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New post 15 Feb 2010, 08:58
thanks bunnel, you are right. S2 isnt sufficient, there was no need to solve, visualizing the way you did was the best way.
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New post 15 Feb 2010, 09:32
Thanks Bunuel!

You made my day.... I learnt something new today..

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Re: good one [#permalink]

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New post 07 Aug 2010, 15:40
Bunuel wrote:
bml wrote:
Bunuel wrote:
First of all I want to say that trigonometry is not tested in GMAT. You should be able to solve all GMAT geometry problems without it.

Now statement (1) is clearly sufficient as shown in previous posts.

As for statement (2): consider BA to be the base. \(Area=30=\frac{1}{2}*base*height\). If angle B is right angle, then \(height=CB=12\) BUT if angle B is acute and CA is reflected symmetricly about vertical line, then CA still will be 13, height still will be the same and area still will be 30, though in this case CB will be much greater than 12.

So statement (2) is not sufficient.

Answer: A.



Area = 30 units 1/2*base *height
going by your explanation: Area is same, height same => base has to be same.
We cannot form two triangles with two sides the same, same height and area yet the third side different.
If one side is13 units, base is 5 units and area of triangle is 30 units, the third side has to be 12 units, perpendicular to the base.
IMO D


We can form two different triangles with same base, same height, same second side and same area.

I found this drawing in the web:
Attachment:
Triangles.jpg

Hope it helps.[/quote


Hello Bunuel,

I think the answer to this question should be D. Because if you have the area of the Triangle, and two sides you can use Heron's formula to figure out the third side

A = \sqrt{S(S-a)(S-b)(S-c)}
Where A = Area of Triangle
S = (a+b+c)/2
a,b,c are the sides of the triangle.

So for us in the above formula all we are missing is the value of c.

Let me know if you think I made some silly assumption in this solution
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New post 07 Aug 2010, 15:56
arundas wrote:
Hello Bunuel,

I think the answer to this question should be D. Because if you have the area of the Triangle, and two sides you can use Heron's formula to figure out the third side

A = \sqrt{S(S-a)(S-b)(S-c)}
Where A = Area of Triangle
S = (a+b+c)/2
a,b,c are the sides of the triangle.

So for us in the above formula all we are missing is the value of c.

Let me know if you think I made some silly assumption in this solution



The problem with your approach is that you'll have more than one solution for c. So statement 2 is not sufficient.
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Manhattan DS - Triangles - Error? [#permalink]

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I am not sure about this one. I believe that using the second equation too you can find the area of the triangle. Using the formula \(sqrt(s(s-a)(s-b)(s-c))\). You have 1 equation and 1 variable. Well it might be a tough 4th degree equation but I am hoping it does not give me 2 positive roots.

Bunuel, Shrouded?

Let me know what I am missing.

Thank you.

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Re: Manhattan DS - Triangles - Error? [#permalink]

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New post 24 Oct 2010, 07:09
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ManG.JPG

I am not sure about this one. I believe that using the second equation too you can find the area of the triangle. Using the formula \(sqrt(s(s-a)(s-b)(s-c))\). You have 1 equation and 1 variable. Well it might be a tough 4th degree equation but I am hoping it does not give me 2 positive roots.

Bunuel, Shrouded?

Let me know what I am missing.

Thank you.


As you can see in the diagrams in the discussion, (2) can lead to more than one triangle.

The problem in arguing the way you did is exactly as you pointed out. You get an equation in the fourth degree which can have upto four roots, and there is no easy way to know and conclude how many of these solutions will be positive numbers and hence define valid triangles. Uniqueness of solution is not easy to guarantee.
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Re: What is the length of segment BC? [#permalink]

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Re: What is the length of segment BC? [#permalink]

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New post 22 Jun 2013, 04:18
Hi Bunnel,

Regarding B: I think it is sufficient to find y
my approach:
Drop a perpendicular from AD to opposite side - this will divide the side into x and 13-x

1/2xh+1/2(13-x)h=30
13/2h=30
=>h=60/13
so x can be found out as =>x=(25-h^2)^(1/2)
therefore 13-x could be found out.
and therefore y could be found out.

Hence answer should be D.
Please correct me if I am wrong.
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Re: What is the length of segment BC? [#permalink]

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New post 02 Dec 2017, 18:10
Hi gurpreetsingh, alexBLR
Statement 2 is not sufficient. please see my explanation.

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gurpreetsingh wrote:
,

s2 is also sufficient.
you have 2 sides of the triangle and the area
\(area = \frac{1}{2}* bc SinA\)

\(30 = \frac{1}{2} * 5 * 13 SinA\)

=> Sin A = 12/13
=> A must be 90 degree but those who are still not sure..check the cosine rule.

\(cos A = \frac{5}{13}\) using \(cos^2 A = 1- sin^2 A\)

\(cos A =\frac{(b^2 + c^2 - a^2)}{2bc}\)

This will solve for a
=> \(\frac{5}{13} = \frac{(25 + 169 -a^2)}{2*5*13}\)
=> 50 = 25 + 169 - a^2 ==> a^2 = 144 thus a = 12

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Re: What is the length of segment BC?   [#permalink] 02 Dec 2017, 18:10
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