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# Find the attachment. I tried adding the image in the post;

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Find the attachment. I tried adding the image in the post; [#permalink]  16 May 2011, 11:25
00:00

Difficulty:

(N/A)

Question Stats:

29% (02:14) correct 71% (01:48) wrong based on 8 sessions
Find the attachment. I tried adding the image in the post; however, all in vain. So, attached the complete question.

Attachment:

1.jpg [ 25.68 KiB | Viewed 932 times ]

Attachments

question.doc [54 KiB]

Last edited by gurpreetsingh on 16 May 2011, 11:56, edited 1 time in total.
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Re: co-ordinate geometry question [#permalink]  16 May 2011, 12:01
1
KUDOS
Statement 1 : Not sufficient.
We know the position of B, but not the position of A. Thus we can not calculate the value of AB.

Statement 2: Sufficient.
We know the position of C
=> we know the perpendicular distance from the line AB
=> we know the altitude of the triangle
=> we have the area.

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Re: co-ordinate geometry question [#permalink]  16 May 2011, 12:04
Attachment:

Area_Of_Equilateral_triangle.PNG [ 15.53 KiB | Viewed 923 times ]

(1)
A can be (0, 0.000000000001)
Or
A can be (0, 5)
Both resulting in different area for ABC.
Not Sufficient.

(2)
We know the altitude of the triangle.
Distance between y-axis and coordinate C or 6.
Thus, the area can be found.

$$Altitude=\frac{\sqrt{3}}{2}*(Side)=6$$

$$Area = \frac{\sqrt{3}}{4}(side)^2$$

Sufficient.

Ans: "B"
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Re: co-ordinate geometry question [#permalink]  16 May 2011, 12:29
Stmt1 . We know only one co-ordinate. we don't know altitude, base or one side of triangle. Hence insufficient.

Stmt2: We know altitude when draw from C to opposite base is 6. Also this altitude will bisect the angle ACB making each angle 30.
So cos30 = altitude/CA {imagine a line from C to AB}
$$\sqrt{3}/2=6/CA$$

$$CA= 12/\sqrt{3}.$$ We know one side of EQUILATERAL triangle.

$$Apply formula Area of equilateral triangle= s^2 * \sqrt{3}/4 = (12/\sqrt{3})^2* \sqrt{3}/4 = 12\sqrt{3}$$
Hence sufficient.

OA B.
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Last edited by jamifahad on 16 May 2011, 12:53, edited 1 time in total.
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Re: co-ordinate geometry question [#permalink]  16 May 2011, 12:40
So sin30 = altitude/CA {imagine a line from C to AB}

I know it doesn't change the answer. But, is it sin{30} or cos{30}?
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Re: co-ordinate geometry question [#permalink]  16 May 2011, 12:55
oh yes cos30. i was timing my solution. got mixed up. thanks.
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Re: co-ordinate geometry question [#permalink]  16 May 2011, 19:11
a the distance from origin to A is not known.Hence side of the equilateral triangle cannot be calculated. Not sufficient.

b C has x coordinate = 6 = altitude of equilateral triangle
y coordinate = 3* 3^(1/2) also does not account for the distance from origin to A.

however altitude a = 6 means side = 6*2/3^(1/2) as in a triangle with angle 60 deg sides are in the ratio 3^(1/2)(perpendicular):1 (base):2(hypotenuse)

thus area can be calculated.

B
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Re: co-ordinate geometry question [#permalink]  17 May 2011, 06:19
Thanks guys, indeed the OA is B.
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Re: co-ordinate geometry question [#permalink]  17 May 2011, 09:39
Yup B is sufficient

With point C there is only one possible equilateral triangle. Hence we can find it area
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Re: co-ordinate geometry question [#permalink]  18 May 2011, 03:48
(1)
Let Co-ord of A = (0,y)

5root(3) - y = Side

Insufficient

(2)

A median drawn from C will bisect AB

and there will be 30-60-90 triangle

The length of perpendicular is known

so other sides can be found, and this the area can be found

Sufficient

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Re: co-ordinate geometry question   [#permalink] 18 May 2011, 03:48
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