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# 1) What is a the greatest possible area of a triangular

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Manager
Joined: 24 Jun 2006
Posts: 61
1) What is a the greatest possible area of a triangular [#permalink]

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25 Dec 2006, 16:59
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

1) What is a the greatest possible area of a triangular region with the vertex at the center of a circle with radius 1 and other two vertices on the circle
1) (sqrt 3)/4
2) 1/2
3) pi/4
4) 1
5) sqrt 2

2) A certain circular area has its center at point P and has radius 4 and Points X & Y lie in the same plane as the circular area. Does point y lie outside the circular area. (Data Sufficiency)
1) Dist between X & P = 4.5
2) Disat between X & Y = 9
VP
Joined: 28 Mar 2006
Posts: 1369

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25 Dec 2006, 17:40
mitul wrote:
1) What is a the greatest possible area of a triangular region with the vertex at the center of a circle with radius 1 and other two vertices on the circle
1) (sqrt 3)/4
2) 1/2
3) pi/4
4) 1
5) sqrt 2

It should be a isoceles right triangle with base=ht=1 So area =1/2

2) A certain circular area has its center at point P and has radius 4 and Points X & Y lie in the same plane as the circular area. Does point y lie outside the circular area. (Data Sufficiency)
1) Dist between X & P = 4.5
NOT SUFF.
Is radius =4 and P is 4.5 from centre then Y should be between 0.5 and 8.5 ('cos thrid side is always less than sum of other 2 sides and greater than the difference).

2) Distance between X & Y = 9

SUFF. The third side length is between 5(9-4) and 13(9+4) which is outside the circle

So B
Director
Joined: 28 Dec 2005
Posts: 919

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25 Dec 2006, 17:48
mitul wrote:
1) What is a the greatest possible area of a triangular region with the vertex at the center of a circle with radius 1 and other two vertices on the circle
1) (sqrt 3)/4
2) 1/2
3) pi/4
4) 1
5) sqrt 2

2) A certain circular area has its center at point P and has radius 4 and Points X & Y lie in the same plane as the circular area. Does point y lie outside the circular area. (Data Sufficiency)
1) Dist between X & P = 4.5
2) Disat between X & Y = 9

1) I am not sure this is so, can someone please confirm, that such a triangle of largest area is an equilateral one?

If that is the case, the answer is sqrt(3)/4.

Using special triangle properties the side of the triangle are in the ratio: 1:sqrt(3):2. [30:60:90]. Thus hyp is 1, so the height is sqrt(3)/2 and the base of the complete triangle is 1/2*2 = 1.

Thus the area is 1/2*1*sqrt(3)/2 = sqrt(3)/4.

2)

The only way to do this is by drawing the circle, the center and the points X and Y.

1) X is obviously outside the circle because it is more than the radius away from the center. But doesnt say anything about Y. INSUFF.
2)Distance between X and Y is given, but this also insufficient since we have no indication about the location wrt the circle. INSUFF.

Together, it becomes clear that Y will always be outside the circle since the only way Y could have been at least on the circle would have been if X and Y were (4.5+4 = 8.5).
Senior Manager
Joined: 23 Jun 2006
Posts: 387

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25 Dec 2006, 18:08
for question 1 answer is 1/2 (C)... i.e. a triangle with 90 degree angle.

an alternative formula for a triangle's area is a*b*sin(x)/2 where a and b are two sides of the triangle and x is the angle between them.
in our case, a=b=1 regardless of angle. so maximum is attained when
sin(x) is maximal which happens when x is 90 degrees. area is 0.5 in that case.

question 2 is indeed C.
Director
Joined: 28 Dec 2005
Posts: 919

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25 Dec 2006, 18:15
hobbit wrote:
for question 1 answer is 1/2 (C)... i.e. a triangle with 90 degree angle.

an alternative formula for a triangle's area is a*b*sin(x)/2 where a and b are two sides of the triangle and x is the angle between them.
in our case, a=b=1 regardless of angle. so maximum is attained when
sin(x) is maximal which happens when x is 90 degrees. area is 0.5 in that case.

question 2 is indeed C.

Thanks Hobbit. Makes sense.
Intern
Joined: 13 Jun 2005
Posts: 28

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25 Dec 2006, 18:20
Prob 1.

If one vertex is on the circle, and the other two sides are on the circle ( two sides are equal). Isosceles triangle.

If it is right angled at the center, area = 1/2 x 1 x1 = 1/2

Tried.....
The third side is >0 and <2 and tried to find the hyp but it works out to be less 1/2

I'm going with (2) Ans 1/2

Prob 2.

1) X is 4.5 from P, outside the circle but where is Y (insuff)
2) distance b/w X & Y is 9
X can be inside the circle or Y can be inside the circle or both can be outside and we can still have a distance of 9 (insuff)

together
diameter of circle is 8 .
X is 4.5 from center and is outside the circle
Since distance between X and Y is 9, Y needs to be outside (SUFF)

Ans: C
VP
Joined: 28 Mar 2006
Posts: 1369

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25 Dec 2006, 19:14
hobbit wrote:
for question 1 answer is 1/2 (C)... i.e. a triangle with 90 degree angle.

an alternative formula for a triangle's area is a*b*sin(x)/2 where a and b are two sides of the triangle and x is the angle between them.
in our case, a=b=1 regardless of angle. so maximum is attained when
sin(x) is maximal which happens when x is 90 degrees. area is 0.5 in that case.

question 2 is indeed C.

Why isnt it B enough for question (2)? Please see my reasoning above.

Thanks!
Intern
Joined: 24 Feb 2006
Posts: 48

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25 Dec 2006, 19:19
2) A certain circular area has its center at point P and has radius 4 and Points X & Y lie in the same plane as the circular area. Does point y lie outside the circular area. (Data Sufficiency)
1) Dist between X & P = 4.5
2) Disat between X & Y = 9

1) insuff as no info about Y
2) insuff as no info relates to the circle.

together : 1 gives X is outside of the circle by (4.5 - 4) = 0.5
If Y were in the circle max distance between X and Y will be
diameter + 0.5 = 4*2 + 0.5 = 8.5
but dist between X & Y = 9 >8.5 so Y is outside of the circle.

ans is (C)
25 Dec 2006, 19:19
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