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m22#30

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m22#30 [#permalink] New post 26 Nov 2008, 20:27
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A rectangle is inscribed in a circle of radius 5. Is the area of the rectangle bigger than 48 ?

1. The ratio of the lengths of sides of the rectangle is 3:4
2. The difference between the lengths of sides of the rectangle is smaller than 3

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A

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Re: m22#30 [#permalink] New post 26 Nov 2008, 20:59
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ventivish wrote:
A rectangle is inscribed in a circle of radius 5. Is the area of the rectangle bigger than 48 ?

1. The ratio of the lengths of sides of the rectangle is 3:4
2. The difference between the lengths of sides of the rectangle is smaller than 3


A. very tricky.

1. The sides of the rectangles has to be 6 and 8 cus the diagnol of the rectangle is 10, which is the diameater of the circle.
If we change (decrease) a side of the rectangle, another side will also be changed (increase) and vice versa.

2. really doesnot help.
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Re: m22#30 [#permalink] New post 26 Nov 2008, 21:07
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Thanks alot GMATTIGER for your answers.
However, the question does not mention that the diagonal of the rectangle passes through the centre of the circle.
Is this something that must be assumed for a rectangle inscribed in a circle?
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Re: m22#30 [#permalink] New post 26 Nov 2008, 21:17
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ventivish wrote:
Thanks alot GMATTIGER for your answers.
However, the question does not mention that the diagonal of the rectangle passes through the centre of the circle.
Is this something that must be assumed for a rectangle inscribed in a circle?



It has to be. You cannot make any rectanlge that is inscribed in a circle with a diagnal not equal tyo the diameter of the circle. No matter however you draw a rectangle, its diagnol is the diameter of the circle.
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Re: m22#30 [#permalink] New post 09 Mar 2010, 05:49
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If the diameter = 10 and radius is 5, then the hypo of the mini triangle is 3:4:5 (you need to draw to visualise).

Hence, the side of the triangle is 6 (3X2) and 8 (4X2). This is sufficient to answer the question..
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Re: m22#30 [#permalink] New post 09 Mar 2010, 08:08
ventivish wrote:
A rectangle is inscribed in a circle of radius 5. Is the area of the rectangle bigger than 48 ?

1. The ratio of the lengths of sides of the rectangle is 3:4
2. The difference between the lengths of sides of the rectangle is smaller than 3

[Reveal] Spoiler: OA
A

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1 is enough,
(3x)^2+(4x)^2 = 10^2
x=2
area = 6*8 = 48 hence sufficient to answer

2 is not enough,
l^2+(l-3)^2=10^2
2l^2-6*l+9 = 100
so not enough to solve.

Hence A.
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Re: m22#30 [#permalink] New post 09 Mar 2010, 08:47
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Suppose sides of the ractangle are a and b.

Then, given: a^2 + b^2 = 10^2 (The diagonal of the ractangle is same as the diameter of the circle.)
=> a^2 + b^2 = 100

Question: Is ab > 48 ?

S1: a = 3k and b = 4k
(3k)^2 + (4k)^2 = 100 => k=2
Thus, a = 6 and b = 8 => ab = 48
Therefore, ab > 48 is not true.

Result: S1 is sufficient to answer the question.

S2: b - a < 3
Squaring both the sides of the inequality, gives:
a^2 + b^2 - 2ab < 9 => 100 - 2ab < 9
Solving the inequality gives: ab > 45.5
Therefore, ab > 48 may or may not be true.

Result: S2 is NOT sufficient to answer the question.

My answer is A.
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Re: m22#30 [#permalink] New post 11 Mar 2010, 17:00
chuckberry007 wrote:
If the diameter = 10 and radius is 5, then the hypo of the mini triangle is 3:4:5 (you need to draw to visualise).

Hence, the side of the triangle is 6 (3X2) and 8 (4X2). This is sufficient to answer the question..



I initially agreed with you but on 2nd thought...


the pythagorean triples only hold for integers. The stimulus never states that the sides of the rectangle were integers.

if a & b are the sides of the rectangle and c is the diagonal

then the pyth theorem states

a^2 + b^2 = c^2

c = 10
lets set a =1

now the equation becomes:

1 + b^2 = 100
b^2 = 99
b = sqrt(99) = aproximately 9.95

the area of the rectangle is a*b = 1*9.95 = 9.95, which is less than 48

we can also use the triples 6,8,10 to get the area = 48.

Therefore we need more information than just the stimulus to answer the question.


1) if the ratio of the sides are 3:4 and the diagnol is 10, then the sides must equal 6 & 8 and the area equals 48

sufficient

2) try 6,8,10 :
8-6 = 2 which meets the criteria. Area = 48

try 7, x, 10:
x^2 = 100-49 = 51
x=sqrt(51)= a lil bit greater than 7

area = 7 * x = 7 * (a lil greater than 7) = something greater than 48.

We have shown that the area can equal 48 as well as be greater than 48 with the difference between the sides of the rectangle less than 3.

Insufficient.
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Re: m22#30 [#permalink] New post 13 Mar 2010, 16:46
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a rectangle with the max area is possible only when diagonal is the diameter. but we can have a square as well with diagonal as 10, resulting in each side with 5root2 and area = 50. So we just need to make sure that it is not a square, since the A says that sides are in 3:4 ratio, so the largest rectangle can be with sides 6,8 and dia 10. and area = 48.

B can conclude in both sides with same length and area = 50..so not enough...

Ans is A.
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Re: m22#30 [#permalink] New post 19 Mar 2010, 07:45
ventivish wrote:
A rectangle is inscribed in a circle of radius 5. Is the area of the rectangle bigger than 48 ?

1. The ratio of the lengths of sides of the rectangle is 3:4
2. The difference between the lengths of sides of the rectangle is smaller than 3

[Reveal] Spoiler: OA
A

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hypotenuse of rectangle is 10 i.e. diameter

stmt1: 3:4 3x and 4x are the sides and 9x^2+16x^2 = 100 =>x^2 = 4 => x = 2 => 6 and 8 are the sides and area is 48 so suff
stmt2: l-b<3 and we have l^2+b^2 = 100 cannot find l or b so insuff
hence A
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Re: m22#30 [#permalink] New post 09 May 2010, 12:27
GMAT TIGER wrote:
ventivish wrote:
Thanks alot GMATTIGER for your answers.
However, the question does not mention that the diagonal of the rectangle passes through the centre of the circle.
Is this something that must be assumed for a rectangle inscribed in a circle?



It has to be. You cannot make any rectanlge that is inscribed in a circle with a diagnal not equal tyo the diameter of the circle. No matter however you draw a rectangle, its diagnol is the diameter of the circle.



The reason is that rectangle has all it's angles equal to 90degree. And you must know that angle inscribed in a semicircle is always 90degrees. This means, the diagonal is exactly the diameter of the circle.
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Re: m22#30 [#permalink] New post 12 May 2010, 01:31
Hello all,

Just trying to confirm this very question ?

Is the area of the rectangle within the corcle bigger than 48?

For both statements 1 ) AREA = 48 !! Not > 48
2) Not sufficient

Please kindly let me know that i am right please .

IMO: E
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Re: m22#30 [#permalink] New post 19 Jun 2010, 23:26
I have gone through the previous posts, and I agree with them:
That the diagonal of an inscribed rectangle is always the diameter of the circle.
(1) 3k:4k:10 => (3k)^2 + (4k)^2 = 100…k=2; 6^2+8^2(area of rec.) = 48…Suff.

(2) 6x8(difference is 2) = 48; 7xsqrt(51) (difference is < 1) > 48…Insuff.

OR a – b < 3 (given)…where a^2 + b^2 = 10^2
=> a^2+b^2 – 2ab < 9
=> 100 – 2ab < 9 i.e ab >45.5. hence, ab may or may not be > 48… Insuff
clearly understood...OA is thus A.
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Re: m22#30 [#permalink] New post 23 Jun 2010, 13:22
If you change the 2nd stmt as follows
2. The difference between the lengths of sides of the rectangle is smaller than 2
then the area will be bigger than 48.
So, the 1st stmt is suff to determine that the area is 48 and not bigger, the second stmt is suff to determine that the area is bigger than 48.
What is the answer in this case?
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Re: m22#30 [#permalink] New post 07 Sep 2010, 22:26
The explanation for A:

1) l:b is defined so l,b values can be calculated.
2) l-b<3 => l^2 + b^2 - 2lb < 9 or lb> 45.5 So unsufficient to say whether lb > 48 or not.
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Re: m22#30 [#permalink] New post 13 Sep 2010, 03:41
Answer is D (Correct me if I am wrong)

Statement 1 : Everybody knows that this is sufficient.
Statement 2 : It says that the difference between the lengths of sides of the rectangle is smaller than 3,so that means l-b < 3..Also it has a fact that the diagonal of the rectangle is 10 (since radius of circle is 5).Now,considering these values (l^2 + b^2 = 100 and l-b<3),we have only one possibility when the diagonal can be of 10 units i.e. when we have sides 8 and 6.Hence,this statement is also sufficient to answer.

Let me know what you guys think....
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Re: m22#30 [#permalink] New post 13 Sep 2010, 21:07
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sahilkhurana06 wrote:
Answer is D (Correct me if I am wrong)

Statement 1 : Everybody knows that this is sufficient.
Statement 2 : It says that the difference between the lengths of sides of the rectangle is smaller than 3,so that means l-b < 3..Also it has a fact that the diagonal of the rectangle is 10 (since radius of circle is 5).Now,considering these values (l^2 + b^2 = 100 and l-b<3),we have only one possibility when the diagonal can be of 10 units i.e. when we have sides 8 and 6.Hence,this statement is also sufficient to answer.

Let me know what you guys think....


Statement (2) is not sufficient, see algebraic solutions on previous page or consider the following examples:

If a=8 and b=6 (a-b=2<3 and a^2+b^2=100) then area=ab=48 and the answer to the question "is area>48" is NO;

If a=b=\sqrt{50} (a-b=0<3 and a^2+b^2=100) then area=ab=50 and the answer to the question "is area>48" is YES. Note that in this case inscribed figure is square, which is a special type of rectangle.

Two different answers not sufficient.

Answer: A.

I think that the problem with your solution is that you assumed with no ground for it that the lengths of the sides of the rectangle are integers.

Hope it's clear.
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Re: m22#30 [#permalink] New post 15 Mar 2012, 05:14
I am new on this forum and would like to know if the GMAT Club Tests were providing 'official' answers explanations.
Having 10-15 posts to answer a question provides some interesting back and forth and sometimes multiple angles to solve it
but it can unfortunately be confusing as well.
Adding an 'official' explanation to the correct answer - maybe as a second show/ hide section - would in my opinion provide added value.
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Re: m22#30 [#permalink] New post 19 Mar 2012, 10:03
1. The ratio of the lengths of sides of the rectangle is 3:4:- This gives area of rectangle equal to 48 hence sufficient
2. The difference between the lengths of sides of the rectangle is smaller than 3: If a,b are the two sides then a:b=3:4 or a=3k,b=4k where k is any constant.
Difference =4k-3k=k<3 hence k=0,1,2 k cannot be equal to 0,1 as if k=1 then a=3,b=4 which is not possible as diameter of circle=10. if k=2 then a=6,b=8 and area=48 hence sufficient.

so IMO D
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Re: m22#30 [#permalink] New post 25 Apr 2012, 06:15
Bunuel wrote:
sahilkhurana06 wrote:
Answer is D (Correct me if I am wrong)

Statement 1 : Everybody knows that this is sufficient.
Statement 2 : It says that the difference between the lengths of sides of the rectangle is smaller than 3,so that means l-b < 3..Also it has a fact that the diagonal of the rectangle is 10 (since radius of circle is 5).Now,considering these values (l^2 + b^2 = 100 and l-b<3),we have only one possibility when the diagonal can be of 10 units i.e. when we have sides 8 and 6.Hence,this statement is also sufficient to answer.

Let me know what you guys think....


Statement (2) is not sufficient, see algebraic solutions on previous page or consider the following examples:

If a=8 and b=6 (a-b=2<3 and a^2+b^2=100) then area=ab=48 and the answer to the question "is area>48" is NO;

If a=b=\sqrt{50} (a-b=0<3 and a^2+b^2=100) then area=ab=50 and the answer to the question "is area>48" is YES. Note that in this case inscribed figure is square, which is a special type of rectangle.

Two different answers not sufficient.

Answer: A.

I think that the problem with your solution is that you assumed with no ground for it that the lengths of the sides of the rectangle are integers.

Hope it's clear.


Bunuel....Can u please explain the solution..( both options A and B)... i am quite confused about this problem of how is A sufficient and how is B insufficient
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