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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
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.
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?
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. _________________
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.
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...
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 _________________
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.
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. _________________
KUDOS me if you feel my contribution has helped you.
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?
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.
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.
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. Greg _________________
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.
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 _________________
Regards, Harsha
Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat