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, I understand the solution given in your other explanation. Thanks for the help. My question is on this first explination. How do we get \(a=b=\sqrt{50}\)? Is it simply because \(a^2+b^2=100\)...I think I figured it out as I was typing this, but is that correct? I was originally trying to draw a connection to the radius and the sides of the square (25+25...) Anyway.