Sorry for posting a image file as i couldn't find the text anywhere. Could anyone please solve it mathematically. No idea how to approach. Thanks

sui generis wrote:Sorry for posting a image file as i couldn't find the text anywhere. Could anyone please solve it mathematically. No idea how to approach. Thanks

A rectangle is inscribed in a circle of radius r. If the rectangle is not a square, which of the following could be equal to the perimeter of the rectangle? A. 2r(√3) B. 2r(√3 + 1) C. 4r(√2) D. 4r(√3) E. 4r(√3 + 1) When shapes overlap, look for what they have IN COMMON. When a rectangle is inscribed in a circle, the DIAGONAL of the rectangle is also the DIAMETER of the circle. Almost every answer choice here includes √3. √3 implies a 30-60-90 triangle. The sides of a 30-60-90 triangle are in the following ratio: 1 - √3 - 2. Let r=1, implying a diameter of 2. Draw the following figure: The perimeter of the rectangle above = 2 + 2√3. This is our target. Now we plug r=1 into the answers to see whether one of them yields our target of 2 + 2√3. Answer choice B: 2r(√3 + 1) = 2(1)(√3 + 1) = 2√3 + 2. Success! The correct answer is B.

Awesome solutions guys. Learnt a new concept today. Thanks Aneesh and Mitch.

The solutions above obtained the right answer by guessing there is 30-60-90 triangle involved. That is a good approach for a COULD BE problem because once you found an answer that works, it is the answer. On the other hand, it remained unclear why the other answers do not work. This is actually an optimization problem in disguise. The perimeter of the rectangle has a range between two extremal values: The first extreme case is when one of the sides is completely squashed into a dot (side a = 0, side b = 2r), which gives a perimeter p = 4r. The other extreme case is when the other side is completely squashed into a dot (side a = 2r, side b = 0), giving the same extremal value of p = 4r. Changing the rectangle by increasing one side and decreasing the other changed the perimeter from 4r to 4r. Somewhere in between, there must be another extremal value. By symmetry it happens exactly when the rectangle is a square (side a = b = r*root(2)), giving a perimeter of r*root(2)*4. The bottom line is the perimeter is between 4r and 4r*root(2), with the lower limit corresponding to a rectangle squashed into a line, and the upper limit corresponding to a square. That excludes all the answers except one: A. approximately 3.4r - too small, reject B. approximately 5.4r - passes C. this is the case of a square - excluded by the problem D. too big - reject E. too big - reject