The figure above shows a rectangle inscribed within a square. Now many

Question

https://gmatclub.com/forum/download/file.php?id=41678 The figure above shows a rectangle inscribed within a square. Now many times greater is the perimeter of the square than the perimeter of the Inscribed rectangle? A. \sqrt{2} B. \frac{2 + \sqrt{2}}{2} C. 2 D. 2 \sqrt{2} E. It cannot be determined from the information given. square.jpg

vp101 · Answer

I think the answer is A

pr3mk5 · Answer

https://i63.tinypic.com/egtroj.png

Lets Assume the side of Square to be 'A'
and the breadth of rectangle to be 'X'

So each side would be X + A-X.

And in square each angle is 90 degrees.

perimeter of Square would be 4A and for rectangle we need to find the sides.

It'd for a 45-45-90 triangle. So we can use 1:1:\sqrt{2} formulae.

Since both side are 'X' breadth would be \sqrt{2} X

and length would be \sqrt{2} (A-X)

So perimeter would be 2(\sqrt{2} X + \sqrt{2}(A-X))

2\sqrt{2}(A)

Therefore perimeter of Square would be equal to \sqrt{2} times 2\sqrt{2}A.

OA - A

EgmatQuantExpert · Answer

Solution

Given:  
 • A rectangle is inscribed within a square as shown

To find:  
 • The perimeter of the square is how many times greater than that of rectangle

Approach and Working:

https://e-gmat.com/blogs/wp-content/uploads/2018/06/square-rectangle.png

Following the diagram, we can say
 • Length of the rectangle =  \sqrt{2} b
• Breadth of the rectangle =  \sqrt{2} a
• Hence, perimeter of the rectangle = 2 ( \sqrt{2} a+  \sqrt{2} b) = 2 \sqrt{2}  (a + b)

Also, for the square,
 • Perimeter = 4 (a + b)

Hence, the required ratio =  \frac{4 (a + b)}{2√2 (a + b)}  =  \sqrt{2}

Hence, the correct answer is option A.

Answer: A

GyMrAT · Answer

the rectangle divides each side of the square into a small segment & a large segment.

Let x be the length of the large segment & y be the length of the small segment.

Hence each side of square is (x + y) units

Perimeter of square = 4(x + y) units.

Now the large segments on perpendicular sides form a 45-45-90 isosceles right triangle with hypotenuse as the length of the rectangle.

Hence Length of rectangle =  \sqrt{2} x

Similarly the small segments on perpendicular sides form a 45-45-90 isosceles right triangle with hypotenuse as the width of the rectangle.

Hence Width of rectangle =  \sqrt{2} y

we have Perimeter of rectangle =  2\sqrt{2} (x + y)

Required ratio =  4 (x + y)/ 2\sqrt{2} (x + y) =  \sqrt{2}

Answer A.

Thanks,
GyM

Tuanguyen248 · Answer

Put a side of the square into 2 smaller: x+y => perimeter of the square = 4(x+y) (1)

each side of rectangle will be:  x\sqrt{2}  and  y\sqrt{2}  = > perimeter of rec =  2\sqrt{2}(x+y])  (2).

(1)/(2) =  \sqrt{2}

Answer: A

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The figure above shows a rectangle inscribed within a square. Now many

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