Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

68% (01:15) correct
32% (01:52) wrong based on 140 sessions

HideShow timer Statistics

If a square is inscribed in a circle that, in turn, is inscribed in a larger square, what is the ratio of the perimeter of the larger square to that of the smaller square?

A. \(\frac{1}{2}\) B. \(\frac{1}{\sqrt{2}}\) C. \(\sqrt{2}\) D. \(2\) E. \(3.14\)

If a square is inscribed in a circle that, in turn, is inscribed in a larger square, what is the ratio of the perimeter of the larger square to that of the smaller square?

A. \(\frac{1}{2}\) B. \(\frac{1}{\sqrt{2}}\) C. \(\sqrt{2}\) D. \(2\) E. \(3.14\)

Consider the diagram below:

The side of a large square is \(a\), thus its perimeter is \(4a\);

The side of a small square is \(\sqrt{(\frac{a}{2})^2+(\frac{a}{2})^2}=\frac{a}{\sqrt{2}}\), thus its perimeter is \(\frac{4a}{\sqrt{2}}\);

Hence the ratio is \(\frac{(4a)}{(\frac{4a}{\sqrt{2}})}=\sqrt{2}\).

Let the side of the smaller square be root 2, hence the diagonal of circle = 2. The diag of circle a the same time equals the side of the larger square. Here is the fraction.

Isn't it also possible to use a smart number here?

Say the larger square's side equals 4 then two sides of the triangle would equal 2. The hypotenuse (and one side of the smaller square) would equal 2root2. The larger square would have a perimeter of 16 and smaller square would have a perimeter of 8root2. At that point I feel like this approach should work but I think I missed something because I can't get to root2 as an answer.

(S) is the side of the smaller square and (L) is the side of the larger square; ratio of the perimeters = (4L)/(4S).

***The 4s will cancel leaving us with (L/S) -> remember that this is the answer we are looking for.

When you draw the figure with the smaller square inscribed in a circle, which is inscribed in a larger square, you will see that the diagonal of the smaller square = side of the larger square (L)

***Diagonal of the smaller square = the side of the larger square.

Diagonal of smaller square = S\(\sqrt{2}\) Side of Larger Square (L) = S\(\sqrt{2}\) -> Plug this into the equation (L/S)

(S\(\sqrt{2}\))/(S) -> the S cancels leaving us with \(\sqrt{2}\) as the answer (C) _________________

Really didn´t know how to solve it... just thought that the ratio of the larger to the smaller should be larger than 1 (thus eliminating A and B), and shouldn´t be twice as large, so the only answer possible was C..

For the smaller square inscribed in the circle, assume that the side of each square is X. Using the properties of squares, we know that the diagonal is x(rt 2). The perimeter of the smaller is square is 4*x = 4x.

Since the diagonal of the square is also the diameter of the circle, therefore the diameter also equals x(rt 2).

For the circle to be inscribed on the larger square, its diameter must equal the length of one of the sides of the square. So the length of the sides of the larger square are x(rt 2).

Therefore the perimeter of the large square is 4 * x(rt 2) = 4x(rt 2).

The question asks for the ratio of the perimeter of the larger square to the smaller square so we arrange:

Assuming each side of the large square is 2. Then the perimeter of the large sqaure is 8

The radius of the circle is half the Perimeter of the large sqaure which is 1 The radius of the circle is also half the diagonal of the small square inside the circle. The full diagonal of the small square is therefore 2.

We can make the small square inside the circle into two 45-45-90 triangles where the hypotnuse equals the diagonal which is 2

Therefore one side of the small sqaure is 2/SQRT(2) = 2* SQRT(2) / 2 = SQRT(2) The perimeter of 4 sides of the small sqaure is 4 * SQRT(2)

Comparing this to the large sqaure would be 8 : 4SQRT(2) which simplifies to 1 : 1/2 SQRT(2)

However, the answer is just SQRT(2), not 1/2 SQRT(2)

Assuming each side of the large square is 2. Then the perimeter of the large sqaure is 8

The radius of the circle is half the Perimeter of the large sqaure which is 1 The radius of the circle is also half the diagonal of the small square inside the circle. The full diagonal of the small square is therefore 2.

We can make the small square inside the circle into two 45-45-90 triangles where the hypotnuse equals the diagonal which is 2

Therefore one side of the small sqaure is 2/SQRT(2) = 2* SQRT(2) / 2 = SQRT(2) The perimeter of 4 sides of the small sqaure is 4 * SQRT(2)

Comparing this to the large sqaure would be 8 : 4SQRT(2) which simplifies to 1 : 1/2 SQRT(2)

However, the answer is just SQRT(2), not 1/2 SQRT(2)

Assuming each side of the large square is 2. Then the perimeter of the large sqaure is 8

The radius of the circle is half the Perimeter of the large sqaure which is 1 The radius of the circle is also half the diagonal of the small square inside the circle. The full diagonal of the small square is therefore 2.

We can make the small square inside the circle into two 45-45-90 triangles where the hypotnuse equals the diagonal which is 2

Therefore one side of the small sqaure is 2/SQRT(2) = 2* SQRT(2) / 2 = SQRT(2) The perimeter of 4 sides of the small sqaure is 4 * SQRT(2)

Comparing this to the large sqaure would be 8 : 4SQRT(2) which simplifies to 1 : 1/2 SQRT(2)

However, the answer is just SQRT(2), not 1/2 SQRT(2)

Side of smaller square = s Perimeter of smaller square = 4s Diagonal of smaller square, and hence the diameter of the circle = s*root2 The diameter of the circle = length of the side of the larger square Perimeter of larger square = 4s*root2 Ratio of perimeter of larger square to perimeter of smaller square = 4s*root2/4s = root2 Answer C

If ratio of sides of 2 Parallelograms (squares are parallelograms) = A/B , then ratio of areas will be \(\frac{A^2}{B^2}\)

Area of Square inscribed in a circle (Smaller square)= \(πa^2/4\) ----------Equation 1 Area of Square with a circle inscribed in it (Bigger square)= \(πa^2/2\)-----Equation 2

Now we have ratio of Areas as: \(\frac{Equation 2}{Equation 1}\) i.e. 2

Hence ratios of perimeters or sides = Square root of 2. \(C\)