M17-23

Brunel,

Could you please explain how you derived the area of a small square ?

Forgive my ignorance but after spending 10 mins i am still lost

Thanks in Advance

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.

(2*4) / (4*root2)

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.

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..

The answer is option C.

The conventional method

Assume the radius of the circle to be r. The length of the smaller square is r*sqrt(2).
The length of the larger square is 2*r.

Now the ratio is (2*r)/(r*sqrt(2))

This is sqrt(2). Hence option 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:

4x(rt 2) / 4x.

The 4x's cancel out so we get rt 2 as the answer.

Where did I go wrong? I used smart numbers

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)

Check below:

\(\frac{8}{4\sqrt{2}}=\frac{2}{\sqrt{2}}=\frac{\sqrt{2}*\sqrt{2}}{\sqrt{2}}=\sqrt{2}\).

Hope it helps.

Oh my goodness, the question asked for ratio of big square to small. Silly mistake on my part. Thanks for the help!

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

We could do it this way:

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. \(C\)

Hello, how do we know that that is how the diagram will look? I knew that the diagonal of the smaller square would need to be the same as the circumference of the circle, but I got stuck after that. Any help would be appreciated. Thank you.

Bunuel KarishmaB avigutman