In the diagram, a circle is inscribed in a large square and a smaller square is inscribed in the circle. If the area of the large square is 36 , the area of the smaller square is
(A) 15
(B) 12
(C) 9
(D) 24
(E) 18

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-077.jpg?height=401&width=398&top_left_y=1022&top_left_x=1292)
Solution 1

Since the area of the large square is 36 , then the side length of the large square is 6 .

Therefore, the diameter of the circle must be 6 as well, and so its radius is 3 .

Label the four vertices of the small square as $A, B, C$, and $D$.

Join $A$ to $C$ and $B$ to $D$.

Since $A B C D$ is a square, then $A C$ and $B D$ are perpendicular, crossing at point $O$, which by symmetry is the centre of the circle.

Therefore, $A O=B O=C O=D O=3$, the radius of the circle.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-346.jpg?height=399&width=425&top_left_y=814&top_left_x=904)

But square $A B C D$ is divided into four identical isosceles right-angled triangles.

The area of each of these triangles is $\frac{1}{2} b h=\frac{1}{2}(3)(3)=\frac{9}{2}$ so the area of the square is $4 \times \frac{9}{2}=18$.

## Solution 2

Rotate the smaller square so that its four corners are at the four points where the circle touches the large square.

Next, join the top and bottom points where the large square and circle touch, and join the left and right points.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-346.jpg?height=404&width=415&top_left_y=1744&top_left_x=904)

By symmetry, these two lines divide the large square into four sections (each of which is square) of equal area.

But the original smaller square occupies exactly one-half of each of these four sections, since each edge of the smaller square is a diagonal of one of these sections.

Therefore, the area of the smaller square is exactly one-half of the area of the larger square, or 18.

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