Circle and Squares Math Puzzle (Pre-Algebra Level)

[Otis]

The diagram shows a square inscribed inside a circle inscribed inside a square. If the area of the larger square is 64 square units, what is the area of the smaller square?




[imath]\;[/imath]

 

[AvgStudent]

The first thing that came into my mind is to use the Pythagorean theorem, but I don't recall learning this in pre-algebra. Is there a pre-algebra way?

Spoiler

Since the area of the large square is 64 units^2, then its sides are 8 units => the diameter of the circle = diagonal of the smaller square = 8 units.
By 45-45-90 triangle or the Pythagorean theorem, the side of the smaller square is 8/sqrt(2)=4sqrt(2) -> Area of the smaller square is (4*sqrt(2))^2=32 units^2

 

[Dr.Peterson]

The diagram shows a square inscribed inside a circle inscribed inside a square. If the area of the larger square is 64 square units, what is the area of the smaller square?



View attachment 31676
[imath]\;[/imath]

Spoiler

If we rotate the inner square, we can see that its corners will touch the sides of the large square. Drawing in its diagonals, we can see that this square consists of 4 out of 8 equal parts of the large square (triangles), so its area is 1/2 of the whole, or 32 square units.

 

[Otis]

Is there a pre-algebra way?

There is -- in some school districts, anyway -- and your solution steps pretty much match mine. Good Job!

In grade school, I learned about square roots and rationalizing denominators, as well as some introductory geometry formulas and relationships (eg: the connection between a square's side and its diagonal).

Spoiler

We see that the circle's diameter, the big square's side length,
and the small square's diagonal are all the same length.

Any square's Diagonal measures: √2 × Side Length,
so Side Length is always [imath]\frac{1}{\sqrt{2}}\text{th}[/imath] of the Diagonal.

Area of small square: [imath]\frac{8}{\sqrt{2}}×\frac{8}{\sqrt{2}}[/imath]

[imath]\;[/imath]