3 squares

Cut the lower half and the upper one into respectively 4 and 9 triangles of equal areas, as below.

Then $A=\frac{1}{2}$ and $B=\frac{4}{9}$ , so $\frac{A}{B}=\frac{1}{2}\frac{9}{4}=\frac{9}{8}$

For the larger square we can say it has a side length of s.

For square A to fit and still be a square, it will need to have half the length of the original, or a side length of $\frac{1}{2}$ s

so the area of square A is $\frac{1}{4}$ $s^{2}$

For square B to fit and still be a square, we can split the diagonal of the big square into 3 equal pieces. To find the length of the diagonal we can use Pythagorean theorem or the 45-45-90 special right triangle rule, we can say that the diagonal has a length $\sqrt{2}$ s so the side length of square B is $\frac{\sqrt{2}}{3}$ s

This means the area of square B is $\frac{2}{9}$ $s^{2}$

so the ratio of the areas $\frac{A}{B}$ is $\frac{\frac{1}{4}s^{2}}{\frac{2}{9}s^{2}}$ which simplifies to $\boxed{\frac{9}{8}}$

Without even using numbers, we see that Square A can be formed from the two triangles above and below it (this can be proven using the facts that a diagonal of a square bisects the 90 degree angles and that one of the legs of each triangle is also one of the sides of Square A) and leave no remaining parts, whereas Square B can be formed from the triangles beside it (this can be proven in the same way as explained for Square A) but leaves a remaining triangular part. So A is larger than B and the only valid answer is the third choice: 2A=.5 2B+c=.5 (where c is the remaining triangular part) 2A=2B+c»»»A=B+(c/2) A>B

Here is a little visualization: Because the upper and lower sum represent both a half of the triangle, it follows: $2.25B=2A\Rightarrow \boldsymbol{\frac{A}{B}=\frac{2.25}{2}=\frac{9}{8}}$

Firstly, we assume that the square pictured is a unit square. Then, notice that square $A$ has side lengths of $\frac{1}{2}$ . Given that information, we seek to find square $B$ 's side length, which we will label $x$ . Additionally, we will label the length of the top left corner to the vertex of square $B$ along the top edge $y$ . The last piece of preliminary information we will use is that the length of half the diagonal, or from a corner to the vertex on square $A$ along the diagonal, is $\frac{\sqrt{2}}{2}$ . All triangles in this diagram are similar by angle-angle similarity because they are all $45$ - $45$ - $90$ triangles. Therefore, by triangle similarity, $\frac{1-y}{x} = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}}$ $y = 1-\sqrt{2}x$ Then plugging the previous equation into another similarity proportion and we get $\frac{1-\sqrt{2}x}{x} = \frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}}$ $x = \frac{\sqrt{2}}{3}$ Thus, the area of square $A$ is $\frac{1}{4}$ and the area of square $B$ is $\frac{2}{9}$ . The ratio of $\frac{A}{B}$ is then $\frac{\frac{1}{4}}{\frac{2}{9}} = \boxed{\frac{9}{8}}$ .

The statement said that A > B therefore A / B> 1 so that the only possible solution was 9/8. I know that it is not the most elegant solution, I have not done it either.

From the figure we can just observe that A is larger than B

I found the solution using triangle congruencies and defining the sides of A as X. By using the triangle congruencies, I found that the big triangle that square A is in has side length of 2X, and therefore all the sides of the big square are 2x. I will attach a picture of my work (Sorry its a bit messy):

Let the sides of box A be of length a, and the sides of box B be of length b. Observe that A's side divides evenly the side of the square, so that the square's side is of length 2a. So then the diagonal of the square has a squared length of 8A. That same diagonal has a length of 3b, or it's square 9B. So 8A = 9B, and A/B = 9/8

let L=the length of the largest square,

A=( $\frac{L}{2}$ )^2= $\frac{L^2}{4}$

We can cut the top left side of the diagonal into 9 identical triangles to show that sqrt(B) is $\frac{diagonal}{3}$

B=( $\frac{diagonal}{3}$ )^2=( $\frac{sqrt(2L^2)}{3}$ )^2= $\frac{2L^2}{9}$

A* $\frac{1}{B}$ = $\frac{L^2*9}{4*2L^2}$ = $\frac{9}{8}$

:-)

The length of one side of the square in the lower half(A) is half of the size of the length of the outer square. Therefore area of square A is $\frac{a^2}{4}$ . Length of one side of the square in the upper half (B) is one third of the diagonal of the outer square. Therefore area of square B is $\frac{2a^2}{9}$ . Therefore $\frac{A}{B}$ equals $\frac{9}{8}$ .

The whole square (let's call it $S$ ) is split into two equal halves. Taking the half with the square $A$ , we notice that it is comprised of $A$ as well as two other sections, $t_{1}$ and $t_{2}$ . $A$ is a square extending from the corner of $S$ to its diagonal, and therefore it is a quarter of $S$ , and thus half of the right triangle made by the diagonal. Triangles $t_{1}$ and $t_{2}$ , then, also together form half the right triangle. So we can say $t_{1}+t_{2}=\frac{1}{4}S=A$ . Together, the left triangle half can be written as $A+t_{1}+t_{2}=A+A$ , or $2A$ .

As for the right half, it is comprised of square $B$ and three triangles: $t_{3}$ and $t_{4}$ , which are identical, and $t_{5}$ in the corner. Noticing that $t_{5}$ can be folded down onto the intersection of the diagonals of $B$ informs us that $t_{5}=\frac{1}{4}B$ , because it can only cover a fourth of the area. Likewise, triangles $t_{3}$ and $t_{4}$ can be folded across $B$ , and each take up half of the area of $B$ . So $t_{3}=t_{4}=\frac{1}{2}B$ . The right half of square $S$ , then, is written $B+t_{3}+t_{4}+t_{5}=B+\frac{1}{2}B+\frac{1}{2}B+\frac{1}{4}B$ , which simplifies to $\frac{9}{4}B$ .

Finally, since each half of $S$ is equivalent, we can set the left half, $2A$ , equal to the right half, $\frac{9}{4}B$ . So, $2A=\frac{9}{4}B$ , and rearranging gives us $\frac{A}{B}=\frac{9}{8}$ .

A can be formed using A1 and A2, and B can be formed using B1 and B2. Since the two halfs of the square are equal, we have

$\begin{aligned} A + A1 + A2 &= B + B1 + B2 + C\\ 2A &= 2B + C\\ A &= B + \frac{C}{2} > B\\ A/B &> 1 \end{aligned}$

(where the labels represent the areas of the geometric object)