Insufficient Information in Geometry?

Note that $\left[BAF \right]$ and $\left[DFE \right]$ are similar. Since $\left[BF\right] : \left[FD \right] = 4:3$ , their area ratio are $16:9$ .

$\begin{aligned} 100 &= 16x + 9x \\ 100 &= 25x \\ x&=4\\ \end{aligned}$

This will gives us $\left[BAF \right] = 16x = 16 \times 4 = 64$ .

Since $\left[BF \right] : \left[FD \right] = 4:3$ , $\left[DFA \right] = \frac{3}{4} \times 64 = 48$ .

Hence, $\left[ABD\right] = \left[DFA \right] + \left[BAF\right] = 64+48 = 112$

Since $\left[ABD\right]$ is half of the square, hence the square area is $2\times 112 = \color{#D61F06}{\boxed{224}}$ .

From the text, we know that $FD : BF = 3 : 4 \space cm$ . And the ratio between two similar triangles is the ratio between the square of the sides of the triangles.

We have $\dfrac{\left[ DEF\right] }{ \left[ BAF \right] } = \dfrac{FD^{2}}{ BF^{2}} = \dfrac{9}{16}$ .

Now, we can find the area of $\left[ DEF \right]$ and $\left[ BAF \right]$ , since we have $\left[ DEF \right] + \left[ BAF \right] = 100 \space cm^2$ .

$\left[ DEF \right] = \dfrac{9}{9+16} \times 100 = 36 \space cm^2$ .

$\left[ BAF \right] = 100 - 36 = 64 \space cm^2$ .

We can draw a line $IJ$ which is parallel to $AD$ , and a line $KL$ which is parallel to $AB$ and prependicular to $IJ$ as you can see in the figure.

Let the side of the square is $a$ and $CE = x \space cm$ , which means $DE = (a-x) \space cm$ . We let the height of $\triangle DEF = t_1 \space cm$ and the height of $\triangle BAF = t_2 \space cm$ .

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We can see that $\begin{aligned} \left[ BAF \right] & = \color{#D61F06} \dfrac{at_2}{2} \\ \color{#D61F06} 64 & = \color{#D61F06} \dfrac{at_2}{2} \\ \color{#D61F06} at_2 & = \color{#D61F06}128 \space cm^2 \end{aligned}$ .

So, we have $\color{#D61F06} \left[ ABLK \right] = \color{#D61F06} 128 \space cm^2$ .

We can see that $\triangle DEF$ and $\triangle BAF$ are similar to each other. Then we have

$\dfrac{a-x}{a} = \dfrac{FD}{BF} = \dfrac{3}{4}$

$3a = 4a - 4x \Rightarrow x = \dfrac{1}{4}a$ .

We know that $\left[ DEF \right] = \dfrac{DE \times t_1}{2} \Rightarrow 36 = \dfrac{(a-x)t_1}{2}$ .

$\begin{aligned} \color{#3D99F6} 36 & = \color{#3D99F6} \dfrac{(a-x)t_1}{2} \\ \color{#3D99F6} 72 & = \color{#3D99F6} (a - \dfrac{1}{4})t_1 \\ \color{#3D99F6} \dfrac{3}{4}at_1 & = \color{#3D99F6} 72 \\ \color{#3D99F6} at_1 & = \color{#3D99F6} 96 \space cm^2 \end{aligned}$ .

Now, we have $\color{#3D99F6} \left[ DKLC \right] = \color{#3D99F6} 96 \space cm^2$ .

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Hence, we have $\begin{aligned} \left[ ABCD \right] & = \color{#D61F06} \left[ ABLK \right] \color{#333333} + \color{#3D99F6} \left[ DKLC \right] \\ & = \color{#D61F06} 128 \color{#333333} + \color{#3D99F6} 96 \color{#333333} = \color{#69047E} 224 \space cm^2 \end{aligned}$

Let $x$ be the side length of the square. By pythagorean theorem, $BD = x\sqrt{2}$ . Then, $BF=\frac{4}{7}x\sqrt{2}$ and $FD=\frac{3}{7}\sqrt{2}$ . Since $\bigtriangleup$ $DFE$ is similar to $\bigtriangleup$ $BFA$ , $\frac{DE}{x}=\frac{DF}{FB}$ $\implies$ $DE=\frac{3}{4}x$ .

Solving for the areas of the blue region, we have

$A_1 = \frac{1}{2}(x)(\frac{4}{7})x\sqrt{2}sin~45=0.285714285x^2$

$A_2=\frac{1}{2}(\frac{3}{4}x)(\frac{3}{7})x\sqrt{2}sin~45=0.160714285x^2$

Adding the areas, we have

$0.285714285x^2 + 0.160714285x^2 = 100$

$0.44642857x^2 = 100$

$\boxed{\large\color{#D61F06}x^2 = 224~cm^2}$

Let a be the length of sides of the square.

Obviously ΔABF~ΔDEF . Therefore, BF:FD=4:3=AB:DE=H:h (where, H is height of ΔABF with base AB & h is height of ΔDEF with base DE). Note that H+h=a

We can write, DE=(3/4)*AB=3a/4, H=4a/7 & h=3a/7

Now, Combined Area of Blue Triangles =Ar(ΔABF)+Ar(ΔDEF)=(1/2)×(AB×H+DE×h)=(1/2) {(a) (4a/7)+(3a/4)×(3a/7)}=(2/7+9/56) a^2= (25/56) a^2 =100 sq.cm (given)

Hence, Area of square ABCD =a^2=100*56/25= 224 sq.cm