Simple Geometric Principle

It is easy to see that the diagonal line is a 45 degree angle (because it passes from corner to corner through the square). Because of this, its horizontal and vertical components are equal. Using the Pythagorean theorem, $x^2+y^2=h^2$

$2y^2=(7 \sqrt{2})^2$

$y^2=49$

$y=7$

The vertical component of AH is the height of the big square, so the area of this square is 49.

Given That Area of Square $ABCD=4$

Let Side of Square $ABCD=x$

So, $x^2=4$

$x=\pm 2$

Since Side cannot Be Negative, Therefore Side of Square $=2=AB=BC$

Now, By Using Pythagoras Theoram In $\triangle ABC$

$AC^2=AB^2+BC^2$

$AC^2=2^2+2^2$

$AC^2=4+4$

$AC^2=8$

$AC=\pm2\sqrt{2}$

Since $\overline{AC} \neq-2$ ,Therefore $\overline{AC}=2$

$AH=7\sqrt{2}\ldots$ (Given)

$CH=AH-AC$

$CH=7\sqrt{2}-2\sqrt{2}$

$CH=5\sqrt{2}$

Now, Since $\triangle ABC$ is a Right Angle Isoceles Triangle

Therefore, $\angle ACB=45^\circ$

$\angle ACB=\angle GCH=45^\circ\ldots$ (VOA)

Now, In $\triangle CGH$

$\angle CGH+\angle GHC+\angle GCH=180^\circ$

$90^\circ+45^\circ+\angle GHC=180^\circ$

$\angle GHC=45^\circ$

Hence $\triangle CGH$ is a Right Angle Isoceles Triangle

Now Let Side of $\triangle CGH$ Be $=y=CG=GH$

By Using Pythagoras Theorm

$CH^2=CG^2+GH^2$

$(5\sqrt{2})^2=y^2+y^2$

$(5\sqrt{2})^2=2y^2$

$(5\sqrt{2})^2=(\sqrt{2}y)^2$

$5\sqrt{2}=\sqrt{2}y$

$y=5$

Therefore, $\overline CG=5$

Now $BG=BC+CG$

$BG=2+5$

$BG=7$

Now, In Square $BEFG$

Area Of Square $BEFG=BG^2$

Area Of Square $BEFG=7^2$

Area Of Square $BEFG=\boxed{49}$

Since the area of the square ABCD is 4 then its side length equal 2 so the diagonal AC equal 2√2 then CH equal 5√2 then GC equal 5 so GB length equal 7 then area equal 49

Simply, if we bisect the right angle of the square, we get of course a 45 degree angle. The special right triangle gives a hypotenuse of $x\cdot\sqrt{2}$ for any x. If we extend this to the next square, the hypotenuse length will be equal to the difference between $\overline{AB}$ and $\overline{BE}$ , $5\sqrt{2}$ . Since we know the area is 4, $\overline{AB}=2$ . Thus, we know that for the square BEFG, the line $BE$ equals $\frac{\overline{AH}}{\sqrt{2}}=7$ . So the area is $\boxed{49}$

49 was the only reasonable factor with a result 7

Length of AC is not needed, BAC is 45' and triangle with hypo 7sqrt2 with 45' sin is 7

The big square looks like it can house 4 small squares,4*4=16 and 49is close to 16ish