Heron? Shoelace? Varignon's? Brahmagupta's?

Let each side of the square ABED be $x$ .

Therefore, the area of the square ABED = $x^2$ .

As we know that the Area of a triangle = $\Delta =\frac{1}{2} Base*Height$

$\Delta ABC = \frac{1}{2} AC*AB = 7 \Rightarrow AC=\frac{14}{x}$ $\Rightarrow DC=x-\frac{14}{x}$ $\Delta BEF = \frac{1}{2} BE*EF = 9 \Rightarrow EF=\frac{18}{x}$ $\Rightarrow DF=x-\frac{18}{x}$ $\Delta CDF = \frac{1}{2} DC*DF = 11$ $\Rightarrow (x-\frac{14}{x})(x-\frac{18}{x})=22$ $\Rightarrow x^4-54{x}^2+252=0$ The above is a quadratic equation in $x^2$ , therefore, $x^2=27 \pm 3\sqrt{53}$ The Area of the Required Shaded Region is the Area of the Square - Sum of the Area of the 3 Given Triangles.

Therefore, Required Area of Shaded Region $= x^2-27 =27 \pm 3\sqrt{53} -27=\pm 3\sqrt{53} = +3\sqrt{53} \equiv \boxed{21.84}$

The negative sign is abandoned as Area cannot be Negative. :D

Lots of algebra, but it's all straighforward.

Let $a$ be the side of the square, $b$ the horizontal side of the 7-triangle, and $c$ the vertical side of the 9-triangle. We write equations representing the given areas:

$ab=14,\quad{ac}=18, \quad(a-b)(a-c)=a^2-ac-ab+bc=22$ The last equation simplifies to $a^2-18-14+\frac{14\times18}{a^2}=22,\quad so\quad a^4-54a^2+252=0.$ We find $a^2=27\pm3\sqrt{53}$ , but we reject the solution with the negative sign as being too small; we need $a^2>7+9+11$ .

The area we seek is $(27+3\sqrt{53})-7-9-11=\boxed{3\sqrt{53}}\approx21.84$

I solved it in this photo i hope that can help you :]

There is a wonderful generalization of the solution for this problem: A = √[(a + b + c)^2 - 4ac] Where “a,” “b,” and “c” are the known areas, and “A” is the unknown area. Caveat: “a” and “c’ must be the triangles whose long legs are the side of the square, in this case 7 and 9. A = √[27x27 - 4(9x7)] = √(729 - 252) = 21.84

I solve this problem by using C programming and answer is 21.840330