In the given square $ABCD$, you are asked to find the area of the square. The given information includes:

• $AE = 6$
• $EF = 8$
• $CF = 10$

Step 1: Apply the Pythagorean Theorem to Triangle $ECF$

In triangle $ECF$, we know that $EF = 8$ and $CF = 10$, and angle $\angle EFC = 90^\circ$, forming a right-angled triangle. We can use the Pythagorean theorem to find $EC$.

The Pythagorean theorem states: $EC^2 + EF^2 = CF^2$ Substitute the given values: $EC^2 + 8^2 = 10^2$ $EC^2 + 64 = 100$ $EC^2 = 36$ $EC = 6$

Step 2: Recognize that $EC = AE = 6$

From the result, we see that $EC = AE = 6$. This means triangle $AEC$ is isosceles.

Step 3: Find the Side Length of the Square

Since both triangles $AEC$ and $EFC$ form part of the geometry inside square $ABCD$, and they share certain relationships, you can deduce that the diagonal $AC$ of the square is composed of these right triangles.

Using the Pythagorean theorem again for the diagonal $AC$, which is the hypotenuse of triangle $ABC$: $AC = \sqrt{AE^2 + CF^2} = \sqrt{6^2 + 10^2} = \sqrt{36 + 100} = \sqrt{136}$

Now, the side length $s$ of the square is related to the diagonal by the formula: $AC = s\sqrt{2}$ So, $s\sqrt{2} = \sqrt{136}$ $s = \frac{\sqrt{136}}{\sqrt{2}} = \sqrt{\frac{136}{2}} = \sqrt{68}$

Step 4: Find the Area of the Square

The area $A$ of the square is: $A = s^2 = (\sqrt{68})^2 = 68$

It seems there was a small error in one of the images, and the correct area might actually be an option from the choices provided. Let me recheck the final side lengths and confirm if there are further details