To solve the problem, we will start by setting up the equations based on the given information.

Let's denote:

• $a$ as the side length of the larger square.
• $b$ as the side length of the smaller square.

We are given two key pieces of information:

1. The perimeter of the entire figure is $86$ cm.
2. The sum of the areas of the two squares is $386$ cm$^2$.

Step 1: Set up the equation for the perimeter

Since the squares share a vertex and one entire side of the smaller square overlaps part of a side of the larger square, the overall perimeter can be calculated as follows:

• The side of the larger square contributes $2a$ to the perimeter.
• The side of the smaller square that does not overlap contributes $2b$.
• Two sides of the smaller square are already counted as part of $a$.

So the total perimeter is: $P = 2a + 2b + 2a - 2b = 2a + 2a = 4a$

Given the total perimeter is $86$ cm: $4a = 86$ $a = \frac{86}{4}$ $a = 21.5$

This value is not an integer, which suggests there may be a need for a different perspective. Reconsider the perimeter formula to: $2a + 2(a - b) + 2b = 86$ $4a = 86$ This still leads us to $a = 21.5$, suggesting reconsidering the equation itself.

Step 2: Correct perimeter formulation

For correct overlap calculation, we split into correct contribution: $a - b + a - b + b + a = a + a + a + a + b + b = 3a + b$ Equates correctly with total: $3a + b = 86$ $3a + b = 86$

This more accurate representation properly links into: $P = 2a + 2(a-b) + 2b$

Step 3: Solve areas contribution

The sum of the areas of the two squares is: $a^2 + b^2 = 386$

Thus, we need:

3a + b = 86 \\ a^2 + b^2 = 386 \end{cases}$$ We solve the system by substituting $$b$$: $$b = 86 - 3a$$ Replace in the area equation: $$a^2 + (86 - 3a)^2 = 386$$ Expand and simplify: $$a^2 + (86 - 3a)^2 = 386$$ $$a^2 + 7396 - 516a + 9a^2 = 386$$ $$10a^2 - 516a + 7396 = 386$$ $$10a^2 - 516a + 7010 = 0$$ ### Step 4: Quadratic solving Solving the quadratic: $$10a^2 - 516a + 7010 = 0$$ Using quadratic formula: $$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Where $$a = 10$$, $$b = -516$$, $$c = 7010$$. Calculates: $$a = \frac{516 \pm \sqrt{(516)^2 - 4 \cdot 10 \cdot 7010}}{20}$$ $$a = \frac{516 \pm \sqrt{266256 - 28040}}{20}$$ $$a = \frac{516 \pm \sqrt{238216}}{20}$$ $$a = \frac{516 \pm 488}{20}$$ Thus, it provides: $$a = \frac{1004}{20} = 25$$ $$a = \frac{28}{20} = 1.4$$ Only integral solution: $$a = 25$$ So, $$b = 86 - 3 \cdot 25$$ $$b = 86 - 75$$ $$b = 11$$ Thus, $$a + b$$: $$\boxed{36}$$ Would you like details on steps or have any questions? Here are further questions to explore: 1. Verify with different perimeter formula. 2. Crosscheck root of quadratic. 3. Other overlap instances. 4. Explore geometric transformations. 5. Investigate if overlap doesn't disrupt. 6. Area linking without perimeter given. 7. Volume interpretations. 8. Physics implying dimensions. **Tip**: Carefully rethink the given equation assumptions if initial results counter physical logical constraints.