Let's denote the side length of Joaquin's original square treasure map as \( x \) inches. According to the problem, he increases each side by 6 inches, resulting in a new side length of \( x + 6 \) inches.
The area of the larger map is given as 121 square inches. The area of a square is calculated using the formula: \[ \text{Area} = \text{side length}^2 \] Therefore, for the larger map, we have: \[ (x + 6)^2 = 121 \]
To find \( x + 6 \), we take the square root of both sides: \[ x + 6 = \sqrt{121} = 11 \]
Now, we can solve for \( x \): \[ x = 11 - 6 = 5 \]
Thus, the dimensions of the original smaller map are: \[ 5 \text{ inches by } 5 \text{ inches} \]
To verify, we can check the dimensions of the larger map: \[ x + 6 = 5 + 6 = 11 \text{ inches} \] The area of the larger map would then be: \[ 11^2 = 121 \text{ square inches} \] This confirms that our calculation is correct.
Therefore, the dimensions of Joaquin's original smaller map are: \[ \boxed{5 \text{ inches by } 5 \text{ inches}} \]