To solve the problem, we need to find the radius of the circle inscribed in a triangle formed by any three vertices of a square with a given center and side length. Here’s how we can approach the problem step by step:
### Step 1: Determine the vertices of the square
Given that the center of the square is at (3, 4) and each side has a length of 4 units, we can find the coordinates of the vertices.
The half-length of the side is 2 units (since 4/2 = 2). The vertices can be calculated as follows:
- Top right vertex: (3 + 2, 4 + 2) = (5, 6)
- Top left vertex: (3 - 2, 4 + 2) = (1, 6)
- Bottom left vertex: (3 - 2, 4 - 2) = (1, 2)
- Bottom right vertex: (3 + 2, 4 - 2) = (5, 2)
Thus, the vertices of the square are (5, 6), (1, 6), (1, 2), and (5, 2).
### Step 2: Identify the triangle formed by any three vertices
We can choose any three vertices to form a triangle. Let's choose the vertices (5, 6), (1, 6), and (1, 2).
### Step 3: Calculate the lengths of the sides of the triangle
Using the distance formula, we can find the lengths of the sides of the triangle:
- Length of side AB (from (5, 6) to (1, 6)):
\[
AB = \sqrt{(5 - 1)^2 + (6 - 6)^2} = \sqrt{4^2} = 4
\]
- Length of side AC (from (5, 6) to (1, 2)):
\[
AC = \sqrt{(5 - 1)^2 + (6 - 2)^2} = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}
\]
- Length of side BC (from (1, 6) to (1, 2)):
\[
BC = \sqrt{(1 - 1)^2 + (6 - 2)^2} = \sqrt{4^2} = 4
\]
### Step 4: Use the formula for the radius of the inscribed circle
The radius \( r \) of the inscribed circle (inradius) of a triangle can be calculated using the formula:
\[
r = \frac{A}{s}
\]
where \( A \) is the area of the triangle and \( s \) is the semi-perimeter.
#### Step 4.1: Calculate the semi-perimeter \( s \)
\[
s = \frac{AB + AC + BC}{2} = \frac{4 + 4\sqrt{2} + 4}{2} = 4 + 2\sqrt{2}
\]
#### Step 4.2: Calculate the area \( A \)
Using the formula for the area of a triangle with vertices at coordinates:
\[
A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the vertices (5, 6), (1, 6), and (1, 2):
\[
A = \frac{1}{2} \left| 5(6 - 2) + 1(2 - 6) + 1(6 - 6) \right| = \frac{1}{2} \left| 5 \cdot 4 + 1 \cdot (-4) + 0 \right| = \frac{1}{2} \left| 20 - 4 \right| = \frac{1}{2} \cdot 16 = 8
\]
### Step 5: Calculate the inradius \( r \)
Now substituting \( A \) and \( s \) into the inradius formula:
\[
r = \frac{A}{s} = \frac{8}{4 + 2\sqrt{2}}
\]
To simplify this, we can multiply the numerator and denominator by the conjugate of the denominator:
\[
r = \frac{8(4 - 2\sqrt{2})}{(4 + 2\sqrt{2})(4 - 2\sqrt{2})} = \frac{32 - 16\sqrt{2}}{16 - 8} = \frac{32 - 16\sqrt{2}}{8} = 4 - 2\sqrt{2}
\]
### Final Answer
The radius of the circle inscribed in the triangle formed by any three vertices is:
\[
r = 2\sqrt{2}(\sqrt{2} - 1)
\]
