Small Circle Radius in 10cm Square Geometry

Answer

The radius of the smaller circle is calculated by modeling the diagonal distance from the corner of the square to the center of the large circle. The exact radius is $15 - 10\sqrt{2} \text{ cm}$, which is approximately $0.858 \text{ cm}$.

Image Analysis

The image displays a square with a side length of $10\text{ cm}$. Inside this square, there is a large circle inscribed such that it touches all four sides. In the upper-left corner, a smaller circle is placed so that it is tangent to two sides of the square and also externally tangent to the large circle.

Explanation

1. Find the radius of the large circle Since the large circle is inscribed in a square of side $s = 10\text{ cm}$, its diameter is equal to the side of the square. Therefore, the radius $R$ is half of the side length. $R = \frac{10}{2} = 5\text{ cm}$ The radius of a circle inscribed in a square is exactly half the length of the square's side.

2. Calculate the diagonal distance from the corner to the center of the large circle Let the top-left corner be the origin $(0,0)$ if we consider the square in a coordinate plane. The center of the large circle is at $(5, 5)$. The distance $D$ from the corner to the center is found using the Pythagorean theorem for a diagonal of a square with side $R$. $D = \sqrt{5^2 + 5^2} = 5\sqrt{2}\text{ cm}$ The distance from a corner to the center of the large circle is the hypotenuse of an isosceles right triangle with legs equal to the large radius.

3. Identify the components of the diagonal Let $r$ be the radius of the smaller circle. The diagonal distance $D$ can be broken into three segments: the distance from the corner to the center of the small circle ($d_r$), the radius of the small circle ($r$), and the radius of the large circle ($R$). ⚠️ This step is required on exams: establishing the geometric relationship along the diagonal line of symmetry. $d_r = r\sqrt{2}$ This is the distance from the corner of the square to the center of the small circle, following the same logic as step 2.

4. Set up the geometric equation The sum of the segments from the corner to the center of the large circle must equal $D$. $r\sqrt{2} + r + R = D$ The total diagonal distance is the sum of the small diagonal, the small radius, and the large radius due to the circles being tangent.

5. Substitute and solve for r Substitute $R = 5$ and $D = 5\sqrt{2}$ into the equation. $r(\sqrt{2} + 1) + 5 = 5\sqrt{2}$ We group the terms containing $r$ to isolate the variable. $r(\sqrt{2} + 1) = 5\sqrt{2} - 5$ Subtracting 5 from both sides shifts the constant to the right. $r = \frac{5\sqrt{2} - 5}{\sqrt{2} + 1}$ The radius is found by dividing the remaining distance by the factor $(\sqrt{2} + 1)$.

6. Rationalize the denominator To simplify, multiply the numerator and denominator by the conjugate $(\sqrt{2} - 1)$. $r = \frac{5(\sqrt{2} - 1)(\sqrt{2} - 1)}{1}$ $r = 5(2 - 2\sqrt{2} + 1) = 5(3 - 2\sqrt{2})$ $r = 15 - 10\sqrt{2}$ Rationalizing removes the radical from the denominator to providing a standard form answer.

Final Answer

The radius of the smaller circle is: $\boxed{15 - 10\sqrt{2} \text{ cm}}$

Common Mistakes

• Incorrect Diagonal Calculation: Students often forget that the distance from the corner to the center of the small circle is $r\sqrt{2}$, not just $r$.
• Arithmetic Errors in Rationalization: Forgetting to square both terms in the conjugate multiplication, which usually leads to a denominator other than 1.
• Confusion of Radii: Mixing up the diameter and the radius of the large circle (using 10 instead of 5).