Solving the Geometry Problem: Area of the Smaller Square

This problem involves a geometric figure consisting of a large square, an inscribed circle, a rectangle, and a smaller square. We are asked to find the area of the smaller square located at corner R of the large square PQRS.

Understanding the Geometric Setup

We have a square PQRS. A circle is inscribed within this square, meaning it touches all four sides. A rectangle with dimensions 4 cm \(\times\) 2 cm is placed at corner P, and a smaller square is placed at corner R. The image shows that a vertex of the rectangle (opposite to P) lies on the circle, and a vertex of the smaller square (opposite to R) also lies on the circle.

Finding the Side Length of the Large Square (PQRS)

Let the side length of the large square PQRS be \(s\). Since the circle is inscribed, its diameter is equal to the side length of the square, so the radius is \(s/2\).

Let's place the corner P at the origin \((0,0)\) of a coordinate system. Then the vertices of the square are P\((0,0)\), Q\((s,0)\), R\((s,s)\), and S\((0,s)\). The center of the square, and thus the center of the inscribed circle, is at \((s/2, s/2)\). The equation of the circle is given by:

\[ \left(x - \frac{s}{2}\right)^2 + \left(y - \frac{s}{2}\right)^2 = \left(\frac{s}{2}\right)^2 \]

The rectangle at corner P\((0,0)\) has dimensions 4 cm and 2 cm. The vertex of this rectangle opposite to P is at coordinates \((4,2)\). According to the problem description and the figure, this vertex lies on the inscribed circle.

Substitute the coordinates \((4,2)\) into the circle's equation:

\[ \left(4 - \frac{s}{2}\right)^2 + \left(2 - \frac{s}{2}\right)^2 = \left(\frac{s}{2}\right)^2 \]

Expand the terms:

\[ \left(16 - 2 \times 4 \times \frac{s}{2} + \frac{s^2}{4}\right) + \left(4 - 2 \times 2 \times \frac{s}{2} + \frac{s^2}{4}\right) = \frac{s^2}{4} \]

\[ \left(16 - 4s + \frac{s^2}{4}\right) + \left(4 - 2s + \frac{s^2}{4}\right) = \frac{s^2}{4} \]

Combine like terms:

\[ 20 - 6s + \frac{2s^2}{4} = \frac{s^2}{4} \]

\[ 20 - 6s + \frac{s^2}{2} = \frac{s^2}{4} \]

Multiply the entire equation by 4 to remove fractions:

\[ 4(20) - 4(6s) + 4\left(\frac{s^2}{2}\right) = 4\left(\frac{s^2}{4}\right) \]

\[ 80 - 24s + 2s^2 = s^2 \]

Rearrange into a quadratic equation:

\[ s^2 - 24s + 80 = 0 \]

We can solve this quadratic equation for \(s\) using the quadratic formula \(s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\[ s = \frac{-(-24) \pm \sqrt{(-24)^2 - 4(1)(80)}}{2(1)} \]

\[ s = \frac{24 \pm \sqrt{576 - 320}}{2} \]

\[ s = \frac{24 \pm \sqrt{256}}{2} \]

\[ s = \frac{24 \pm 16}{2} \]

This gives two possible values for \(s\):

• \(s_1 = \frac{24 + 16}{2} = \frac{40}{2} = 20\) cm
• \(s_2 = \frac{24 - 16}{2} = \frac{8}{2} = 4\) cm

Looking at the figure, the side length of the large square \(s\) must be greater than the dimensions of the rectangle at corner P (which are 4 and 2). Therefore, \(s = 20\) cm is the correct side length for the large square PQRS.

Finding the Side Length of the Smaller Square

The smaller square is located at corner R. Let its side length be \(a\). With P at \((0,0)\), the coordinates of R are \((s,s)\), which is \((20,20)\). The smaller square at R has vertices \((20,20)\), \((20-a, 20)\), \((20, 20-a)\), and \((20-a, 20-a)\). The vertex of this smaller square opposite to R is at \((20-a, 20-a)\). This vertex lies on the inscribed circle.

Substitute the coordinates \((20-a, 20-a)\) into the circle's equation with \(s=20\) (so \(s/2 = 10\)):

\[ \left((20-a) - 10\right)^2 + \left((20-a) - 10\right)^2 = 10^2 \]

\[ \left(10 - a\right)^2 + \left(10 - a\right)^2 = 100 \]

\[ 2(10 - a)^2 = 100 \]

Divide by 2:

\[ (10 - a)^2 = 50 \]

Take the square root of both sides:

\[ 10 - a = \pm \sqrt{50} \]

\[ 10 - a = \pm \sqrt{25 \times 2} \]

\[ 10 - a = \pm 5\sqrt{2} \]

Now, solve for \(a\):

\[ a = 10 \mp 5\sqrt{2} \]

This gives two possible values for \(a\):

• \(a_1 = 10 + 5\sqrt{2}\)
• \(a_2 = 10 - 5\sqrt{2}\)

The smaller square is at corner R. Its side length \(a\) must be less than the radius of the circle \(s/2 = 10\) because the circle curves away from the corner. \(10 + 5\sqrt{2} \approx 10 + 5(1.414) = 10 + 7.07 = 17.07\), which is greater than 10. \(10 - 5\sqrt{2} \approx 10 - 7.07 = 2.93\), which is less than 10.

Therefore, the side length of the smaller square is \(a = 10 - 5\sqrt{2}\) cm.

Calculating the Area of the Smaller Square

The area of the smaller square is \(a^2\).

\[ \text{Area} = (10 - 5\sqrt{2})^2 \]

Expand the square using the formula \((p-q)^2 = p^2 - 2pq + q^2\):

\[ \text{Area} = 10^2 - 2(10)(5\sqrt{2}) + (5\sqrt{2})^2 \]

\[ \text{Area} = 100 - 100\sqrt{2} + (25 \times 2) \]

\[ \text{Area} = 100 - 100\sqrt{2} + 50 \]

\[ \text{Area} = 150 - 100\sqrt{2} \]

Factor out 50:

\[ \text{Area} = 50(3 - 2\sqrt{2}) \text{ square cm} \]

Conclusion

The area of the smaller square is \(50(3 - 2\sqrt{2})\) square cm.

Revision Table: Key Geometric Concepts

Additional Information: Solving Geometry Problems

When tackling geometry problems involving coordinates and curves, it's often helpful to:

• Draw or carefully examine the figure.
• Set up a coordinate system at a convenient point (like a corner or center).
• Write down the equations of the shapes involved (lines, circles, etc.).
• Use the given information (like points on a curve) to create equations.
• Solve the system of equations to find unknown values.
• Always check if the obtained solutions make sense in the context of the geometry (e.g., lengths must be positive, side lengths must fit within the figure).
• Be proficient with algebraic manipulations, including solving quadratic equations and simplifying radicals.