1. **Problem Statement:** We have a large square, and inside it, a smaller square is inscribed by connecting the midpoints of the sides of the larger square. This process is repeated indefinitely, creating a sequence of inscribed squares. We want to find: a) The area of the square inscribed the $n^{th}$ time. b) The area of one right triangle formed in the $n^{th}$ inscribed square. 2. **Understanding the problem and formula:** When a square is inscribed inside another by joining midpoints, the side length of the smaller square is $\frac{\sqrt{2}}{2}$ times the side length of the larger square. If the original square has side length $s_0$, then the side length of the $n^{th}$ inscribed square, $s_n$, is: $$ s_n = s_0 \left( \frac{\sqrt{2}}{2} \right)^n $$ The area of a square is side length squared, so the area of the $n^{th}$ square, $A_n$, is: $$ A_n = s_n^2 = s_0^2 \left( \frac{\sqrt{2}}{2} \right)^{2n} $$ Simplify the term inside the power: $$ \left( \frac{\sqrt{2}}{2} \right)^2 = \frac{2}{4} = \frac{1}{2} $$ So, $$ A_n = s_0^2 \left( \frac{1}{2} \right)^n = s_0^2 \cdot 2^{-n} $$ 3. **Answer for part (a):** The area of the $n^{th}$ inscribed square is: $$ A_n = s_0^2 \cdot 2^{-n} $$ where $s_0$ is the side length of the original square. 4. **Finding the area of one right triangle formed in the $n^{th}$ inscribed square:** Each inscribed square is formed by connecting midpoints, creating four right triangles around it inside the larger square. The area of the $n^{th}$ square is $A_n$, and the original square area is $A_0 = s_0^2$. The right triangles formed are congruent and each has an area equal to half the difference between the areas of the $(n-1)^{th}$ and $n^{th}$ squares divided by 4 (since 4 triangles are formed). But more simply, each right triangle in the $n^{th}$ inscribed square has area equal to half the area of one of the four triangles formed by the midpoints. Alternatively, the right triangle formed in the $n^{th}$ inscribed square has legs equal to half the side length of the $(n-1)^{th}$ square: Leg length = $\frac{s_{n-1}}{2} = \frac{s_0}{2} \left( \frac{\sqrt{2}}{2} \right)^{n-1}$ Area of one right triangle: $$ T_n = \frac{1}{2} \times \left( \frac{s_{n-1}}{2} \right)^2 = \frac{1}{2} \times \frac{s_{n-1}^2}{4} = \frac{s_{n-1}^2}{8} $$ Using $s_{n-1}^2 = s_0^2 \cdot 2^{-(n-1)}$ from above: $$ T_n = \frac{s_0^2 \cdot 2^{-(n-1)}}{8} = s_0^2 \cdot \frac{1}{8} \cdot 2^{-(n-1)} = s_0^2 \cdot 2^{-(n+2)} $$ 5. **Answer for part (b):** The area of one right triangle formed in the $n^{th}$ inscribed square is: $$ T_n = s_0^2 \cdot 2^{-(n+2)} $$ --- **Summary:** - Area of $n^{th}$ inscribed square: $$ A_n = s_0^2 \cdot 2^{-n} $$ - Area of one right triangle in the $n^{th}$ inscribed square: $$ T_n = s_0^2 \cdot 2^{-(n+2)} $$