Square $A B C D$ has an area of 4 . $E$ is the midpoint of $A B$. Similarly, $F, G, H$, and $I$ are the midpoints of $D E, C F, D G$, and $C H$, respectively. The area of $\triangle I D C$ is
(A) $\frac{1}{4}$
(B) $\frac{1}{8}$
(C) $\frac{1}{16}$
(D) $\frac{1}{32}$
(E) $\frac{1}{64}$
![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-135.jpg?height=836&width=462&top_left_y=1658&top_left_x=1316)

## Part C: Each correct answer is worth 8.
Square $A B C D$ has an area of 4 . $E$ is the midpoint of $A B$. Similarly, $F, G, H$, and $I$ are the midpoints of $D E, C F$, $D G$, and $C H$, respectively. The area of $\triangle I D C$ is
(A) $\frac{1}{4}$
(B) $\frac{1}{8}$
(C) $\frac{1}{16}$
(D) $\frac{1}{32}$
(E) $\frac{1}{64}$

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-222.jpg?height=412&width=371&top_left_y=518&top_left_x=1495)

## Solution

We start by joining $E$ to $C$ and noting that $\triangle D E C$ has half the area of the square or an area of 2 . From here, we observe $\triangle D E C$ and the dividing line, $C F$. Since $F$ is the midpoint of $D E$, triangles $D F C$ and $E F C$ have the same height and the same base and hence the same area. Therefore, $\triangle D F C$ will have an area of 1 unit. If we continue using this same idea, $\triangle D G C$ has an area of $\frac{1}{2}$, $\triangle C H D$ has an area of $\frac{1}{4}$ and $\triangle I D C$ had area of $\frac{1}{8}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-222.jpg?height=415&width=369&top_left_y=974&top_left_x=1493)

ANSWER: (B)

## Part C