2004 AMC 10A Problem 9

The question asks for a difference of two areas, which hints that some unknown piece might cancel.

1. Find the two big areas. Both are right triangles with the leg $AB = 4$:
$$\text{area of } ABE = \tfrac12 \times 4 \times 8 = 16, \qquad \text{area of } ABC = \tfrac12 \times 4 \times 6 = 12.$$
2. Look at how each big triangle is cut up by point $D$.
- Triangle $ABE$ is made of triangle $ADE$ and triangle $ABD$.
- Triangle $ABC$ is made of triangle $BDC$ and the same triangle $ABD$.
3. Call the area of the shared triangle $ABD$ by the letter $t$. Then
$$\text{area of } ADE = 16 - t, \qquad \text{area of } BDC = 12 - t.$$
4. Subtract. The unknown $t$ disappears:
$$(16 - t) - (12 - t) = 16 - 12 = 4.$$

The answer is $\boxed{\textbf{(B)}\ 4}$.

If two shapes share a common piece, removing that piece from both does not change the difference of their areas. Whenever a problem asks for a difference, check whether a shared region cancels. Here it means you never need to find where $D$ is.

Find each area directly. Sides $AE$ and $BC$ are both perpendicular to $AB$, so they are parallel. That makes triangles $ADE$ and $CDB$ the same shape, with sides in the ratio $8 : 6 = 4 : 3$.

1. Their heights from $D$ (measured along the direction of $AB$) are in the same ratio $4 : 3$ and add up to $AB = 4$. So they are $\tfrac{16}{7}$ and $\tfrac{12}{7}$.
2. Area of $ADE = \tfrac12 \times 8 \times \tfrac{16}{7} = \tfrac{64}{7}$, and area of $BDC = \tfrac12 \times 6 \times \tfrac{12}{7} = \tfrac{36}{7}$.
3. The difference is $\tfrac{64 - 36}{7} = \tfrac{28}{7} = 4$.

Both large triangles are right triangles with leg $AB = 4$:
$$[ABE] = \tfrac12 \cdot 4 \cdot 8 = 16, \qquad [ABC] = \tfrac12 \cdot 4 \cdot 6 = 12.$$

Segment $AD$ splits $\triangle ABE$ into $\triangle ADE$ and $\triangle ABD$, and segment $BD$ splits $\triangle ABC$ into $\triangle BDC$ and the same $\triangle ABD$. Writing $t = [ABD]$,
$$[ADE] = 16 - t, \qquad [BDC] = 12 - t.$$

Subtracting, the unknown $t$ cancels:
$$[ADE] - [BDC] = 16 - 12 = 4.$$

The answer is $\boxed{\textbf{(B)}\ 4}$.