2004 AMC 10A Problem 9

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 $\boxed{\textbf{(B)}\ 4}$.

When two overlapping figures share a common piece, the difference of their areas equals the difference of the non-shared parts. Asking for a difference rather than an individual area is the hint that the intersection point $D$ is a red herring; you never need to locate it.

Since $AE \parallel BC$, triangles $ADE$ and $CDB$ are similar with ratio $8 : 6 = 4 : 3$, so the heights from $D$ to $AE$ and to $BC$ split the distance $4$ in the ratio $4 : 3$: they are $\tfrac{16}{7}$ and $\tfrac{12}{7}$. Then $[ADE] = \tfrac12 \cdot 8 \cdot \tfrac{16}{7} = \tfrac{64}{7}$ and $[BDC] = \tfrac12 \cdot 6 \cdot \tfrac{12}{7} = \tfrac{36}{7}$, whose difference is $\tfrac{28}{7} = 4$.

• Locating D with similar triangles and computing each small area separately, which is slow and invites fraction errors.
• Using $AE$ and $BC$ as the two legs of a single triangle, or swapping the lengths $6$ and $8$.