2004 AMC 10A Problem 9

P1-10~2 minPrint

Triangles: Area & Pythagorean Cut the figure into known shapes (triangles, rectangles, sectors)

In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ , $\angle{EAB}$ and $\angle{ABC}$ are right angles, $AB=4$ , $BC=6$ , $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$ ?

Show solution without answering

Key insight

Both big right triangles contain the same piece, triangle ABD. Take it away from each and the difference stays the same: 16 - 12 = 4.

Solution

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

Why this works

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.

Alternative approach

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 $8 : 6 = 4 : 3$ .

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 $\tfrac{12}{7}$ .
Area of $ADE = \tfrac12 \times 8 \times \tfrac{16}{7} = \tfrac{64}{7}$ , and area of $BDC = \tfrac12 \times 6 \times \tfrac{12}{7} = \tfrac{36}{7}$ .
The difference is $\tfrac{64 - 36}{7} = \tfrac{28}{7} = 4$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)
Introduction to Geometry, Chapter 11: Circles (§11.3)
Introduction to Geometry, Chapter 10: Geometric Inequalities (§10.3)