2018 AMC 10A Problem 9

P1-10~2 minPrint

All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$ ?

Show solution without answering

Key insight

Four small triangles fit side by side along DE, so triangle ADE is a small one scaled up 4 times, with area 4 times 4 = 16.

Solution

The trapezoid $DBCE$ is the big triangle $ABC$ with the top triangle $ADE$ cut off. We know $ABC$ has area $40$ , so we only need the area of $ADE$ .

Measure $DE$ in small triangles. Look along the segment $DE$ . Four of the small triangles point upward and sit with their bottom sides on $DE$ ; the other three point downward and fill the gaps. The four bottom sides cover $DE$ exactly, so $DE$ is $4$ times as long as the base of one small triangle.
Compare $ADE$ with a small triangle. Every triangle in the picture has the same shape. So $ADE$ is a small triangle blown up $4$ times in every direction.
How does area grow? If a shape is stretched $4$ times as wide and $4$ times as tall, its area grows by $4 \cdot 4 = 16$ . A small triangle has area $1$ , so $ADE$ has area $16$ .
Subtract. The trapezoid has area $40 - 16 = 24$ .

The answer is $\boxed{\textbf{(E)}\ 24}$ .

Why this works

For two shapes of the same shape, if lengths are $k$ times as long, areas are $k^2$ times as big. The picture gives the length ratio for free by lining small triangles up along $DE$ . After that, cutting off a known triangle from a known triangle is just a subtraction.

Alternative approach

Count pieces inside $ADE$ . The $7$ small triangles form a strip of area $7$ . Its top edge is $3$ small bases long, so the triangle sitting above the strip is a small triangle blown up $3$ times, with area $3 \cdot 3 = 9$ . So $ADE$ has area $7 + 9 = 16$ , and the trapezoid again has area $40 - 16 = 24$ .

Concise solution

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

The trapezoid is what remains of $\triangle ABC$ after removing $\triangle ADE$ , so we need the area of $\triangle ADE$ .

Along $\overline{DE}$ sit four of the small triangles with their bases on $DE$ (the other three point downward and fill the gaps). Those four bases exactly cover $DE$ , so $DE$ is $4$ times the base of a small triangle.

Since $\triangle ADE$ is similar to each small triangle (all triangles in the figure are similar to $\triangle ABC$ ), its area is the small area scaled by $4^2$ :
$[ADE] = 16 \cdot 1 = 16.$

Therefore
$[DBCE] = [ABC] - [ADE] = 40 - 16 = 24.$

The answer is $\boxed{\textbf{(E)}\ 24}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 3: Congruent Triangles
Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Geometry, Chapter 5: Similar Triangles
Introduction to Geometry, Chapter 9: Polygons (§9.3)