2018 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 9 Easy

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? (2018 AMC 10A Problem, Question#9)

  • A.

    16

  • B.

    18

  • C.

    20

  • D.

    22

  • E.

    24

Answer:E

Let x be the area of ADE. Note that x is comprised of the 7 small isosceles triangles and a triangle similar to ADE with side length ratio 3:4 (so an area ratio of 9:16). Thus, we have x=7+\frac{9}{16}x.This gives x=16, so the area of DBCE=40-x=\boxed{\rm (E)~24}.

Let the base length of the small triangle be x. Then, there is a triangle ADE encompassing the 7 small triangles and sharing the top angle with a base length of 4x. Because the area is proportional to the square of the side, let the base BC be \sqrt{40}x. Then triangle ADE has an area of 16. So the area is 40-16=\boxed{24}.

Notice [DBCE]=[ABC]-[ADE]. Let the base of the small triangles of area 1 be x, then the base length of \triangle ADE=4x. Notice, \left(\frac{DE}{BC}\right)^2=\frac{1}{40}\Rightarrow \frac{x}{BC}=\frac{1}{\sqrt{40}},

then 4x=\frac{4BC}{\sqrt{40}}\Rightarrow [ADE]=\left(\frac{4}{\sqrt{40}}\right)^2\cdot [ABC]=\frac{2}{5}[ABC] Thus, [DBCE]=[ABC]-[ADE]=[ABC]\left(1-\frac{2}{5}\right)=\frac{3}{5}\cdot 40=\boxed{24}.

The area of ADE is 16 times the area of the small triangle, as they are similar and their side ratio is 4:1. Therefore the area of the trapezoid is 40-16=\boxed{24}.

You can see that we can create a "stack" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be 7+5+3+1=16, so to find the area of such trapezoid BCED, we just take 40-16=\boxed{24}, like so.

The combined area of the small triangles is 7 , and from the fact that each small triangle has an area of 1, we can deduce that the larger triangle above has an area of 9 (as the sides of the triangles are in a proportion of \frac{1}{3}, so will their areas have a proportion that is the square of the proportion of their sides, or \frac{1}{9}). Thus, the combined area of the top triangle and the trapezoid immediately below is 7+9=16. The area of trapezoid BCED is thus the area of triangle ABC-16=\boxed{24}.