2018 AMC 10A Problem 9

Attempt Problem 9 of the 2018 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 10A solutions, or check the answer key.

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9.

All of the triangles in the diagram below are similar to isosceles triangle ABC,ABC, in which AB=AC.AB=AC. Each of the 77 smallest triangles has area 1,1, and ABC\triangle ABC has area 40.40. What is the area of trapezoid DBCE?DBCE?

1616

1818

2020

2222

2424

Answer: E
Concepts:similarityarea ratio
Difficulty rating: 1420
Solution:

We know that the side length of the smaller triangles is 140\sqrt{\frac{1}{40}} times the length of the larger triangle from similar triangles.

Then the side length of ADE\triangle ADE is 41404\sqrt{\frac{1}{40}} times the length of the side length of the larger triangle.

This makes the ratio of the areas (4140)2=16140=25. \left(4\sqrt{\dfrac{1}{40}}\right)^2 = 16 \cdot \dfrac{1}{40} = \dfrac{2}{5}.

Therefore, the area of ADE\triangle ADE is 2540=16.\frac{2}{5} \cdot 40 = 16. The area of the trapezoid is then 4016=24.40 - 16 = 24.

Thus, E is the correct answer.

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