2023 AMC 10B Problem 7

Attempt Problem 7 of the 2023 AMC 10B 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 2023 AMC 10B solutions, or check the answer key.

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7.

Square ABCDABCD is rotated 20∘20^\circ clockwise about its center to obtain square EFGH,EFGH, as shown below. What is the degree measure of ∠EAB?\angle EAB?

24∘24^\circ

35∘35^\circ

30∘30^\circ

32∘32^\circ

20∘20^\circ

Answer: B
Concepts:angle chasingisosceles triangletransformation
Difficulty rating: 1310
Small Hint:

Let OO be the center; then OA=OEOA = OE and ∠AOE=20∘\angle AOE = 20^\circ

Big Hint:

Triangle OAEOAE is isosceles, and the diagonal AOAO makes a 45∘45^\circ angle with ABAB

Solution:

Let OO be the shared center. The rotation carries AA to E,E, so OA=OEOA = OE and ∠AOE=20∘.\angle AOE = 20^\circ. That makes triangle OAEOAE isosceles, with base angles ∠OAE=180∘−20∘2=80∘.\angle OAE = \frac{180^\circ - 20^\circ}{2} = 80^\circ. The diagonal ACAC splits the right angle at A,A, so ∠OAB=45∘.\angle OAB = 45^\circ. Subtracting, ∠EAB=80∘−45∘=35∘.\angle EAB = 80^\circ - 45^\circ = 35^\circ. Thus, B is the correct answer.

Problem 6#6
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