AMC 10 Daily Practice Round 1

Complete problem set with solutions and individual problem pages

Problem 7 Easy

As shown in the figure, right triangle \triangle ABC is rotated clockwise around point A by a certain angle, resulting in right triangle \triangle ADE. Point B is mapped to point D, which lies exactly on side BC. Given that DE = 12 and \angle B = 60^\circ, what is the distance between point E and point C?

  • A.

    12

  • B.

    6\sqrt{3}

  • C.

    6

  • D.

    6\sqrt{2}

  • E.

    8

Answer:B

​Since right triangle \triangle ABC is rotated clockwise around point A, the resulting triangle is \triangle ADE. We know that DE = BC = 12, and because AD = AB and AC = AE, and the angles \angle DAB and \angle EAC are equal, the triangles are similar. Given \angle B = 60^\circ, it follows that \angle ACB = 30^\circ. Therefore, AB = \frac{1}{2}BC = 6, and AC = \sqrt{3} \times AB = 6\sqrt{3}. Since AD = AB and \triangle ABD is equilateral (as \angle B = 60^\circ), we know \angle DAB = 60^\circ = \angle EAC, meaning \triangle ACE is also equilateral. Thus, AC = AE = EC = 6\sqrt{3}, so the distance between points E and C is 6\sqrt{3}. The answer is \boxed{9}.

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