2005 AMC 10B Problem 10

Attempt Problem 10 of the 2005 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 2005 AMC 10B solutions, or check the answer key.

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10.

In ABC,\triangle ABC, we have AC=BC=7AC = BC = 7 and AB=2.AB = 2. Suppose that DD is a point on line ABAB such that BB lies between AA and DD and CD=8.CD = 8. What is BD?BD?

33

232\sqrt{3}

44

55

424\sqrt{2}

Answer: A
Concepts:isosceles trianglealtitudePythagorean Theorem
Difficulty rating: 1370
Solution:

Let HH be the foot of the altitude from CC to line AB.AB. Since AC=BC,AC = BC, HH is the midpoint of AB,AB, so BH=1BH = 1 and CH2=7212=48.CH^2 = 7^2 - 1^2 = 48.

Applying the Pythagorean theorem in CHD,\triangle CHD, where HD=BH+BD=1+BD,HD = BH + BD = 1 + BD, gives 82=48+(1+BD)2, 8^2 = 48 + (1 + BD)^2, so (1+BD)2=16.(1 + BD)^2 = 16.

Then 1+BD=4,1 + BD = 4, so BD=3.BD = 3.

Thus, A is the correct answer.

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