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2009 AMC 10A Problem 10

Problem 10 of 25EasierGeometry

Triangle ABCABC has a right angle at B.B. Point DD is the foot of the altitude from B,B, AD=3,AD = 3, and DC=4.DC = 4. What is the area of △ABC?\triangle ABC?

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Solution

For the altitude from the right angle to the hypotenuse, BD2=AD⋅DC=3⋅4=12,BD^2 = AD \cdot DC = 3 \cdot 4 = 12, so BD=23.BD = 2\sqrt{3}. The hypotenuse is AC=3+4=7,AC = 3 + 4 = 7, so the area is 12⋅7⋅23=73.\dfrac12 \cdot 7 \cdot 2\sqrt3 = 7\sqrt3. Thus, the correct answer is B.

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Concepts: right triangle · altitude · triangle area

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.