AMC 10 · 2009 · #10

Grade 8 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
similar-trianglesarea-trianglespythagorean-theorem identify-subproblems ↑ Prerequisites: similar-triangles
📏 Medium solution 💡 2 insights 📊 Diagram
Problem
Right triangle ABC has its right angle at B. D is the foot of the altitude from B to the hypotenuse AC, and the two pieces of the hypotenuse are AD = 3 and DC = 4. Find the area of triangle ABC.

Pick an answer.

(A)
$\ 4\sqrt3$
(B)
$\ 7\sqrt3$
(C)
$\ 21$
(D)
$\ 14\sqrt3$
(E)
$\ 42$

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

The area needs a base and a matching height. AC is an easy base since AC = AD + DC = 7, so the real subproblem is the height BD. The altitude from the right angle cuts the big right triangle into two smaller right triangles that are copies of the whole, and that similarity pins down BD. Solve that piece, then combine.

1STEP 1

Split with the altitude

The altitude BD splits ABC into two right triangles, and the shared acute angles make triangle ABD similar to triangle BCD.

△ ABD ∼ △ BCD
2STEP 2

Similar triangles give BD squared

Matching sides give ADBD=BDDC\frac{AD}{BD}=\frac{BD}{DC}, so cross-multiplying yields BD2=3⋅4=12BD^2 = 3 \cdot 4 = 12.

AD/BD = BD/DC → BD² = AD · DC = 3 · 4 = 12
3STEP 3

Take the square root

A length is positive and 12=4⋅312 = 4 \cdot 3, so pulling out the perfect square gives BD=12=23BD = \sqrt{12} = 2\sqrt{3}.

BD = √(12) = √(4) √(3) = 2√(3)
4STEP 4

Compute the area

Base AC=3+4=7AC = 3 + 4 = 7 with height BD gives area 12⋅7⋅23=73\frac{1}{2} \cdot 7 \cdot 2\sqrt{3} = 7\sqrt{3}, choice (B).

Area = 1/2 · AC · BD = 1/2 · 7 · 2√(3) = 7√(3)
Answer
7√3
The height BD = 2 root 3 is about 3.46, and the base AC is 7, so the area is about one half times 7 times 3.46, roughly 12.1. The answer 7 root 3 is about 12.12, which matches. Checking the choices: 4 root 3 (about 6.9) and 14 root 3 (about 24.2) and 42 are too far off, and 21 would need a height near 6, which is impossible since BD sits inside a triangle only 7 wide. Only 7 root 3 fits.
💡Key takeaway

When a right angle drops a height onto the far side, that height is the geometric mean of the two pieces it lands between.

  • Split with the altitude
  • Similar triangles give BD squared
  • Take the square root
  • Compute the area

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