AMC 10 · 2014 · #9

Grade 8 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
area-trianglespythagorean-theorem identify-subproblems ↑ Prerequisites: area-trianglespythagorean-theorem
📏 Medium solution 💡 2 insights
Problem
A right triangle has its two legs measuring 2√3 and 6. In a right triangle each leg is perpendicular to the other, so each leg is also an altitude, and these two given lengths are two of the triangle's three altitudes. Find the length of the third altitude, the one drawn from the right angle to the hypotenuse.

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

The third altitude cannot be read off directly, so Tool #7 (Identify Subproblems) breaks the task into three reachable pieces: find the area from the two legs, find the hypotenuse, then find the altitude that pairs with that hypotenuse. Tool #1 (Draw a Diagram) makes the key fact visible — the two perpendicular legs already form a base-and-height pair, so the area is easy. Tool #4 (Introduce a Variable) names the unknown altitude h so the fixed area can be written a second way and solved. The hinge is that the area is the same whichever side is the base.

1STEP 1

Read the legs as base and height

The legs meet at the right angle, so one is the base and the other the height: area = 1/2 · 2√3 · 6 = 6√3.

area = 1/2 · 2√3 · 6 = 6√3
2STEP 2

Find the hypotenuse squared

The hypotenuse faces the right angle, so by the Pythagorean theorem c² = (2√3)² + 6² = 12 + 36 = 48.

c² = (2√3)² + 6² = 12 + 36 = 48
3STEP 3

Simplify the hypotenuse

Take the square root, pulling out the perfect square: c = √(48) = √(16 · 3) = 4√3, the base for the last altitude.

c = √(48) = √(16 · 3) = 4√3
4STEP 4

Solve for the third altitude

With the hypotenuse as base and h the altitude on it, 1/2 · 4√3 · h = 6√3, so h = 3 — choice (C).

2√3 h = 6√3 → h = 3 (C)
Answer
3
Plug back: with base 4√3 and height 3 the area is 1/2 · 4√3 · 3 = 6√3, matching the area from the legs. The value is also sensible because the altitude to the hypotenuse must be shorter than either leg (it is the shortest way from the right angle to the opposite side), and indeed 3 ＜ 2√3 ≈ 3.46 and 3 ＜ 6. That rules out choices (D) 4 and (E) 5, both larger than a leg. So 3, choice (C), fits.
💡Key takeaway

The area stays the same for every base, so find the area from the two legs, find the hypotenuse with Pythagoras, then divide back out to get the last height of 3.

  • Read the legs as base and height
  • Find the hypotenuse squared
  • Simplify the hypotenuse
  • Solve for the third altitude

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