AMC 10 · 2014 · #9
Grade 8 geometry-2dThe two legs of a right triangle, which are altitudes, have lengths 23 and 6. How long is the third altitude of the triangle?
Pick an answer.
AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A right triangle has its two legs measuring $2\sqrt3$ 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.
Givens: The triangle has a right angle, and its two legs have lengths $2\sqrt3$ and $6$; Each leg is perpendicular to the other, so the two legs are two of the triangle's three altitudes; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Unknowns: The length of the third altitude — the perpendicular segment from the right angle down to the hypotenuse
Understand
Restated: A right triangle has its two legs measuring $2\sqrt3$ 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.
Givens: The triangle has a right angle, and its two legs have lengths $2\sqrt3$ and $6$; Each leg is perpendicular to the other, so the two legs are two of the triangle's three altitudes; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #4 Introduce a Variable
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.
Execute — Answer: C
6.G.A.1 Step 1 Read the legs as base and height
- The two legs meet at the right angle, so they are perpendicular: one is the height while the other is the base.
- That makes the area easy: $\text{area} = \tfrac12 \cdot 2\sqrt3 \cdot 6 = 6\sqrt3$.
💡 In a right triangle the two legs are already perpendicular, so one is the height the moment the other is the base — no altitude has to be drawn.
8.G.B.7 Step 2 Find the hypotenuse squared
- The hypotenuse is the side opposite the right angle.
- By the Pythagorean theorem its square is the sum of the squares of the two legs: $c^2 = (2\sqrt3)^2 + 6^2 = 12 + 36 = 48$.
💡 The Pythagorean theorem is the direct bridge from the two legs to the hypotenuse in any right triangle.
8.EE.A.2 Step 3 Simplify the hypotenuse
- Take the square root and pull out the perfect-square factor: $c = \sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt3$.
- This is the base for the third altitude.
💡 Pulling a perfect square out of the root turns a messy $\sqrt{48}$ into a clean $4\sqrt3$ that is easy to compute with.
6.EE.B.7 Step 4 Solve for the third altitude
- Let $h$ be the third altitude, the one to the hypotenuse.
- The area written with this base is $\tfrac12 \cdot 4\sqrt3 \cdot h = 2\sqrt3\,h$.
- Since the area is still $6\sqrt3$, set them equal: $2\sqrt3\,h = 6\sqrt3$, so $h = 3$.
- The third altitude has length $3$, which is choice (C).
💡 The area never changes, so setting the same area equal using the new base immediately solves for its matching height.
6.G.A.1 The two legs meet at the right angle, so they are perpendicular: one is the heig 8.G.B.7 The hypotenuse is the side opposite the right angle. By the Pythagorean theorem 8.EE.A.2 Take the square root and pull out the perfect-square factor: $c = \sqrt{48} = \s 6.EE.B.7 Let $h$ be the third altitude, the one to the hypotenuse. The area written with Review
Reasonableness: Plug back: with base $4\sqrt3$ and height $3$ the area is $\tfrac12 \cdot 4\sqrt3 \cdot 3 = 6\sqrt3$, 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 $3 < 2\sqrt3 \approx 3.46$ and $3 < 6$. That rules out choices (D) $4$ and (E) $5$, both larger than a leg. So $3$, choice (C), fits.
Alternative: Use the shortcut that in a right triangle the altitude to the hypotenuse equals the product of the legs divided by the hypotenuse: $h = \dfrac{2\sqrt3 \cdot 6}{4\sqrt3} = \dfrac{12\sqrt3}{4\sqrt3} = 3$. (This is just the two area formulas combined into one line.) The legs $2\sqrt3$ and $6$ also reveal a 30-60-90 triangle, and its altitude to the hypotenuse is $3$, confirming the answer.
CCSS standards used (min grade 8)
6.G.A.1Find the area of right triangles, other triangles, and polygons by composing or decomposing shapes (Computing the area $6\sqrt3$ from the two perpendicular legs used as base and height.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the hypotenuse squared, $c^2 = 12 + 36 = 48$, from the two legs.)8.EE.A.2Use square root symbols to represent solutions and evaluate square roots of small perfect squares (Simplifying $\sqrt{48}$ to $4\sqrt3$ to get the hypotenuse length.)6.EE.B.7Solve real-world and mathematical problems by writing and solving equations of the form px = q (Solving $2\sqrt3\,h = 6\sqrt3$ for the third altitude $h = 3$.)
⭐ 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 $3$.
⭐ 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 $3$.
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