AMC 10 · 2002 · #13
Grade 8 geometry-2dGiven a triangle with side lengths 15, 20, and 25, find the triangle's shortest altitude.
Pick an answer.
AMC 10 2002 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 triangle has sides $15$, $20$, and $25$. Among its three altitudes (heights), find the length of the shortest one.
Givens: The three side lengths are $15$, $20$, and $25$; An altitude is the perpendicular height drawn from a vertex to the side across from it; Answer choices: (A) $6$, (B) $12$, (C) $12.5$, (D) $13$, (E) $15$
Unknowns: The length of the triangle's shortest altitude
Understand
Restated: A triangle has sides $15$, $20$, and $25$. Among its three altitudes (heights), find the length of the shortest one.
Givens: The three side lengths are $15$, $20$, and $25$; An altitude is the perpendicular height drawn from a vertex to the side across from it; Answer choices: (A) $6$, (B) $12$, (C) $12.5$, (D) $13$, (E) $15$
Plan
Primary tool: #14 Extreme Principle
Secondary: #5 Look for a Pattern, #7 Identify Subproblems
The area of a triangle is $\tfrac{1}{2}\times\text{base}\times\text{height}$, and that area is fixed no matter which side you call the base. So for each side $s$ with altitude $h$, the product $s\times h$ always equals twice the area — a constant. Tool #14 (Extreme Principle) reads that directly: if $s\times h$ is fixed, then to make $h$ as small as possible you must draw it to the largest side $s$. So the shortest altitude is the one dropped to the longest side. First find the area (easy, because #5 spots that $15,20,25$ is a right triangle), then divide twice the area by the longest side.
Execute — Answer: B
8.G.B.6 Step 1 Spot the right triangle
- Check whether the sides fit the Pythagorean relationship: $15^2+20^2 = 225+400 = 625$, and $25^2 = 625$.
- Since the squares of the two shorter sides add up to the square of the longest side, the converse of the Pythagorean theorem says the triangle is right-angled.
- The legs (the two sides meeting at the right angle) are $15$ and $20$, and the hypotenuse is $25$.
- This is just the $3\text{-}4\text{-}5$ triple scaled up by $5$.
💡 When the two smaller squares add up to the biggest square, the corner between the short sides is a perfect right angle.
6.G.A.1 Step 2 Find the area
- In a right triangle the two legs are perpendicular, so one leg is a base and the other is its height.
- The area is $\tfrac{1}{2}\times 15\times 20 = \tfrac{1}{2}\times 300 = 150$.
- Because the legs meet at $90^\circ$, no extra height calculation is needed — the legs do the job for free.
💡 The two legs of a right triangle are already perpendicular, so they act as base and height without any extra work.
6.G.A.1 Step 3 Match the smallest altitude to the longest side
- For any side $s$ used as base, the area formula gives $150 = \tfrac{1}{2}\,s\,h$, so $h = \dfrac{2\times 150}{s} = \dfrac{300}{s}$.
- The top number $300$ never changes, so $h$ shrinks as $s$ grows.
- The longest side is the hypotenuse $25$, so the altitude drawn to it is the shortest of the three.
💡 Base times height is fixed, so stretching the base as long as possible squeezes the height down to its smallest.
6.NS.B.2 Step 4 Divide to get the height
- Put the longest side into the formula: $h = \dfrac{300}{25} = 12$.
- So the shortest altitude has length $12$, which is choice (B).
💡 Dividing twice the area by the base you chose reads off the exact height to that base.
8.G.B.6 Check whether the sides fit the Pythagorean relationship: $15^2+20^2 = 225+400 = 6.G.A.1 In a right triangle the two legs are perpendicular, so one leg is a base and the 6.G.A.1 For any side $s$ used as base, the area formula gives $150 = \tfrac{1}{2}\,s\,h$ 6.NS.B.2 Put the longest side into the formula: $h = \dfrac{300}{25} = 12$. So the shorte Review
Reasonableness: Compute all three altitudes to be sure the smallest one was found: to side $15$, $h=\tfrac{300}{15}=20$; to side $20$, $h=\tfrac{300}{20}=15$; to side $25$, $h=\tfrac{300}{25}=12$. The three heights are $20, 15, 12$, and $12$ is indeed the smallest, matching choice (B). It also makes sense that the altitude to the hypotenuse ($12$) is shorter than either leg ($15$ and $20$), since a leg is itself the altitude to the other leg. Choice (E) $15$ is the trap for anyone who stops at a leg instead of the hypotenuse.
Alternative: Use the geometric-mean (altitude-on-hypotenuse) relationship for right triangles: the altitude to the hypotenuse equals the product of the legs divided by the hypotenuse, $h = \dfrac{15\times 20}{25} = \dfrac{300}{25} = 12$. Same answer (B), reached without separately computing the area.
CCSS standards used (min grade 8)
8.G.B.6Explain a proof of the Pythagorean theorem and its converse (Using the converse ($15^2+20^2=25^2$) to confirm the triangle is right-angled with legs $15,20$ and hypotenuse $25$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the area $\tfrac{1}{2}\times15\times20=150$ and rearranging $\text{Area}=\tfrac{1}{2}sh$ into $h=\tfrac{300}{s}$.)6.NS.B.2Fluently divide multi-digit numbers using the standard algorithm (Evaluating $\tfrac{300}{25}=12$ (and the check values $300/15=20$, $300/20=15$).)
⭐ Since base times height is a fixed number (twice the area), the shortest height always drops to the longest side — so find the area, then divide by the biggest side.
⭐ Since base times height is a fixed number (twice the area), the shortest height always drops to the longest side — so find the area, then divide by the biggest side.
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