AMC 10 · 2013 · #15
Grade 7 geometry-2dTwo sides of a triangle have lengths 10 and 15. The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?
Pick an answer.
AMC 10 2013 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 two sides of length $10$ and $15$. Inside a triangle, each side has an altitude drawn to it (the perpendicular height when that side is the base). The altitude drawn to the third side equals the average of the altitudes drawn to the sides of length $10$ and $15$. Find the length of the third side.
Givens: Two side lengths are $10$ and $15$; Every side of a triangle has an altitude (height) drawn to it; The altitude to the third side $=$ the average of the altitude to the side of length $10$ and the altitude to the side of length $15$; Answer choices: (A) $6$, (B) $8$, (C) $9$, (D) $12$, (E) $18$
Unknowns: The length $x$ of the third side
Understand
Restated: A triangle has two sides of length $10$ and $15$. Inside a triangle, each side has an altitude drawn to it (the perpendicular height when that side is the base). The altitude drawn to the third side equals the average of the altitudes drawn to the sides of length $10$ and $15$. Find the length of the third side.
Givens: Two side lengths are $10$ and $15$; Every side of a triangle has an altitude (height) drawn to it; The altitude to the third side $=$ the average of the altitude to the side of length $10$ and the altitude to the side of length $15$; Answer choices: (A) $6$, (B) $8$, (C) $9$, (D) $12$, (E) $18$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #13 Convert to Algebra
The altitudes are never given as numbers, so the useful move is to name the one quantity that ties them all together: the triangle's area. That is Tool #4 (Introduce a Variable) — call the area $S$. The single triangle has one area, and $S = \frac{1}{2}\,(\text{base})(\text{height})$ for each of the three bases, so every altitude can be written as $\frac{2S}{\text{side}}$. Once each altitude is an expression in $S$, Tool #13 (Convert to Algebra) turns the words "is the average of" into an equation. The reason this works so cleanly is that the unknown area $S$ appears in every term and cancels, leaving a short equation in $x$ alone.
Execute — Answer: D
6.G.A.1 Step 1 Name the shared area
- The triangle has one fixed area, and that area is the same no matter which of the three sides you treat as the base.
- Call the area $S$.
- Writing the area formula $\text{area} = \frac{1}{2}\,(\text{base})(\text{height})$ once for each base ties all three altitudes to the single number $S$.
💡 One triangle has one area, so the same $S$ works with every side-and-height pair.
6.EE.A.2 Step 2 Write each altitude
- Solve each copy of the area formula for the altitude.
- Dividing $S = \frac{1}{2}(\text{base})(\text{height})$ by the base gives $\text{height} = \frac{2S}{\text{base}}$.
- Doing this for all three bases turns each altitude into an expression in $S$.
💡 A bigger base needs a shorter height to keep the same area, so height is $2S$ divided by the side.
6.SP.B.5 Step 3 Translate "average"
- The altitude to the third side equals the average of the other two altitudes.
- The average of two quantities is their sum divided by $2$.
- Turn that sentence directly into an equation.
💡 "Average of two numbers" is just add them and cut in half.
5.NF.A.1 Step 4 Cancel the area
- Substitute the expressions from the earlier step.
- Every term contains the factor $2S$, so it divides out of the whole equation, leaving only the sides.
- What remains is a statement about $\frac{1}{x}$, and the fractions on the right add cleanly using the common denominator $30$.
💡 Because the mystery area sits in every term, it disappears and only the side lengths are left to decide the answer.
7.EE.B.4 Step 5 Solve for the side
- Finish the arithmetic on the right, then take the reciprocal.
- Halving $\frac{1}{6}$ gives $\frac{1}{12}$, so $\frac{1}{x} = \frac{1}{12}$ forces $x = 12$.
- The third side has length $12$, which is choice (D).
💡 If two positive numbers have equal reciprocals, the numbers themselves are equal.
6.G.A.1 The triangle has one fixed area, and that area is the same no matter which of th 6.EE.A.2 Solve each copy of the area formula for the altitude. Dividing $S = \frac{1}{2}( 6.SP.B.5 The altitude to the third side equals the average of the other two altitudes. Th 5.NF.A.1 Substitute the expressions from the earlier step. Every term contains the factor 7.EE.B.4 Finish the arithmetic on the right, then take the reciprocal. Halving $\frac{1}{ Review
Reasonableness: Plug in a concrete area to test. Give the triangle area $S = 30$: then $h_{10} = \frac{2\cdot 30}{10} = 6$ and $h_{15} = \frac{2\cdot 30}{15} = 4$, whose average is $5$. The altitude to a side of length $12$ would be $h_{x} = \frac{2\cdot 30}{12} = 5$ — it matches the required average exactly. The sides $10, 12, 15$ also pass the triangle inequality ($10 + 12 > 15$), so a real triangle exists. Both checks confirm $x = 12$.
Alternative: Tool #3 (Eliminate Possibilities): the third altitude is the average of the altitudes to the sides of length $10$ and $15$, so it lies strictly between those two altitudes. Since a longer side always carries a shorter altitude (their product is the fixed $2S$), the third side must lie strictly between $10$ and $15$. Among the choices $6, 8, 9, 12, 18$, only $12$ falls between $10$ and $15$, so the answer is (D) — reachable without any equation.
CCSS standards used (min grade 7)
6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Writing the triangle's area as $\frac{1}{2}(\text{base})(\text{height})$ for each of the three sides, so all three altitudes share one common area $S$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Rewriting each altitude as the expression $\frac{2S}{\text{side}}$ in terms of the named area $S$.)6.SP.B.5Summarize numerical data sets by reporting number of observations and measures (Turning "the average of the two altitudes" into the mean expression $\frac{h_{10} + h_{15}}{2}$.)5.NF.A.1Add and subtract fractions with unlike denominators (Adding $\frac{1}{10} + \frac{1}{15} = \frac{5}{30} = \frac{1}{6}$ using the common denominator $30$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Setting up and solving $\frac{1}{x} = \frac{1}{12}$ after the shared area cancels, giving $x = 12$.)
⭐ A triangle's area is the same whichever side you call the base, so each altitude equals twice the area divided by its side — name the area once, watch it cancel, and a clean equation for the missing side is all that is left.
⭐ A triangle's area is the same whichever side you call the base, so each altitude equals twice the area divided by its side — name the area once, watch it cancel, and a clean equation for the missing side is all that is left.
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