AMC 10 · 2005 · #14
Grade 8 geometry-2dEquilateral △ABC has side length 2, M is the midpoint of AC, and C is the midpoint of BD. What is the area of △CDM?
Pick an answer.
AMC 10 2005 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: An equilateral triangle $ABC$ has every side equal to $2$. The point $M$ is the midpoint of side $AC$. The point $D$ lies on ray $BC$ so that $C$ is the midpoint of segment $BD$, which puts $D$ on the far side of $C$ from $B$. Find the area of triangle $CDM$.
Givens: Triangle $ABC$ is equilateral with side length $2$, so $AB=BC=CA=2$; $M$ is the midpoint of $\overline{AC}$; $C$ is the midpoint of $\overline{BD}$, so $B$, $C$, $D$ lie on one straight line with $BC=CD$; Answer choices: (A) $\tfrac{\sqrt{2}}{2}$, (B) $\tfrac{3}{4}$, (C) $\tfrac{\sqrt{3}}{2}$, (D) $1$, (E) $\sqrt{2}$
Unknowns: The area of triangle $CDM$
Understand
Restated: An equilateral triangle $ABC$ has every side equal to $2$. The point $M$ is the midpoint of side $AC$. The point $D$ lies on ray $BC$ so that $C$ is the midpoint of segment $BD$, which puts $D$ on the far side of $C$ from $B$. Find the area of triangle $CDM$.
Givens: Triangle $ABC$ is equilateral with side length $2$, so $AB=BC=CA=2$; $M$ is the midpoint of $\overline{AC}$; $C$ is the midpoint of $\overline{BD}$, so $B$, $C$, $D$ lie on one straight line with $BC=CD$; Answer choices: (A) $\tfrac{\sqrt{2}}{2}$, (B) $\tfrac{3}{4}$, (C) $\tfrac{\sqrt{3}}{2}$, (D) $1$, (E) $\sqrt{2}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
Tool #1 (Draw a Diagram) is the natural move: sketch the equilateral triangle sitting on the line through $B$, $C$, $D$, and the picture immediately shows that side $CD$ of the target triangle lies flat on that line. That makes $CD$ an obvious base, so the area is just $\tfrac{1}{2}\cdot\text{base}\cdot\text{height}$. Tool #7 (Identify Subproblems) then splits the work into two clean pieces: find the base $CD$, and find the height, which is how high $M$ sits above the line. The height needs the triangle's altitude (one Pythagorean step) and the fact that $M$ is only halfway up.
Execute — Answer: C
6.G.A.1 Step 1 Line up B, C, D and read off the base
- Because $C$ is the midpoint of $\overline{BD}$, the points $B$, $C$, $D$ sit on one straight line and $CD=CB$.
- Side $CB$ of the equilateral triangle is $2$, so $CD=2$.
- In triangle $CDM$, take this side $CD$ as the base: it lies flat on line $BD$, and its length is $2$.
💡 A midpoint cuts a segment into two equal halves, so $CD$ is just a copy of the side $CB$.
8.G.B.7 Step 2 Find the triangle's altitude
- To measure heights above line $BD$, first find how high the apex $A$ sits.
- Drop a perpendicular from $A$ to line $BC$, meeting it at foot $F$.
- In equilateral triangle $ABC$ this foot is the midpoint of $BC$, so $BF=1$.
- Right triangle $AFB$ has hypotenuse $AB=2$ and leg $BF=1$, so by the Pythagorean theorem $AF=\sqrt{2^2-1^2}=\sqrt{3}$.
- The apex $A$ stands $\sqrt{3}$ above the line.
💡 The altitude and half the base are the two legs of a right triangle whose hypotenuse is the slanted side.
8.G.A.4 Step 3 Halve the altitude to reach M
- The height of triangle $CDM$ is how high $M$ sits above line $BD$.
- Drop a perpendicular from $M$ to the line, foot $G$.
- Right triangles $CGM$ and $CFA$ share the angle at $C$ and both have a right angle, so they are similar.
- Since $M$ is the midpoint of $AC$, the ratio $CM:CA = 1:2$, so every matching length shrinks by half.
- Therefore $MG=\tfrac{1}{2}AF=\tfrac{\sqrt{3}}{2}$.
💡 Halfway up the side means halfway up in height, because the height climbs evenly as you slide from $C$ toward $A$.
6.G.A.1 Step 4 Multiply base by height
- Triangle $CDM$ has base $CD=2$ and height $MG=\tfrac{\sqrt{3}}{2}$.
- The area of a triangle is $\tfrac{1}{2}\cdot\text{base}\cdot\text{height}$, so the area is $\tfrac{1}{2}\cdot 2\cdot\tfrac{\sqrt{3}}{2}=\tfrac{\sqrt{3}}{2}$.
- That is choice (C).
💡 With the base flat on the line, the height is just how high the third corner floats above it.
6.G.A.1 Because $C$ is the midpoint of $\overline{BD}$, the points $B$, $C$, $D$ sit on 8.G.B.7 To measure heights above line $BD$, first find how high the apex $A$ sits. Drop 8.G.A.4 The height of triangle $CDM$ is how high $M$ sits above line $BD$. Drop a perpen 6.G.A.1 Triangle $CDM$ has base $CD=2$ and height $MG=\tfrac{\sqrt{3}}{2}$. The area of Review
Reasonableness: The value $\tfrac{\sqrt{3}}{2}\approx 0.87$ is comfortably positive and matches a base of $2$ paired with a smallish height of about $0.87$. The most tempting trap is to use the full altitude $\sqrt{3}$ as the height, which would give $\tfrac{1}{2}\cdot 2\cdot\sqrt{3}=\sqrt{3}\approx 1.73$ — but $M$ is only halfway up side $AC$, so its height is half of that. Choice (D) $1$ and choice (E) $\sqrt{2}$ are the kinds of near-miss values that show up if the height or base is mis-set; the careful base-times-height gives $\tfrac{\sqrt{3}}{2}$.
Alternative: Compare areas instead of computing a height. Since $C$ is the midpoint of $\overline{BD}$, triangles $MBC$ and $MDC$ share the apex $M$ and have equal bases $BC=CD$ on the same line, so they have equal areas. Now $BM$ is a median of equilateral triangle $ABC$ (it runs from $B$ to the midpoint $M$ of $AC$), and a median splits a triangle into two equal areas, so $[MBC]=\tfrac{1}{2}[ABC]$. The area of the equilateral triangle is $\tfrac{\sqrt{3}}{4}\cdot 2^2=\sqrt{3}$, so $[MBC]=\tfrac{\sqrt{3}}{2}$, and therefore $[MDC]=\tfrac{\sqrt{3}}{2}$ as well — the same answer.
CCSS standards used (min grade 8)
6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Choosing $CD=2$ as the base of triangle $CDM$ and finishing with area $=\tfrac{1}{2}\cdot\text{base}\cdot\text{height}=\tfrac{\sqrt{3}}{2}$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the equilateral triangle's altitude $AF=\sqrt{2^2-1^2}=\sqrt{3}$ from right triangle $AFB$.)8.G.A.4Understand that a two-dimensional figure is similar to another using transformations (Using the similar right triangles $CGM$ and $CFA$ (ratio $1:2$ from the midpoint $M$) to get the height $MG=\tfrac{\sqrt{3}}{2}$.)
⭐ When one side of a triangle lies flat on a line, use it as the base — then the area is just half that base times how high the last corner floats above the line.
⭐ When one side of a triangle lies flat on a line, use it as the base — then the area is just half that base times how high the last corner floats above the line.
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