AMC 10 · 2013 · #16
Grade 8 geometry-2dIn triangle ABC, medians AD and CE intersect at P, PE=1.5, PD=2, and DE=2.5. What is the area of AEDC?
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: In triangle $ABC$, the medians $AD$ and $CE$ meet at point $P$. The three small segments near $P$ measure $PE=1.5$, $PD=2$, and $DE=2.5$. Find the area of quadrilateral $AEDC$ (the four corners $A$, $E$, $D$, $C$ in order).
Givens: $AD$ and $CE$ are medians of triangle $ABC$, so $D$ is the midpoint of $BC$ and $E$ is the midpoint of $AB$; The medians cross at $P$ with $PE=1.5$, $PD=2$, and $DE=2.5$; $AEDC$ is the quadrilateral with vertices $A$, $E$, $D$, $C$; Answer choices: (A) $13$, (B) $13.5$, (C) $14$, (D) $14.5$, (E) $15$
Unknowns: The area of quadrilateral $AEDC$
Understand
Restated: In triangle $ABC$, the medians $AD$ and $CE$ meet at point $P$. The three small segments near $P$ measure $PE=1.5$, $PD=2$, and $DE=2.5$. Find the area of quadrilateral $AEDC$ (the four corners $A$, $E$, $D$, $C$ in order).
Givens: $AD$ and $CE$ are medians of triangle $ABC$, so $D$ is the midpoint of $BC$ and $E$ is the midpoint of $AB$; The medians cross at $P$ with $PE=1.5$, $PD=2$, and $DE=2.5$; $AEDC$ is the quadrilateral with vertices $A$, $E$, $D$, $C$; Answer choices: (A) $13$, (B) $13.5$, (C) $14$, (D) $14.5$, (E) $15$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #5 Look for a Pattern, #1 Draw a Diagram
The figure looks busy, so tool #7 (Identify Subproblems) breaks $AEDC$ into small, familiar pieces: first find the missing half-lengths of the medians, then check the angle at $P$, then build the area. Tool #5 (Look for a Pattern) spots that the little triangle $PED$ has sides $1.5, 2, 2.5$ — a scaled $3$-$4$-$5$ right triangle — which reveals a right angle. Tool #1 (Draw a Diagram) keeps track of which segments are the diagonals of $AEDC$ and how they cross at $P$.
Execute — Answer: B
6.RP.A.3 Step 1 Use the centroid 2:1 split
- Two medians of a triangle always meet at one point, the centroid $P$.
- The centroid cuts each median so the part reaching the vertex is twice the part reaching the midpoint.
- Median $AD$ is split into $AP$ and $PD$, and median $CE$ into $CP$ and $PE$.
- Since $PD=2$ and $PE=1.5$ are the short (midpoint) pieces, the long (vertex) pieces are $AP=2\cdot PD=4$ and $CP=2\cdot PE=3$.
💡 The centroid balances a triangle, and that balance point always sits twice as far from the midpoint as from the opposite vertex.
8.G.B.6 Step 2 Spot the right angle at P
- Look at the small triangle $PED$ with the given sides $PE=1.5$, $PD=2$, and $ED=2.5$.
- These are just $3, 4, 5$ each cut in half.
- Because $1.5^2+2^2 = 2.25+4 = 6.25 = 2.5^2$, the converse of the Pythagorean theorem says the angle across from the longest side is a right angle.
- That angle is $\angle EPD$, so the two medians cross at right angles at $P$.
💡 If three sides fit $a^2+b^2=c^2$, the triangle must contain a right angle, no picture-measuring needed.
7.NS.A.3 Step 3 Name the diagonals of AEDC
- Go around $AEDC$ in order $A \to E \to D \to C$.
- The diagonals join the non-touching corners: $A$ to $D$ and $E$ to $C$.
- But those are exactly the two medians, and they cross at $P$.
- Their full lengths add the two pieces from Step 1: $AD = AP+PD = 4+2 = 6$ and $EC = EP+PC = 1.5+3 = 4.5$.
- From Step 2 these diagonals are perpendicular.
💡 The two diagonals of $AEDC$ are just the medians themselves, so their lengths are the pieces you already found, added back together.
6.G.A.1 Step 4 Area from perpendicular diagonals
- The point $P$ cuts $AEDC$ into four right triangles that all meet at $P$, because the diagonals cross there at $90^\circ$.
- Adding the four triangle areas is the same as the shortcut for any quadrilateral with perpendicular diagonals: area $= \tfrac{1}{2}\,d_1\,d_2$.
- Using the diagonal lengths $6$ and $4.5$ gives $\tfrac{1}{2}\cdot 6\cdot 4.5 = 13.5$, which is choice (B).
💡 When the diagonals meet at a right angle, the quadrilateral is four right triangles, and their areas sum to half the product of the diagonals.
6.RP.A.3 Two medians of a triangle always meet at one point, the centroid $P$. The centro 8.G.B.6 Look at the small triangle $PED$ with the given sides $PE=1.5$, $PD=2$, and $ED= 7.NS.A.3 Go around $AEDC$ in order $A \to E \to D \to C$. The diagonals join the non-touc 6.G.A.1 The point $P$ cuts $AEDC$ into four right triangles that all meet at $P$, becaus Review
Reasonableness: The answer $13.5$ sits right in the middle of the choices $13$ to $15$, so it is a sensible size. Quick cross-check: the four right triangles have legs $(4,3)$, $(4,1.5)$, $(2,3)$, $(2,1.5)$, giving areas $6$, $3$, $3$, $1.5$, which total $13.5$ — matching the $\tfrac{1}{2}\,d_1 d_2$ shortcut exactly. The right angle at $P$ is guaranteed by the $3$-$4$-$5$ pattern, so no measuring is needed.
Alternative: Skip the shortcut and add the four triangles directly. With the right angle at $P$, each of $\triangle APC$, $\triangle APE$, $\triangle DPC$, $\triangle DPE$ is a right triangle whose legs are pieces of the two medians. Their areas are $\tfrac12(4)(3)=6$, $\tfrac12(4)(1.5)=3$, $\tfrac12(2)(3)=3$, and $\tfrac12(2)(1.5)=1.5$. Summing gives $6+3+3+1.5=13.5$.
CCSS standards used (min grade 8)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Applying the centroid's $2:1$ split to get $AP=2\cdot PD=4$ and $CP=2\cdot PE=3$.)8.G.B.6Explain a proof of the Pythagorean theorem and its converse (Using the converse of the Pythagorean theorem on $1.5, 2, 2.5$ to show $\angle EPD=90^\circ$, so the medians are perpendicular.)7.NS.A.3Solve real-world problems involving the four operations with rational numbers (Adding the median pieces to get the diagonal lengths $AD=4+2=6$ and $EC=1.5+3=4.5$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Decomposing $AEDC$ into four right triangles and using $\tfrac12 d_1 d_2$ to get area $13.5$.)
⭐ The two medians are the diagonals of $AEDC$; they cross at a right angle, so the area is just half the product of the two full median lengths.
⭐ The two medians are the diagonals of $AEDC$; they cross at a right angle, so the area is just half the product of the two full median lengths.
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