AMC 10 · 2013 · #12
Grade 8 geometry-2dIn △ABC, AB=AC=28 and BC=20. Points D,E, and F are on sides AB, BC, and AC, respectively, such that DE and EF are parallel to AC and AB, respectively. What is the perimeter of parallelogram ADEF?
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 isosceles triangle $ABC$ with $AB=AC=28$ and $BC=20$, point $D$ is on $\overline{AB}$, point $E$ is on $\overline{BC}$, and point $F$ is on $\overline{AC}$. Segment $\overline{DE}$ is parallel to $\overline{AC}$ and segment $\overline{EF}$ is parallel to $\overline{AB}$, so $ADEF$ is a parallelogram. Find the perimeter of that parallelogram.
Givens: $\triangle ABC$ is isosceles with $AB = AC = 28$; $BC = 20$; $D$ is on $\overline{AB}$, $E$ is on $\overline{BC}$, $F$ is on $\overline{AC}$; $\overline{DE}\parallel\overline{AC}$ and $\overline{EF}\parallel\overline{AB}$; In the figure, $ADEF$ has $AD$ along $AB$, $AF$ along $AC$, and the two drawn segments $DE$, $EF$ inside the triangle; Answer choices: (A) $48$, (B) $52$, (C) $56$, (D) $60$, (E) $72$
Unknowns: The perimeter of parallelogram $ADEF$
Understand
Restated: In isosceles triangle $ABC$ with $AB=AC=28$ and $BC=20$, point $D$ is on $\overline{AB}$, point $E$ is on $\overline{BC}$, and point $F$ is on $\overline{AC}$. Segment $\overline{DE}$ is parallel to $\overline{AC}$ and segment $\overline{EF}$ is parallel to $\overline{AB}$, so $ADEF$ is a parallelogram. Find the perimeter of that parallelogram.
Givens: $\triangle ABC$ is isosceles with $AB = AC = 28$; $BC = 20$; $D$ is on $\overline{AB}$, $E$ is on $\overline{BC}$, $F$ is on $\overline{AC}$; $\overline{DE}\parallel\overline{AC}$ and $\overline{EF}\parallel\overline{AB}$; In the figure, $ADEF$ has $AD$ along $AB$, $AF$ along $AC$, and the two drawn segments $DE$, $EF$ inside the triangle; Answer choices: (A) $48$, (B) $52$, (C) $56$, (D) $60$, (E) $72$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems
The figure is the key. Working from the marked diagram (Tool #1), the perimeter of parallelogram $ADEF$ is $AD+DE+EF+FA$, and opposite sides are equal, so it collapses to $2(AD+AF)$. The parallel line $DE\parallel AC$ cuts a small triangle $BDE$ off the corner at $B$; reading the equal angles straight off the figure shows that small triangle is isosceles, which pins down one of the lengths. To see that the exact spot of $D$ does not matter, name $BD=x$ (Tool #4) and watch it cancel. Breaking the perimeter into the little triangle plus the parallelogram is Tool #7, Identify Subproblems.
Execute — Answer: C
8.G.A.5 Step 1 Write the perimeter from the figure
- Since $EF\parallel AB$ and $DE\parallel AC$, figure $ADEF$ is a parallelogram: side $AD$ lies on $AB$ and is parallel to $EF$, while side $AF$ lies on $AC$ and is parallel to $DE$.
- Opposite sides of a parallelogram are equal, so $EF = AD$ and $DE = AF$.
- The perimeter is $AD + DE + EF + FA = 2(AD + AF)$.
- So the whole job is to find $AD + AF$.
💡 A parallelogram has two pairs of equal sides, so its perimeter is just twice the sum of two neighboring sides.
8.G.A.5 Step 2 Spot the isosceles corner triangle
- Look at the small triangle $BDE$ cut off at corner $B$.
- Because $\triangle ABC$ has $AB = AC$, its base angles are equal: $\angle B = \angle C$.
- Now $DE\parallel AC$ with side $BC$ acting as a transversal, so $\angle DEB = \angle ACB = \angle C$.
- That makes $\angle DBE = \angle DEB$ inside triangle $BDE$.
- Equal base angles mean triangle $BDE$ is isosceles, and the sides opposite those equal angles are equal, so $BD = DE$.
💡 A line parallel to one side of an isosceles triangle copies its equal base angles, so the little triangle it cuts off is isosceles too.
6.EE.A.2 Step 3 Name BD and let it cancel
- Let $BD = x$.
- From the last step $DE = x$, and from Step 1 $AF = DE = x$.
- Point $D$ sits on $\overline{AB}$, so $AD = AB - BD = 28 - x$.
- Add the two sides of the parallelogram: $AD + AF = (28 - x) + x = 28$.
- The $x$ disappears, which means it never mattered where $D$ was placed on $AB$.
💡 The length taken from $AB$ is handed straight back as the opposite side, so the two pieces always rebuild the full side $AB$.
6.EE.A.3 Step 4 Double it for the perimeter
- From Step 1 the perimeter is $2(AD + AF)$, and $AD + AF = 28$, so $P = 2 \cdot 28 = 56$.
- Notice $BC = 20$ was never used; the perimeter is always twice the equal side $AB$.
- The answer is $\textbf{(C)}\ 56$.
💡 Two copies of the two sides give the full loop around the parallelogram, and those two sides always add up to $AB$.
8.G.A.5 Since $EF\parallel AB$ and $DE\parallel AC$, figure $ADEF$ is a parallelogram: s 8.G.A.5 Look at the small triangle $BDE$ cut off at corner $B$. Because $\triangle ABC$ 6.EE.A.2 Let $BD = x$. From the last step $DE = x$, and from Step 1 $AF = DE = x$. Point 6.EE.A.3 From Step 1 the perimeter is $2(AD + AF)$, and $AD + AF = 28$, so $P = 2 \cdot 2 Review
Reasonableness: The parallelogram lives inside the triangle, so its perimeter should be less than the triangle's perimeter $28+28+20 = 76$; indeed $56 < 76$. The result $2\cdot AB = 56$ uses only the equal side $28$ and ignores $BC=20$, which fits the fact that $x$ cancelled: the answer cannot depend on where $D$ sits. Every answer choice is even, matching $P = 2(AD+AF)$, and $56$ is on the list as (C).
Alternative: Since the position of $D$ does not matter, put $D$, $E$, $F$ at the midpoints of the sides. Then $ADEF$ is a smaller figure with $AD = 14$ and each drawn segment equal to $14$, so all four sides are $14$ and the perimeter is $4 \cdot 14 = 56$, the same answer. This midpoint shortcut is a quick way to confirm the general result.
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments to establish facts about angles of triangles and the angles created when parallel lines are cut by a transversal (Reading equal angles off the figure: the isosceles triangle's equal base angles plus the parallel line $DE\parallel AC$ show triangle $BDE$ is isosceles, giving $BD = DE$, and that opposite sides of the parallelogram are equal.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Naming $BD = x$ and writing $AD = 28 - x$ and $AF = x$ so the two adjacent sides can be added symbolically.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Simplifying $(28 - x) + x$ to $28$ so the variable cancels, then doubling to get the perimeter $2 \cdot 28 = 56$.)
⭐ A line drawn parallel to a side of an isosceles triangle cuts off another isosceles triangle, and the parallelogram's two sides always add back to the full slant side, so the perimeter is just twice that side.
⭐ A line drawn parallel to a side of an isosceles triangle cuts off another isosceles triangle, and the parallelogram's two sides always add back to the full slant side, so the perimeter is just twice that side.
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