AMC 10 · 2005 · #23
Grade 8 geometry-2dIn trapezoid ABCD we have AB parallel to DC, E as the midpoint of BC, and F as the midpoint of DA. The area of ABEF is twice the area of FECD. What is AB/DC?
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: In trapezoid $ABCD$ the top side $AB$ is parallel to the bottom side $DC$. Point $E$ is the midpoint of leg $BC$ and point $F$ is the midpoint of leg $DA$, so segment $FE$ cuts the trapezoid into an upper piece $ABEF$ and a lower piece $FECD$. The upper piece has twice the area of the lower piece. Find the ratio $AB/DC$.
Givens: $ABCD$ is a trapezoid with $\overline{AB}\parallel\overline{DC}$; $E$ is the midpoint of leg $\overline{BC}$ and $F$ is the midpoint of leg $\overline{DA}$; Area of $ABEF$ is twice the area of $FECD$; Answer choices: (A) $2$, (B) $3$, (C) $5$, (D) $6$, (E) $8$
Unknowns: The ratio $AB/DC$
Understand
Restated: In trapezoid $ABCD$ the top side $AB$ is parallel to the bottom side $DC$. Point $E$ is the midpoint of leg $BC$ and point $F$ is the midpoint of leg $DA$, so segment $FE$ cuts the trapezoid into an upper piece $ABEF$ and a lower piece $FECD$. The upper piece has twice the area of the lower piece. Find the ratio $AB/DC$.
Givens: $ABCD$ is a trapezoid with $\overline{AB}\parallel\overline{DC}$; $E$ is the midpoint of leg $\overline{BC}$ and $F$ is the midpoint of leg $\overline{DA}$; Area of $ABEF$ is twice the area of $FECD$; Answer choices: (A) $2$, (B) $3$, (C) $5$, (D) $6$, (E) $8$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
The picture has no numbers, only two unknown bases and a middle line, so Tool #4 (Introduce a Variable) is the natural start: call the top base $a=AB$ and the bottom base $b=DC$ and hunt for one equation relating them. Tool #1 (Draw a Diagram) makes the key feature visible — the segment $FE$ joining the two leg-midpoints is the midsegment, which is parallel to both bases, equals their average, and sits exactly halfway up. Tool #7 (Identify Subproblems) then breaks the job into two clean pieces: first pin down the length of $FE$ and the fact that the two smaller trapezoids have the same height, then turn the 'twice the area' condition into a single linear equation in $a$ and $b$ and read off the ratio.
Execute — Answer: C
6.EE.B.6 Step 1 Name the two bases
- Let $a$ stand for the top base $AB$ and let $b$ stand for the bottom base $DC$.
- The question asks for $AB/DC = a/b$, so the whole job is to find one equation linking $a$ and $b$.
- Everything about the areas will be written using these two letters and the trapezoid's height.
💡 When the figure gives no numbers, giving the unknown lengths names lets you compute with them as if they were known.
8.G.A.4 Step 2 The middle line is the average, halfway up
- Segment $FE$ joins the midpoints of the two legs, so it is the midsegment of the trapezoid.
- Draw diagonal $DB$: in triangle $DAB$, segment $F\!X$ from the midpoint of $DA$ to the midpoint of $DB$ is half of $AB$, and in triangle $BCD$ the piece from the midpoint of $DB$ to $E$ is half of $DC$.
- Adding them, $FE=\tfrac{1}{2}(AB+DC)=\tfrac{a+b}{2}$, and $FE$ is parallel to both bases.
- Because $E$ and $F$ are the midpoints, $FE$ lies exactly halfway between $AB$ and $DC$, so trapezoids $ABEF$ and $FECD$ each have height equal to half the full height $h$.
💡 The line through the midpoints of the slanted sides is the average of the top and bottom, and it splits the height evenly.
6.G.A.1 Step 3 Equal heights turn area into a sum of bases
- A trapezoid's area is $\tfrac{1}{2}(\text{sum of the two parallel sides})\times(\text{height})$.
- Both pieces share the same height $\tfrac{h}{2}$, so their areas are proportional to the sum of their parallel sides.
- The upper piece $ABEF$ has parallel sides $a$ and $\tfrac{a+b}{2}$, and the lower piece $FECD$ has parallel sides $\tfrac{a+b}{2}$ and $b$.
- Clearing the common factor $\tfrac{1}{2}\cdot\tfrac{h}{2}$, the areas compare as $a+\tfrac{a+b}{2}$ to $\tfrac{a+b}{2}+b$, which simplify to $\tfrac{3a+b}{2}$ and $\tfrac{a+3b}{2}$.
💡 When two trapezoids are the same height, whichever has the longer pair of parallel sides has the proportionally bigger area.
8.EE.C.7 Step 4 Use 'twice as big' and solve
- The upper area is twice the lower area, so $\tfrac{3a+b}{2}=2\cdot\tfrac{a+3b}{2}$.
- The halves cancel, giving $3a+b=2(a+3b)=2a+6b$.
- Subtracting $2a$ and $b$ from both sides leaves $a=5b$.
- Dividing by $b$, the ratio is $\tfrac{a}{b}=5$, so $AB/DC=5$.
- That is choice (C).
💡 Turning the area condition into one linear equation lets the unknown heights and constants cancel, leaving a clean ratio.
6.EE.B.6 Let $a$ stand for the top base $AB$ and let $b$ stand for the bottom base $DC$. 8.G.A.4 Segment $FE$ joins the midpoints of the two legs, so it is the midsegment of the 6.G.A.1 A trapezoid's area is $\tfrac{1}{2}(\text{sum of the two parallel sides})\times( 8.EE.C.7 The upper area is twice the lower area, so $\tfrac{3a+b}{2}=2\cdot\tfrac{a+3b}{2 Review
Reasonableness: A quick sanity test: with $a=5$ and $b=1$ the midsegment is $\tfrac{5+1}{2}=3$. The upper piece then behaves like $5+3=8$ and the lower like $3+1=4$, and $8$ really is twice $4$ — the condition checks out and gives $AB/DC=5$. The ratio also makes sense in the picture: the top must be much wider than the bottom for its half of the trapezoid to carry twice the area. A common trap is to forget that the midsegment sits halfway and instead compare only the bases $a:b$, or to set the lower piece to be the doubled one; both mis-steps land on the decoy answers like $2$ or $3$.
Alternative: Skip the midsegment length and use coordinates. Put $D=(0,0)$, $C=(b,0)$, and place $A=(p,h)$, $B=(p+a,h)$ so $AB=a$ and $DC=b$. Then $F=\big(\tfrac{p}{2},\tfrac{h}{2}\big)$ and $E=\big(\tfrac{p+a+b}{2},\tfrac{h}{2}\big)$, so $FE=\tfrac{a+b}{2}$ at height $\tfrac{h}{2}$ — the same facts as before. The area equation $\tfrac{3a+b}{2}=2\cdot\tfrac{a+3b}{2}$ again gives $a=5b$, confirming $AB/DC=5$.
CCSS standards used (min grade 8)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the unknown bases $a=AB$ and $b=DC$ so the areas can be written and compared algebraically.)8.G.A.4Understand that a two-dimensional figure is similar to another using transformations (Using the trapezoid midsegment fact (via triangle midsegments on diagonal $DB$) that $FE=\tfrac{a+b}{2}$ and lies halfway up, so both pieces share height $\tfrac{h}{2}$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Applying the trapezoid area formula and reducing each equal-height piece to the sum of its parallel sides, $\tfrac{3a+b}{2}$ and $\tfrac{a+3b}{2}$.)8.EE.C.7Solve linear equations in one variable (Solving $3a+b=2(a+3b)$ to get $a=5b$ and hence $AB/DC=5$.)
⭐ The line joining the midpoints of a trapezoid's slanted sides is the average of the two bases and sits halfway up, so the two halves have the same height and their areas compare just by the length of their parallel sides.
⭐ The line joining the midpoints of a trapezoid's slanted sides is the average of the two bases and sits halfway up, so the two halves have the same height and their areas compare just by the length of their parallel sides.
More like this
Same archetype — closest grade level first.
