AMC 8 · 2018 · #22
Grade 6 geometry-2dProblem
Point E is the midpoint of side CD in square ABCD, and BE meets diagonal AC at F. The area of quadrilateral AFED is 45. What is the area of ABCD?
Pick an answer.
AMC 8 2018 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 square $ABCD$, point $E$ is the midpoint of side $\overline{CD}$, and segment $\overline{BE}$ crosses diagonal $\overline{AC}$ at point $F$. The quadrilateral $AFED$ (bounded by $\overline{AF}$, $\overline{FE}$, $\overline{ED}$, $\overline{DA}$) has area $45$. Find the area of the whole square.
Givens: $ABCD$ is a square, so all four sides are equal and the diagonals split it into congruent triangles; $E$ is the midpoint of $\overline{CD}$, so $CE = ED = \tfrac{1}{2} \cdot CD$; $F$ is the intersection of diagonal $\overline{AC}$ and segment $\overline{BE}$; Area of quadrilateral $AFED = 45$; Answer choices: (A) $100$, (B) $108$, (C) $120$, (D) $135$, (E) $144$
Unknowns: The area of square $ABCD$
Understand
Restated: In square $ABCD$, point $E$ is the midpoint of side $\overline{CD}$, and segment $\overline{BE}$ crosses diagonal $\overline{AC}$ at point $F$. The quadrilateral $AFED$ (bounded by $\overline{AF}$, $\overline{FE}$, $\overline{ED}$, $\overline{DA}$) has area $45$. Find the area of the whole square.
Givens: $ABCD$ is a square, so all four sides are equal and the diagonals split it into congruent triangles; $E$ is the midpoint of $\overline{CD}$, so $CE = ED = \tfrac{1}{2} \cdot CD$; $F$ is the intersection of diagonal $\overline{AC}$ and segment $\overline{BE}$; Area of quadrilateral $AFED = 45$; Answer choices: (A) $100$, (B) $108$, (C) $120$, (D) $135$, (E) $144$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem
The figure is already drawn, but Tool #1 (Draw a Diagram) tells us to *add to it* — mark $E$ as a midpoint, label $CE = ED = s/2$, and notice the small triangle $\triangle FCE$ tucked inside the larger right triangle $\triangle ACD$. That picture suggests Tool #7 (Identify Subproblems): the awkward quadrilateral $AFED$ equals the easy half-square triangle $\triangle ACD$ minus the small triangle $\triangle FCE$, so we only need to find the small triangle. To find $F$ without algebra, we use Tool #9 (Solve an Easier Related Problem): pick a concrete easy side length (say $s=6$, since $E$ is a midpoint and $\triangle ACD$ is right-angled) and discover how far below $\overline{AB}$ the point $F$ sits. The ratio we find scales to any $s$.
Execute — Answer: B
5.G.A.1 Step 1 - Pick a friendly square.
- By Tool #9, set $s = 6$ so $CE = ED = 3$ and arithmetic stays whole.
- Place $D = (0,0)$, $C = (6,0)$, $B = (6,6)$, $A = (0,6)$, $E = (3,0)$ — matching the given picture.
💡 Putting the square on a coordinate grid turns geometry into easy counting — a Grade 5 coordinate-plane skill.
5.G.A.2 Step 2 - Find point $F$, where diagonal $\overline{AC}$ meets segment $\overline{BE}$.
- Going from $A(0,6)$ down to $C(6,0)$, you drop $1$ unit for every $1$ unit right, so the diagonal is the set of points where $x + y = 6$.
- Going from $B(6,6)$ down to $E(3,0)$, you drop $2$ units for every $1$ unit left, so $\overline{BE}$ rises $2$ units in $y$ for each $1$ unit in $x$ starting at $E$: $y = 2(x - 3)$.
- Setting $y = 6 - x = 2(x-3)$ gives $6 - x = 2x - 6$, so $3x = 12$, $x = 4$ and $y = 2$.
- Thus $F = (4, 2)$ — the exact point labeled in the picture.
💡 Marking $F$'s coordinates is just locating an intersection point on the grid — a Grade 5 graphing problem.
3.MD.C.7 Step 3 - Use Tool #7 to split the picture.
- Diagonal $\overline{AC}$ cuts the square into two congruent right triangles; the half containing $D$ is $\triangle ACD$, and $AFED$ is what's left after removing the tiny corner triangle $\triangle FCE$ from $\triangle ACD$.
- So $[AFED] = [\triangle ACD] - [\triangle FCE]$.
💡 Adding and subtracting rectangle/triangle areas to handle a compound shape is exactly the Grade 3 area-as-addition idea.
6.G.A.1 Step 4 - Compute the two pieces with the $s = 6$ test square.
- $\triangle ACD$ has legs $AD = 6$ and $DC = 6$, so its area is $\tfrac{1}{2} \cdot 6 \cdot 6 = 18$.
- For $\triangle FCE$, take base $CE = 3$ on the bottom side; the height is the vertical distance from $F$ to $\overline{CD}$, which is $y_F = 2$.
- So $[\triangle FCE] = \tfrac{1}{2} \cdot 3 \cdot 2 = 3$.
💡 Finding triangle areas with $\tfrac{1}{2} \cdot \text{base} \cdot \text{height}$ and combining them is the Grade 6 area-of-polygons standard.
6.RP.A.3 Step 5 - Scale back to the real problem.
- With $s = 6$ the square has area $36$ and the quadrilateral $AFED$ has area $15$, so $[AFED]$ is the fraction $\tfrac{15}{36} = \tfrac{5}{12}$ of the whole square.
- That ratio is independent of $s$.
- The problem says $[AFED] = 45$, so the area of the square satisfies $\tfrac{5}{12} \cdot [\square ABCD] = 45$, giving $[\square ABCD] = 45 \cdot \tfrac{12}{5} = 108$.
💡 Using a ratio found from an easy case and scaling it to the real number is Grade 6 ratio reasoning.
5.G.A.1 Pick a friendly square. By Tool #9, set $s = 6$ so $CE = ED = 3$ and arithmetic 5.G.A.2 Find point $F$, where diagonal $\overline{AC}$ meets segment $\overline{BE}$. Go 3.MD.C.7 Use Tool #7 to split the picture. Diagonal $\overline{AC}$ cuts the square into 6.G.A.1 Compute the two pieces with the $s = 6$ test square. $\triangle ACD$ has legs $A 6.RP.A.3 Scale back to the real problem. With $s = 6$ the square has area $36$ and the qu Review
Reasonableness: $AFED$ takes up roughly a third of the square in the picture (a bit less than the half-square $\triangle ACD$), so the square's area should be a bit less than $3 \times 45 = 135$. The computed value $108$ is in that range and matches $\tfrac{12}{5} \times 45$. Also, $108$ is the only answer choice equal to $\tfrac{12}{5}$ of $45$, since $\tfrac{12}{5} \cdot 45 = 12 \cdot 9 = 108$ — a clean whole number, as expected for a contest problem.
Alternative: Tool #3 (Eliminate Possibilities) on the answer choices: for each candidate area $A$, the side is $s = \sqrt{A}$ and $[AFED] = \tfrac{5}{12} A$ once the $5/12$ ratio is known. Only $A = 108$ gives $\tfrac{5}{12} \cdot 108 = 45$. Choices $100, 120, 135, 144$ give $\tfrac{125}{3}, 50, 56.25, 60$ — none equal $45$, so $(B)$ is forced.
CCSS standards used (min grade 6)
5.G.A.1Use a pair of perpendicular number lines forming a coordinate system (Placing the square on a coordinate grid with $D=(0,0)$, $C=(6,0)$, $B=(6,6)$, $A=(0,6)$, $E=(3,0)$ so distances become coordinate differences.)5.G.A.2Represent real-world and mathematical problems by graphing points (Locating the intersection point $F = (4,2)$ where diagonal $\overline{AC}$ meets segment $\overline{BE}$ on the grid.)3.MD.C.7Relate area to multiplication and addition operations (Recognizing that the compound area $[AFED]$ equals the larger triangle area minus the smaller triangle area.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing $[\triangle ACD] = 18$ and $[\triangle FCE] = 3$ with the $\tfrac{1}{2} \cdot \text{base} \cdot \text{height}$ formula and combining them into $[AFED] = 15$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Scaling the ratio $\tfrac{[AFED]}{[\square ABCD]} = \tfrac{5}{12}$ found in the easy case ($s=6$) up to the real condition $[AFED] = 45$ to get $[\square ABCD] = 108$.)
⭐ This AMC 8 problem only needs Grade 6 ratio reasoning and the triangle area formula you already know!
⭐ This AMC 8 problem only needs Grade 6 ratio reasoning and the triangle area formula you already know!
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