AMC 8 · 2018 · #22

Grade 6 geometry-2d

Archetype Compound Area by Decomposition / Difference

similar-trianglesarea-trianglesratio-proportionarea-rectangles area-differenceidentify-subproblemsconvert-to-algebra ↑ Prerequisites: similar-trianglesarea-triangles

📏 Long solution 💡 4 insights 📊 Diagram

Problem

Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$

$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$

Pick an answer.

(A)

100

(B)

108

(C)

120

(D)

135

(E)

144

View mode:

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

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

#9 Solve an Easier Related Problem 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.

$$s = 6,\ CE = ED = 3$$

💡 Putting the square on a coordinate grid turns geometry into easy counting — a Grade 5 coordinate-plane skill.

#1 Draw a Diagram 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.

$F = (4, 2)$, height of $F$ above $\overline{CD}$ is $y_F = 2 = \tfrac{s}{3}$

💡 Marking $F$'s coordinates is just locating an intersection point on the grid — a Grade 5 graphing problem.

#7 Identify Subproblems 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 $[AFED] = [\triangle ACD] - [\triangle FCE]$.

$$[AFED] = [\triangle ACD] - [\triangle FCE]$$

💡 Adding and subtracting rectangle/triangle areas to handle a compound shape is exactly the Grade 3 area-as-addition idea.

#7 Identify Subproblems 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 $[\triangle FCE] = \tfrac{1}{2} \cdot 3 \cdot 2 = 3$.

$$[\triangle ACD] = 18,\ [\triangle FCE] = 3,\ [AFED] = 18 - 3 = 15$$

💡 Finding triangle areas with $\tfrac{1}{2} \cdot \text{base} \cdot \text{height}$ and combining them is the Grade 6 area-of-polygons standard.

#9 Solve an Easier Related Problem 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 $[\square ABCD] = 45 \cdot \tfrac{12}{5} = 108$.

$$\dfrac{[AFED]}{[\square ABCD]} = \dfrac{15}{36} = \dfrac{5}{12} \;\Rightarrow\; [\square ABCD] = 45 \cdot \dfrac{12}{5} = 108 \;\Rightarrow\; \textbf{(B)}$$

💡 Using a ratio found from an easy case and scaling it to the real number is Grade 6 ratio reasoning.

[1] #9 5.G.A.1 Pick a friendly square. By Tool #9, set $s = 6$ so $CE = ED = 3$ and arithmetic

[2] #1 5.G.A.2 Find point $F$, where diagonal $\overline{AC}$ meets segment $\overline{BE}$. Go

[3] #7 3.MD.C.7 Use Tool #7 to split the picture. Diagonal $\overline{AC}$ cuts the square into

[4] #7 6.G.A.1 Compute the two pieces with the $s = 6$ test square. $\triangle ACD$ has legs $A

[5] #9 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.1 Use 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.2 Represent 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.7 Relate 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.1 Find 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.3 Use 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

More like this