AMC 8 · 2004 · #24

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

area-rectanglesarea-trianglespythagorean-theoremformula-substitution area-differenceidentify-subproblems ↑ Prerequisites: area-rectanglesarea-trianglespythagorean-theorem

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

In the figure, $ABCD$ is a rectangle and $EFGH$ is a parallelogram. Using the measurements given in the figure, what is the length $d$ of the segment that is perpendicular to $\overline{HE}$ and $\overline{FG}$ ?

Pick an answer.

(A)

6.8

(B)

7.1

(C)

7.6

(D)

7.8

(E)

8.1

View mode:

Toolkit + CCSS Solution

Understand

Restated: Rectangle $ABCD$ has a parallelogram $EFGH$ inscribed in it. On each side of the rectangle, the parallelogram's vertex splits the side into two labeled pieces: $AE=4, EB=6$ on top; $BF=5, FC=3$ on the right; $CG=4, GD=6$ on the bottom; $DH=5, HA=3$ on the left. Find the distance $d$ between the two parallel sides $\overline{HE}$ and $\overline{FG}$ of the parallelogram.

Givens: $ABCD$ is a rectangle with width $AE+EB = 4+6 = 10$ and height $BF+FC = 5+3 = 8$; $EFGH$ is a parallelogram inscribed in the rectangle, one vertex on each side; Side pieces: $AE=4$, $EB=6$, $BF=5$, $FC=3$, $CG=4$, $GD=6$, $DH=5$, $HA=3$; $d$ is the perpendicular distance between the parallel sides $\overline{HE}$ and $\overline{FG}$; Answer choices: (A) $6.8$, (B) $7.1$, (C) $7.6$, (D) $7.8$, (E) $8.1$

Unknowns: The length $d$ of the segment perpendicular to both $\overline{HE}$ and $\overline{FG}$

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #7 Identify Subproblems, #1 Draw a Diagram

The parallelogram is tilted, so measuring it directly is painful. Tool #16 (Count the Complement) flips the question: instead of measuring $EFGH$, measure the four corner right triangles that the rectangle gives up to form it, then subtract from the rectangle's area. That gives the parallelogram's area without any tilted measurements. Tool #7 (Identify Subproblems) splits the job in two — first find the area, then find the base $FG$ — and Tool #1 (Draw a Diagram) lets us read each corner triangle's two leg lengths straight off the labeled sides. With the area and the base in hand, $d$ comes from area $=$ base $\times$ height.

Execute — Answer: C

#1 Draw a Diagram 4.G.A.1 Step 1

Read the four corner right triangles off the figure.
Each cut corner is a right triangle whose legs run along two sides of the rectangle, so the legs are exactly the labeled pieces.
At $A$: legs $AE=4$ and $AH=3$.
At $B$: legs $BE=6$ and $BF=5$.
At $C$: legs $CF=3$ and $CG=4$.
At $D$: legs $DG=6$ and $DH=5$.

$$\triangle AEH:\;4,3 \quad \triangle BEF:\;6,5 \quad \triangle CFG:\;3,4 \quad \triangle DGH:\;6,5$$

💡 Each corner sits at a $90^\circ$ angle of the rectangle, so the labeled pieces on the two adjacent sides are exactly the legs — Grade 4 "identify right angles in figures".

#7 Identify Subproblems 6.G.A.1 Step 2

Compute the four corner triangle areas using $\tfrac{1}{2} \times \text{leg} \times \text{leg}$.
Notice the four triangles come in two matching pairs: $\triangle AEH \cong \triangle CFG$ (both $3$-$4$ right triangles) and $\triangle BEF \cong \triangle DGH$ (both $5$-$6$ right triangles).

$$2 \cdot \tfrac{1}{2}(4)(3) + 2 \cdot \tfrac{1}{2}(6)(5) = 12 + 30 = 42$$

💡 Two pairs of congruent right triangles let you double the area of one from each pair — a Grade 6 right-triangle-area shortcut.

#16 Change Focus / Count the Complement 6.G.A.1 Step 3

Apply the complement.
The parallelogram is what is left after the four corner triangles are removed from the rectangle.

$$[EFGH] = [ABCD] - 42 = (10)(8) - 42 = 80 - 42 = 38$$

💡 "Whole minus the easy pieces" gives the tilted parallelogram's area without measuring any tilted side.

#7 Identify Subproblems 8.G.B.7 Step 4

Subproblem 2 — find the base $FG$ using the Pythagorean theorem on $\triangle CFG$.
This is a $3$-$4$ right triangle (legs $CF=3$ and $CG=4$), so its hypotenuse $FG$ is the classic $3$-$4$-$5$ length.

$$FG = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$$

💡 Grade 8 Pythagorean theorem on a $3$-$4$-$5$ triangle gives an integer side — the cleanest possible base for the parallelogram.

#16 Change Focus / Count the Complement 6.G.A.1 Step 5

Use the parallelogram area formula with $FG$ as the base and $d$ as the height.
Set the two expressions for the area equal and solve for $d$.

$$[EFGH] = FG \cdot d \;\Rightarrow\; 38 = 5d \;\Rightarrow\; d = \tfrac{38}{5} = 7.6 \;\Rightarrow\; \textbf{(C)}$$

💡 One area, two expressions: equating them turns the unknown height $d$ into a one-step division.

[1] #1 4.G.A.1 Read the four corner right triangles off the figure. Each cut corner is a right

[2] #7 6.G.A.1 Compute the four corner triangle areas using $\tfrac{1}{2} \times \text{leg} \ti

[3] #16 6.G.A.1 Apply the complement. The parallelogram is what is left after the four corner tr

[4] #7 8.G.B.7 Subproblem 2 — find the base $FG$ using the Pythagorean theorem on $\triangle CF

[5] #16 6.G.A.1 Use the parallelogram area formula with $FG$ as the base and $d$ as the height.

Review

Reasonableness: Sanity-check the size of $d$. The rectangle is $10$ wide and $8$ tall, so any segment that stays inside it has length at most $\sqrt{10^2 + 8^2} \approx 12.8$. The segment $d$ goes from near the top side to near the bottom side, so it should be a bit longer than the height $8$ but not much — a value in the $7$-$8$ range looks right. Among the choices, $7.6$ fits cleanly; $6.8$ is too short to reach across, and $7.8$ or $8.1$ would push the foot of the perpendicular outside $\overline{FG}$. Also, the parallelogram area $38$ and base $FG = 5$ give $d = 38/5 = 7.6$ exactly — no rounding needed.

Alternative: Tool #13 (Convert to Algebra) with coordinates: set $D=(0,0)$, $C=(10,0)$, $B=(10,8)$, $A=(0,8)$. Then $E=(4,8)$, $F=(10,3)$, $G=(6,0)$, $H=(0,5)$. Vector $\overrightarrow{HE} = (4, 3)$ has length $5$, confirming $HE = FG = 5$. The area of the parallelogram equals $|\overrightarrow{HE} \times \overrightarrow{HG}|$ where $\overrightarrow{HG} = (6, -5)$, giving $|4(-5) - 3(6)| = |-20 - 18| = 38$. Then $d = 38 / 5 = 7.6$. Same answer (C).

CCSS standards used (min grade 8)

4.G.A.1 Draw points, lines, line segments, rays, angles, and perpendicular and parallel lines, and identify these in two-dimensional figures (Recognizing that each corner of the rectangle is a right angle, so the labeled side pieces are the legs of the four corner right triangles.)
6.G.A.1 Find the area of triangles, quadrilaterals, and polygons by composing into rectangles or decomposing into triangles (Computing the four corner triangle areas via $\tfrac{1}{2} \times \text{leg} \times \text{leg}$, subtracting from the rectangle area $80$ to get the parallelogram area $38$, and then using base $\times$ height $= 5d$ to solve for $d$.)
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Finding the parallelogram's base $FG$ as the hypotenuse of the $3$-$4$-$5$ right triangle $\triangle CFG$.)

⭐ The tilted parallelogram is hard to measure directly, but the four corner triangles the rectangle gives up are easy: subtract their combined area $42$ from the rectangle's $80$ to get area $38$, then divide by the base $FG = 5$ (a $3$-$4$-$5$ triangle) to land on $d = 7.6$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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