AMC 8 · 2004 · #24

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation
area-rectanglesarea-trianglespythagorean-theoremformula-substitution area-differenceidentify-subproblems ↑ Prerequisites: area-rectanglesarea-trianglespythagorean-theorem
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
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 HE and FG of the parallelogram.

Pick an answer.

(A)
6.8
(B)
7.1
(C)
7.6
(D)
7.8
(E)
8.1

AMC 8 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

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 × height.

1STEP 1

Each cut-off corner is a right triangle whose two legs are the labeled pieces — A: 4,3; B: 6,5; C: 3,4; D: 6,5.

△ AEH: 4,3 △ BEF: 6,5 △ CFG: 3,4 △ DGH: 6,5
2STEP 2

The four corners pair up as two 3-4 and two 5-6 right triangles, so their combined area is 42.

2 · 12\frac{1}{2}(4)(3) + 2 · 12\frac{1}{2}(6)(5) = 12 + 30 = 42
3STEP 3

The parallelogram is what remains after the corners are removed, so its area is 80 - 42 = 38.

[EFGH] = [ABCD] - 42 = (10)(8) - 42 = 80 - 42 = 38
4STEP 4

Triangle CFG is a 3-4-5 right triangle, so the base FG equals its hypotenuse 5.

FG = √(3² + 4²) = √(25) = 5
5STEP 5

Set the two area expressions equal: FG × d = 38, so 5d = 38 and d = 7.6.

[EFGH] = FG · d → 38 = 5d → d = 385\frac{38}{5} = 7.6 → (C)
Answer
7.6
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 √(10² + 8²) ≈ 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 FG. Also, the parallelogram area 38 and base FG = 5 give d = 385\frac{38}{5} = 7.6 exactly — no rounding needed.
💡Key takeaway

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.

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