AMC 8 · 2020 · #18

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

pythagorean-theoremarea-rectanglesarea-circles identify-subproblems ↑ Prerequisites: pythagorean-theoremarea-rectangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A rectangle ABCD sits inside a semicircle so that side DA lies on the diameter FE, with D and A between F and E. The two upper corners B and C touch the curved part of the semicircle. We are told DA = 16 and FD = AE = 9. Find the area of rectangle ABCD.

Pick an answer.

(A)

240

(B)

248

(C)

256

(D)

264

(E)

272

AMC 8 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The picture already exists, but the key insight is hidden until we add one more thing to it (Tool #1): mark the center O of the semicircle and draw the radius OC. That single extra segment turns the problem into a right triangle △ ODC that we can attack directly. Tool #7 (Identify Subproblems) then breaks the area question into three clean pieces: (1) find the radius from the diameter, (2) find the height CD from a right triangle, (3) multiply length × height to get the area. Each subproblem uses only one idea at a time.

1STEP 1

Mark the center O; the diameter FE = 9 + 16 + 9 = 34, so the radius is r = 17, and by symmetry O is the midpoint of DA.

FE = 9 + 16 + 9 = 34, r =

\frac{34}{2}

= 17

2STEP 2

O is the midpoint of DA, so OD = 16 ÷ 2 = 8 — the short leg of the right triangle we will use.

OD =

\frac{DA}{2}

\frac{16}{2}

= 8

3STEP 3

Draw radius OC = 17; since ∠ODC is a right angle, △ODC is a right triangle with hypotenuse OC.

OC = 17, ∠ ODC = 90^°

4STEP 4

Pythagoras on △ODC: 8² + CD² = 17², so CD² = 289 - 64 = 225 and CD = 15.

8² + CD² = 17² → CD² = 289 - 64 = 225 → CD = 15

5STEP 5

Multiply width × height: 16 × 15 = 240, which matches choice (A).

Area = DA × CD = 16 × 15 = 240 → (A)

Answer

240

The radius is 17 and the rectangle's half-width OD = 8, so the height must be less than 17 (it cannot poke out of the semicircle). We got CD = 15, which fits — and it lands on the Pythagorean triple (8, 15, 17), a well-known triple. The area 16 × 15 = 240 is comfortably the smallest answer choice, which makes sense: the rectangle is wide but not very tall, so its area should be on the smaller end of the listed options.

💡Key takeaway

This AMC 8 problem only needs Grade 8 Pythagorean theorem — a² + b² = c² on a right triangle — you already know!

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