Competition · AMC preparation · step 4 of 4

AMC 8 · 2014 · #20

Grade 7 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid

area-rectanglesarea-circlesestimation area-differenceidentify-subproblems ↑ Prerequisites: area-rectanglesarea-circles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A 3 × 5 rectangle ABCD has three circles centered at three of its corners: radius 1 at A, radius 2 at B, radius 3 at C. Only the parts of the circles that fall inside the rectangle matter. Find the area of the rectangle that is not covered by any circle, then pick the closest answer choice.

Pick an answer.

(A)

3.5

(B)

4.0

(C)

4.5

(D)

5.0

(E)

5.5

AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The shaded region is the rectangle with three corner pieces removed, so the natural move is Tool #7 (Subproblems): compute the rectangle's area, compute each quarter-circle's area, then subtract. Tool #1 (Diagram) keeps us honest about which part of each circle sits inside the rectangle — at a corner with a 90° angle, exactly one quarter does — and lets us check on the picture that the quarter-circles do not overlap before we add them up.

1STEP 1

Find the rectangle's area

The rectangle's area is DA × CD = 5 × 3 = 15 — the whole we carve corners from.

Area_rect = 5 × 3 = 15

2STEP 2

Take a quarter of each circle

Each 90° corner traps exactly a quarter-circle, so apply the area rule $\frac{1}{4}$ π r² to each circle.

Area_A = 1/4π(1)² = 0.25π, Area_B = 1/4π(2)² = π, Area_C = 1/4π(3)² = 2.25π

3STEP 3

Check the pieces don't overlap

Overlap check: 1+2=3=AB and 2+3=5=BC, so the quarter-circles only touch and add with no double-count.

1 + 2 = 3 = AB, 2 + 3 = 5 = BC

Splitting the corner pieces into separate disjoint subproblems is only legal if they don't overlap — the side-length check confirms it.

Because the three quarter-circles inside the rectangle never overlap, the whole area the circles cover there is exactly the sum of the three separate quarter-circle areas (0.25π + π + 2.25π = 3.5π).

▸ Why?

The part of each corner circle that lies inside the rectangle is exactly one quarter of that circle, because the two sides meeting at the corner make a right angle and carve out a 90-degree wedge — one of the four equal quarters of the full 360-degree disk.

▸ Why?

Two perpendicular cuts through the center split a disk into four wedges, and a quarter turn about the center drops each wedge exactly onto the next, so all four cover equal area and the corner's wedge is one of them.

Rigid motion preserves length and angleSliding, turning, or flipping a shape lays it exactly onto a copy, so every length and angle stays the same (folding is the flip case).

▸ Why?

The three quarter-circles meet only at single points and share no area, so setting them side by side and adding their three areas counts every covered spot once and only once.

▸ Why?

Along side AB the circle at A reaches 1 unit and the circle at B reaches 2 units, and 1+2=3 is the whole length of AB, so their edges just touch; the same holds on BC where 2+3=5=BC, and the circles at A and C sit too far across the rectangle to reach each other.

▸ Why?

Every point of a circle is exactly its radius from the center and no farther, so the circle at A covers only points within 1 of A and the circle at B only points within 2 of B — their reaches along AB add up to the side and stop.

A circle's radius is constantEvery point on a circle sits exactly one radius from the center, so all radii of a circle are equal.

▸ Why?

Pieces that do not overlap add up: the area of all of them together is just the sum of the separate areas, with nothing counted twice.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

7.G.B.6Identify Subproblems

4STEP 4

Add the quarter-circle areas

Add the three disjoint quarter-circles: 0.25π + π + 2.25π = 3.5π covered.

Total covered = 0.25π + π + 2.25π = 3.5π

5STEP 5

Subtract and compare the choices

Uncovered = 15 - 3.5π ≈ 15 - 10.99 = 4.01, closest to choice (B) 4.0.

15 - 3.5π ≈ 15 - 3.5(3.14) = 15 - 10.99 = 4.01 → (B) 4.0

Answer

4.0

Sanity-check the magnitude. The rectangle has area 15. The three quarter-circles together are 3.5π ≈ 11, which is a bit more than two-thirds of the rectangle — visually that matches the figure, where the big circle at C alone eats up a large corner. So the leftover should be a small single-digit number, and 15 - 11 = 4 lands right in the middle of the choice range (3.5, 4.0, 4.5, 5.0, 5.5). Choice (B) is the only one consistent with 3.5π very close to 11.

💡Key takeaway

This AMC 8 problem only needs Grade 7 circle-area know-how plus the "break the shape into pieces and subtract" idea you already use for L-shaped figures.

Find the rectangle's area
Take a quarter of each circle
Check the pieces don't overlap
Add the quarter-circle areas
Subtract and compare the choices

A parent dashboard for the family lives at sensimlab.com.

Find the rectangle's area

Take a quarter of each circle

Check the pieces don't overlap

Add the quarter-circle areas

Subtract and compare the choices

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