AMC 8 · 2000 · #19

Grade 7 geometry-2d

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

area-circlesarea-trianglesspatial-visualizationreflection-symmetry area-differenceidentify-subproblems ↑ Prerequisites: area-circlesarea-triangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A region is bounded by three circular arcs, each of radius 5. Two of them, arcs AB and AD, are quarter-circles that bulge inward at the bottom corners; the third, arc BCD, is a semicircle that bulges outward at the top. Find the area of the region.

Pick an answer.

(A)

(B)

$10+5\pi$

(C)

(D)

$50+5\pi$

(E)

$25\pi$

AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The figure is a curvy blob, but the asy coordinates pin down a clean rectangle in the middle: B=(-5,5) and D=(5,5) sit directly above A=(0,0). Tool #1 (Draw a Diagram) makes the trick visible — cut the region with the segment BD. Above BD is exactly the semicircle on diameter BD. Below BD is a 10 × 5 rectangle with two quarter-circles scooped out of its bottom corners. Tool #7 (Break Into Subproblems) then turns the area into three easy circle/rectangle areas. The payoff: the semicircle added on top has the same area as the two quarter-circles scooped out of the bottom ( $\frac{25π}{2}$ = 2 · $\frac{25π}{4}$ ), so all the π terms cancel and only the rectangle is left.

1STEP 1

Draw segment BD. From B=(-5,5) and D=(5,5) it is horizontal with length 10, splitting the region into a top and a bottom piece.

BD = 5 - (-5) = 10

2STEP 2

Top piece: arc BCD is a semicircle of diameter 10, radius 5, so its area is 25π/2.

A_upper =

\frac{1}{2}

π(5)² =

\frac{25π}{2}

3STEP 3

Bottom piece: a 10×5 rectangle (area 50) with a radius-5 quarter-circle scooped from each bottom corner, so its area is 50 - 25π/2.

A_lower = 10 × 5 - 2 ·

\frac{1}{4}

π(5)² = 50 -

\frac{25π}{2}

4STEP 4

Add them: the + $\frac{25π}{2}$ on top and the - $\frac{25π}{2}$ below cancel, leaving just the rectangle — area 50.

A_total =

\frac{25π}{2}

+ (50 -

\frac{25π}{2}

) = 50 → (C)

Answer

The two quarter-circles cut out at the bottom together form half of a radius-5 circle, the same shape as the semicircle glued on top. Mentally pick up each scooped quarter and slide it up to fill its matching half of the semicircle: the curvy region rearranges into the 10 × 5 rectangle exactly, area 50. That rules out (A) 25 (too small — even the rectangle alone is 50), (E) 25π ≈ 78.5 (too big — the region fits inside the 10 × 10 bounding square of area 100 but is clearly less than the bounding square), and (B), (D), which would mean the π terms do not cancel. The clean integer answer 50 matches (C).

💡Key takeaway

Cut the region with segment BD — the semicircle glued on top has the same area as the two quarter-circles scooped out of the rectangle below, so the π terms cancel and the answer is just the rectangle, 10 × 5 = 50.

More like this