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AMC 8 · 2000 · #19

Grade 7 geometry-2d
Archetype Compound Area by Decomposition / Difference
area-circlesarea-trianglesspatial-visualizationreflection-symmetry area-differenceidentify-subproblems ↑ Prerequisites: area-circlesarea-triangles
📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Three circular arcs of radius 55 units bound the region shown. Arcs ABAB and ADAD are quarter-circles, and arc BCDBCD is a semicircle. What is the area, in square units, of the region?

Pick an answer.

(A)
25
(B)
$10+5\pi$
(C)
50
(D)
$50+5\pi$
(E)
$25\pi$
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Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: All three arcs have radius $5$; Arc $AB$ is a quarter-circle centered at $(-5, 0)$; Arc $AD$ is a quarter-circle centered at $(5, 0)$; Arc $BCD$ is a semicircle on diameter $BD$; From the asy code: $A=(0,0)$, $B=(-5,5)$, $C=(0,10)$, $D=(5,5)$; Answer choices: (A) $25$, (B) $10+5\pi$, (C) $50$, (D) $50+5\pi$, (E) $25\pi$

Unknowns: The area of the region bounded by the three arcs

Understand

Restated: 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.

Givens: All three arcs have radius $5$; Arc $AB$ is a quarter-circle centered at $(-5, 0)$; Arc $AD$ is a quarter-circle centered at $(5, 0)$; Arc $BCD$ is a semicircle on diameter $BD$; From the asy code: $A=(0,0)$, $B=(-5,5)$, $C=(0,10)$, $D=(5,5)$; Answer choices: (A) $25$, (B) $10+5\pi$, (C) $50$, (D) $50+5\pi$, (E) $25\pi$

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Break Into Subproblems

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 $10 \times 5$ 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 ($\tfrac{25\pi}{2} = 2 \cdot \tfrac{25\pi}{4}$), so all the $\pi$ terms cancel and only the rectangle is left.

Execute — Answer: C

#1 Draw a Diagram 6.NS.C.8 Step 1
  • Draw the segment $BD$ and read coordinates off the figure.
  • The asy code places $A=(0,0)$, $B=(-5,5)$, $C=(0,10)$, $D=(5,5)$.
  • So $BD$ is the horizontal segment at height $y=5$ from $x=-5$ to $x=5$, length $10$.
  • The segment $BD$ splits the region into two pieces: an upper piece bounded by arc $BCD$ and segment $BD$, and a lower piece bounded by arcs $AB$, $AD$ and segment $BD$.
$$BD = 5 - (-5) = 10$$

💡 Horizontal length on the coordinate plane is just the difference of $x$-coordinates — a Grade 6 distance-on-axis move.

#7 Break Into Subproblems 7.G.B.4 Step 2
  • Find the area of the upper piece.
  • Arc $BCD$ is a semicircle with diameter $BD = 10$, so its radius is $5$.
  • Its area is half of $\pi r^2$.
$$A_{\text{upper}} = \tfrac{1}{2}\pi(5)^2 = \dfrac{25\pi}{2}$$

💡 $A = \pi r^2$ is the Grade 7 circle-area formula; a semicircle is half of that.

#7 Break Into Subproblems 7.G.B.4 Step 3
  • Find the area of the lower piece.
  • Place a rectangle with corners $(-5,0), (5,0), (5,5), (-5,5)$.
  • It has width $10$ and height $5$, area $50$.
  • The arcs $AB$ and $AD$ are quarter-circles of radius $5$ centered at $(-5,0)$ and $(5,0)$, and they curve inward, scooping a quarter-circle out of each bottom corner of the rectangle.
  • Each scoop has area $\tfrac{1}{4}\pi(5)^2 = \tfrac{25\pi}{4}$.
$$A_{\text{lower}} = 10 \times 5 - 2 \cdot \tfrac{1}{4}\pi(5)^2 = 50 - \dfrac{25\pi}{2}$$

💡 Rectangle area minus the two scooped corners. Each scoop is a quarter of a radius-$5$ circle.

#7 Break Into Subproblems 6.EE.A.3 Step 4
  • Add the upper and lower pieces.
  • The $\dfrac{25\pi}{2}$ added by the semicircle on top is exactly the $\dfrac{25\pi}{2}$ subtracted by the two scooped quarter-circles below, so they cancel and only the rectangle's area is left.
$$A_{\text{total}} = \dfrac{25\pi}{2} + \left(50 - \dfrac{25\pi}{2}\right) = 50 \;\Rightarrow\; \textbf{(C)}$$

💡 Combining like terms: the $\pi$ pieces are opposites and add to zero, leaving the plain number $50$.

[1] #1 6.NS.C.8 Draw the segment $BD$ and read coordinates off the figure. The asy code places $
[2] #7 7.G.B.4 Find the area of the upper piece. Arc $BCD$ is a semicircle with diameter $BD =
[3] #7 7.G.B.4 Find the area of the lower piece. Place a rectangle with corners $(-5,0), (5,0),
[4] #7 6.EE.A.3 Add the upper and lower pieces. The $\dfrac{25\pi}{2}$ added by the semicircle o

Review

Reasonableness: 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 $10 \times 5$ rectangle exactly, area $50$. That rules out (A) $25$ (too small — even the rectangle alone is $50$), (E) $25\pi \approx 78.5$ (too big — the region fits inside the $10 \times 10$ bounding square of area $100$ but is clearly less than the bounding square), and (B), (D), which would mean the $\pi$ terms do not cancel. The clean integer answer $50$ matches (C).

Alternative: Tool #11 (Find an Invariant): the total area is the bounding rectangle's area plus the net signed area contributed by the arcs. The semicircle adds $\tfrac{25\pi}{2}$ and the two quarter-circles each remove $\tfrac{25\pi}{4}$, so the net $\pi$ contribution is $\tfrac{25\pi}{2} - 2\cdot\tfrac{25\pi}{4} = 0$. The $\pi$ terms must cancel by symmetry — so the answer cannot contain $\pi$. Only choices (A) $25$ and (C) $50$ are $\pi$-free, and (A) is clearly too small.

CCSS standards used (min grade 7)

  • 6.NS.C.8 Solve real-world and mathematical problems by graphing points in the coordinate plane, including finding distances between points with the same first or second coordinate (Reading $BD = 10$ as the difference of $x$-coordinates of $B=(-5,5)$ and $D=(5,5)$ on the horizontal segment.)
  • 6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Combining $\tfrac{25\pi}{2} + \left(50 - \tfrac{25\pi}{2}\right)$ by cancelling the $\pi$ terms to leave $50$.)
  • 7.G.B.4 Know the formulas for the area and circumference of a circle and use them to solve problems (Computing the semicircle area $\tfrac{1}{2}\pi(5)^2 = \tfrac{25\pi}{2}$ and each quarter-circle area $\tfrac{1}{4}\pi(5)^2 = \tfrac{25\pi}{4}$ from $A = \pi r^2$.)
  • 7.G.B.6 Solve real-world and mathematical problems involving area of two-dimensional objects composed of triangles, quadrilaterals, and other shapes (Decomposing the curvy region into a $10 \times 5$ rectangle, a semicircle, and two quarter-circles, then combining their areas.)

⭐ 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 $\pi$ terms cancel and the answer is just the rectangle, $10 \times 5 = 50$.

⭐ 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 $\pi$ terms cancel and the answer is just the rectangle, $10 \times 5 = 50$.

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