AMC 10 · 2004 · #9
Grade 7 geometry-2dA square has sides of length 10, and a circle centered at one of its vertices has radius 10. What is the area of the union of the regions enclosed by the square and the circle?
Pick an answer.
AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A square has side length $10$. A circle of radius $10$ is centered at one corner of that square. Find the area of the union of the two regions — every point that lies inside the square, the circle, or both.
Givens: The square has side length $10$; The circle has radius $10$ and is centered at one vertex of the square; Answer choices: (A) $200+25\pi$, (B) $100+75\pi$, (C) $75+100\pi$, (D) $100+100\pi$, (E) $100+125\pi$
Unknowns: The total area covered by the square and the circle together (their union)
Understand
Restated: A square has side length $10$. A circle of radius $10$ is centered at one corner of that square. Find the area of the union of the two regions — every point that lies inside the square, the circle, or both.
Givens: The square has side length $10$; The circle has radius $10$ and is centered at one vertex of the square; Answer choices: (A) $200+25\pi$, (B) $100+75\pi$, (C) $75+100\pi$, (D) $100+100\pi$, (E) $100+125\pi$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
A picture is the whole game here (#1 Draw a Diagram): once you place the circle on a corner you can see that only a corner wedge of the circle sits inside the square. Then the union splits into three clean pieces to compute — square area, circle area, and the overlap — which is Identify Subproblems (#7), combined by the union rule so the shared part is not double-counted. Finally Eliminate Possibilities (#3) checks the result against the five choices, since the traps come from mishandling the overlap.
Execute — Answer: B
3.MD.C.7 Step 1 Draw it and spot the overlap
- Put the square in the corner of a grid with the circle's center at the origin, so the square covers $0\le x\le 10$, $0\le y\le 10$.
- The two sides of the square that meet at the center run along the positive $x$- and $y$-axes, so they open a right angle — a quarter turn — at that corner.
- The only part of the circle that falls inside the square is the wedge of the disk sitting in that right-angle corner.
- Because the radius $10$ equals the side, that wedge is exactly a quarter of the full circle.
💡 The square only opens a $90^\circ$ corner at the center, so it can grab just a quarter of the circle.
7.G.B.4 Step 2 Area of the square and the circle
- The square's area is side times side.
- The circle's area is $\pi$ times radius squared.
- Keep them separate for now — the next step handles the part they share.
💡 Break the messy shape into two areas you already know the formulas for.
7.G.B.4 Step 3 Measure the shared corner wedge
- The overlap is the quarter of the disk that lies in the square's right-angle corner.
- Every point of that quarter-disk is within distance $10$ of the corner and has both coordinates between $0$ and $10$, so it really does stay inside the $10\times10$ square — the whole quarter-circle fits.
- Its area is one fourth of the circle's area.
💡 A quarter of the circle sits inside the square, so the shared region is a quarter-circle.
7.G.B.6 Step 4 Combine without double-counting
- The union is the square plus the circle, minus the piece counted in both.
- Adding $100$ and $100\pi$ counts the corner wedge twice, so subtract that $25\pi$ once.
- The total is $100+100\pi-25\pi=100+75\pi$, which is choice (B).
- The traps come from forgetting to subtract the overlap (giving (D) $100+100\pi$) or from subtracting the wrong shape.
💡 Add both areas, then remove the shared wedge once so it is counted a single time.
3.MD.C.7 Put the square in the corner of a grid with the circle's center at the origin, s 7.G.B.4 The square's area is side times side. The circle's area is $\pi$ times radius sq 7.G.B.4 The overlap is the quarter of the disk that lies in the square's right-angle cor 7.G.B.6 The union is the square plus the circle, minus the piece counted in both. Adding Review
Reasonableness: The answer should be a bit less than the naive sum $100+100\pi$, because the square and circle overlap; $100+75\pi$ is exactly $25\pi$ smaller, which matches the quarter-circle they share. It should also be more than the circle alone ($100\pi$) minus that quarter — indeed $100+75\pi\approx 100+235.6=335.6$, larger than the square ($100$) and larger than three-quarters of the circle ($75\pi\approx235.6$), as a union of the two must be. Choice (D) ignores the overlap; (A), (C), (E) use the wrong overlap fraction. Only (B) fits.
Alternative: Count the union as (circle) plus (the part of the square outside the circle). The square minus its quarter-circle corner is $100-25\pi$, and adding the full circle gives $100-25\pi+100\pi=100+75\pi$ — the same result, reached by attaching the leftover square-corner to the whole circle instead of subtracting the overlap.
CCSS standards used (min grade 7)
3.MD.C.7Relate area to multiplication; find area of rectangles by multiplying side lengths (Computing the square's area as $10\cdot10=100$ and recognizing the right-angle corner where the square meets the circle's center.)7.G.B.4Know and use the formula for the area of a circle (Finding the circle's area $\pi\cdot10^2=100\pi$ and taking a quarter of it, $25\pi$, for the shared wedge.)7.G.B.6Solve problems involving area of two-dimensional figures composed of other shapes (Combining the square, circle, and overlap into the union with $100+100\pi-25\pi$ so the shared region is counted once.)
⭐ When two shapes overlap, add both areas and subtract the shared piece once — here the square grabs only a quarter of the circle, so the union is $100+100\pi-25\pi=100+75\pi$.
⭐ When two shapes overlap, add both areas and subtract the shared piece once — here the square grabs only a quarter of the circle, so the union is $100+100\pi-25\pi=100+75\pi$.
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