AMC 10 · 2006 · #19
Grade 8 geometry-2dA circle of radius 2 is centered at O. Square OABC has side length 1. Sides AB and CB are extended past B to meet the circle at D and E, respectively. What is the area of the shaded region in the figure, which is bounded by BD, BE, and the minor arc connecting D and E?
Pick an answer.
AMC 10 2006 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 circle of radius $2$ sits at the origin $O$, and a unit square $OABC$ has one corner at $O$. Side $AB$ is pushed up past $B$ until it hits the circle at $D$, and side $CB$ is pushed right past $B$ until it hits the circle at $E$. Find the area of the region penned in by segment $BD$, segment $BE$, and the short arc from $D$ to $E$.
Givens: A circle centered at $O$ with radius $2$; A square $OABC$ of side length $1$, so $O=(0,0)$, $A=(1,0)$, $B=(1,1)$, $C=(0,1)$; $D$ is where line $AB$ (the vertical line $x=1$) crosses the circle above $B$; $E$ is where line $CB$ (the horizontal line $y=1$) crosses the circle right of $B$
Unknowns: The area of the region bounded by $BD$, $BE$, and the minor arc $DE$
Understand
Restated: A circle of radius $2$ sits at the origin $O$, and a unit square $OABC$ has one corner at $O$. Side $AB$ is pushed up past $B$ until it hits the circle at $D$, and side $CB$ is pushed right past $B$ until it hits the circle at $E$. Find the area of the region penned in by segment $BD$, segment $BE$, and the short arc from $D$ to $E$.
Givens: A circle centered at $O$ with radius $2$; A square $OABC$ of side length $1$, so $O=(0,0)$, $A=(1,0)$, $B=(1,1)$, $C=(0,1)$; $D$ is where line $AB$ (the vertical line $x=1$) crosses the circle above $B$; $E$ is where line $CB$ (the horizontal line $y=1$) crosses the circle right of $B$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities
The shaded region has a curved side, so no single area formula fits. Tool #1 (Draw a Diagram) pins everything to coordinates so $D$ and $E$ become computable. Tool #7 (Identify Subproblems) is the key move: split the awkward region into a plain triangle $BDE$ plus the circular segment that bulges out to the arc, and get the segment as a sector minus a triangle. Tool #3 (Eliminate Possibilities) uses the picture — the region is clearly small — to sanity-check which choice can be right.
Execute — Answer: A
8.G.B.7 Step 1 Locate D and E with coordinates
- Put $O$ at the origin, so the square gives $A=(1,0)$, $B=(1,1)$, $C=(0,1)$.
- Line $AB$ is the vertical line $x=1$; where it meets the circle $x^2+y^2=4$, we get $1+y^2=4$, so $y=\sqrt{3}$ and $D=(1,\sqrt{3})$.
- By the mirror symmetry across the line $y=x$, line $CB$ (which is $y=1$) meets the circle at $E=(\sqrt{3},1)$.
💡 Every point on the circle obeys $x^2+y^2=4$, so fixing $x=1$ leaves the Pythagorean relation to hand you $y$.
6.G.A.1 Step 2 Split the region into two pieces
- The region is bounded by the straight segments $BD$ and $BE$ and the outward-bulging arc $DE$.
- Cut it with the straight chord $DE$: what is left toward $B$ is the plain triangle $BDE$, and what pokes out to the arc is the circular segment sitting on chord $DE$.
- So the shaded area equals (area of triangle $BDE$) plus (area of segment $DE$).
💡 A region with one curved side becomes easy once you slice off the bulge along a straight chord.
6.G.A.1 Step 3 Area of triangle BDE
- Triangle $BDE$ has its right angle at $B=(1,1)$.
- The leg $BD$ runs straight up from $y=1$ to $y=\sqrt{3}$, and the leg $BE$ runs straight right from $x=1$ to $x=\sqrt{3}$, so both legs have length $\sqrt{3}-1$.
- A right triangle's area is half the product of its legs.
💡 The two extended sides are equal by symmetry, so the triangle is just half of a small square.
7.G.B.4 Step 4 Area of the circular segment DE
- The segment equals the sector $ODE$ minus the triangle $ODE$.
- The radius to $D=(1,\sqrt3)$ makes a $60^\circ$ angle with the $x$-axis and the radius to $E=(\sqrt3,1)$ makes a $30^\circ$ angle, so the central angle $\angle DOE = 60^\circ-30^\circ = 30^\circ$.
- That sector is $\tfrac{30}{360}=\tfrac{1}{12}$ of the whole circle of area $\pi(2)^2=4\pi$, giving $\tfrac{4\pi}{12}=\tfrac{\pi}{3}$.
- Triangle $ODE$ has area $\tfrac12\,|x_D\,y_E-x_E\,y_D|=\tfrac12\,|1\cdot1-\sqrt3\cdot\sqrt3|=\tfrac12\,|1-3|=1$.
💡 The bulge between a chord and its arc is exactly what is left when you scoop the flat triangle out of the pie slice.
6.G.A.1 Step 5 Add the pieces and pick the choice
- Combine the triangle and the segment: $[\text{shaded}]=(2-\sqrt3)+\left(\tfrac{\pi}{3}-1\right)=\tfrac{\pi}{3}+1-\sqrt3$.
- Numerically this is about $1.047+1-1.732\approx 0.315$ — a small sliver, matching the picture.
- This is choice (A).
💡 Assembling the flat triangle with the curved bulge gives the whole region in one clean sum.
8.G.B.7 Put $O$ at the origin, so the square gives $A=(1,0)$, $B=(1,1)$, $C=(0,1)$. Line 6.G.A.1 The region is bounded by the straight segments $BD$ and $BE$ and the outward-bul 6.G.A.1 Triangle $BDE$ has its right angle at $B=(1,1)$. The leg $BD$ runs straight up f 7.G.B.4 The segment equals the sector $ODE$ minus the triangle $ODE$. The radius to $D=( 6.G.A.1 Combine the triangle and the segment: $[\text{shaded}]=(2-\sqrt3)+\left(\tfrac{\ Review
Reasonableness: The region is a thin sliver just outside the corner $B$, so its area must be small and positive. Choice (A) gives $\tfrac{\pi}{3}+1-\sqrt3\approx0.315$, which is tiny — perfect for the picture. Choices (B) $\tfrac{\pi}{2}(2-\sqrt3)\approx0.42$ and (C) $\pi(2-\sqrt3)\approx0.84$ are noticeably bigger, and (D) $\approx1.89$ is far too large; (E) $\tfrac{\pi}{3}-1+\sqrt3\approx1.78$ is also too big. Only (A) is small enough, confirming the answer.
Alternative: Subtract directly instead of adding. The sector $ODE$ (area $\tfrac{\pi}{3}$) is bounded by radii $OD$, $OE$ and the arc; the shaded region is exactly that sector with the four-sided piece $ODBE$ removed. Quadrilateral $ODBE$ has vertices $O=(0,0)$, $D=(1,\sqrt3)$, $B=(1,1)$, $E=(\sqrt3,1)$; the shoelace formula gives its area as $\tfrac12|2-2\sqrt3|=\sqrt3-1$. So the shaded area $=\tfrac{\pi}{3}-(\sqrt3-1)=\tfrac{\pi}{3}+1-\sqrt3$, the same result (A).
CCSS standards used (min grade 8)
8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Solving $1^2+y^2=2^2$ to locate $D=(1,\sqrt3)$ and, by symmetry, $E=(\sqrt3,1)$ on the circle.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Decomposing the shaded region into triangle $BDE$ plus a segment, computing $[\triangle BDE]=2-\sqrt3$, and adding the pieces.)7.G.B.4Know the formulas for area and circumference of a circle (Taking the $30^\circ$ sector $ODE$ as $\tfrac{1}{12}$ of the circle $\pi(2)^2=4\pi$ to get $\tfrac{\pi}{3}$.)7.G.B.5Use facts about supplementary, complementary, vertical, and adjacent angles (Finding the central angle $\angle DOE = 60^\circ-30^\circ=30^\circ$ from the two radius directions.)
⭐ When a region has one curved side, slice it along a straight chord into a plain triangle plus a pie-slice-minus-triangle bulge, then add the parts.
⭐ When a region has one curved side, slice it along a straight chord into a plain triangle plus a pie-slice-minus-triangle bulge, then add the parts.
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