AMC 10 · 2007 · #11
Grade 8 geometry-2dA circle passes through the three vertices of an isosceles triangle that has two sides of length 3 and a base of length 2. What is the area of this circle?
Pick an answer.
AMC 10 2007 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: An isosceles triangle has two equal sides of length $3$ and a base of length $2$. A circle passes through all three vertices of this triangle. Find the area of that circle.
Givens: The triangle is isosceles with two legs of length $3$; The base of the triangle has length $2$; One circle passes through all three vertices (the circumscribed circle); Answer choices: (A) $2\pi$, (B) $\tfrac52\pi$, (C) $\tfrac{81}{32}\pi$, (D) $3\pi$, (E) $\tfrac72\pi$
Unknowns: The area of the circle through the three vertices
Understand
Restated: An isosceles triangle has two equal sides of length $3$ and a base of length $2$. A circle passes through all three vertices of this triangle. Find the area of that circle.
Givens: The triangle is isosceles with two legs of length $3$; The base of the triangle has length $2$; One circle passes through all three vertices (the circumscribed circle); Answer choices: (A) $2\pi$, (B) $\tfrac52\pi$, (C) $\tfrac{81}{32}\pi$, (D) $3\pi$, (E) $\tfrac72\pi$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems
The area of the circle is $\pi R^2$, so everything hinges on the radius $R$. Drawing the triangle with coordinates (Tool #1) and putting the base flat with its midpoint at the origin exposes the triangle's mirror symmetry, which forces the circle's center onto the vertical axis. Naming the center's height $k$ (Tool #4) turns 'the center is equidistant from all three vertices' into equations. The job then splits into small subproblems (Tool #7): first the triangle's height, then the center's position, then the radius, then the area.
Execute — Answer: C
8.G.B.7 Step 1 Set coordinates and find the height
- Place the base flat on the $x$-axis with its midpoint at the origin, so the base vertices are $B=(-1,0)$ and $C=(1,0)$ (this makes the base length $2$).
- The apex $A$ sits directly above the origin at $(0,h)$.
- Each equal side has length $3$, so the distance from $A$ to $C$ is $3$.
- That distance is the hypotenuse of a right triangle with horizontal leg $1$ and vertical leg $h$, so the Pythagorean theorem gives $1^2+h^2=3^2$, hence $h^2=8$ and $h=2\sqrt2$.
💡 Splitting the isosceles triangle down its middle makes a right triangle, so the Pythagorean theorem hands over the height.
8.G.B.8 Step 2 Put the center on the axis of symmetry
- The circle's center is the same distance $R$ from all three vertices.
- Because $B$ and $C$ are mirror images across the $y$-axis, the center must lie on that axis, at some point $O=(0,k)$.
- Write the radius two ways using distances.
- The distance from $O$ to $B=(-1,0)$ has horizontal part $1$ and vertical part $k$, so $R^2=1^2+k^2$.
- The distance from $O$ to the apex $A=(0,2\sqrt2)$ is straight up the axis, so $R=2\sqrt2-k$, giving $R^2=(2\sqrt2-k)^2$.
💡 The center is equally far from both base corners, so it can only sit on the line of mirror symmetry.
8.EE.C.7 Step 3 Solve for the center's height
- Both expressions equal $R^2$, so set them equal: $1+k^2=(2\sqrt2-k)^2$.
- Expanding the right side gives $8-4\sqrt2\,k+k^2$.
- The $k^2$ terms cancel from both sides, leaving a linear equation $1=8-4\sqrt2\,k$.
- Solving, $4\sqrt2\,k=7$, so $k=\dfrac{7}{4\sqrt2}$.
💡 The squared terms match on both sides and cancel, collapsing a scary-looking equation into a one-step linear solve.
7.G.B.4 Step 4 Compute the radius squared and the area
- The area only needs $R^2$, and $R^2=1+k^2$.
- Square $k$: $k^2=\left(\dfrac{7}{4\sqrt2}\right)^2=\dfrac{49}{16\cdot2}=\dfrac{49}{32}$.
- Then $R^2=1+\dfrac{49}{32}=\dfrac{32}{32}+\dfrac{49}{32}=\dfrac{81}{32}$.
- The area of the circle is $\pi R^2=\dfrac{81}{32}\pi$, which is choice $\textbf{(C)}$.
💡 Since area is $\pi R^2$, we never need $R$ itself — just $R^2$, which drops out directly.
8.G.B.7 Place the base flat on the $x$-axis with its midpoint at the origin, so the base 8.G.B.8 The circle's center is the same distance $R$ from all three vertices. Because $B 8.EE.C.7 Both expressions equal $R^2$, so set them equal: $1+k^2=(2\sqrt2-k)^2$. Expandin 7.G.B.4 The area only needs $R^2$, and $R^2=1+k^2$. Square $k$: $k^2=\left(\dfrac{7}{4\s Review
Reasonableness: Sanity-check the size: $R^2=\tfrac{81}{32}\approx2.53$, so $R\approx1.59$. The circle must be big enough to reach every vertex; the apex is $2\sqrt2\approx2.83$ above the base, and the center at height $k=\tfrac{7}{4\sqrt2}\approx1.24$ leaves $2.83-1.24\approx1.59$ up to the apex and $\sqrt{1+1.24^2}\approx1.59$ out to a base corner — the two distances agree, confirming a single consistent radius. The area $\tfrac{81}{32}\pi\approx2.53\pi$ lands neatly between choices $2\pi$ and $\tfrac52\pi$, matching only (C).
Alternative: Use the circumradius formula $R=\dfrac{abc}{4K}$, where $a,b,c$ are the side lengths and $K$ is the triangle's area. The area is $K=\tfrac12\cdot\text{base}\cdot\text{height}=\tfrac12\cdot2\cdot2\sqrt2=2\sqrt2$. With sides $3,3,2$, $R=\dfrac{3\cdot3\cdot2}{4\cdot2\sqrt2}=\dfrac{18}{8\sqrt2}=\dfrac{9}{4\sqrt2}$, so $R^2=\dfrac{81}{32}$ and the area is $\dfrac{81}{32}\pi$, confirming (C).
CCSS standards used (min grade 8)
8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the triangle's height $h=2\sqrt2$ from the leg $3$ and half-base $1$.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Writing the radius as a distance from the center $(0,k)$ to a base vertex and to the apex.)8.EE.C.7Solve linear equations in one variable (Solving $1=8-4\sqrt2\,k$ for the center's height after the $k^2$ terms cancel.)7.G.B.4Know the formulas for area and circumference of a circle (Turning the radius into the circle's area via $\pi R^2$.)
⭐ Put the shape on a grid, use symmetry to pin the circle's center on the mirror line, and remember the area only needs $R^2$, not $R$ itself.
⭐ Put the shape on a grid, use symmetry to pin the circle's center on the mirror line, and remember the area only needs $R^2$, not $R$ itself.
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