AMC 10 · 2006 · #8
Grade 8 geometry-2dA square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle?
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 square whose area is $40$ sits inside a semicircle: its base lies on the flat diameter and its two upper corners touch the curved arc. Find the area of the semicircle.
Givens: The square has area $40$; The square's base rests on the diameter of the semicircle, centered on it; The two top corners of the square lie on the semicircle's arc; Answer choices: (A) $20\pi$, (B) $25\pi$, (C) $30\pi$, (D) $40\pi$, (E) $50\pi$
Unknowns: The area of the semicircle
Understand
Restated: A square whose area is $40$ sits inside a semicircle: its base lies on the flat diameter and its two upper corners touch the curved arc. Find the area of the semicircle.
Givens: The square has area $40$; The square's base rests on the diameter of the semicircle, centered on it; The two top corners of the square lie on the semicircle's arc; Answer choices: (A) $20\pi$, (B) $25\pi$, (C) $30\pi$, (D) $40\pi$, (E) $50\pi$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems
The area formula for a semicircle needs the radius, and the radius is hidden inside the picture, so Tool #1 (Draw a Diagram) marks the center at the middle of the diameter and draws the straight line from that center out to a top corner of the square. Tool #4 (Introduce a Variable) names the square's half-side so the corner's position can be written down, and Tool #7 (Identify Subproblems) splits the job into three clean pieces: turn the area into a side length, use a right triangle to get the radius, then plug the radius into the circle-area formula.
Execute — Answer: B
6.G.A.3 Step 1 Mark the center and a half-side
- Put the center $O$ of the semicircle at the middle of the diameter.
- Because the square is centered, its base is split into two equal halves by $O$.
- Let $a$ be the length of one of those halves, so the whole side of the square is $2a$.
- A bottom corner is then a horizontal distance $a$ from $O$, and the matching top corner is straight above it, a height $2a$ up.
💡 Placing the center at the middle turns the symmetric picture into two equal halves, so one small length $a$ describes the whole square.
6.EE.B.7 Step 2 Turn the area into a fact about $a$
- The area of a square is the side times itself.
- Here the side is $2a$, so the area is $(2a)^2 = 4a^2$.
- That must equal $40$, giving $4a^2 = 40$.
- Divide both sides by $4$ to get $a^2 = 10$.
- Notice we only need $a^2$, not $a$ itself, so no square roots are required.
💡 Keeping the answer as $a^2$ instead of $a$ avoids messy roots, because the radius will also come out squared.
8.G.B.7 Step 3 Get the radius with a right triangle
- Draw the line from the center $O$ to the top corner on the arc; that line is a radius $r$.
- It is the hypotenuse of a right triangle whose horizontal leg is $a$ (across to the bottom corner) and whose vertical leg is $2a$ (up to the top corner).
- By the Pythagorean Theorem, $r^2 = a^2 + (2a)^2 = a^2 + 4a^2 = 5a^2$.
- Substitute $a^2 = 10$ to get $r^2 = 5 \times 10 = 50$.
💡 The radius, the half-base, and the side form a right triangle, so the Pythagorean Theorem links them directly.
7.G.B.4 Step 4 Apply the circle-area formula
- A full circle of radius $r$ has area $\pi r^2$; a semicircle is exactly half of that, so its area is $\tfrac{1}{2}\pi r^2$.
- With $r^2 = 50$, the semicircle's area is $\tfrac{1}{2}\pi(50) = 25\pi$.
- That matches choice (B).
💡 Because $r^2$ is already known, the area formula needs no square root at all.
6.G.A.3 Put the center $O$ of the semicircle at the middle of the diameter. Because the 6.EE.B.7 The area of a square is the side times itself. Here the side is $2a$, so the are 8.G.B.7 Draw the line from the center $O$ to the top corner on the arc; that line is a r 7.G.B.4 A full circle of radius $r$ has area $\pi r^2$; a semicircle is exactly half of Review
Reasonableness: A quick sanity check: the square of area $40$ fits snugly inside the semicircle, and a semicircle must have more room than the square it contains. The semicircle's area is $25\pi \approx 78.5$, comfortably larger than $40$, which fits. The full circle would be $50\pi$, so the half is $25\pi$ — choice (E) $50\pi$ is the trap of forgetting to halve, and choice (D) $40\pi$ is the trap of confusing $r^2$ with the square's area $40$.
Alternative: Use coordinates instead of a bare triangle. Put $O$ at the origin with the diameter on the $x$-axis. The corners of the square are $(\pm a, 0)$ and $(\pm a, 2a)$ with $a^2 = 10$. A top corner $(a, 2a)$ lies on the circle $x^2 + y^2 = r^2$, so $r^2 = a^2 + (2a)^2 = 5a^2 = 50$, and the semicircle area is $\tfrac12\pi(50) = 25\pi$. Same answer (B).
CCSS standards used (min grade 8)
6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Placing the center at the middle of the diameter and locating the square's corners relative to it.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Turning the area statement $4a^2 = 40$ into $a^2 = 10$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding $r^2 = a^2 + (2a)^2 = 50$ from the right triangle joining the center to a top corner.)7.G.B.4Know the formulas for area and circumference of a circle (Computing the semicircle area as half of $\pi r^2$, giving $25\pi$.)
⭐ To find a circle's size, hunt for a right triangle that reaches from the center to a point on the edge — the Pythagorean Theorem hands you the radius squared, which is all the area formula needs.
⭐ To find a circle's size, hunt for a right triangle that reaches from the center to a point on the edge — the Pythagorean Theorem hands you the radius squared, which is all the area formula needs.
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