AMC 8 · 2013 · #20
Grade 8 geometry-2dProblem
A 1×2 rectangle is inscribed in a semicircle with the longer side on the diameter. What is the area of the semicircle?
Pick an answer.
AMC 8 2013 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 $1 \times 2$ rectangle sits inside a semicircle so that its longer side (length $2$) lies flat on the diameter. Find the area of the semicircle.
Givens: The rectangle has dimensions $1 \times 2$; The longer side ($2$) lies on the diameter of the semicircle; The two upper vertices of the rectangle touch the semicircular arc; Answer choices: (A) $\tfrac{\pi}{2}$, (B) $\tfrac{2\pi}{3}$, (C) $\pi$, (D) $\tfrac{4\pi}{3}$, (E) $\tfrac{5\pi}{3}$
Unknowns: The area of the semicircle
Understand
Restated: A $1 \times 2$ rectangle sits inside a semicircle so that its longer side (length $2$) lies flat on the diameter. Find the area of the semicircle.
Givens: The rectangle has dimensions $1 \times 2$; The longer side ($2$) lies on the diameter of the semicircle; The two upper vertices of the rectangle touch the semicircular arc; Answer choices: (A) $\tfrac{\pi}{2}$, (B) $\tfrac{2\pi}{3}$, (C) $\pi$, (D) $\tfrac{4\pi}{3}$, (E) $\tfrac{5\pi}{3}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
The problem is pure 2D geometry, so Tool #1 (Draw a Diagram) is the right opener: a quick sketch — semicircle, rectangle, center of diameter — exposes a right triangle whose hypotenuse is the radius. Tool #7 (Identify Subproblems) then splits the work into two clean steps: (1) find $r^2$ from that right triangle (Pythagorean theorem), and (2) plug $r^2$ into the semicircle area formula. We do not need algebra (Tool #13) because the rectangle's symmetry hands us the right triangle directly.
Execute — Answer: C
5.G.A.2 Step 1 - Draw the picture and place coordinates.
- Put the center of the diameter at the origin $(0,0)$.
- By symmetry the rectangle is centered on the origin, so its bottom corners are at $(-1, 0)$ and $(1, 0)$, and its top corners (which sit on the arc) are at $(-1, 1)$ and $(1, 1)$.
💡 Plotting the four corners on a coordinate grid is exactly the Grade 5 "graph points in the coordinate plane" skill, and it turns the geometry into something we can measure.
5.G.A.2 Step 2 - Spot the right triangle.
- The radius from the center $(0,0)$ to the top corner $(1,1)$ is the hypotenuse of a right triangle with horizontal leg $1$ (from $(0,0)$ to $(1,0)$) and vertical leg $1$ (from $(1,0)$ to $(1,1)$).
💡 Breaking the figure into the rectangle plus a right triangle with the radius is the Tool #7 "subproblem" move — the radius is hidden until we draw it in.
8.G.B.7 Step 3 - Use the Pythagorean theorem on that right triangle to get $r^2$ directly.
- We do not even need $r$ itself; the area formula uses $r^2$.
💡 Applying $a^2 + b^2 = c^2$ to a right triangle is the Grade 8 Pythagorean theorem standard, used here on a triangle whose legs were given by the rectangle's sides.
7.G.B.4 Step 4 - Plug $r^2 = 2$ into the semicircle area formula.
- A semicircle is half of a full circle, so its area is $\tfrac{1}{2} \pi r^2$.
💡 Using $A = \pi r^2$ for a circle (and halving it for a semicircle) is the Grade 7 area-of-a-circle standard; the $r^2$ from Step 3 drops straight in.
5.G.A.2 Draw the picture and place coordinates. Put the center of the diameter at the or 5.G.A.2 Spot the right triangle. The radius from the center $(0,0)$ to the top corner $( 8.G.B.7 Use the Pythagorean theorem on that right triangle to get $r^2$ directly. We do 7.G.B.4 Plug $r^2 = 2$ into the semicircle area formula. A semicircle is half of a full Review
Reasonableness: Sanity check the size. The diameter is $2r = 2\sqrt{2} \approx 2.83$, comfortably bigger than the rectangle's base of $2$ — good, the rectangle fits. The semicircle's area $\pi \approx 3.14$ must be larger than the rectangle's area $1 \times 2 = 2$, and it is ($3.14 > 2$). It must also be smaller than the bounding $2\sqrt{2} \times \sqrt{2} \approx 4$ rectangle that would contain the semicircle, and it is ($3.14 < 4$). The answer (C) $\pi$ sits in the right window.
Alternative: Tool #3 (Eliminate Possibilities) on the choices: every choice is a multiple of $\pi$, so really we are picking the value of $\tfrac{1}{2} r^2$. Choice (A) $\tfrac{\pi}{2}$ needs $r^2 = 1$ (radius $1$, but the rectangle alone is $2$ wide — impossible). Choice (E) $\tfrac{5\pi}{3}$ needs $r^2 = \tfrac{10}{3}$, far larger than the snug fit suggests. Only (C) $r^2 = 2$ matches a radius that just reaches the corner $(1,1)$.
CCSS standards used (min grade 8)
5.G.A.2Represent real world and mathematical problems by graphing points in the first quadrant of the coordinate plane (Placing the diameter's center at the origin and reading off the rectangle's corners as coordinate points $(\pm 1, 0)$ and $(\pm 1, 1)$, which exposes the hidden right triangle.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Computing $r^2 = 1^2 + 1^2 = 2$ from the right triangle whose legs are half the rectangle's base and the rectangle's height.)7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Applying $\tfrac{1}{2} \pi r^2$ with $r^2 = 2$ to get the semicircle's area $= \pi$.)
⭐ Draw the picture, find the right triangle hiding inside, and the Pythagorean theorem plus the circle-area formula do the rest — a Grade 8 idea applied to a tidy Grade 5 setup.
⭐ Draw the picture, find the right triangle hiding inside, and the Pythagorean theorem plus the circle-area formula do the rest — a Grade 8 idea applied to a tidy Grade 5 setup.
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