AMC 8 · 2013 · #23
Grade 8 geometry-2dProblem
Angle ABC of △ABC is a right angle. The sides of △ABC are the diameters of semicircles as shown. The area of the semicircle on AB equals 8π, and the arc of the semicircle on AC has length 8.5π. What is the radius of the semicircle on BC?
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: Right triangle $\triangle ABC$ has its right angle at $B$. Each side is the diameter of a semicircle drawn outward. The semicircle on $\overline{AB}$ has area $8\pi$, and the semicircle on $\overline{AC}$ has arc length $8.5\pi$. Find the radius of the semicircle on $\overline{BC}$.
Givens: $\angle ABC = 90^\circ$, so $\overline{AC}$ is the hypotenuse; Each side of $\triangle ABC$ is the diameter of a semicircle on that side; Area of the semicircle on $\overline{AB}$ $= 8\pi$; Arc length of the semicircle on $\overline{AC}$ $= 8.5\pi$; Answer choices: (A) $7$, (B) $7.5$, (C) $8$, (D) $8.5$, (E) $9$
Unknowns: The radius of the semicircle on $\overline{BC}$
Understand
Restated: Right triangle $\triangle ABC$ has its right angle at $B$. Each side is the diameter of a semicircle drawn outward. The semicircle on $\overline{AB}$ has area $8\pi$, and the semicircle on $\overline{AC}$ has arc length $8.5\pi$. Find the radius of the semicircle on $\overline{BC}$.
Givens: $\angle ABC = 90^\circ$, so $\overline{AC}$ is the hypotenuse; Each side of $\triangle ABC$ is the diameter of a semicircle on that side; Area of the semicircle on $\overline{AB}$ $= 8\pi$; Arc length of the semicircle on $\overline{AC}$ $= 8.5\pi$; Answer choices: (A) $7$, (B) $7.5$, (C) $8$, (D) $8.5$, (E) $9$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
The problem describes a figure, so Tool #1 (Draw a Diagram) — sketch the right triangle and label each side with its semicircle — makes the chain $\text{semicircle measurement} \to \text{radius} \to \text{diameter} \to \text{side of triangle}$ visible. Tool #7 (Identify Subproblems) then splits the work into three clean pieces: (1) recover $AB$ from the area, (2) recover $AC$ from the arc length, (3) use the Pythagorean theorem to get $BC$ and halve it for the radius. No algebra heavier than $r^2 = 16$ is needed.
Execute — Answer: B
7.G.B.4 Step 1 - Sketch the figure: right triangle $\triangle ABC$ with the right angle at $B$, plus a semicircle on each side.
- Label the radius of each semicircle $r_{AB}$, $r_{AC}$, $r_{BC}$.
- Since each side IS the diameter, the side length equals $2r$.
💡 Drawing the diagram and writing each side as $2r$ makes it obvious what we need from each semicircle: just its radius.
7.G.B.4 Step 2 - Use the semicircle area formula on $\overline{AB}$.
- A full circle has area $\pi r^2$, so a semicircle has area $\tfrac{1}{2}\pi r^2$.
- Setting this equal to $8\pi$ gives $r_{AB}^2 = 16$, so $r_{AB} = 4$ and $AB = 2 \cdot 4 = 8$.
💡 Inverting the Grade 7 circle-area formula recovers the radius, and doubling gives the side length — a one-step subproblem.
7.G.B.4 Step 3 - Use the semicircle arc-length formula on $\overline{AC}$.
- A full circle's circumference is $2\pi r$, so a semicircle's arc is half of that: $\pi r$.
- Setting $\pi r_{AC} = 8.5\pi$ gives $r_{AC} = 8.5$ and $AC = 2 \cdot 8.5 = 17$.
💡 Same Grade 7 formula family — arc length instead of area — gives the hypotenuse directly.
8.G.B.7 Step 4 - Apply the Pythagorean theorem with $AB$ and $AC$ in hand.
- $\overline{AC}$ is the hypotenuse (it sits opposite the right angle at $B$), so $AB^2 + BC^2 = AC^2$.
- With $AB = 8$ and $AC = 17$ this is the famous $8\text{-}15\text{-}17$ Pythagorean triple, giving $BC = 15$.
💡 Recognizing $8$-$15$-$17$ as a Pythagorean triple skips the square-root step — pattern recognition at Grade 8 level.
7.G.B.4 Step 5 - Finally, halve $BC$ to get the radius of its semicircle.
- Since $BC$ is the diameter, $r_{BC} = BC/2 = 15/2 = 7.5$.
💡 The very last subproblem — diameter divided by two — closes the loop we set up in Step 1.
7.G.B.4 Sketch the figure: right triangle $\triangle ABC$ with the right angle at $B$, p 7.G.B.4 Use the semicircle area formula on $\overline{AB}$. A full circle has area $\pi 7.G.B.4 Use the semicircle arc-length formula on $\overline{AC}$. A full circle's circum 8.G.B.7 Apply the Pythagorean theorem with $AB$ and $AC$ in hand. $\overline{AC}$ is the 7.G.B.4 Finally, halve $BC$ to get the radius of its semicircle. Since $BC$ is the diame Review
Reasonableness: Check the triangle sides are consistent: $8^2 + 15^2 = 64 + 225 = 289 = 17^2$ ✓ — the famous $8\text{-}15\text{-}17$ right triangle. The semicircle on $\overline{BC}$ should sit between the other two in size: $r_{AB} = 4$, $r_{BC} = 7.5$, $r_{AC} = 8.5$, and indeed $4 < 7.5 < 8.5$. Choice (B) lands in the expected range and equals exactly half of $15$.
Alternative: Tool #3 (Eliminate Possibilities) on the choices: each candidate $r_{BC}$ gives $BC = 2r_{BC}$, so we test whether $AB^2 + BC^2 = AC^2$ holds with $AB = 8$, $AC = 17$. (A) $r=7$ gives $BC=14$, $8^2 + 14^2 = 260 \neq 289$. (B) $r=7.5$ gives $BC=15$, $8^2 + 15^2 = 289 = 17^2$ ✓. (C)-(E) give $BC = 16, 17, 18$, all too large. Only (B) satisfies the Pythagorean check.
CCSS standards used (min grade 8)
7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Inverting the semicircle area formula $\tfrac{1}{2}\pi r^2 = 8\pi$ to get $r_{AB} = 4$, inverting the semicircle arc length $\pi r = 8.5\pi$ to get $r_{AC} = 8.5$, and halving $BC$ to get $r_{BC}$.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Solving $8^2 + BC^2 = 17^2$ for $BC = 15$, the missing leg of the right triangle (the classic $8$-$15$-$17$ Pythagorean triple).)
⭐ Each side is a diameter, so each semicircle's measurement (area or arc length) hands you a side of the triangle — then the Pythagorean theorem finishes the job. It's a tidy Grade 8 idea built from two Grade 7 circle formulas.
⭐ Each side is a diameter, so each semicircle's measurement (area or arc length) hands you a side of the triangle — then the Pythagorean theorem finishes the job. It's a tidy Grade 8 idea built from two Grade 7 circle formulas.
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