AMC 10 · 2023 · #3
Grade 7 geometry-2dProblem
A 3−4−5 right triangle is inscribed in circle A, and a 5−12−13 right triangle is inscribed in circle B. What is the ratio of the area of circle A to the area of circle B?
Pick an answer.
AMC 10 2023 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 $3$-$4$-$5$ right triangle is inscribed in circle $A$, and a $5$-$12$-$13$ right triangle is inscribed in circle $B$. Find $\frac{\text{area of } A}{\text{area of } B}$.
Givens: Triangle in circle $A$: right triangle with sides $3, 4, 5$; Triangle in circle $B$: right triangle with sides $5, 12, 13$; Each triangle is *inscribed* — all three vertices lie on its circle; Answer choices: (A) $\frac{9}{25}$, (B) $\frac{1}{9}$, (C) $\frac{1}{5}$, (D) $\frac{25}{169}$, (E) $\frac{4}{25}$
Unknowns: The ratio of the two circle areas
Understand
Restated: A $3$-$4$-$5$ right triangle is inscribed in circle $A$, and a $5$-$12$-$13$ right triangle is inscribed in circle $B$. Find $\frac{\text{area of } A}{\text{area of } B}$.
Givens: Triangle in circle $A$: right triangle with sides $3, 4, 5$; Triangle in circle $B$: right triangle with sides $5, 12, 13$; Each triangle is *inscribed* — all three vertices lie on its circle; Answer choices: (A) $\frac{9}{25}$, (B) $\frac{1}{9}$, (C) $\frac{1}{5}$, (D) $\frac{25}{169}$, (E) $\frac{4}{25}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
A quick sketch (Tool #1) of each circle with its right triangle makes the key fact pop out: the $90^\circ$ vertex sits on the circle and the hypotenuse goes straight across, so the hypotenuse is a diameter. Once that is seen, the problem breaks into three easy subproblems (Tool #7): (a) read off each diameter, (b) write each area in terms of its diameter, (c) form the ratio and cancel. No algebra heavier than squaring a fraction is needed.
Execute — Answer: D
7.G.B.4 Step 1 - Sketch circle $A$ with the $3$-$4$-$5$ triangle inscribed.
- The right angle is opposite the longest side.
- A standard inscribed-angle fact says any inscribed right angle's opposite side is a diameter — so the hypotenuse of length $5$ is the diameter of circle $A$.
💡 Drawing the triangle on the circle shows the hypotenuse stretches from one side of the circle to the other — that's the diameter.
7.G.B.4 Step 2 - Same picture for circle $B$ with the $5$-$12$-$13$ triangle.
- The hypotenuse $13$ is again the diameter, by the same inscribed-right-angle reasoning.
💡 Same drawing argument, different right triangle — the hypotenuse becomes the diameter again.
7.G.B.4 Step 3 - Write each area in terms of its diameter.
- Since $r = d/2$, the area $\pi r^2$ equals $\pi (d/2)^2 = \frac{\pi d^2}{4}$.
- Now form the ratio: $\pi$ and the $4$ cancel, leaving the square of the diameter ratio.
💡 When two circles' areas are compared, only the diameter ratio matters — the $\pi$ and the $4$ are the same on both.
6.EE.A.1 Step 4 Substitute the two diameters and square.
💡 Squaring $\tfrac{5}{13}$ squares the top and bottom separately — a Grade 6 exponent rule.
7.G.B.4 Sketch circle $A$ with the $3$-$4$-$5$ triangle inscribed. The right angle is op 7.G.B.4 Same picture for circle $B$ with the $5$-$12$-$13$ triangle. The hypotenuse $13$ 7.G.B.4 Write each area in terms of its diameter. Since $r = d/2$, the area $\pi r^2$ eq 6.EE.A.1 Substitute the two diameters and square. Review
Reasonableness: Sanity-check magnitudes. Diameter $5$ vs diameter $13$ means circle $A$ is much smaller than $B$, so the ratio should be well below $1$ — and $\frac{25}{169} \approx 0.148$ fits. Among the choices, the only ones below $\frac{1}{4}$ are (B) $\frac{1}{9} \approx 0.111$, (C) $\frac{1}{5} = 0.2$, (D) $\frac{25}{169} \approx 0.148$, (E) $\frac{4}{25} = 0.16$. Only (D) matches the exact squared diameter ratio.
Alternative: Tool #13 (Convert to Algebra) computing radii directly: $r_A = 5/2$, $r_B = 13/2$. Areas: $\pi (5/2)^2 = \frac{25\pi}{4}$ and $\pi (13/2)^2 = \frac{169\pi}{4}$. Ratio: $\frac{25\pi/4}{169\pi/4} = \frac{25}{169}$ — same answer with a touch more arithmetic.
CCSS standards used (min grade 7)
6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Squaring the diameter ratio $\left(\frac{5}{13}\right)^2 = \frac{25}{169}$.)7.G.B.4Know the formulas for area and circumference of a circle (Writing each area as $\frac{\pi d^2}{4}$ and forming the area ratio.)
⭐ This AMC 10 problem only needs the Grade 7 circle-area formula — a right triangle's hypotenuse is the circle's diameter, so the area ratio is just $\left(\tfrac{5}{13}\right)^2$.
⭐ This AMC 10 problem only needs the Grade 7 circle-area formula — a right triangle's hypotenuse is the circle's diameter, so the area ratio is just $\left(\tfrac{5}{13}\right)^2$.