AMC 10 · 2023 · #3
Grade 7 geometry-2dPick an answer.
A quick sketch (Tool #1) of each circle with its right triangle makes the key fact pop out: the 90° 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.
The first circle's diameter
The hypotenuse is the diameter.
Drawing the triangle on the circle shows the hypotenuse stretches from one side of the circle to the other — that's the diameter.
A right triangle drawn on a circle has its hypotenuse stretching across as a diameter.
▸ Why?
The midpoint of the hypotenuse is equally far from all three corners, so it is the centre.
▸ Why?
That equal distance is exactly half the hypotenuse, which the right angle guarantees.
The second circle's diameter
The second works the same way.
Same drawing argument, different right triangle — the hypotenuse becomes the diameter again.
7.G.B.4Draw A DiagramWrite the area ratio
The area ratio is the square of the diameter ratio.
When two circles' areas are compared, only the diameter ratio matters — the π and the 4 are the same on both.
7.G.B.4Identify SubproblemsCompute it
Computing gives twenty-five over one hundred sixty-nine.
Squaring 5/13 squares the top and bottom separately — a Grade 6 exponent rule.
6.EE.A.1Identify SubproblemsThis AMC 12 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 (5/13)².
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