AMC 10 · 2018 · #4

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

chord-perpendicular-from-centerpythagorean-theoremarea-circlesisosceles-right-triangle identify-subproblemssymmetry-argument ↑ Prerequisites: pythagorean-theoremarea-circles

📏 Short solution 💡 2 insights

Problem

Inside some circle sits a chord — a straight segment whose two endpoints are on the circle — and that chord is 10 units long. The shortest distance from the circle's centre to that chord is 5 units. Neither the radius nor the diameter is given. From just those two lengths, find the area of the whole circle.

Pick an answer.

(A)

$25\pi$

(B)

$50\pi$

(C)

$75\pi$

(D)

$100\pi$

(E)

$125\pi$

How to solve

Strategy Draw a Diagram

Tool #1 (Draw a Diagram): the two given numbers describe segments that are not drawn yet. Sketching the circle, the chord, the perpendicular from the center, and the two radii to the chord's endpoints turns the word problem into a picture where a right triangle appears on its own — that triangle is the entire problem. Tool #7 (Identify Subproblems): the target is the area, but the area formula needs the radius, so the real work splits into a small first job (get the radius from the right triangle) and a trivial second job (plug into π r²). Tool #16 (Change Focus): the area formula uses r², not r, so we can stop the moment the Pythagorean theorem hands us r² and never bother taking a square root.

1STEP 1

Draw the picture the words describe

Draw the right triangle the words describe.

AB=10, OM=5, OM ⊥ AB, OA=OB=r

2STEP 2

The perpendicular cuts the chord in half

The perpendicular cuts the chord in half.

AM=BM=10/2=5

3STEP 3

Pythagoras on the half-picture

Pythagoras gives the radius squared.

r² = OM² + AM² = 5² + 5² = 25 + 25 = 50

4STEP 4

Stop at r² and take the area

Stopping there, the area is fifty pi.

Area = π r² = π · 50 = 50π → (B)

Answer

50π

Sanity-check the size. The chord alone spans 10, so the diameter is at least 10 and the radius is at least 5; that already rules out any area below 25π, and 25π itself would mean r=5, which would put the chord through the center at distance 0, not 5. On the other side, the chord's endpoint is 5 across and 5 up from the center, so r=√(50)≈ 7.07, comfortably between 5 and the 10 that would give 100π. So 50π sits in the only plausible range. A second check: because OM=AM=5, △ OMA is an isosceles right triangle, so r = 5√(2) and r² = (5√(2))² = 25 · 2 = 50 — the same 50 from a different route. The answer is (B) 50π.

💡Key takeaway

Half the chord, the distance from the center, and the radius form a right triangle — so r² = 5²+5² = 50, and the area π r² is 50π without ever finding r itself.

Draw the picture the words describe
The perpendicular cuts the chord in half
Pythagoras on the half-picture
Stop at r² and take the area

Draw the picture the words describe

The perpendicular cuts the chord in half

Pythagoras on the half-picture

Stop at r² and take the area

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