AMC 10 · 2017 · #22

Grade 8 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
similar-trianglespythagorean-theoreminscribed-anglearea-triangles identify-subproblemsconvert-to-algebra ↑ Prerequisites: similar-triangles
📏 Long solution 💡 3 insights
Problem
A circle has radius 2, so its diameter AB has length 4. The diameter is extended past B to a point D with BD=3. A point E sits so that segment ED has length 5 and is perpendicular to line AD. Segment AE crosses the circle at a point C between A and E. Find the area of △ ABC.

Pick an answer.

(A)
$\frac{120}{37}$
(B)
$\frac{140}{39}$
(C)
$\frac{145}{39}$
(D)
$\frac{140}{37}$
(E)
$\frac{120}{31}$

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Tool #1 (Draw a Diagram): the problem stacks a circle, a stretched diameter, and a perpendicular segment, so a labeled picture is the only way to see how the pieces line up. Tool #7 (Identify Subproblems): I split the work into 'find the big right triangle ADE', 'notice the small right triangle ACB hiding inside it', and 'turn the matching shapes into lengths'. Tool #4 (Introduce a Variable): I name the two legs AC and CB of the target triangle and let the similar triangles pin their lengths.

1STEP 1

Set up the figure

Radius 2 gives diameter AB=4; extend past B by BD=3 so AD=7, then draw ED=5 perpendicular to AD, marking the right angle at D.

AD = AB + BD = 4 + 3 = 7, ∠ ADE = 90°
2STEP 2

Find AE

In right triangle ADE with legs AD=7 and ED=5, Pythagoras gives AE=√(49+25)=√(74).

AE = √(AD² + ED²) = √(49 + 25) = √(74)
3STEP 3

Spot the similar triangles

AB is a diameter, so ∠ACB=90° like ∠ADE; sharing ∠A gives △ ACB ∼ △ ADE by AA.

∠ ACB = ∠ ADE = 90°, ∠ A shared → △ ACB ∼ △ ADE
4STEP 4

Scale the legs

The similarity ratio ABAE\frac{AB}{AE}=4√(74)\frac{4}{√(74)} scales the legs: AC=28√(74)\frac{28}{√(74)} and CB=20√(74)\frac{20}{√(74)}.

AC = 28√(74)\frac{28}{√(74)}, CB = 20√(74)\frac{20}{√(74)}
5STEP 5

Compute the area

The right angle at C makes AC, CB the base and height, so area=½·28√(74)\frac{28}{√(74)}·20√(74)\frac{20}{√(74)}=14037\frac{140}{37}, choice (D).

[△ ABC] = 12\frac{1}{2}·28√(74)\frac{28}{√(74)}·20√(74)\frac{20}{√(74)} = 560148\frac{560}{148} = 14037\frac{140}{37} → (D)
Answer
14037\frac{140}{37}
The two legs AC=28√(74)\frac{28}{√(74)}≈ 3.25 and CB=20√(74)\frac{20}{√(74)}≈ 2.32 are both shorter than the hypotenuse AB=4, as they must be inside a circle of diameter 4. Their product halved gives about 3.78, and 14037\frac{140}{37}≈ 3.78, so the size is sensible. The legs also satisfy AC²+CB² = (784+400)74\frac{(784+400)}{74} = 118474\frac{1184}{74} = 16 = AB², confirming the right angle at C.
💡Key takeaway

Any triangle whose longest side is a diameter has a right angle, so the little triangle ABC is just a scaled copy of the big right triangle ADE — match the sides, then take half of leg times leg to get area 14037\frac{140}{37}, choice (D).

  • Set up the figure
  • Find AE
  • Spot the similar triangles
  • Scale the legs
  • Compute the area

More like this

Same archetype — closest grade level first.