AMC 10 · 2022 · #16

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

pythagorean-theoremsimilar-trianglescoordinate-geometryarea-trianglesarea-rectangles identify-subproblemsarea-differencecomplementary-counting ↑ Prerequisites: pythagorean-theoremsimilar-triangles

📏 Long solution 💡 3 insights 📊 Diagram

Problem

A 4 × 8 rectangle contains a square of side 5. Three corners of the square sit on three different sides of the rectangle (bottom, right, top). Find the area of the region that lies inside both shapes.

Pick an answer.

(A)

$15\dfrac{1}{8}$

(B)

$15\dfrac{3}{8}$

(C)

$15\dfrac{1}{2}$

(D)

$15\dfrac{5}{8}$

(E)

$15\dfrac{7}{8}$

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Diagram): drop the picture on coordinates so the rectangle is [0,8] × [0,4]. Tool #7 (Subproblems): the two right triangles formed by the square's slanted sides and the bottom/right edges of the rectangle are both 3-4-5 — that fixes every vertex coordinate. Tool #16 (Complement): instead of computing the overlap directly (a pentagon), compute the whole square (25) and subtract the small triangle that pokes above the rectangle's top edge.

1STEP 1

Place the rectangle as (0,0)–(8,4); the square's on-side corners are A (top), B (bottom), C (right), right angle at B, so AB = BC = 5.

AB = BC = 5, ∠ ABC = 90°

2STEP 2

Drop a perpendicular from A to the bottom edge: vertical leg 4, hypotenuse 5, so the horizontal leg is 3 — a 3-4-5 triangle.

A'B = √(25 - 16) = 3

3STEP 3

By symmetry the twin triangle at B is also 3-4-5, so A = (1, 4), B = (4, 0), C = (8, 3).

A = (1, 4), B = (4, 0), C = (8, 3)

4STEP 4

Because ABCD is a square, D = (5, 7); its height 7 pokes 3 above the top edge y = 4.

D = (1+4, 4+3) = (5, 7)

5STEP 5

Side CD (from C(8,3) to D(5,7), slope - $\frac{4}{3}$ ) meets the top edge y = 4 at G = ( $\frac{29}{4}$ , 4).

G = (

\frac{29}{4}

, 4)

6STEP 6

The escaped piece is triangle A-G-D: base $\frac{25}{4}$ , height 3, area = $\frac{1}{2}$ · $\frac{25}{4}$ · 3 = $\frac{75}{8}$ .

outside area =

\frac{1}{2}

\frac{25}{4}

· 3 =

\frac{75}{8}

7STEP 7

Complement: overlap = square - escaped triangle = 25 - $\frac{75}{8}$ = $\frac{125}{8}$ = 15 $\frac{5}{8}$ , choice (D).

25 -

\frac{75}{8}

\frac{125}{8}

= 15

\frac{5}{8}

Answer

\frac{5}{8}

The overlap must be less than the square (25) and less than the rectangle (32). 15

\frac{5}{8}

sits comfortably below both. The spilled triangle has area

\frac{75}{8}

≈ 9.4, which feels right for a triangle of base ≈ 6.25 and height 3. Also, the answer choices are tightly bunched between 15

\frac{1}{8}

and 15

\frac{7}{8}

— only careful arithmetic on the eighth (

\frac{75}{8}

) picks the right one.

💡Key takeaway

Two hidden 3-4-5 right triangles fix every corner of the square. The square's top corner (5, 7) pokes above the rectangle, so chop off that little triangle — area $\frac{75}{8}$ — and the leftover is 25 - $\frac{75}{8}$ = 15 $\frac{5}{8}$ , choice (D).

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