AMC 8 · 2018 · #22

Grade 6 geometry-2d

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

similar-trianglesarea-trianglesratio-proportionarea-rectangles area-differenceidentify-subproblemsconvert-to-algebra ↑ Prerequisites: similar-trianglesarea-triangles

📏 Long solution 💡 4 insights 📊 Diagram

Problem

In square ABCD, point E is the midpoint of side CD, and segment BE crosses diagonal AC at point F. The quadrilateral AFED (bounded by AF, FE, ED, DA) has area 45. Find the area of the whole square.

Pick an answer.

(A)

100

(B)

108

(C)

120

(D)

135

(E)

144

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The figure is already drawn, but Tool #1 (Draw a Diagram) tells us to add to it — mark E as a midpoint, label CE = ED = $\frac{s}{2}$ , and notice the small triangle △ FCE tucked inside the larger right triangle △ ACD. That picture suggests Tool #7 (Identify Subproblems): the awkward quadrilateral AFED equals the easy half-square triangle △ ACD minus the small triangle △ FCE, so we only need to find the small triangle. To find F without algebra, we use Tool #9 (Solve an Easier Related Problem): pick a concrete easy side length (say s=6, since E is a midpoint and △ ACD is right-angled) and discover how far below AB the point F sits. The ratio we find scales to any s.

1STEP 1

Drop the square on a grid (Tool #9): with s = 6 every coordinate is whole — E=(3,0), so CE = ED = 3.

s = 6, CE = ED = 3

2STEP 2

Find F where AC meets BE: the diagonal is x + y = 6 and line BE is y = 2(x - 3); solving gives F = (4, 2).

F = (4, 2), height of F above CD is y_F = 2 =

\frac{s}{3}

3STEP 3

Tool #7: AC halves the square into △ ACD, and AFED is that triangle with only the corner triangle △ FCE removed.

[AFED] = [△ ACD] - [△ FCE]

4STEP 4

In the s = 6 square, [△ ACD] = $\frac{1}{2}$ ·6·6 = 18 and [△ FCE] = $\frac{1}{2}$ ·3·2 = 3, so [AFED] = 15.

[△ ACD] = 18, [△ FCE] = 3, [AFED] = 18 - 3 = 15

5STEP 5

In the test square $\frac{[AFED]}{[□ ABCD]}$ = $\frac{15}{36}$ = $\frac{5}{12}$ , a ratio independent of s; so $\frac{5}{12}$ · area = 45 gives area = 108 → (B).

\frac{[AFED]}{([□ ABCD])}

\frac{15}{36}

\frac{5}{12}

→ [□ ABCD] = 45 ·

\frac{12}{5}

= 108 → (B)

Answer

108

AFED takes up roughly a third of the square in the picture (a bit less than the half-square △ ACD), so the square's area should be a bit less than 3 × 45 = 135. The computed value 108 is in that range and matches

\frac{12}{5}

× 45. Also, 108 is the only answer choice equal to

\frac{12}{5}

of 45, since

\frac{12}{5}

· 45 = 12 · 9 = 108 — a clean whole number, as expected for a contest problem.

💡Key takeaway

This AMC 8 problem only needs Grade 6 ratio reasoning and the triangle area formula you already know!

More like this