Competition · AMC preparation · step 4 of 4

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 = 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

Pick a friendly side length

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 where the lines meet

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 = s/3

Marking F's coordinates is just locating an intersection point on the grid — a Grade 5 graphing problem.

The point F, where diagonal AC crosses segment BE, sits at height s/3 above the bottom side CD — one third of the way up the square.

▸ Why?

F is the shared tip where two triangles meet point-to-point — the big triangle FAB hanging from the top side AB and the small triangle FCE standing on the bottom piece CE — and these two triangles are similar, so their matching lengths, including the two heights measured from F to the two bases, are locked into one and the same ratio.

▸ Why?

The two triangles have equal angles: AB and CE lie on the square's parallel top and bottom sides, so the diagonal AC and the segment BE cross them as transversals and make equal alternate angles.

A transversal of parallel lines makes equal anglesA line crossing two parallel lines makes matching corresponding and alternate angles equal.

▸ Why?

Two triangles with equal angles are similar, and that similarity locks every pair of corresponding lengths — the full bases AB and CE and the two heights dropped from F alike — into a single fixed ratio k.

Similar triangles have proportional sidesTriangles with equal angles are similar: their corresponding sides are in a fixed ratio k, and their areas in ratio k².

▸ Why?

That shared ratio is 2 to 1, because the big triangle's top side AB is a whole side of the square while the small triangle's base CE is only half of the bottom side — E is its midpoint.

▸ Why?

CE is half of the bottom side because the midpoint E splits CD into two equal pieces that together make the whole side, and in a square that side is the same length as AB.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

With the heights from F in the ratio 2 to 1, F divides the square's full height into 2 equal parts above it and 1 equal part below, so F sits one part — a third of the height, s/3 — up from CD.

▸ Why?

The height of the big triangle and the height of the small triangle exactly fill the square's side from CD up to AB with no gap or overlap, so those 2 parts and 1 part are 3 equal parts of one whole side.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

5.G.A.2Draw A Diagram

3STEP 3

Split the region in two

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

Compute the two pieces

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

Scale back to a ratio

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

[AFED]/([□ ABCD]) = 15/36 = 5/12 → [□ ABCD] = 45 · 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!

Pick a friendly side length
Find where the lines meet
Split the region in two
Compute the two pieces
Scale back to a ratio

A parent dashboard for the family lives at sensimlab.com.

Pick a friendly side length

Find where the lines meet

Split the region in two

Compute the two pieces

Scale back to a ratio

More like this