AMC 8 · 2002 · #20

Grade 7 geometry-2d

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

area-trianglessimilar-trianglessimilar-figuresfraction-arithmetic area-differenceidentify-subproblems ↑ Prerequisites: area-trianglesfraction-arithmetic

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Triangle XYZ has area 8 square inches. A is the midpoint of XY, B is the midpoint of XZ, and the altitude XC from X bisects YZ. Find the area of the shaded quadrilateral bounded by YC on the bottom, XY on the left, the segment from A to the midpoint D of XC on top, and XC on the right.

Pick an answer.

(A)

$1\frac{1}2$

(B)

(C)

$2\frac{1}2$

(D)

(E)

$3\frac{1}2$

AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The shaded quadrilateral is awkward to attack head-on, so Tool #7 (Identify Subproblems) splits the job into two clean pieces: first the left half-triangle △ XYC, then the small unshaded top triangle △ XAD that sits inside it. Tool #1 (Draw a Diagram) tracks the symmetry: the altitude cuts △ XYZ into two congruent halves, and the midpoint segment AB crosses XC at its midpoint D. Tool #16 (Count the Complement) finishes the job: the shaded area equals the half-triangle minus the small top triangle.

1STEP 1

XC is an altitude that also bisects YZ, so △ XYZ is isosceles and XC halves it into two congruent right triangles, each of area 4.

Area(△ XYC) =

\frac{1}{2}

· 8 = 4 sq in

2STEP 2

AB joins the midpoints, so it is parallel to YZ and meets XC at its midpoint D; the unshaded piece of △ XYC is the small top triangle △ XAD.

XA =

\frac{1}{2}

XY, XD =

\frac{1}{2}

3STEP 3

△ XAD and △ XYC share the X-angle with sides in ratio 1:2, so the area scales by ( $\frac{1}{2}$ )² = $\frac{1}{4}$ and △ XAD has area 1 sq in.

Area(△ XAD) =

\frac{1}{4}

· 4 = 1 sq in

4STEP 4

The shaded quadrilateral is △ XYC minus the small top triangle △ XAD, giving 4 - 1 = 3 sq in.

Shaded area = Area(△ XYC) - Area(△ XAD) = 4 - 1 = 3 → (D)

Answer

Sanity check with coordinates: place Y=(0,0), Z=(10,0), X=(5,4) so YZ=10, height =4, and area =

\frac{1}{2}

· 10 · 4 = 20. To match the problem's area of 8, scale all areas by

\frac{8}{20}

\frac{2}{5}

. In these coordinates A=(2.5,2), C=(5,0), D=(5,2), so the shaded quadrilateral has vertices Y=(0,0), A=(2.5,2), D=(5,2), C=(5,0) — a trapezoid with parallel sides YC=5 and AD=2.5 and height 2, giving area

\frac{1}{2}

(5+2.5)(2) = 7.5. Scaling: 7.5 ·

\frac{2}{5}

= 3 square inches. Matches (D). Also, the small unshaded triangle is

\frac{1}{4}

of the half-triangle, so the shaded region is

\frac{3}{4}

of 4 = 3 — the fraction

\frac{3}{8}

of the whole triangle's area of 8, which is again 3.

💡Key takeaway

Half the triangle, minus a small corner — the line of symmetry gives the 4, the half-scale similar triangle gives the 1, and 4 - 1 = 3 is the shaded area.

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