Competition · AMC preparation · step 4 of 4

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)
2
(C)
$2\frac{1}2$
(D)
3
(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

Halve along the altitude

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) = 1/2 · 8 = 4 sq in
2STEP 2

Locate the small triangle

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 = 1/2 XY, XD = 1/2 XC
3STEP 3

Find the small triangle's area

△ XAD and △ XYC share the X-angle with sides in ratio 1:2, so the area scales by (12\frac{1}{2})² = 14\frac{1}{4} and △ XAD has area 1 sq in.

Area(△ XAD) = 1/4 · 4 = 1 sq in
4STEP 4

Subtract the small triangle

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
3
Sanity check with coordinates: place Y=(0,0), Z=(10,0), X=(5,4) so YZ=10, height =4, and area = 12\frac{1}{2} · 10 · 4 = 20. To match the problem's area of 8, scale all areas by 820\frac{8}{20} = 25\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 12\frac{1}{2}(5+2.5)(2) = 7.5. Scaling: 7.5 · 25\frac{2}{5} = 3 square inches. Matches (D). Also, the small unshaded triangle is 14\frac{1}{4} of the half-triangle, so the shaded region is 34\frac{3}{4} of 4 = 3 — the fraction 38\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.

  • Halve along the altitude
  • Locate the small triangle
  • Find the small triangle's area
  • Subtract the small triangle

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