AMC 8 · 2002 · #20
Grade 7 geometry-2dProblem
The area of triangle XYZ is 8 square inches. Points A and B are midpoints of congruent segments XY and XZ. Altitude XC bisects YZ. The area (in square inches) of the shaded region is
Pick an answer.
AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Triangle $XYZ$ has area $8$ square inches. $A$ is the midpoint of $\overline{XY}$, $B$ is the midpoint of $\overline{XZ}$, and the altitude $\overline{XC}$ from $X$ bisects $\overline{YZ}$. Find the area of the shaded quadrilateral bounded by $\overline{YC}$ on the bottom, $\overline{XY}$ on the left, the segment from $A$ to the midpoint $D$ of $\overline{XC}$ on top, and $\overline{XC}$ on the right.
Givens: Area of $\triangle XYZ$ is $8$ square inches; $A$ is the midpoint of $\overline{XY}$ and $B$ is the midpoint of $\overline{XZ}$ (so $\overline{XY} \cong \overline{XZ}$); $\overline{XC}$ is an altitude from $X$ to $\overline{YZ}$ and $C$ is the midpoint of $\overline{YZ}$; Answer choices: (A) $1\tfrac{1}{2}$, (B) $2$, (C) $2\tfrac{1}{2}$, (D) $3$, (E) $3\tfrac{1}{2}$
Unknowns: The area of the shaded quadrilateral on the left half of the triangle
Understand
Restated: Triangle $XYZ$ has area $8$ square inches. $A$ is the midpoint of $\overline{XY}$, $B$ is the midpoint of $\overline{XZ}$, and the altitude $\overline{XC}$ from $X$ bisects $\overline{YZ}$. Find the area of the shaded quadrilateral bounded by $\overline{YC}$ on the bottom, $\overline{XY}$ on the left, the segment from $A$ to the midpoint $D$ of $\overline{XC}$ on top, and $\overline{XC}$ on the right.
Givens: Area of $\triangle XYZ$ is $8$ square inches; $A$ is the midpoint of $\overline{XY}$ and $B$ is the midpoint of $\overline{XZ}$ (so $\overline{XY} \cong \overline{XZ}$); $\overline{XC}$ is an altitude from $X$ to $\overline{YZ}$ and $C$ is the midpoint of $\overline{YZ}$; Answer choices: (A) $1\tfrac{1}{2}$, (B) $2$, (C) $2\tfrac{1}{2}$, (D) $3$, (E) $3\tfrac{1}{2}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #16 Change Focus / Count the Complement
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 $\triangle XYC$, then the small unshaded top triangle $\triangle XAD$ that sits inside it. Tool #1 (Draw a Diagram) tracks the symmetry: the altitude cuts $\triangle XYZ$ into two congruent halves, and the midpoint segment $\overline{AB}$ crosses $\overline{XC}$ at its midpoint $D$. Tool #16 (Count the Complement) finishes the job: the shaded area equals the half-triangle minus the small top triangle.
Execute — Answer: D
4.G.A.3 Step 1 - Use the altitude as a line of symmetry.
- Because $\overline{XC}$ is an altitude that also bisects $\overline{YZ}$, triangle $XYZ$ is isosceles and $\overline{XC}$ splits it into two congruent right triangles, $\triangle XYC$ and $\triangle XZC$.
- Each has half the area.
💡 Recognizing the line of symmetry that splits a figure into two congruent halves is the Grade 4 symmetry idea.
7.G.A.1 Step 2 - Locate the small unshaded triangle inside $\triangle XYC$.
- Let $D$ be the point where $\overline{AB}$ meets the altitude $\overline{XC}$.
- Since $A$ and $B$ are midpoints of $\overline{XY}$ and $\overline{XZ}$, segment $\overline{AB}$ is parallel to $\overline{YZ}$ and runs at half the height — so $D$ is the midpoint of $\overline{XC}$.
- Inside $\triangle XYC$, the unshaded piece is the small top triangle $\triangle XAD$, with $A$ a midpoint of $\overline{XY}$ and $D$ a midpoint of $\overline{XC}$.
💡 Naming the small top triangle $\triangle XAD$ turns the shaded quadrilateral into "big minus small," a Grade 7 scale-drawing setup.
7.G.A.1 Step 3 - Get the area of $\triangle XAD$ using the side-length ratio.
- $\triangle XAD$ and $\triangle XYC$ share the angle at $X$ and have sides in ratio $1:2$, so they are similar with linear scale factor $\tfrac{1}{2}$.
- The area scales by the square of the side ratio, so the small triangle has $\bigl(\tfrac{1}{2}\bigr)^2 = \tfrac{1}{4}$ the area of $\triangle XYC$.
💡 Halving every length quarters the area — the Grade 7 "scale factor squared" rule for similar figures.
7.G.B.6 Step 4 - Subtract the small triangle from the half-triangle.
- The shaded region is everything inside $\triangle XYC$ except the small top triangle $\triangle XAD$, so subtract its area.
💡 Computing a composite region as "whole minus the hole" is the Grade 7 area-of-composite-shape move.
4.G.A.3 Use the altitude as a line of symmetry. Because $\overline{XC}$ is an altitude t 7.G.A.1 Locate the small unshaded triangle inside $\triangle XYC$. Let $D$ be the point 7.G.A.1 Get the area of $\triangle XAD$ using the side-length ratio. $\triangle XAD$ and 7.G.B.6 Subtract the small triangle from the half-triangle. The shaded region is everyth Review
Reasonableness: Sanity check with coordinates: place $Y=(0,0)$, $Z=(10,0)$, $X=(5,4)$ so $\overline{YZ}=10$, height $=4$, and area $= \tfrac{1}{2}\cdot 10 \cdot 4 = 20$. To match the problem's area of $8$, scale all areas by $\tfrac{8}{20} = \tfrac{2}{5}$. 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 $\tfrac{1}{2}(5+2.5)(2) = 7.5$. Scaling: $7.5 \cdot \tfrac{2}{5} = 3$ square inches. Matches (D). Also, the small unshaded triangle is $\tfrac{1}{4}$ of the half-triangle, so the shaded region is $\tfrac{3}{4}$ of $4 = 3$ — the fraction $\tfrac{3}{8}$ of the whole triangle's area of $8$, which is again $3$.
Alternative: Tool #5 (Look for a Pattern) via fractions of the whole. The midpoint segment $\overline{AB}$ cuts $\triangle XYZ$ into a small top triangle $\triangle XAB$ (area $\tfrac{1}{4}$ of the whole, so $2$) and a trapezoid $AYZB$ (area $\tfrac{3}{4}$, so $6$). The altitude $\overline{XC}$ splits that trapezoid into two congruent pieces by symmetry, so the shaded half is $6 \div 2 = 3$. Same answer (D).
CCSS standards used (min grade 7)
4.G.A.3Recognize a line of symmetry for a two-dimensional figure as a line across the figure such that the figure can be folded along the line into matching parts (Using altitude $\overline{XC}$ as a line of symmetry of the isosceles $\triangle XYZ$ to split it into two congruent halves, so $\triangle XYC$ has area $4$.)7.G.A.1Solve problems involving scale drawings of geometric figures, including computing actual lengths and areas from a scale drawing and reproducing a scale drawing at a different scale (Recognizing $\triangle XAD$ as a scaled copy of $\triangle XYC$ with side ratio $1{:}2$, so area ratio $1{:}4$ gives area $1$.)7.G.B.6Solve real-world and mathematical problems involving area, volume and surface area of two- and three-dimensional objects composed of triangles, quadrilaterals, polygons, cubes, and right prisms (Computing the shaded quadrilateral as the half-triangle minus the small top triangle: $4 - 1 = 3$.)
⭐ 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.
⭐ 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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