AMC 8 · 2002 · #20

Grade 7 geometry-2d

Archetype Compound Area by Decomposition / Difference

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

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$ . Altitude $\overline{XC}$ bisects $\overline{YZ}$ . The area (in square inches) of the shaded region is

Pick an answer.

(A)

$1rac{1}2$

(B)

(C)

$2rac{1}2$

(D)

(E)

$3rac{1}2$

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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

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

#1 Draw a Diagram 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.

$$\text{Area}(\triangle XYC) = \tfrac{1}{2} \cdot 8 = 4 \text{ sq in}$$

💡 Recognizing the line of symmetry that splits a figure into two congruent halves is the Grade 4 symmetry idea.

#7 Identify Subproblems 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 $\overline{XC}$.

$$XA = \tfrac{1}{2} XY, \quad XD = \tfrac{1}{2} XC$$

💡 Naming the small top triangle $\triangle XAD$ turns the shaded quadrilateral into "big minus small," a Grade 7 scale-drawing setup.

#7 Identify Subproblems 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 $\triangle XYC$.

$$\text{Area}(\triangle XAD) = \tfrac{1}{4} \cdot 4 = 1 \text{ sq in}$$

💡 Halving every length quarters the area — the Grade 7 "scale factor squared" rule for similar figures.

#16 Change Focus / Count the Complement 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.

$$\text{Shaded area} = \text{Area}(\triangle XYC) - \text{Area}(\triangle XAD) = 4 - 1 = 3 \;\Rightarrow\; \textbf{(D)}$$

💡 Computing a composite region as "whole minus the hole" is the Grade 7 area-of-composite-shape move.

[1] #1 4.G.A.3 Use the altitude as a line of symmetry. Because $\overline{XC}$ is an altitude t

[2] #7 7.G.A.1 Locate the small unshaded triangle inside $\triangle XYC$. Let $D$ be the point

[3] #7 7.G.A.1 Get the area of $\triangle XAD$ using the side-length ratio. $\triangle XAD$ and

[4] #16 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.3 Recognize 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.1 Solve 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.6 Solve 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 7)

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