AMC 8 · 2000 · #6

Grade 3 geometry-2d

Archetype Compound Area by Decomposition / Difference

area-rectanglesspatial-visualizationline-symmetry area-differenceidentify-subproblems ↑ Prerequisites: area-rectangles

📏 Short solution 💡 2 insights 📊 Diagram

Problem

Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded $L$ -shaped region is

Pick an answer.

(A)

(B)

(C)

12.5

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: A $5 \times 5$ square $ABCD$ contains three smaller squares with sides $1$, $3$, $1$ arranged so that the $3 \times 3$ square sits in a corner and the two unit squares sit in adjacent corners. Find the area of the shaded L-shaped region that wraps around the $3 \times 3$ square.

Givens: Outer square $ABCD$ has side $1 + 3 + 1 = 5$; One small square has side $1$ in the top-left strip; The middle square has side $3$; Another small square has side $1$ in the bottom-right strip; Answer choices: (A) $7$, (B) $10$, (C) $12.5$, (D) $14$, (E) $15$

Unknowns: The area of the shaded L-shaped region

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Break Into Subproblems

The labeled lengths $1, 3, 1$ pin down every coordinate on the figure, so Tool #1 (Draw a Diagram) lets us read all the side lengths of the L-shape directly off the grid — no algebra needed. Once the L is drawn with its corner coordinates known, Tool #7 (Break Into Subproblems) handles it cleanly: slice the L into two rectangles along a single horizontal or vertical cut, find each rectangle's area, then add. This avoids the larger "big square minus three smaller squares, then divide by $2$" route, which is correct but does extra work the diagram itself makes unnecessary.

Execute — Answer: A

#1 Draw a Diagram 3.MD.C.7 Step 1

Set up the grid.
The outer square has side $1 + 3 + 1 = 5$, so place it on a grid with corners at $(0,0)$ and $(5,5)$.
The shaded L wraps around the $3 \times 3$ middle square: it runs along the left edge with width $1$ from $y = 0$ up to $y = 4$, and along the bottom edge with height $1$ from $x = 0$ to $x = 4$.

$$\text{outer side} = 1 + 3 + 1 = 5$$

💡 Reading the figure on a grid is the Grade 3 "tile a rectangle" move — every length is a whole number of unit squares.

#7 Break Into Subproblems 3.MD.C.7 Step 2

Cut the L into two rectangles with one horizontal slice at $y = 1$.
The top piece is a tall rectangle along the left edge, from $(0,1)$ to $(1,4)$, so it is $1$ wide and $3$ tall.
The bottom piece is a wide rectangle along the bottom edge, from $(0,0)$ to $(4,1)$, so it is $4$ wide and $1$ tall.
The two pieces meet at the cut but do not overlap.

$$\text{top piece}: 1 \times 3, \quad \text{bottom piece}: 4 \times 1$$

💡 Splitting an L into two rectangles with one straight cut is the Grade 3 "decompose into rectangles" idea — same total area, easier shapes.

#7 Break Into Subproblems 3.MD.C.7 Step 3

Find each rectangle's area, then add.
Area of the top piece is $1 \times 3 = 3$.
Area of the bottom piece is $4 \times 1 = 4$.
Add them for the L's area.

$$\text{L area} = 1 \times 3 + 4 \times 1 = 3 + 4 = 7 \;\Rightarrow\; \textbf{(A)}$$

💡 Adding the two rectangle areas is the Grade 3 "area is additive" property.

[1] #1 3.MD.C.7 Set up the grid. The outer square has side $1 + 3 + 1 = 5$, so place it on a gri

[2] #7 3.MD.C.7 Cut the L into two rectangles with one horizontal slice at $y = 1$. The top piec

[3] #7 3.MD.C.7 Find each rectangle's area, then add. Area of the top piece is $1 \times 3 = 3$.

Review

Reasonableness: Cross-check with the "big minus inner" view. The lower-left $4 \times 4$ sub-square (corners $(0,0)$ to $(4,4)$) contains both the L-region and the $3 \times 3$ middle square exactly, with no leftover. So L $= 4 \times 4 - 3 \times 3 = 16 - 9 = 7$. Same answer. A unit-square count also works: drawing the full $5 \times 5$ grid and counting shaded cells gives $4$ along the bottom strip plus $3$ along the left strip $= 7$ — matches (A). Choices (D) $14$ and (E) $15$ are tempting only if a solver forgets the L wraps around just one $3 \times 3$ square and instead subtracts all three small squares from $25$, getting $25 - 9 - 1 - 1 = 14$; that counts the L-region plus its mirror image, not one L alone.

Alternative: Tool #9 (Solve an Easier Related Problem): instead of measuring the L directly, surround it. The lower-left $4 \times 4$ sub-square is the smallest axis-aligned square that contains the L plus exactly the $3 \times 3$ inner square — no leftover. So L $= 4^2 - 3^2 = 16 - 9 = 7$, which again gives (A). This trades the two-rectangle add for a single subtraction once you spot the right enclosing square.

CCSS standards used (min grade 3)

3.MD.C.7 Relate area to multiplication and to addition (Computing each rectangle's area as length $\times$ width ($1 \times 3$ and $4 \times 1$) and using "area is additive" to add the two pieces for the L's total area.)
3.OA.D.8 Solve two-step word problems using the four operations (Combining the two rectangle areas $3$ and $4$ in one short calculation $3 + 4 = 7$ to land on answer (A).)

⭐ Slice the L into two rectangles along one straight cut. One piece is $1 \times 3 = 3$, the other is $4 \times 1 = 4$, and the L's area is $3 + 4 = 7$ — answer (A).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 3)

More like this