AMC 8 · 2006 · #6

Grade 3 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Rectilinear Frame / Box Border

perimeterarea-rectanglesspatial-visualization area-differenceidentify-subproblems ↑ Prerequisites: perimetermulti-digit-arithmetic

📏 Short solution 💡 2 insights 📊 Diagram

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Problem

The letter T is formed by placing two $2 \times 4$ inch rectangles next to each other, as shown. What is the perimeter of the T, in inches?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2006 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.

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Toolkit + CCSS Solution

Understand

Restated: Two $2 \times 4$ inch rectangles are placed against each other to form the letter T — one rectangle laid flat across the top, the other standing upright underneath, centered. What is the perimeter of the resulting T shape?

Givens: Each rectangle measures $2$ inches by $4$ inches; The two rectangles together form the letter T; The top rectangle is $4$ inches wide and $2$ inches tall; the vertical rectangle is $2$ inches wide and $4$ inches tall, attached centered to the bottom of the top one; Answer choices: (A) $12$, (B) $16$, (C) $20$, (D) $22$, (E) $24$

Unknowns: The perimeter of the T shape, in inches

Understand

Plan

Primary tool: #1 Draw a Picture

Secondary: #7 Break into Subproblems

Tool #1 (Draw a Picture) is the natural first move for any perimeter question on an irregular shape: sketch the T, label every side length, then walk around the boundary once and add the segments. The T has eight straight edges, each of which is either $1$, $2$, or $4$ inches, all readable straight from the figure. Tool #7 (Break into Subproblems) gives a clean check: treat the T as the two rectangles, add their perimeters, then subtract the part of the boundary that gets hidden where they meet.

Execute — Answer: C

#1 Draw a Picture 3.MD.D.8 Step 1

Sketch the T and label the side lengths.
The top rectangle is $4$ inches wide and $2$ inches tall.
The vertical rectangle is $2$ inches wide and $4$ inches tall, attached to the middle of the top rectangle's bottom edge.
Because the vertical piece is $2$ inches wide and is centered under a $4$-inch top, there are $\dfrac{4 - 2}{2} = 1$ inch of overhang on each side.

$$\text{overhang on each side} = \dfrac{4 - 2}{2} = 1 \text{ inch}$$

💡 Drawing the shape and writing each length next to its edge turns the problem into simple addition along the outline.

#1 Draw a Picture 3.MD.D.8 Step 2

Walk around the boundary of the T, starting at the top-left corner and going clockwise.
List every edge in order: across the top, down the right side of the top rectangle, left along the overhang, down the right side of the vertical rectangle, across the bottom, up the left side of the vertical rectangle, right along the left overhang, and up the left side of the top rectangle.

$$\text{edges in order} = 4,\; 2,\; 1,\; 4,\; 2,\; 4,\; 1,\; 2 \text{ inches}$$

💡 Going in one direction around the shape guarantees each outside edge is counted exactly once.

#1 Draw a Picture 3.MD.D.8 Step 3

Add the eight edge lengths to get the perimeter.

$$P = 4 + 2 + 1 + 4 + 2 + 4 + 1 + 2 = 20 \text{ inches} \;\Rightarrow\; \textbf{(C)}$$

💡 Grouping the edges makes the arithmetic clean: top + bottom $= 4 + 2 = 6$; the two long stem sides $= 4 + 4 = 8$; the two short sides of the horizontal bar $= 2 + 2 = 4$; the two overhangs $= 1 + 1 = 2$. Total $= 6 + 8 + 4 + 2 = 20$.

[1] #1 3.MD.D.8 Sketch the T and label the side lengths. The top rectangle is $4$ inches wide an

[2] #1 3.MD.D.8 Walk around the boundary of the T, starting at the top-left corner and going clo

[3] #1 3.MD.D.8 Add the eight edge lengths to get the perimeter.

Review

Reasonableness: Two separate $2 \times 4$ rectangles would each have perimeter $2(2+4) = 12$, for a combined $24$ inches. When the rectangles are pushed together to form the T, a $2$-inch segment of each rectangle's edge is hidden inside the figure, so the perimeter loses $2 + 2 = 4$ inches. That gives $24 - 4 = 20$ inches, matching the boundary-trace answer. The result $20$ is one of the offered choices and sits between $16$ and $24$, exactly where it should be for a shape made of two joined rectangles.

Alternative: Tool #7 (Break into Subproblems): compute the perimeter of each $2 \times 4$ rectangle separately, then remove the shared edge twice. Each rectangle has perimeter $2(2+4) = 12$, so together they contribute $24$ inches. The two pieces meet along a $2$-inch segment (the top of the vertical rectangle meets the middle of the top rectangle's bottom edge), and that segment is on the boundary of both rectangles but inside the T. Removing it once from each gives $24 - 2 \cdot 2 = 20$ inches.

CCSS standards used (min grade 3)

3.MD.D.8 Solve real world and mathematical problems involving perimeters of polygons (Adding the lengths of the eight outside edges of the T to compute its perimeter.)
3.OA.D.8 Solve two-step word problems using the four operations (Computing the $1$-inch overhang from $(4 - 2) \div 2$ before adding all edge lengths.)

⭐ For a perimeter question on an irregular shape, draw the figure, label every edge, and walk around the outside once — adding $4+2+1+4+2+4+1+2$ gives $20$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 3)

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