AMC 8 · 2012 · #6

Grade 4 geometry-2d

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

area-rectanglesperimeter area-differenceidentify-subproblems ↑ Prerequisites: multi-digit-arithmeticarea-rectangles

📏 Short solution 💡 2 insights

Problem

A rectangular photograph is placed in a frame that forms a border two inches wide on all sides of the photograph. The photograph measures $8$ inches high and $10$ inches wide. What is the area of the border, in square inches?

Pick an answer.

(A)

$hspace{.05in}36$

(B)

$hspace{.05in}40$

(C)

$hspace{.05in}64$

(D)

$hspace{.05in}72$

(E)

$hspace{.05in}88$

AMC 8 2012 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: A rectangular photograph that is $8$ inches tall and $10$ inches wide sits inside a frame whose border is $2$ inches wide on every side. Find the area, in square inches, of just the border (the frame region, not including the photo).

Givens: Photo dimensions: $8$ inches tall, $10$ inches wide; Border width: $2$ inches on every side (top, bottom, left, right); Answer choices: (A) $36$, (B) $40$, (C) $64$, (D) $72$, (E) $88$ (square inches)

Unknowns: Area of the border region (the frame ring around the photo), in square inches

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram

The border is a ring — an awkward shape to measure directly. Tool #7 (Identify Subproblems) turns it into two easy rectangles: find the area of the big outer rectangle (photo $+$ border), find the area of the inner photo, then subtract. Tool #1 (Draw a Diagram) is the supporting move: sketch the photo inside the frame and label every side, so the $2$-inch border gets added to BOTH ends of each dimension (not just one), which is the most common mistake on this problem.

Execute — Answer: E

#1 Draw a Diagram 4.MD.A.3 Step 1

Draw the picture.
Sketch the photo ($8 \times 10$) and around it draw the frame border $2$ inches thick on every side.
Label the outer rectangle's height and width by adding $2$ inches at the top AND $2$ inches at the bottom, $2$ inches at the left AND $2$ inches at the right.

$$\text{outer height} = 2 + 8 + 2 = 12 \text{ in}, \quad \text{outer width} = 2 + 10 + 2 = 14 \text{ in}$$

💡 Drawing the labeled diagram forces you to add the border twice per dimension (both sides), which is the whole trick of the problem.

#7 Identify Subproblems 4.MD.A.3 Step 2

Find the area of the outer rectangle (photo $+$ border).
Use the rectangle area formula: area $=$ length $\times$ width.

$$\text{outer area} = 12 \times 14 = 168 \text{ in}^2$$

💡 The outer rectangle is the first subproblem — a clean rectangle whose area is the standard length-times-width formula.

#7 Identify Subproblems 3.MD.C.7 Step 3

Find the area of the inner rectangle (the photo itself).

$$\text{photo area} = 8 \times 10 = 80 \text{ in}^2$$

💡 The photo is the second subproblem — another clean rectangle, computed the same way.

#7 Identify Subproblems 3.NBT.A.2 Step 4

Subtract the photo from the outer rectangle.
What is left is exactly the border ring.

$$\text{border area} = 168 - 80 = 88 \text{ in}^2 \;\Rightarrow\; \textbf{(E)}$$

💡 "Big shape minus inside shape $=$ the ring around it" is the decompose-and-subtract move at the heart of Tool #7.

[1] #1 4.MD.A.3 Draw the picture. Sketch the photo ($8 \times 10$) and around it draw the frame

[2] #7 4.MD.A.3 Find the area of the outer rectangle (photo $+$ border). Use the rectangle area

[3] #7 3.MD.C.7 Find the area of the inner rectangle (the photo itself).

[4] #7 3.NBT.A.2 Subtract the photo from the outer rectangle. What is left is exactly the border

Review

Reasonableness: Sanity check the size: an $8 \times 10$ photo has area $80$. A border $2$ inches wide on every side adds roughly the perimeter ($2 \times (8+10) = 36$) times the border width ($2$), which is about $72$, plus the four corner squares ($4 \times 2 \times 2 = 16$), totaling $88$. That matches answer (E) exactly and is in the right ballpark — bigger than $72$, smaller than the photo itself.

Alternative: Tool #7 can also split the border into $4$ corners $+$ $4$ side strips. Four corner squares: $4 \times (2 \times 2) = 16$. Top and bottom strips above and below the photo: $2 \times (2 \times 10) = 40$. Left and right strips beside the photo: $2 \times (2 \times 8) = 32$. Total: $16 + 40 + 32 = 88$ in$^2$, again answer (E). Same answer from a totally different decomposition is strong confirmation.

CCSS standards used (min grade 4)

3.MD.C.7 Relate area to multiplication and to addition (Computing the photo's area as $8 \times 10 = 80$ in$^2$ using the rectangle area formula.)
4.MD.A.3 Apply the area and perimeter formulas for rectangles in real world and mathematical problems (Building the outer rectangle's dimensions ($12 \times 14$) by adding the $2$-inch border to BOTH ends of each side, then computing its area as $168$ in$^2$.)
3.NBT.A.2 Fluently add and subtract within $1000$ using strategies and algorithms (Subtracting the photo area from the outer area: $168 - 80 = 88$ in$^2$ to isolate the border.)

⭐ A border around a rectangle is just "big rectangle minus the photo inside" — Grade 4 area facts are all you need!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 4)

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