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AMC 8 · 2012 · #6

Grade 4 geometry-2d
Archetype Compound Area by Decomposition / Difference
area-rectanglesperimeter area-differenceidentify-subproblems ↑ Prerequisites: multi-digit-arithmeticarea-rectangles
📏 Short solution 💡 2 insights
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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 88 inches high and 1010 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)
$hspace{.05in}88$
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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

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)

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!

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

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