← Sensim Lab Sensim Math

AMC 8 · 2012 · #6

Easy mode Grade 4
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Problem

Imagine a photo that is 88 inches tall and 1010 inches wide. The photo sits inside a frame.

The frame adds a border of width 22 inches on every side of the photo.

What is the area of just 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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