AMC 8 · 2019 · #2

Easy mode Grade 4

Problem

Picture three identical small rectangles. They are arranged to form one big rectangle, called $ABCD$ , like in the picture below.

The short side of each small rectangle is $5$ feet long.

What is the area of the big rectangle $ABCD$ , in square feet?

$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$

Pick an answer.

(A)

(B)

(C)

100

(D)

125

(E)

150

View mode:

Toolkit + CCSS Solution

Understand

Restated: Three identical small rectangles are arranged to make one big rectangle $ABCD$. The figure shows two of the small rectangles stacked on the left (each lying with its long side horizontal) and one standing on the right with its long side vertical. The shorter side of each small rectangle is $5$ feet. Find the area, in square feet, of the big rectangle $ABCD$.

Givens: Three small rectangles are identical (same short side $s$ and same long side $l$); Short side of each small rectangle: $s = 5$ feet; On the left: two small rectangles are stacked one above the other (each with its long side horizontal); On the right: one small rectangle stands upright with its long side vertical; Answer choices: (A) $45$, (B) $75$, (C) $100$, (D) $125$, (E) $150$ (square feet)

Unknowns: The area of the big rectangle $ABCD$ in square feet

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

The figure is the whole story, so Tool #1 (Draw a Diagram) is the natural entry: label every short side $s$ and every long side $l$ on the picture, and the constraint that the two columns must reach the same height of $ABCD$ pops out as the equation $2s = l$. From there Tool #7 (Identify Subproblems) splits the work into three clean pieces — (1) deduce $l$ from $s$, (2) compute the width and height of $ABCD$, (3) multiply for the area — instead of trying to attack the area in one move. Reaching for algebra (Tool #13) would be overkill: a single labeled drawing makes the relationship visible.

Execute — Answer: E

#1 Draw a Diagram 3.G.A.1 Step 1

Label every short side of a small rectangle as $s$ and every long side as $l$ on the figure.
Looking at the LEFT column (the two stacked rectangles) the height of $ABCD$ is two short sides, $s + s = 2s$.
Looking at the RIGHT column (the one upright rectangle) the height of $ABCD$ is one long side, $l$.
Both columns must reach the same top, so $2s = l$.

$$2s = l$$

💡 Recognizing that the same height of $ABCD$ has to equal both expressions is a Grade 3 'shapes share attributes' move — same length, two names.

#7 Identify Subproblems 3.OA.A.3 Step 2

Plug in the given $s = 5$ feet to find the long side of every small rectangle.

$$l = 2 \times 5 = 10 \text{ ft}$$

💡 Doubling a one-digit number is a Grade 3 multiplication word-problem step.

#7 Identify Subproblems 3.G.A.1 Step 3

Find the dimensions of the big rectangle $ABCD$.
The HEIGHT is $l = 10$ ft (the right-column rectangle stands $l$ tall, and equivalently the two stacked left rectangles are $s + s = 10$ ft tall).
The WIDTH is the long side of a stacked left rectangle ($l = 10$ ft, horizontal) plus the short side of the upright right rectangle ($s = 5$ ft, horizontal), giving $10 + 5 = 15$ ft.

$$\text{height} = 10 \text{ ft}, \quad \text{width} = 10 + 5 = 15 \text{ ft}$$

💡 Adding the widths of the two side-by-side pieces to get the whole width is the same compose-and-decompose-shapes idea from Grade 3 geometry.

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

Apply the rectangle-area formula $\text{area} = \text{length} \times \text{width}$ to $ABCD$ to get the final answer in square feet.

$$\text{area} = 15 \times 10 = 150 \text{ sq ft} \;\Rightarrow\; \textbf{(E)}$$

💡 Using the rectangle area formula on a real-world measurement problem is exactly the Grade 4 standard 4.MD.A.3.

[1] #1 3.G.A.1 Label every short side of a small rectangle as $s$ and every long side as $l$ on

[2] #7 3.OA.A.3 Plug in the given $s = 5$ feet to find the long side of every small rectangle.

[3] #7 3.G.A.1 Find the dimensions of the big rectangle $ABCD$. The HEIGHT is $l = 10$ ft (the

[4] #7 4.MD.A.3 Apply the rectangle-area formula $\text{area} = \text{length} \times \text{width

Review

Reasonableness: Cross-check by counting copies: each small rectangle has area $s \times l = 5 \times 10 = 50$ sq ft, and the big rectangle is made of exactly $3$ of them, so the total area must be $3 \times 50 = 150$ sq ft. This matches the formula answer and lands on choice (E). The magnitudes (a $15 \times 10$ ft floor area) also make physical sense for 'rectangle made of three small rectangles'.

Alternative: Tool #10 (Physical Representation) — cut three identical $5 \times 10$ paper rectangles, lay two flat-stacked on the left and one upright on the right, and the outer shape is a $15 \times 10$ rectangle. Since the big rectangle is exactly the three tiles glued together with no gaps, $\text{area} = 3 \times (5 \times 10) = 150$ sq ft — same answer with no formula at all.

CCSS standards used (min grade 4)

3.G.A.1 Understand that shapes in different categories share attributes (Recognizing that the height of $ABCD$ can be measured along either column (giving $2s$ on the left and $l$ on the right) and that these must describe the same length, and that the width of $ABCD$ is the sum of the widths of its two side-by-side parts.)
3.OA.A.3 Solve multiplication and division word problems within 100 (Computing $l = 2 \times 5 = 10$ ft, doubling the given short side to find the long side of each small rectangle.)
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems (Applying $\text{area} = \text{length} \times \text{width} = 15 \times 10 = 150$ sq ft to the big rectangle $ABCD$.)

⭐ This AMC 8 problem only needs the Grade 4 rectangle area formula (length times width) you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 4)

More like this