Sensim Math Original · sm-5
Easy mode Grade 4Problem
Picture a rectangular community-garden plot. It is 15 meters wide (from east to west) and 11 meters tall (from north to south).
Inside this plot, two things take up space. First, a tool shed sits in the southwest corner. The shed is a 4 meter by 3 meter rectangle, pushed snugly into that corner. Second, a straight gravel footpath also sits inside the plot. The path is a 9 meter by 2 meter rectangle, and it does not touch the shed.
The rest of the plot — every spot that is not the shed and not the path — will be planted with vegetables.
What is the area of the vegetable land, in square meters?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A rectangular community-garden plot is $15$ m by $11$ m. Two non-overlapping rectangles sit inside the plot: a $4$ m by $3$ m tool shed flush in one corner, and a $9$ m by $2$ m gravel footpath somewhere else inside the plot. The remaining area — what's left of the plot after removing both the shed and the footpath — is the planting area. Find that planting area in square meters.
Givens: Outer rectangular plot: $15$ m by $11$ m; Tool shed: $4$ m by $3$ m, in a corner of the plot; Gravel footpath: $9$ m by $2$ m, entirely inside the plot; The shed and the footpath do NOT overlap each other; Answer choices: (A) $126$, (B) $132$, (C) $135$, (D) $147$, (E) $153$
Unknowns: The area (in square meters) of the planting region: plot $-$ shed $-$ footpath
Understand
Restated: A rectangular community-garden plot is $15$ m by $11$ m. Two non-overlapping rectangles sit inside the plot: a $4$ m by $3$ m tool shed flush in one corner, and a $9$ m by $2$ m gravel footpath somewhere else inside the plot. The remaining area — what's left of the plot after removing both the shed and the footpath — is the planting area. Find that planting area in square meters.
Givens: Outer rectangular plot: $15$ m by $11$ m; Tool shed: $4$ m by $3$ m, in a corner of the plot; Gravel footpath: $9$ m by $2$ m, entirely inside the plot; The shed and the footpath do NOT overlap each other; Answer choices: (A) $126$, (B) $132$, (C) $135$, (D) $147$, (E) $153$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities
The planting region is not a single nice shape, but everything we remove from it (the shed and the footpath) IS a nice rectangle. Tool #7 (Identify Subproblems) is the natural move: split the work into three one-line area calculations — the whole plot, the shed, and the footpath — then subtract. Tool #1 (Draw a Diagram) — a quick sketch of the plot with a shed in one corner and a footpath strip elsewhere — confirms the two interior pieces don't overlap, so simple subtraction is valid (no inclusion-exclusion needed). Tool #3 (Eliminate Possibilities) matches the final number to one of (A)–(E) and checks the near-miss distractors (forgot the shed, forgot the path) we'd expect a hurried student to make.
Execute — Answer: C
2.G.A.1 Step 1 - Sketch the situation: a $15 \times 11$ rectangle for the plot, a small $4 \times 3$ box wedged into one corner for the shed, and a $9 \times 2$ strip floating somewhere else inside for the gravel footpath.
- Because the problem says the shed and footpath do NOT overlap, the planting region is simply what is left after we erase both interior rectangles from the plot.
💡 Recognizing the three pieces as plain rectangles is the Grade 2 "shapes with given attributes" skill.
4.MD.A.3 Step 2 - Compute the plot area using length $\times$ width.
- The plot is $15$ m by $11$ m, so its area is $15 \times 11 = 165$ square meters.
💡 Applying the rectangle area formula $\ell \times w$ to a real-world plot is exactly the Grade 4 area-formula standard.
3.MD.C.7 Step 3 Compute the shed area the same way: $4$ m by $3$ m gives $4 \times 3 = 12$ square meters.
💡 A $4 \times 3$ rectangle is the canonical Grade 3 area picture — three rows of four unit squares.
3.MD.C.7 Step 4 Compute the footpath area: $9$ m by $2$ m gives $9 \times 2 = 18$ square meters.
💡 Same Grade 3 area-of-a-rectangle idea, just with different side lengths.
3.NBT.A.2 Step 5 - Subtract both interior rectangles from the plot to get the planting area.
- We can either subtract them one at a time, or add them first and subtract once: $12 + 18 = 30$, then $165 - 30 = 135$.
- Either order gives the same answer.
💡 Adding and subtracting whole numbers under $1000$ is fluent Grade 3 arithmetic.
3.NBT.A.2 Step 6 - Check against the answer list $126, 132, 135, 147, 153$.
- Our value $135$ matches choice **(C)** exactly.
- Spot-check the near-misses: forgetting the footpath gives $165 - 12 = 153$ — that's (E).
- Forgetting the shed gives $165 - 18 = 147$ — that's (D).
- So (D) and (E) are exactly the "missed one piece" traps, and only (C) accounts for both removed pieces.
💡 Comparing $135$ to the five three-digit choices is Grade 3 add-and-subtract fluency.
2.G.A.1 Sketch the situation: a $15 \times 11$ rectangle for the plot, a small $4 \times 4.MD.A.3 Compute the plot area using length $\times$ width. The plot is $15$ m by $11$ m, 3.MD.C.7 Compute the shed area the same way: $4$ m by $3$ m gives $4 \times 3 = 12$ squar 3.MD.C.7 Compute the footpath area: $9$ m by $2$ m gives $9 \times 2 = 18$ square meters. 3.NBT.A.2 Subtract both interior rectangles from the plot to get the planting area. We can 3.NBT.A.2 Check against the answer list $126, 132, 135, 147, 153$. Our value $135$ matches Review
Reasonableness: Sanity check the magnitude: the full plot is $165$ m$^2$ and we're removing two small interior rectangles totaling $30$ m$^2$ (about $18\%$ of the plot), so the planting region should be a bit more than $80\%$ of $165$, roughly $130$–$140$ m$^2$. Our $135$ m$^2$ sits squarely in that range. Plugging back: $135 + 12 + 18 = 165$, which is the plot area — the three pieces partition the plot, as required. The units are square meters throughout, consistent with what the question asked.
Alternative: An alternative is Tool #13 (Convert to Algebra): write $A = \ell w - a_1 - a_2$ with $\ell = 15,\, w = 11,\, a_1 = 4 \cdot 3,\, a_2 = 9 \cdot 2$, then evaluate. The algebra gives the same $135$ but is heavier than needed — the straight "compute three rectangle areas and subtract" path from Tool #7 (Identify Subproblems) keeps it firmly at the Grade 3–4 level, which matches the product's lowest-grade message.
CCSS standards used (min grade 4)
2.G.A.1Recognize and draw shapes having specified attributes (e.g., rectangles) (Identifying the plot, the shed, and the footpath as three distinct rectangles, and sketching them so the planting region is clearly what remains.)3.MD.C.7Relate area to multiplication and addition operations (area of a rectangle = length × width; area of composite figures by adding/subtracting parts) (Computing the $4 \times 3 = 12$ shed area and the $9 \times 2 = 18$ footpath area as side-times-side, then using composite-area subtraction to get the planting region.)3.NBT.A.2Fluently add and subtract within 1000 (Adding $12 + 18 = 30$ and subtracting $165 - 30 = 135$ to finish the calculation and match against the answer choices.)4.MD.A.3Apply area and perimeter formulas for rectangles in real-world problems (Using $\ell \times w$ on the $15 \times 11$ garden plot (a real-world rectangle context) to get $165$ m$^2$.)
⭐ This AMC 8 problem only needs the Grade 4 rectangle-area formula (length $\times$ width) and a little subtraction you already know!
⭐ This AMC 8 problem only needs the Grade 4 rectangle-area formula (length $\times$ width) and a little subtraction you already know!
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