AMC 10 · 2010 · #2
Easy mode Grade 4The picture shows a shape like the letter L. It has six sides, and every corner is a right angle. Four of the six sides have their lengths marked in the picture; the other two do not. What is the area inside the L?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: The picture shows a big letter $L$ drawn with six sides, every corner a right angle. Only four of the six sides are labeled: the left side is $8$, the bottom is $5$, the short vertical side on the right is $2$, and the top is $2$. Find the area the $L$ encloses.
Givens: The $L$ is a closed six-sided figure and every side is horizontal or vertical; The left side has length $8$; The bottom side has length $5$; The short vertical side on the right has length $2$; The top side has length $2$; Answer choices: (A) $22$, (B) $24$, (C) $26$, (D) $28$, (E) $30$
Unknowns: The area enclosed by the $L$; The lengths of the two sides the picture leaves unlabeled
Understand
Restated: The picture shows a big letter $L$ drawn with six sides, every corner a right angle. Only four of the six sides are labeled: the left side is $8$, the bottom is $5$, the short vertical side on the right is $2$, and the top is $2$. Find the area the $L$ encloses.
Givens: The $L$ is a closed six-sided figure and every side is horizontal or vertical; The left side has length $8$; The bottom side has length $5$; The short vertical side on the right has length $2$; The top side has length $2$; Answer choices: (A) $22$, (B) $24$, (C) $26$, (D) $28$, (E) $30$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #4 Introduce a Variable, #13 Convert to Algebra, #16 Change Focus / Count the Complement
An $L$ has no area formula of its own, but it is built out of rectangles, which do — so Tool #7 (Identify Subproblems) says cut it into rectangles and add. The obstacle comes first. Two of the six sides carry no number, and every cut you could make needs one of them, so nothing can be multiplied until those two lengths are pinned down. Tool #4 (Introduce a Variable) names them and Tool #13 (Convert to Algebra) forces their values from the one structural fact available: the boundary closes. Once the shape is fully measured, Tool #16 (Change Focus / Count the Complement) recomputes the area from the outside in — fill the $L$ out to a full rectangle and subtract the bite that is missing — as a check on the cut-and-add count.
Execute — Answer: A
2.MD.B.5 Step 1 Name the two unlabeled sides
- Trace the boundary.
- The $L$ has six sides, and the picture prints only four of them: left $8$, bottom $5$, short right side $2$, top $2$.
- The two sides that turn the inner corner — the horizontal one running left from the top of the short right side, and the vertical one running up from where it stops — carry no label.
- Call the horizontal one $a$ and the vertical one $b$.
- Every rectangle hiding inside this $L$ has $a$ or $b$ as one of its dimensions, so these two come before any area.
💡 You cannot multiply a length you do not have yet, so the two missing sides are the real first question.
1.OA.D.8 Step 2 Close the loop to find a and b
- Walk once around the boundary.
- Because you finish where you began, the rightward steps must exactly cancel the leftward steps, and the upward steps must exactly cancel the downward steps.
- Going around from the bottom-left corner: right $5$, up $2$, left $a$, up $b$, left $2$, down $8$.
- The horizontal moves give $5 = a + 2$, so $a = 3$.
- The vertical moves give $2 + b = 8$, so $b = 6$.
- This is where $5-2$ and $8-2$ actually come from — not from how the picture looks, but because a closed path made of horizontal and vertical steps has no other option.
💡 A round trip has to undo every step it takes, so the sideways moves cancel each other and so do the up-and-down moves.
3.MD.C.7 Step 3 Cut the L into two rectangles
- Extend the inner vertical side straight down to the bottom edge.
- That single cut splits the $L$ into two pieces that overlap nowhere and together make the whole shape: a tall rectangle on the left, $2$ wide and $8$ tall, and the foot on the right, $a = 3$ wide and $2$ tall.
- Both pieces are genuine rectangles because every side involved is horizontal or vertical.
- Area is additive over non-overlapping pieces, so add them: $2 \times 8 = 16$ and $3 \times 2 = 6$.
💡 Chop a shape into pieces that do not overlap and the areas simply add up.
4.MD.A.3 Step 4 Check by filling out the rectangle
- Now compute it from the outside in.
- The $L$ fits snugly inside a $5$-by-$8$ rectangle of area $5 \times 8 = 40$.
- What is missing is the top-right corner region, and that region really is a rectangle: on the left it is bounded by the line through the inner vertical side, below by the line through the inner horizontal side, and on the other two sides by the big rectangle itself — four horizontal or vertical lines.
- Its width is $a = 3$ and its height is $b = 6$, so it has area $3 \times 6 = 18$.
- Subtracting gives $40 - 18 = 22$, the same total as the cut-and-add count.
- The area is $22$, so the answer is (A).
💡 What is left after a bite is taken out of a rectangle is just the rectangle minus the bite.
2.MD.B.5 Trace the boundary. The $L$ has six sides, and the picture prints only four of t 1.OA.D.8 Walk once around the boundary. Because you finish where you began, the rightward 3.MD.C.7 Extend the inner vertical side straight down to the bottom edge. That single cut 4.MD.A.3 Now compute it from the outside in. The $L$ fits snugly inside a $5$-by-$8$ rect Review
Reasonableness: Bounds alone are not enough here: the $L$ sits inside the $5 \times 8$ rectangle and contains the $2 \times 8$ tall arm, so its area is between $16$ and $40$ — and every answer choice satisfies that. So count unit squares instead of multiplying. The two bottom rows, from height $0$ to $1$ and from $1$ to $2$, are $5$ squares wide, giving $10$ squares. The rows above them, from height $2$ up to $8$, are only $2$ squares wide, and there are $6$ of them, giving $12$. The tally is $10 + 12 = 22$ squares, reached by counting rather than by any area formula. It matches.
Alternative: Put the figure on coordinates and use the shoelace formula, which needs only the six corner points and no rectangle formula at all. The corners are $(0,0)$, $(5,0)$, $(5,2)$, $(2,2)$, $(2,8)$, $(0,8)$, and the shoelace terms are $0, 10, 6, 12, 16, 0$, summing to $44$, so the area is $\frac{1}{2} \times 44 = 22$. Pick's theorem gives a third route that shares nothing with the first two: the boundary passes through $B = 26$ lattice points and $I = 10$ lattice points sit strictly inside, so the area is $I + \frac{B}{2} - 1 = 10 + 13 - 1 = 22$. Three different routes, one answer (A).
CCSS standards used (min grade 4)
2.MD.B.5Solve word problems involving lengths using same units (Spotting that two of the six side lengths are missing and naming them $a$ and $b$ before anything is computed.)1.OA.D.8Determine the unknown whole number in an addition or subtraction equation (Solving $a + 2 = 5$ and $2 + b = 8$, which the closed boundary forces, to get $a = 3$ and $b = 6$.)3.MD.C.7Relate area to multiplication and addition operations (Cutting the $L$ into two non-overlapping rectangles and adding their areas, $16 + 6 = 22$.)4.MD.A.3Apply area and perimeter formulas for rectangles in real-world problems (Computing the $5 \times 8$ bounding rectangle and the $3 \times 6$ missing corner, then subtracting to confirm $22$.)
⭐ When a figure hides some of its side lengths, walk all the way around it — the trip has to close, and that alone hands you the missing lengths.
⭐ When a figure hides some of its side lengths, walk all the way around it — the trip has to close, and that alone hands you the missing lengths.
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