AMC 10 · 2011 · #14
Grade 8 geometry-2dA rectangular parking lot has a diagonal of 25 meters and an area of 168 square meters. In meters, what is the perimeter of the parking lot?
Pick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A rectangle has a diagonal of $25$ meters and an area of $168$ square meters. Find the perimeter of the rectangle.
Givens: The diagonal of the rectangle is $25$ meters; The area of the rectangle is $168$ square meters; Answer choices: (A) $52$, (B) $58$, (C) $62$, (D) $68$, (E) $70$
Unknowns: The perimeter of the rectangle, in meters
Understand
Restated: A rectangle has a diagonal of $25$ meters and an area of $168$ square meters. Find the perimeter of the rectangle.
Givens: The diagonal of the rectangle is $25$ meters; The area of the rectangle is $168$ square meters; Answer choices: (A) $52$, (B) $58$, (C) $62$, (D) $68$, (E) $70$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
Tool #4 (Introduce a Variable) names the two sides $l$ and $w$ so the diagonal and the area become two equations. Tool #1 (Draw a Diagram) shows why the diagonal makes a right triangle, which turns the diagonal fact into $l^2+w^2=625$. Tool #7 (Identify Subproblems) is the key move: the perimeter only needs $l+w$, not $l$ and $w$ on their own, and $(l+w)^2 = l^2+w^2+2lw$ recycles both facts we already have. This dodges the messy work of solving for each side.
Execute — Answer: C
8.G.B.7 Step 1 Cut the rectangle with its diagonal
- The diagonal splits the rectangle into two right triangles.
- The two sides of the rectangle are the legs, and the $25$-meter diagonal is the hypotenuse.
- By the Pythagorean theorem, the squares of the sides add up to the square of the diagonal.
💡 A rectangle's diagonal always makes a right triangle, so its length ties the two sides together through the Pythagorean theorem.
3.MD.C.7 Step 2 Turn the area into an equation
- Call the two side lengths $l$ and $w$.
- The area of a rectangle is length times width, so the area fact becomes a second equation.
- Now two simple equations hold everything the problem tells us.
💡 Naming the sides lets each sentence of the problem become one clean equation.
6.EE.B.6 Step 3 Aim straight at the perimeter
- The perimeter is $2(l+w)$, so all we really need is $l+w$ — not the two sides by themselves.
- Squaring $l+w$ opens up into $l^2 + w^2 + 2lw$, and both of those pieces are already known: $l^2+w^2=625$ and $2lw = 2 \cdot 168 = 336$.
- Add them.
💡 Squaring the sum of the sides mixes in exactly the two facts you were handed, so no side-by-side solving is needed.
8.EE.A.2 Step 4 Undo the square
- Since $l+w$ is a positive length, take the positive square root of $961$.
- Because $31^2 = 961$, the sum of the two sides is $31$ meters.
💡 A square root peels off the square to reveal the sum you were chasing.
6.EE.B.7 Step 5 Double it for the perimeter
- The perimeter is two lengths plus two widths, which is $2(l+w)$.
- Doubling $31$ gives $62$ meters.
- That matches choice (C).
💡 A rectangle's border is just the two sides counted twice, so double their sum.
8.G.B.7 The diagonal splits the rectangle into two right triangles. The two sides of the 3.MD.C.7 Call the two side lengths $l$ and $w$. The area of a rectangle is length times w 6.EE.B.6 The perimeter is $2(l+w)$, so all we really need is $l+w$ — not the two sides by 8.EE.A.2 Since $l+w$ is a positive length, take the positive square root of $961$. Becaus 6.EE.B.7 The perimeter is two lengths plus two widths, which is $2(l+w)$. Doubling $31$ g Review
Reasonableness: Check that real sides exist: $l+w=31$ and $lw=168$ mean the sides solve $x^2-31x+168=0$, which factors as $(x-7)(x-24)=0$, giving sides $7$ and $24$. Those check out: $7^2+24^2 = 49+576 = 625 = 25^2$, and $7 \cdot 24 = 168$. Their perimeter $2(7+24)=62$ confirms choice (C).
Alternative: Use Tool #6 (Guess and Check) with Pythagorean triples. Since $25$ is the hypotenuse, the classic triple $7$-$24$-$25$ jumps out. Its legs give area $7 \cdot 24 = 168$, exactly as required, so the sides are $7$ and $24$ and the perimeter is $2(7+24)=62$.
CCSS standards used (min grade 8)
8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Turning the $25$-meter diagonal into the equation $l^2+w^2=625$ via the right triangle it forms.)3.MD.C.7Relate area to multiplication and addition operations (Writing the rectangle's area as $l \cdot w = 168$.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Expressing the perimeter through $l+w$ and expanding $(l+w)^2 = l^2+w^2+2lw$ to combine the known facts.)8.EE.A.2Use square root and cube root symbols to represent solutions (Taking $\sqrt{961}=31$ to recover the sum of the sides.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Computing the perimeter as $P = 2 \cdot 31 = 62$.)
⭐ When you only need the sum of two sides, square the sum — it reuses the area and diagonal facts you already have, so you never have to find each side.
⭐ When you only need the sum of two sides, square the sum — it reuses the area and diagonal facts you already have, so you never have to find each side.
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