AMC 8 · 2019 · #4
Grade 8 geometry-2dProblem
Quadrilateral ABCD is a rhombus with perimeter 52 meters. The length of diagonal AC is 24 meters. What is the area in square meters of rhombus ABCD?
Pick an answer.
AMC 8 2019 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: Rhombus $ABCD$ has perimeter $52$ meters and one diagonal $\overline{AC}$ of length $24$ meters. Find the area of the rhombus in square meters.
Givens: $ABCD$ is a rhombus (all four sides equal); Perimeter $= 52$ meters; Diagonal $AC = 24$ meters; Answer choices: (A) $60$, (B) $90$, (C) $105$, (D) $120$, (E) $144$
Unknowns: Area of rhombus $ABCD$ in square meters
Understand
Restated: Rhombus $ABCD$ has perimeter $52$ meters and one diagonal $\overline{AC}$ of length $24$ meters. Find the area of the rhombus in square meters.
Givens: $ABCD$ is a rhombus (all four sides equal); Perimeter $= 52$ meters; Diagonal $AC = 24$ meters; Answer choices: (A) $60$, (B) $90$, (C) $105$, (D) $120$, (E) $144$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
The problem is a 2-D geometry question, so Tool #1 (Draw a Diagram) is the natural entry point: sketch the rhombus, add both diagonals, and mark where they cross. Once the diagonals are drawn, Tool #7 (Identify Subproblems) makes the path obvious — the rhombus splits into four congruent right triangles, so the problem decomposes into (i) get the side length from the perimeter, (ii) get the missing half-diagonal from a right triangle, (iii) combine the two diagonals into the area formula.
Execute — Answer: D
3.OA.A.3 Step 1 - Find the side length.
- A rhombus has four equal sides, so dividing the perimeter by $4$ gives one side.
💡 Splitting the total perimeter evenly among $4$ equal sides is a Grade 3 multiplication/division word-problem move.
5.G.B.4 Step 2 - Sketch the rhombus with both diagonals.
- Because the diagonals of a rhombus are perpendicular and bisect each other, they meet at the center $M$ and cut the rhombus into four congruent right triangles.
- Half of diagonal $AC$ is one leg: $AM = 24 / 2 = 12$ meters.
💡 Recognizing the rhombus's diagonals as perpendicular bisectors is part of classifying special quadrilaterals in Grade 5.
8.G.B.7 Step 3 - Zoom in on right triangle $\triangle AMB$.
- The hypotenuse is the rhombus's side ($AB = 13$), one leg is $AM = 12$, and the other leg $BM$ is the unknown half of the second diagonal.
- Apply the Pythagorean theorem to solve for $BM$.
💡 Splitting the rhombus produces a right triangle with two known sides — exactly the Grade 8 Pythagorean-theorem setup.
5.G.B.4 Step 4 Double $BM$ to get the full second diagonal $BD$, since the diagonals bisect each other.
💡 Using the bisecting property of the rhombus's diagonals to recover the full length is still Grade 5 quadrilateral reasoning.
6.G.A.1 Step 5 Apply the rhombus area formula: half the product of the two diagonals.
💡 The half-product-of-diagonals formula for a special quadrilateral is Grade 6 area-of-polygons content.
3.OA.A.3 Find the side length. A rhombus has four equal sides, so dividing the perimeter 5.G.B.4 Sketch the rhombus with both diagonals. Because the diagonals of a rhombus are p 8.G.B.7 Zoom in on right triangle $\triangle AMB$. The hypotenuse is the rhombus's side 5.G.B.4 Double $BM$ to get the full second diagonal $BD$, since the diagonals bisect eac 6.G.A.1 Apply the rhombus area formula: half the product of the two diagonals. Review
Reasonableness: Quick sanity check: the rhombus fits inside the $24 \times 10$ rectangle formed by its diagonals, which has area $240$. The rhombus is exactly half of that rectangle (its four triangle pieces fill half), giving $120$ square meters. The answer also lies between $60$ (a very 'flat' rhombus) and $144$ (the would-be square with side $\sqrt{144}=12$), so $120$ is in the right magnitude. Choice (D) is confirmed.
Alternative: Tool #6 (Guess and Check) on a side-$13$ triangle: the only Pythagorean triple with hypotenuse $13$ and integer legs is $5$–$12$–$13$. Once you spot that, $BM = 5$ instantly without writing the equation. Combined with the diagonal-area formula, you reach $(1/2)(24)(10) = 120$ in just two lines.
CCSS standards used (min grade 8)
3.OA.A.3Solve multiplication and division word problems within 100 (Dividing the perimeter $52$ by the $4$ equal sides of the rhombus to get the side length $13$.)5.G.B.4Classify two-dimensional figures in a hierarchy based on properties (Using the defining properties of a rhombus — equal sides and perpendicular-bisecting diagonals — to label half-diagonals and recover full ones.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Applying the rhombus area formula $\tfrac{1}{2} d_1 d_2$ once both diagonals are known.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the unknown half-diagonal $BM$ from the right triangle with legs $AM = 12$ and hypotenuse $AB = 13$.)
⭐ This AMC 8 problem only needs the Grade 8 Pythagorean theorem (the $5$-$12$-$13$ right triangle hiding inside a rhombus) you already know!
⭐ This AMC 8 problem only needs the Grade 8 Pythagorean theorem (the $5$-$12$-$13$ right triangle hiding inside a rhombus) you already know!
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