Competition · AMC preparation · step 4 of 4
AMC 8 · 2019 · #4
Grade 8 geometry-2d
Pick an answer.
AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Find the side length
A rhombus has four equal sides, so dividing the perimeter by 4 gives one side of 13 meters.
Splitting the total perimeter evenly among 4 equal sides is a Grade 3 multiplication/division word-problem move.
3.OA.A.3Draw A DiagramHalve the given diagonal
Both diagonals split the rhombus into four right triangles meeting at center M, where AM is half of AC = 12 meters.
Recognizing the rhombus's diagonals as perpendicular bisectors is part of classifying special quadrilaterals in Grade 5.
5.G.B.4Draw A DiagramUse the Pythagorean theorem
In right triangle AMB the side AB=13 is the hypotenuse and AM=12 a leg, so the Pythagorean theorem gives BM = 5.
Splitting the rhombus produces a right triangle with two known sides — exactly the Grade 8 Pythagorean-theorem setup.
The unknown half-diagonal BM measures 5 meters.
▸ Why?
In right triangle △ AMB the legs AM and BM and the hypotenuse AB satisfy AM² + BM² = AB², so with AM=12 and AB=13 we get BM² = 13² - 12² = 25, giving BM=5.
▸ Why?
The corner of the triangle at M is a right angle, because the two diagonals of the rhombus cross each other squarely.
▸ Why?
Folding the rhombus along diagonal AC lays side AB exactly onto side AD, so AC is the fold line that meets the other diagonal BD at a right angle at M.
▸ Why?
The leg AM is 12, half of the given diagonal AC=24, because M is the midpoint of AC.
▸ Why?
Folding the rhombus along diagonal BD lays A exactly onto C, so the fold line cuts AC into two equal halves at M.
▸ Why?
The hypotenuse AB is 13, one side of the rhombus, because the perimeter 52 is four equal sides put together.
▸ Why?
The perimeter is 4 equal groups of one side, so sharing 52 into those 4 equal groups gives 13 for each side.
▸ Why?
In every right triangle the square on the hypotenuse equals the sum of the squares on the two legs.
Double to get the other diagonal
Since the diagonals bisect each other, double BM to get the full second diagonal BD = 10 meters.
Using the bisecting property of the rhombus's diagonals to recover the full length is still Grade 5 quadrilateral reasoning.
5.G.B.4Identify SubproblemsApply the rhombus area formula
The rhombus area is half the product of the diagonals: × 24 × 10 = 120 square meters.
The half-product-of-diagonals formula for a special quadrilateral is Grade 6 area-of-polygons content.
6.G.A.1Identify SubproblemsThis AMC 8 problem only needs the Grade 8 Pythagorean theorem (the 5-12-13 right triangle hiding inside a rhombus) you already know!
- Find the side length
- Halve the given diagonal
- Use the Pythagorean theorem
- Double to get the other diagonal
- Apply the rhombus area formula
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