AMC 10 · 2012 · #14
Grade 8 geometry-2dTwo equilateral triangles are contained in square whose side length is 23. The bases of these triangles are the opposite side of the square, and their intersection is a rhombus. What is the area of the rhombus?
Pick an answer.
AMC 10 2012 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 square has side length $2\sqrt{3}$. Two equilateral triangles sit inside it: one has its base on the bottom side of the square, the other has its base on the top side. The region where the two triangles overlap is a rhombus. Find the area of that rhombus.
Givens: The square has side length $2\sqrt{3}$.; Each triangle is equilateral, so all three of its sides are equal and every angle is $60^\circ$.; One triangle's base is the bottom side of the square; the other's base is the top side.; The overlap of the two triangles is a rhombus.
Unknowns: The area of the rhombus formed by the overlap.
Understand
Restated: A square has side length $2\sqrt{3}$. Two equilateral triangles sit inside it: one has its base on the bottom side of the square, the other has its base on the top side. The region where the two triangles overlap is a rhombus. Find the area of that rhombus.
Givens: The square has side length $2\sqrt{3}$.; Each triangle is equilateral, so all three of its sides are equal and every angle is $60^\circ$.; One triangle's base is the bottom side of the square; the other's base is the top side.; The overlap of the two triangles is a rhombus.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #4 Introduce a Variable, #17 Visualize Spatial Relationships
The shapes are all defined by position, so a coordinate diagram pins down every point exactly. Once the square and both triangles have coordinates, the rhombus is just the overlap, and its four corners are points I can compute. I break the work into small pieces: find the triangle height, find the corners, then combine the two diagonals into the area.
Execute — Answer: D
6.G.A.3 Step 1 Place the square on a grid
- Put the square's corners at $(0,0)$, $(2\sqrt{3},0)$, $(2\sqrt{3},2\sqrt{3})$, and $(0,2\sqrt{3})$.
- The bottom triangle has its base along the bottom side and points up; the top triangle has its base along the top side and points down.
- Both bases have length $2\sqrt{3}$.
💡 Coordinates turn a picture into exact numbers I can calculate with.
8.G.B.7 Step 2 Find each triangle's height
- An equilateral triangle with side $s$ has height $\frac{\sqrt{3}}{2}s$, which comes from splitting it into two right triangles and using the Pythagorean theorem: $h^2 + (\tfrac{s}{2})^2 = s^2$.
- With $s = 2\sqrt{3}$ the height is $\frac{\sqrt{3}}{2}\cdot 2\sqrt{3} = 3$.
- So the bottom triangle's apex is at $(\sqrt{3},3)$ and the top triangle's apex is at $(\sqrt{3},2\sqrt{3}-3)$.
💡 Cutting an equilateral triangle down the middle makes a right triangle, and Pythagoras gives its height.
8.EE.C.8 Step 3 Locate the two side corners
- The bottom triangle's left edge runs from $(0,0)$ to the apex $(\sqrt{3},3)$, giving the line $y=\sqrt{3}\,x$.
- The top triangle's left edge runs from $(0,2\sqrt{3})$ to its apex, giving $y=-\sqrt{3}\,x+2\sqrt{3}$.
- Solving the pair, $\sqrt{3}x=-\sqrt{3}x+2\sqrt{3}$, gives $x=1$, so the left corner of the rhombus is $(1,\sqrt{3})$.
- By symmetry the right corner is $(2\sqrt{3}-1,\sqrt{3})$.
💡 Two edges cross where their line equations agree, so solving them together finds the corner.
8.G.B.8 Step 4 Measure the two diagonals
- The rhombus's vertical diagonal joins the two apexes $(\sqrt{3},3)$ and $(\sqrt{3},2\sqrt{3}-3)$, so its length is $3-(2\sqrt{3}-3)=6-2\sqrt{3}$.
- The horizontal diagonal joins the side corners $(1,\sqrt{3})$ and $(2\sqrt{3}-1,\sqrt{3})$, so its length is $(2\sqrt{3}-1)-1=2\sqrt{3}-2$.
💡 Points sharing an $x$ (or $y$) value are a straight distance apart — just subtract the other coordinate.
7.G.B.6 Step 5 Combine into the area
- A rhombus's area is half the product of its diagonals: $\frac{1}{2}d_1 d_2$.
- So the area is $\frac{1}{2}(6-2\sqrt{3})(2\sqrt{3}-2)$.
- Expanding, $(6-2\sqrt{3})(2\sqrt{3}-2)=12\sqrt{3}-12-12+4\sqrt{3}=16\sqrt{3}-24$, and half of that is $8\sqrt{3}-12$.
- The answer is $(D)$.
💡 The two diagonals cut the rhombus into four right triangles, and their half-product gives the whole area.
6.G.A.3 Put the square's corners at $(0,0)$, $(2\sqrt{3},0)$, $(2\sqrt{3},2\sqrt{3})$, a 8.G.B.7 An equilateral triangle with side $s$ has height $\frac{\sqrt{3}}{2}s$, which co 8.EE.C.8 The bottom triangle's left edge runs from $(0,0)$ to the apex $(\sqrt{3},3)$, gi 8.G.B.8 The rhombus's vertical diagonal joins the two apexes $(\sqrt{3},3)$ and $(\sqrt{ 7.G.B.6 A rhombus's area is half the product of its diagonals: $\frac{1}{2}d_1 d_2$. So Review
Reasonableness: Numerically $8\sqrt{3}-12 \approx 1.86$. The whole square has area $(2\sqrt{3})^2 = 12$, and the small overlap rhombus is a thin sliver near the center, so an area under $2$ is sensible. The diagonals $6-2\sqrt{3}\approx 2.54$ and $2\sqrt{3}-2\approx 1.46$ are both positive and fit inside the square, confirming the corners are placed correctly.
Alternative: Instead of coordinates, use the $30$-$60$-$90$ triangles directly. Each triangle has height $3$, but the square is only $2\sqrt{3}\approx 3.46$ tall, so the two heights overlap by $3+3-2\sqrt{3}=6-2\sqrt{3}$; that overlap is the vertical diagonal. The rhombus is four $30$-$60$-$90$ triangles, and finding one triangle's legs from that diagonal gives the same area $8\sqrt{3}-12$.
CCSS standards used (min grade 8)
6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Placing the square and the two triangles on a coordinate grid.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding each equilateral triangle's height of 3.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Solving two edge lines together to locate a side corner of the rhombus.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Measuring the two diagonals from their endpoints' coordinates.)7.G.B.6Solve real-world problems involving area, surface area, and volume (Using the half-product-of-diagonals formula to get the rhombus area.)
⭐ Put the figure on a grid, find the four corners of the overlap, and a rhombus's area is just half the product of its two diagonals.
⭐ Put the figure on a grid, find the four corners of the overlap, and a rhombus's area is just half the product of its two diagonals.
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