AMC 10 · 2006 · #15
Grade 8 geometry-2dRhombus ABCD is similar to rhombus BFDE. The area of rhombus ABCD is 24 and ∠BAD=60∘. What is the area of rhombus BFDE?
Pick an answer.
AMC 10 2006 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 area $24$ and one angle $\angle BAD=60^\circ$. A second rhombus $BFDE$ has the same shape (it is similar to $ABCD$) and shares the diagonal $BD$. Find the area of rhombus $BFDE$.
Givens: $ABCD$ is a rhombus with $\angle BAD=60^\circ$; The area of rhombus $ABCD$ is $24$; Rhombus $BFDE$ is similar to rhombus $ABCD$; $B$ and $D$ are vertices of both rhombi, so segment $BD$ is a diagonal of each; Answer choices: (A) $6$, (B) $4\sqrt{3}$, (C) $8$, (D) $9$, (E) $6\sqrt{3}$
Unknowns: The area of rhombus $BFDE$
Understand
Restated: Rhombus $ABCD$ has area $24$ and one angle $\angle BAD=60^\circ$. A second rhombus $BFDE$ has the same shape (it is similar to $ABCD$) and shares the diagonal $BD$. Find the area of rhombus $BFDE$.
Givens: $ABCD$ is a rhombus with $\angle BAD=60^\circ$; The area of rhombus $ABCD$ is $24$; Rhombus $BFDE$ is similar to rhombus $ABCD$; $B$ and $D$ are vertices of both rhombi, so segment $BD$ is a diagonal of each; Answer choices: (A) $6$, (B) $4\sqrt{3}$, (C) $8$, (D) $9$, (E) $6\sqrt{3}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #4 Introduce a Variable
The whole problem lives in one figure, so Tool #1 (Draw a Diagram) is primary: reading the rhombus tells you that $BD$ is a diagonal of both rhombi and that the $60^\circ$ angle turns triangle $ABD$ into an equilateral triangle. Tool #7 (Identify Subproblems) splits the work into two smaller jobs — first pin down both diagonals of $ABCD$ from its area, then compare the two rhombi. Tool #4 (Introduce a Variable) names the common side length $s$ so every diagonal is written in terms of one letter, which is exactly what makes the shared diagonal $BD$ link the big rhombus to the small one.
Execute — Answer: C
7.G.A.2 Step 1 The 60 degree angle makes an equilateral triangle
- Let each side of rhombus $ABCD$ be $s$.
- Draw diagonal $BD$.
- In triangle $ABD$ the two sides $AB$ and $AD$ are both $s$ and the angle between them is $\angle BAD=60^\circ$.
- An isosceles triangle with a $60^\circ$ apex is equilateral, so the third side $BD$ also equals $s$.
- This $BD$ is the shorter diagonal of $ABCD$ (it joins the two $120^\circ$ corners $B$ and $D$), and it is the same segment that will be a diagonal of the inner rhombus $BFDE$.
💡 Two equal sides plus a 60 degree angle between them force the third side to be equal too, so the short diagonal equals the side.
8.G.B.7 Step 2 Find the long diagonal and the area
- The diagonals of a rhombus cross at right angles and cut each other in half.
- Half of $BD$ is $\tfrac{s}{2}$, and it forms a right triangle with a side $s$ as the hypotenuse and half of the long diagonal $AC$ as the other leg.
- So $\left(\tfrac{AC}{2}\right)^2=s^2-\left(\tfrac{s}{2}\right)^2=\tfrac{3s^2}{4}$, giving $AC=s\sqrt{3}$.
- The area of a rhombus is half the product of its diagonals: $\tfrac12\cdot BD\cdot AC=\tfrac12\cdot s\cdot s\sqrt3=\tfrac{\sqrt3}{2}s^2$.
- Since this equals $24$, we have $\tfrac{\sqrt3}{2}s^2=24$.
💡 Cutting a rhombus along both diagonals makes four right triangles, so the Pythagorean theorem hands you the missing diagonal.
8.G.A.4 Step 3 Locate BD inside the smaller rhombus
- Rhombus $BFDE$ is similar to $ABCD$, so it has the same $60^\circ/120^\circ$ shape and the same rule: short diagonal equals the side, long diagonal equals the side times $\sqrt3$.
- The segment $BD=s$ is a diagonal of $BFDE$ too.
- If $BD$ were the short diagonal of $BFDE$, then $BFDE$ would have side $s$ and be congruent to $ABCD$ — but $BFDE$ is the smaller inner rhombus, so that cannot be.
- Therefore $BD$ is the long diagonal of $BFDE$.
- Letting $t$ be the side of $BFDE$, the long diagonal is $t\sqrt3$, so $t\sqrt3=s$, which gives $t=\tfrac{s}{\sqrt3}$ and $t^2=\tfrac{s^2}{3}$.
💡 The shared diagonal is short for the big rhombus but long for the small one, which is exactly why the inner rhombus shrinks.
7.RP.A.3 Step 4 Compute the area of BFDE
- Rhombus $BFDE$ has the same shape as $ABCD$, so its area follows the same formula $\tfrac{\sqrt3}{2}(\text{side})^2$.
- Using $t^2=\tfrac{s^2}{3}$: $\text{Area}_{BFDE}=\tfrac{\sqrt3}{2}t^2=\tfrac{\sqrt3}{2}\cdot\tfrac{s^2}{3}=\tfrac13\left(\tfrac{\sqrt3}{2}s^2\right)=\tfrac13\cdot 24=8$.
- The area of rhombus $BFDE$ is $8$, which is choice (C).
💡 Since the smaller rhombus is built from the same shape scaled down, its area is a clean fraction of the big one.
7.G.A.2 Let each side of rhombus $ABCD$ be $s$. Draw diagonal $BD$. In triangle $ABD$ th 8.G.B.7 The diagonals of a rhombus cross at right angles and cut each other in half. Hal 8.G.A.4 Rhombus $BFDE$ is similar to $ABCD$, so it has the same $60^\circ/120^\circ$ sha 7.RP.A.3 Rhombus $BFDE$ has the same shape as $ABCD$, so its area follows the same formul Review
Reasonableness: The answer $8$ is exactly one third of the given area $24$, which fits the picture: the inner rhombus $BFDE$ is clearly smaller than $ABCD$, and a third of the area is a believable amount of shrink. It also passes the similarity test — because area scales as the square of the linear ratio and the side ratio is $t/s=1/\sqrt3$, the area ratio is $(1/\sqrt3)^2=1/3$, matching $24\to 8$. The distractors $9$, $6$, $6\sqrt3$, and $4\sqrt3$ do not come from this clean $1/3$ scaling.
Alternative: Decompose the figure directly. Drawing both diagonals of $ABCD$ and the diagonal $AC$ shows that the $60^\circ$ rhombus $ABCD$ splits into two equilateral triangles $ABD$ and $CBD$, and the inner rhombus $BFDE$ is built from congruent smaller triangles. Counting these congruent pieces shows $BFDE$ covers exactly one third of $ABCD$, so its area is $\tfrac13\cdot 24=8$ without computing any side length.
CCSS standards used (min grade 8)
7.G.A.2Draw geometric shapes with given conditions including triangles (Recognizing that an isosceles triangle with a $60^\circ$ angle is equilateral, so diagonal $BD$ equals the side $s$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the long diagonal $AC=s\sqrt3$ from the right triangle formed by the half-diagonals, then the area $\tfrac{\sqrt3}{2}s^2$.)8.G.A.4Understand that a two-dimensional figure is similar to another using transformations (Using the similarity of $BFDE$ to $ABCD$ to conclude $BD$ is the long diagonal of $BFDE$ and $t=s/\sqrt3$.)7.RP.A.3Use proportional relationships to solve multi-step ratio and percent problems (Scaling the area by the factor $t^2/s^2=1/3$ to get $\tfrac13\cdot 24=8$.)
⭐ When two shapes are similar and share a segment, figure out whether that segment is the long or the short diagonal in each — here $BD$ is short for the big rhombus but long for the small one, which makes the small area exactly one third, so $8$.
⭐ When two shapes are similar and share a segment, figure out whether that segment is the long or the short diagonal in each — here $BD$ is short for the big rhombus but long for the small one, which makes the small area exactly one third, so $8$.
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