AMC 10 · 2011 · #9
Grade 8 geometry-2dThe area of △EBD is one third of the area of △ABC. Segment DE is perpendicular to segment AB. What is BD?
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: In the figure, triangle $ABC$ has side lengths $AC=3$, $CB=4$, and $AB=5$. Point $D$ lies on $AB$ and point $E$ is placed so that segment $DE$ is perpendicular to $AB$. The area of $\triangle EBD$ is exactly one third of the area of $\triangle ABC$. Find the length $BD$.
Givens: $\triangle ABC$ has $AC=3$, $CB=4$, $AB=5$; $DE \perp AB$, with $D$ on segment $AB$ (right angle marked at $D$); $[\triangle EBD] = \tfrac13\,[\triangle ABC]$; Answer choices: (A) $\tfrac{4}{3}$, (B) $\sqrt5$, (C) $\tfrac{9}{4}$, (D) $\tfrac{4\sqrt3}{3}$, (E) $\tfrac{5}{2}$
Unknowns: The length of segment $BD$
Understand
Restated: In the figure, triangle $ABC$ has side lengths $AC=3$, $CB=4$, and $AB=5$. Point $D$ lies on $AB$ and point $E$ is placed so that segment $DE$ is perpendicular to $AB$. The area of $\triangle EBD$ is exactly one third of the area of $\triangle ABC$. Find the length $BD$.
Givens: $\triangle ABC$ has $AC=3$, $CB=4$, $AB=5$; $DE \perp AB$, with $D$ on segment $AB$ (right angle marked at $D$); $[\triangle EBD] = \tfrac13\,[\triangle ABC]$; Answer choices: (A) $\tfrac{4}{3}$, (B) $\sqrt5$, (C) $\tfrac{9}{4}$, (D) $\tfrac{4\sqrt3}{3}$, (E) $\tfrac{5}{2}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #4 Introduce a Variable
Tool #1 (Draw a Diagram) is the key: reading the figure shows that $\triangle EBD$ and $\triangle ABC$ share the corner at $B$ and each contain a right angle (at $D$ and at $C$), which is exactly the setup for similar triangles. Tool #7 (Identify Subproblems) breaks the work into three clean pieces — find the area of $\triangle ABC$, turn the area condition into a ratio of sides, then read off $BD$. Tool #4 (Introduce a Variable) names the shared scale factor $k$ so the area-to-side relationship becomes a short equation.
Execute — Answer: D
8.G.B.7 Step 1 Recognize the 3-4-5 right triangle
- The three sides of $\triangle ABC$ are $3$, $4$, and $5$.
- Checking the Pythagorean relation, $3^2+4^2 = 9+16 = 25 = 5^2$, so the triangle is right-angled, with the right angle at $C$ (opposite the longest side $AB=5$).
💡 If the three sides fit $a^2+b^2=c^2$, the corner across from the longest side is a perfect right angle.
6.G.A.1 Step 2 Area of the big triangle
- With the right angle at $C$, the two legs $CA=3$ and $CB=4$ are perpendicular, so they act as base and height.
- The area of $\triangle ABC$ is $\tfrac12\cdot 3\cdot 4 = 6$.
- The target triangle $\triangle EBD$ has one third of this, so its area is $\tfrac13\cdot 6 = 2$.
💡 In a right triangle the two legs are already a base and a matching height, so the area is just half their product.
8.G.A.4 Step 3 Find the similar triangles
- Since $DE\perp AB$, the angle at $D$ in $\triangle BDE$ is $90^\circ$, matching the right angle at $C$ in $\triangle BCA$.
- Both triangles also share the angle at $B$.
- Two equal pairs of angles mean the triangles are similar by AA: $\triangle BDE \sim \triangle BCA$.
- Matching the corners, $BD$ corresponds to $BC$ and $DE$ corresponds to $CA$.
💡 A shared angle plus a matching right angle forces the third angles to agree, so the two triangles are the same shape.
8.G.A.4 Step 4 Area ratio gives the scale factor
- For similar triangles, the ratio of areas equals the square of the ratio of matching sides.
- Call that side ratio $k = \tfrac{BD}{BC}$.
- Then $k^2$ equals the area ratio, which is $\tfrac{[\triangle EBD]}{[\triangle ABC]} = \tfrac{2}{6} = \tfrac13$.
💡 Shrinking every side by a factor $k$ shrinks the area by $k^2$, so the area ratio is the side ratio squared.
8.EE.A.2 Step 5 Solve for BD
- Take the square root: $k = \tfrac{1}{\sqrt3}$.
- Since $k = \tfrac{BD}{BC}$ and $BC=4$, multiply to get $BD = 4\cdot\tfrac{1}{\sqrt3} = \tfrac{4}{\sqrt3}$.
- Rationalizing the denominator gives $\tfrac{4\sqrt3}{3}$, which is choice (D).
💡 Undo the squaring with a square root, then rescale $BC$ by that factor to land on $BD$.
8.G.B.7 The three sides of $\triangle ABC$ are $3$, $4$, and $5$. Checking the Pythagore 6.G.A.1 With the right angle at $C$, the two legs $CA=3$ and $CB=4$ are perpendicular, s 8.G.A.4 Since $DE\perp AB$, the angle at $D$ in $\triangle BDE$ is $90^\circ$, matching 8.G.A.4 For similar triangles, the ratio of areas equals the square of the ratio of matc 8.EE.A.2 Take the square root: $k = \tfrac{1}{\sqrt3}$. Since $k = \tfrac{BD}{BC}$ and $B Review
Reasonableness: $BD = \tfrac{4\sqrt3}{3}\approx 2.31$, which is less than $BC=4$ and less than $AB=5$, so $D$ falls comfortably between $A$ and $B$ — a sensible spot on the segment. Checking the area directly: $DE = CA\cdot k = 3\cdot\tfrac{1}{\sqrt3} = \sqrt3$, so $[\triangle EBD] = \tfrac12\cdot BD\cdot DE = \tfrac12\cdot\tfrac{4}{\sqrt3}\cdot\sqrt3 = 2$, exactly one third of $6$. This confirms choice (D).
Alternative: Skip the area-ratio rule and use the shape directly. In $\triangle BDE$ the legs satisfy $\tfrac{DE}{BD} = \tfrac{CA}{CB} = \tfrac34$, so $DE = \tfrac34\,BD$. Setting the area equal to $2$: $\tfrac12\cdot BD\cdot\tfrac34 BD = \tfrac{3}{8}BD^2 = 2$, giving $BD^2 = \tfrac{16}{3}$ and $BD = \tfrac{4}{\sqrt3} = \tfrac{4\sqrt3}{3}$ — the same answer (D).
CCSS standards used (min grade 8)
8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems (Confirming that the $3$-$4$-$5$ triangle $ABC$ is right-angled at $C$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing $[\triangle ABC] = \tfrac12\cdot 3\cdot 4 = 6$ and the target area $2$.)8.G.A.4Understand that a two-dimensional figure is similar to another if one can be obtained from the other by a sequence of rotations, reflections, translations, and dilations (Establishing $\triangle BDE \sim \triangle BCA$ by AA and using area ratio $=$ (side ratio)$^2$.)8.EE.A.2Use square root and cube root symbols to represent solutions and evaluate square roots of small perfect squares (Taking $k=\sqrt{1/3}=\tfrac{1}{\sqrt3}$ and rationalizing to get $BD=\tfrac{4\sqrt3}{3}$.)
⭐ Same-shape triangles have areas that grow by the square of the side ratio, so match the areas, take the square root, and scale the side you know.
⭐ Same-shape triangles have areas that grow by the square of the side ratio, so match the areas, take the square root, and scale the side you know.
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