AMC 8 · 2017 · #16

Grade 6 geometry-2dalgebra

Archetype Compound Area by Decomposition / Difference

perimeterarea-trianglesratio-proportionpythagorean-theorem identify-subproblemsconvert-to-algebraratio-proportion ↑ Prerequisites: area-trianglespythagorean-theorem

📏 Medium solution 💡 4 insights 📊 Diagram

Problem

In the figure below, choose point $D$ on $\overline{BC}$ so that $\triangle ACD$ and $\triangle ABD$ have equal perimeters. What is the area of $\triangle ABD$ ?

$\textbf{(A) }\frac{3}{4}\qquad\textbf{(B) }\frac{3}{2}\qquad\textbf{(C) }2\qquad\textbf{(D) }\frac{12}{5}\qquad\textbf{(E) }\frac{5}{2}$

Pick an answer.

(A)

$frac{3}{4}$

(B)

$frac{3}{2}$

(C)

(D)

$frac{12}{5}$

(E)

$frac{5}{2}$

View mode:

Toolkit + CCSS Solution

Understand

Restated: A right triangle $ABC$ has legs $AC = 3$ and $AB = 4$ and hypotenuse $BC = 5$. Place a point $D$ somewhere on segment $\overline{BC}$ so that the two smaller triangles formed, $\triangle ACD$ and $\triangle ABD$, have the same perimeter. Find the area of $\triangle ABD$.

Givens: Right triangle with legs $AC = 3$, $AB = 4$, hypotenuse $BC = 5$ (the right angle is at $A$); Point $D$ lies on $\overline{BC}$, so $BD + DC = 5$; Perimeter of $\triangle ACD$ equals perimeter of $\triangle ABD$; Answer choices: (A) $\tfrac{3}{4}$, (B) $\tfrac{3}{2}$, (C) $2$, (D) $\tfrac{12}{5}$, (E) $\tfrac{5}{2}$

Unknowns: The area of $\triangle ABD$

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram, #6 Guess and Check

The big question "area of $\triangle ABD$" breaks cleanly into three smaller questions (Tool #7): (a) where is $D$ on $\overline{BC}$? (b) what is the area of the whole right triangle $\triangle ABC$? (c) what fraction of that area belongs to $\triangle ABD$? Tool #1 (Draw a Diagram) supports (a): on the given figure we label $BD$ and $CD$ and notice the shared side $AD$ cancels from both perimeters, so the perimeter condition becomes a simple statement about $BD$ and $CD$ alone. Tool #6 (Guess and Check) handles "two pieces sum to $5$, differ by $1$" without needing formal algebra — perfect for an elementary solver.

Execute — Answer: D

#1 Draw a Diagram 4.G.A.1 Step 1

Draw the figure and add labels.
Triangle $ABC$ has the right angle at $A$, legs $AC = 3$ and $AB = 4$, hypotenuse $BC = 5$.
Drop point $D$ on $\overline{BC}$ and mark the two unknown pieces $BD$ and $CD$.
Notice that both small triangles share the cevian $AD$.

$$BD + CD = 5$$

💡 Labeling points, segments, and the shared cevian on the diagram is exactly the Grade 4 "identify lines and segments in figures" skill.

#7 Identify Subproblems 3.MD.D.8 Step 2

Write each perimeter and cancel the shared side.
The perimeter of $\triangle ACD$ is $AC + CD + AD$ and the perimeter of $\triangle ABD$ is $AB + BD + AD$.
Setting them equal, $AD$ cancels, leaving $AC + CD = AB + BD$, i.e.
$3 + CD = 4 + BD$, so $CD = BD + 1$.

$$AC + CD + AD = AB + BD + AD \;\Rightarrow\; 3 + CD = 4 + BD \;\Rightarrow\; CD - BD = 1$$

💡 Adding up the three sides of each triangle is Grade 3 polygon-perimeter work; the cancellation is just "same thing on both sides".

#6 Guess and Check 4.OA.A.3 Step 3

Find $BD$ using the two facts $BD + CD = 5$ and $CD = BD + 1$.
Guess and check: if $BD = 2$, then $CD = 3$, and $2 + 3 = 5$ with a difference of $1$ — both conditions hold.
So $BD = 2$ and $CD = 3$.

$$BD = 2, \; CD = 3$$

💡 "Two whole numbers sum to $5$ and differ by $1$" is a one-line multi-step word problem at the Grade 4 level.

#7 Identify Subproblems 6.G.A.1 Step 4

Compute the area of the whole right triangle $\triangle ABC$ using its two legs as base and height.
With $AB = 4$ and $AC = 3$ meeting at a right angle, the area is $\tfrac{1}{2} \cdot 4 \cdot 3 = 6$.

$$\text{Area}(\triangle ABC) = \tfrac{1}{2} \cdot AB \cdot AC = \tfrac{1}{2} \cdot 4 \cdot 3 = 6$$

💡 Half of base times height for a triangle is the Grade 6 area formula.

#7 Identify Subproblems 6.RP.A.3 Step 5

Use the shared-altitude principle.
Triangles $\triangle ABD$ and $\triangle ABC$ both have their apex at $A$ and bases along the same line $BC$, so they share the same altitude from $A$.
When two triangles share an altitude, the ratio of their areas equals the ratio of their bases: $\text{Area}(\triangle ABD) : \text{Area}(\triangle ABC) = BD : BC = 2 : 5$.

$$\dfrac{\text{Area}(\triangle ABD)}{\text{Area}(\triangle ABC)} = \dfrac{BD}{BC} = \dfrac{2}{5}$$

💡 Comparing two area-pieces that share a height is a Grade 6 ratio-reasoning move.

#7 Identify Subproblems 5.NF.B.6 Step 6

Multiply to get the final area: $\text{Area}(\triangle ABD) = \tfrac{2}{5} \cdot 6 = \tfrac{12}{5}$.
This matches answer choice (D).

$$\text{Area}(\triangle ABD) = \dfrac{2}{5} \cdot 6 = \dfrac{12}{5} \;\Rightarrow\; \textbf{(D)}$$

💡 Multiplying a fraction by a whole number ($\tfrac{2}{5}$ of $6$) is a Grade 5 fraction-of-a-whole calculation.

[1] #1 4.G.A.1 Draw the figure and add labels. Triangle $ABC$ has the right angle at $A$, legs

[2] #7 3.MD.D.8 Write each perimeter and cancel the shared side. The perimeter of $\triangle ACD

[3] #6 4.OA.A.3 Find $BD$ using the two facts $BD + CD = 5$ and $CD = BD + 1$. Guess and check:

[4] #7 6.G.A.1 Compute the area of the whole right triangle $\triangle ABC$ using its two legs

[5] #7 6.RP.A.3 Use the shared-altitude principle. Triangles $\triangle ABD$ and $\triangle ABC$

[6] #7 5.NF.B.6 Multiply to get the final area: $\text{Area}(\triangle ABD) = \tfrac{2}{5} \cdot

Review

Reasonableness: Sanity check the size of the answer. The whole right triangle has area $6$. Since $BD = 2$ is less than half of $BC = 5$, triangle $\triangle ABD$ should have less than half the total area, i.e. less than $3$. We got $\tfrac{12}{5} = 2.4$, which is indeed less than $3$ but still a meaningful chunk — exactly $\tfrac{2}{5}$ of $6$. Also $\tfrac{12}{5} + \text{Area}(\triangle ACD) = 6$, so $\text{Area}(\triangle ACD) = \tfrac{18}{5}$, and the ratio $\tfrac{12}{5} : \tfrac{18}{5} = 2 : 3 = BD : CD$, confirming the shared-altitude argument.

Alternative: Tool #13 (Convert to Algebra) gives a direct route: let $BD = x$, then $CD = 5 - x$, and the perimeter condition $3 + (5 - x) = 4 + x$ collapses to $8 - x = 4 + x$, so $x = 2$. Then compute the area exactly as in steps 4-6. Algebra is cleaner here, but the guess-and-check path above is faster for an elementary student and avoids introducing a variable.

CCSS standards used (min grade 6)

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures (Labeling $D$ on $\overline{BC}$ and marking the two new segments $BD$ and $CD$ plus the shared cevian $AD$ on the figure.)
3.MD.D.8 Solve real-world problems involving perimeters of polygons (Writing the perimeters of $\triangle ACD$ and $\triangle ABD$ as sums of three sides and cancelling the shared side $AD$ to get $3 + CD = 4 + BD$.)
4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Finding two whole-number lengths that sum to $5$ and differ by $1$, giving $BD = 2$ and $CD = 3$.)
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Computing the area of right triangle $\triangle ABC$ as $\tfrac{1}{2} \cdot 4 \cdot 3 = 6$.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Setting the area ratio $\text{Area}(\triangle ABD) : \text{Area}(\triangle ABC)$ equal to the base ratio $BD : BC = 2 : 5$ since the two triangles share an altitude from $A$.)
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers (Multiplying $\tfrac{2}{5} \cdot 6 = \tfrac{12}{5}$ to extract the final area of $\triangle ABD$.)

⭐ This AMC 8 problem only needs Grade 6 ratio reasoning — when two triangles share a height, their areas split in the same ratio as their bases — that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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