AMC 8 · 2023 · #24

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

area-trianglessimilar-figuresratio-proportion area-differenceidentify-subproblemsconvert-to-algebra ↑ Prerequisites: area-trianglessimilar-figures

📏 Medium solution 💡 4 insights 📊 Diagram

Problem

Isosceles $\triangle ABC$ has equal side lengths $AB$ and $BC$ . In the figure below, segments are drawn parallel to $\overline{AC}$ so that the shaded portions of $\triangle ABC$ have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of $h$ of $\triangle ABC$ ? (Diagram not drawn to scale.)

$\textbf{(A) } 14.6 \qquad \textbf{(B) } 14.8 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 15.2 \qquad \textbf{(E) } 15.4$

Pick an answer.

(A)

14.6

(B)

14.8

(C)

(D)

15.2

(E)

15.4

View mode:

Toolkit + CCSS Solution

Understand

Restated: An isosceles triangle $ABC$ (with $AB = BC$) has total height $h$ from base $\overline{AC}$ up to the apex $B$. Two copies of the same triangle are drawn. In the first copy, one segment parallel to $\overline{AC}$ is drawn so that a small triangle of height $11$ is left UNshaded at the top, and the trapezoid below it is shaded. In the second copy, a different parallel segment is drawn so that the small triangle near the apex is shaded and the trapezoid at the bottom (height $5$) is UNshaded. The two shaded regions have the same area. Find $h$.

Givens: $\triangle ABC$ is isosceles with $AB = BC$ and total height $h$; Figure 1: small unshaded triangle at the top has height $11$; the rest (a trapezoid) is shaded; Figure 2: small shaded triangle at the top; unshaded trapezoid at the bottom has height $5$, so the shaded triangle has height $h - 5$; $\text{Area(shaded}_1) = \text{Area(shaded}_2)$; Answer choices: (A) $14.6$, (B) $14.8$, (C) $15$, (D) $15.2$, (E) $15.4$

Unknowns: The full height $h$ of $\triangle ABC$

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram, #13 Convert to Algebra

The two figures are independent area calculations that must be made equal — that's a textbook Tool #7 (Subproblems) setup: solve each shaded area separately, then equate. Tool #1 (Diagram) keeps the geometry honest: each small top triangle (heights $11$ and $h-5$) is similar to $\triangle ABC$, so its area is $A_{total} \cdot (\text{height ratio})^2$. Once each shaded area is written symbolically, Tool #13 (Algebra) finishes it: the $A_{total}$ and $h^2$ cancel, leaving a simple linear equation in $h$. Younger tools (Guess & Check on the five answer choices) also work and we mention it in Review.

Execute — Answer: A

#1 Draw a Diagram 8.G.A.4 Step 1

Mark the diagrams.
In Figure 1, the cut parallel to $\overline{AC}$ leaves a small UNshaded triangle of height $11$ at the apex.
In Figure 2, a different cut leaves the trapezoid of height $5$ UNshaded at the bottom, so the shaded triangle at the top has height $h - 5$.
Both small top triangles share angles with $\triangle ABC$ (parallel cut $\Rightarrow$ corresponding angles equal), so they are similar to $\triangle ABC$.

$$\text{Fig 1: top triangle height} = 11. \quad \text{Fig 2: top triangle height} = h - 5.$$

💡 Parallel cuts produce similar triangles — a Grade 8 similarity fact.

#7 Identify Subproblems 8.G.A.4 Step 2

Subproblem A — area of the shaded region in Figure 1.
Let $A_{total}$ be the area of $\triangle ABC$.
The small top triangle is similar to $\triangle ABC$ with height ratio $\tfrac{11}{h}$, so its area ratio is $\left(\tfrac{11}{h}\right)^2 = \tfrac{121}{h^2}$.
The shaded trapezoid is everything else.

$$A_{\text{shaded},1} = A_{total} - A_{total} \cdot \dfrac{121}{h^2} = A_{total}\!\left(1 - \dfrac{121}{h^2}\right)$$

💡 Same-shape figures have area in the square of their length ratio — Grade 8 similarity again.

#7 Identify Subproblems 8.G.A.4 Step 3

Subproblem B — area of the shaded region in Figure 2.
The shaded region IS the small top triangle, similar to $\triangle ABC$ with height ratio $\tfrac{h-5}{h}$.
Squaring this ratio gives its share of the total area.

$$A_{\text{shaded},2} = A_{total} \cdot \left(\dfrac{h-5}{h}\right)^2 = A_{total} \cdot \dfrac{(h-5)^2}{h^2}$$

💡 Squared height ratio gives area ratio — the same Grade 8 similarity tool, applied to the second figure.

#13 Convert to Algebra 6.EE.B.5 Step 4

Combine the subproblems.
The problem says the two shaded areas are equal, so set the two expressions equal.
$A_{total}$ is nonzero (it's a real triangle), so divide it out; then multiply both sides by $h^2$ to clear the denominators.

$$1 - \dfrac{121}{h^2} = \dfrac{(h-5)^2}{h^2} \;\Longrightarrow\; h^2 - 121 = (h-5)^2$$

💡 Writing "same area" as an equation is the Grade 6 idea of solving equations by finding values that make a statement true.

#13 Convert to Algebra 8.EE.C.7 Step 5

Expand the right side and notice the $h^2$ cancels on both sides — what looked like a quadratic equation is actually linear in $h$.

$$h^2 - 121 = h^2 - 10h + 25 \;\Longrightarrow\; -121 = -10h + 25 \;\Longrightarrow\; 10h = 146$$

💡 Solving a linear equation in one variable is the Grade 8 linear-equation standard.

#13 Convert to Algebra 5.NBT.B.7 Step 6

Divide by $10$ to isolate $h$, then match the result against the choices.

$$h = \dfrac{146}{10} = 14.6 \;\Longrightarrow\; \textbf{(A)}$$

💡 Dividing $146$ by $10$ to get the decimal $14.6$ is a Grade 5 decimal-arithmetic move.

[1] #1 8.G.A.4 Mark the diagrams. In Figure 1, the cut parallel to $\overline{AC}$ leaves a sma

[2] #7 8.G.A.4 Subproblem A — area of the shaded region in Figure 1. Let $A_{total}$ be the are

[3] #7 8.G.A.4 Subproblem B — area of the shaded region in Figure 2. The shaded region IS the s

[4] #13 6.EE.B.5 Combine the subproblems. The problem says the two shaded areas are equal, so set

[5] #13 8.EE.C.7 Expand the right side and notice the $h^2$ cancels on both sides — what looked l

[6] #13 5.NBT.B.7 Divide by $10$ to isolate $h$, then match the result against the choices.

Review

Reasonableness: Sanity check the answer. With $h = 14.6$, the small UNshaded triangle in Figure 1 has area-fraction $\tfrac{121}{14.6^2} = \tfrac{121}{213.16} \approx 0.568$, so the shaded trapezoid is about $1 - 0.568 = 0.432$ of the whole triangle. The shaded triangle in Figure 2 has height $14.6 - 5 = 9.6$ and area-fraction $\left(\tfrac{9.6}{14.6}\right)^2 = \tfrac{92.16}{213.16} \approx 0.432$ — the two match. Also $h = 14.6$ is just barely bigger than $11$ (required so the Figure 1 cut sits below the apex), so the geometry is consistent.

Alternative: Tool #6 (Guess & Check) on the five answer choices. For each candidate $h$, compute $h^2 - 121$ and $(h-5)^2$ and see which makes them equal. For (A) $h = 14.6$: $14.6^2 - 121 = 213.16 - 121 = 92.16$ and $(14.6 - 5)^2 = 9.6^2 = 92.16$. They match exactly, confirming (A). Quick checks of (B) $14.8$ give $98.04$ vs $96.04$ — not equal — eliminating it; the remaining choices fall away by the same arithmetic.

CCSS standards used (min grade 8)

5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths (Dividing $146 \div 10 = 14.6$ to express the final height as a decimal.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Translating "the two shaded areas are equal" into the equation $1 - \tfrac{121}{h^2} = \tfrac{(h-5)^2}{h^2}$ as a statement that has a specific value of $h$ as its solution.)
8.G.A.4 Understand that a two-dimensional figure is similar to another using transformations (Recognizing that each small top triangle is similar to $\triangle ABC$ via a dilation about the apex, so its area equals $A_{total}$ times the square of the height ratio.)
8.EE.C.7 Solve linear equations in one variable (After the $h^2$ terms cancel, solving the linear equation $10h = 146$ for the height $h$.)

⭐ This AMC 8 problem only needs Grade 8 similarity (area scales with the square of the height ratio) plus a one-line linear equation you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 8)

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