AMC 8 · 2023 · #24
Grade 8 geometry-2dProblem
Isosceles △ABC has equal side lengths AB and BC. In the figure below, segments are drawn parallel to AC so that the shaded portions of △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 △ABC? (Diagram not drawn to scale.)
Pick an answer.
AMC 8 2023 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: 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
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) $15.4$
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
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 $\triangle ABC$.
💡 Parallel cuts produce similar triangles — a Grade 8 similarity fact.
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.
💡 Same-shape figures have area in the square of their length ratio — Grade 8 similarity again.
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.
💡 Squared height ratio gives area ratio — the same Grade 8 similarity tool, applied to the second figure.
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.
💡 Writing "same area" as an equation is the Grade 6 idea of solving equations by finding values that make a statement true.
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 $h$.
💡 Solving a linear equation in one variable is the Grade 8 linear-equation standard.
5.NBT.B.7 Step 6 Divide by $10$ to isolate $h$, then match the result against the choices.
💡 Dividing $146$ by $10$ to get the decimal $14.6$ is a Grade 5 decimal-arithmetic move.
8.G.A.4 Mark the diagrams. In Figure 1, the cut parallel to $\overline{AC}$ leaves a sma 8.G.A.4 Subproblem A — area of the shaded region in Figure 1. Let $A_{total}$ be the are 8.G.A.4 Subproblem B — area of the shaded region in Figure 2. The shaded region IS the s 6.EE.B.5 Combine the subproblems. The problem says the two shaded areas are equal, so set 8.EE.C.7 Expand the right side and notice the $h^2$ cancels on both sides — what looked l 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.7Add, subtract, multiply, and divide decimals to hundredths (Dividing $146 \div 10 = 14.6$ to express the final height as a decimal.)6.EE.B.5Understand 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.4Understand 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.7Solve 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!
⭐ 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!
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