AMC 8 · 1999 · #14

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

perimeterpythagorean-theoremisosceles-trianglearea-rectangles identify-subproblemsarea-difference ↑ Prerequisites: pythagorean-theoremperimeter

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

In trapezoid $ABCD$ , the sides $AB$ and $CD$ are equal. The perimeter of $ABCD$ is

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Trapezoid $ABCD$ has $AB = CD$ (isosceles). From the figure, the parallel sides are $BC = 8$ on top and $AD = 16$ on the bottom, and the trapezoid's height (perpendicular distance between the parallel sides) is $3$. Find the perimeter of $ABCD$.

Givens: $ABCD$ is a trapezoid with $AB = CD$, so it is isosceles; Top base $BC = 8$; Bottom base $AD = 16$; Height of the trapezoid $= 3$; Answer choices: (A) $27$, (B) $30$, (C) $32$, (D) $34$, (E) $48$

Unknowns: The perimeter $AB + BC + CD + DA$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

The trapezoid's slanted sides are unknown but its height $3$ and the two parallel sides $8$ and $16$ are given. Tool #1 (Draw a Diagram) suggests adding the standard auxiliary lines: drop perpendiculars from $B$ and $C$ down to the bottom base $AD$. That cut splits the figure into a central rectangle and two congruent right triangles (congruent because the trapezoid is isosceles), which is Tool #7 (Identify Subproblems) — the trapezoid problem becomes a Pythagorean-theorem problem on one easy right triangle.

Execute — Answer: D

#1 Draw a Diagram 6.G.A.1 Step 1

Drop perpendiculars from $B$ and $C$ down to the bottom base $AD$, meeting it at $E$ and $F$.
The middle piece $BCFE$ is a rectangle (two pairs of parallel sides, all right angles), and the two outer pieces $\triangle ABE$ and $\triangle CFD$ are right triangles with the right angle at $E$ and $F$.

$$BE = CF = 3, \quad EF = BC = 8$$

💡 Grade 6 "decompose a polygon into simpler shapes": a rectangle plus two right triangles is far easier to measure than the trapezoid itself.

#7 Identify Subproblems 6.G.A.1 Step 2

The bottom base $AD = 16$ is made of three segments: $AE + EF + FD$.
We know $EF = 8$, so the two overhangs $AE$ and $FD$ together must account for the leftover $16 - 8 = 8$.

$$AE + FD = 16 - 8 = 8$$

💡 Splitting the long base into the rectangle's footprint plus two overhangs is the natural Grade 6 length-arithmetic move.

#1 Draw a Diagram 4.G.A.2 Step 3

Because the trapezoid is isosceles, the left and right right triangles are congruent mirror images, so the two overhangs are equal: $AE = FD$.
Splitting the leftover $8$ in half gives each overhang as $4$.

$$AE = FD = \dfrac{8}{2} = 4$$

💡 Grade 4 classifies trapezoids and recognizes that the isosceles version has a vertical line of symmetry — that symmetry forces the two overhangs to match.

#7 Identify Subproblems 8.G.B.7 Step 4

Now $\triangle ABE$ is a right triangle with legs $AE = 4$ and $BE = 3$.
The Pythagorean theorem gives the hypotenuse $AB$, which is one of the slanted sides we need.

$$AB = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$

💡 This is the classic $3$-$4$-$5$ right triangle from Grade 8 Pythagorean practice — no calculator needed.

#7 Identify Subproblems 4.MD.A.3 Step 5

The other triangle is congruent, so $CD = AB = 5$.
Add the four sides of the trapezoid.

$$\text{Perimeter} = AB + BC + CD + DA = 5 + 8 + 5 + 16 = 34 \;\Rightarrow\; \textbf{(D)}$$

💡 Grade 4 "perimeter = sum of side lengths" closes the problem once all four sides are known.

[1] #1 6.G.A.1 Drop perpendiculars from $B$ and $C$ down to the bottom base $AD$, meeting it at

[2] #7 6.G.A.1 The bottom base $AD = 16$ is made of three segments: $AE + EF + FD$. We know $EF

[3] #1 4.G.A.2 Because the trapezoid is isosceles, the left and right right triangles are congr

[4] #7 8.G.B.7 Now $\triangle ABE$ is a right triangle with legs $AE = 4$ and $BE = 3$. The Pyt

[5] #7 4.MD.A.3 The other triangle is congruent, so $CD = AB = 5$. Add the four sides of the tra

Review

Reasonableness: Sanity check the slant length with the triangle inequality and a rough estimate. The slant rises $3$ over a run of $4$, so it must be longer than $4$ but shorter than $4 + 3 = 7$ — and $5$ sits right in that range. Sanity check the total: the two horizontal sides already contribute $8 + 16 = 24$, so the perimeter must exceed $24$, which eliminates choices below $24$ immediately. Adding two slants of length $5$ gives $24 + 10 = 34$, matching (D). Choice (E) $48$ would require each slant to be $12$, far too long for a $3$-tall rise; choice (A) $27$ would force slants of length $1.5$, shorter than the height itself — impossible.

Alternative: Tool #3 (Eliminate Possibilities). The horizontal sides alone sum to $8 + 16 = 24$, so the perimeter equals $24 + 2 \cdot (\text{slant})$. The slant length must exceed the height $3$ (legs are always shorter than the hypotenuse), so the perimeter exceeds $24 + 6 = 30$, ruling out (A) and (B). The horizontal run for each slant is half of $16 - 8 = 8$, namely $4$, so the slant is less than $4 + 3 = 7$ (triangle inequality), making the perimeter less than $24 + 14 = 38$, ruling out (E). Only (C) $32$ and (D) $34$ remain. A single Pythagorean check $\sqrt{3^2 + 4^2} = 5$ then pins down (D).

CCSS standards used (min grade 8)

4.G.A.2 Classify two-dimensional figures based on the presence of parallel or perpendicular lines, or the presence of angles of a specified size (Identifying $ABCD$ as an isosceles trapezoid and using its line of symmetry to conclude that the two bottom overhangs $AE$ and $FD$ are equal.)
4.MD.A.3 Apply the area and perimeter formulas for rectangles in real world and mathematical problems (Computing the trapezoid's perimeter as the sum of its four side lengths $5 + 8 + 5 + 16 = 34$.)
6.G.A.1 Find the area of polygons by composing into rectangles or decomposing into triangles and other shapes (Decomposing the trapezoid into a central rectangle $BCFE$ plus two right triangles $\triangle ABE$ and $\triangle CFD$ by dropping perpendiculars from $B$ and $C$ to the bottom base.)
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems (Computing the slanted side $AB = \sqrt{4^2 + 3^2} = 5$ from the right triangle with legs $3$ (height) and $4$ (overhang).)

⭐ Drop two height lines, split the trapezoid into a rectangle plus two matching right triangles, and a single Pythagorean step on a $3$-$4$-$5$ triangle gives slant $= 5$ — perimeter $5 + 8 + 5 + 16 = 34$, answer (D).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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