AMC 8 · 1999 · #14
Grade 8 geometry-2dProblem
In trapezoid ABCD, the sides AB and CD are equal. The perimeter of ABCD is
Pick an answer.
AMC 8 1999 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: 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
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) $48$
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
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 $F$.
💡 Grade 6 "decompose a polygon into simpler shapes": a rectangle plus two right triangles is far easier to measure than the trapezoid itself.
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 $16 - 8 = 8$.
💡 Splitting the long base into the rectangle's footprint plus two overhangs is the natural Grade 6 length-arithmetic move.
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 $4$.
💡 Grade 4 classifies trapezoids and recognizes that the isosceles version has a vertical line of symmetry — that symmetry forces the two overhangs to match.
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.
💡 This is the classic $3$-$4$-$5$ right triangle from Grade 8 Pythagorean practice — no calculator needed.
4.MD.A.3 Step 5 - The other triangle is congruent, so $CD = AB = 5$.
- Add the four sides of the trapezoid.
💡 Grade 4 "perimeter = sum of side lengths" closes the problem once all four sides are known.
6.G.A.1 Drop perpendiculars from $B$ and $C$ down to the bottom base $AD$, meeting it at 6.G.A.1 The bottom base $AD = 16$ is made of three segments: $AE + EF + FD$. We know $EF 4.G.A.2 Because the trapezoid is isosceles, the left and right right triangles are congr 8.G.B.7 Now $\triangle ABE$ is a right triangle with legs $AE = 4$ and $BE = 3$. The Pyt 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.2Classify 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.3Apply 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.1Find 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.7Apply 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).
⭐ 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).
More like this
Same archetype — closest grade level first.
