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AMC 8 · 2005 · #19

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference
pythagorean-theoremperimeterarea-rectangles identify-subproblems ↑ Prerequisites: pythagorean-theoremperimeter
📏 Medium solution 💡 4 insights 📊 Diagram

Problem

What is the perimeter of trapezoid ABCDABCD?

Pick an answer.

(A)
180
(B)
188
(C)
196
(D)
200
(E)
204
View mode:

Toolkit + CCSS Solution

Understand

Restated: Trapezoid $ABCD$ has slant side $AB = 30$, top base $BC = 50$, slant side $CD = 25$, and a vertical height of $24$ from $B$ down to the base $AD$. Find the perimeter of $ABCD$.

Givens: $AB = 30$ (left slant side); $BC = 50$ (top base, parallel to $AD$); $CD = 25$ (right slant side); Height of the trapezoid $= 24$; Answer choices: (A) $180$, (B) $188$, (C) $196$, (D) $200$, (E) $204$

Unknowns: The perimeter $AB + BC + CD + DA$ of trapezoid $ABCD$; In particular, the length of the bottom base $AD$, the only side not given

Understand

Restated: Trapezoid $ABCD$ has slant side $AB = 30$, top base $BC = 50$, slant side $CD = 25$, and a vertical height of $24$ from $B$ down to the base $AD$. Find the perimeter of $ABCD$.

Givens: $AB = 30$ (left slant side); $BC = 50$ (top base, parallel to $AD$); $CD = 25$ (right slant side); Height of the trapezoid $= 24$; Answer choices: (A) $180$, (B) $188$, (C) $196$, (D) $200$, (E) $204$

Plan

Primary tool: #1 Draw a Diagram

Three of the four sides are given; the missing piece is the bottom base $AD$. Tool #1 (Draw a Diagram) says: add the right auxiliary lines to the picture and let the shape do the work. Dropping a perpendicular from $B$ to $AD$ and from $C$ to $AD$ cuts the trapezoid into a rectangle in the middle and two right triangles on the sides. The middle rectangle has the same length as the top base $BC = 50$. Each side triangle has the height $24$ as one leg and a known slant ($30$ or $25$) as the hypotenuse, so the Pythagorean theorem finishes each foot. The two famous triples $3$-$4$-$5$ and $7$-$24$-$25$ then snap the answer into place — no algebra needed.

Execute — Answer: A

#1 Draw a Diagram 4.G.A.1 Step 1
  • Add auxiliary lines to the picture.
  • Drop a perpendicular from $B$ to $AD$ — call its foot $E$ — and a perpendicular from $C$ to $AD$ — call its foot $F$.
  • The trapezoid now consists of right triangle $ABE$ on the left, rectangle $BCFE$ in the middle, and right triangle $CFD$ on the right.
  • The two perpendiculars are the height of the trapezoid, so $BE = CF = 24$.
$$AD = AE + EF + FD, \quad BE = CF = 24, \quad EF = BC = 50$$

💡 Drawing perpendicular auxiliary lines is the Grade 4 "identify perpendicular lines in plane figures" move that converts a hard quadrilateral into shapes whose sides we know how to measure.

#1 Draw a Diagram 8.G.B.7 Step 2
  • Find $AE$ from the left right triangle.
  • Triangle $ABE$ has hypotenuse $AB = 30$ and one leg $BE = 24$.
  • By the Pythagorean theorem, $AE^2 = 30^2 - 24^2 = 900 - 576 = 324$, so $AE = 18$.
  • (Equivalently, $30 = 6 \times 5$ and $24 = 6 \times 4$, so this is the $3$-$4$-$5$ triple scaled by $6$, giving $AE = 6 \times 3 = 18$.)
$$AE = \sqrt{30^2 - 24^2} = \sqrt{324} = 18$$

💡 Once the picture shows a right triangle with two known sides, the Grade 8 Pythagorean theorem hands you the third. Recognizing the $3$-$4$-$5$ family makes the arithmetic instant.

#1 Draw a Diagram 8.G.B.7 Step 3
  • Find $FD$ from the right right triangle.
  • Triangle $CFD$ has hypotenuse $CD = 25$ and one leg $CF = 24$.
  • By the Pythagorean theorem, $FD^2 = 25^2 - 24^2 = 625 - 576 = 49$, so $FD = 7$.
  • This is the well-known $7$-$24$-$25$ triple.
$$FD = \sqrt{25^2 - 24^2} = \sqrt{49} = 7$$

💡 Same Grade 8 theorem, second triangle. Spotting the $7$-$24$-$25$ triple is the geometry analogue of recognizing $3$-$4$-$5$ — pure pattern recall.

#1 Draw a Diagram 3.MD.D.8 Step 4
  • Assemble the bottom base, then add the four sides.
  • The base splits as $AD = AE + EF + FD = 18 + 50 + 7 = 75$.
  • The perimeter is the sum of the four sides.
$$AD = 18 + 50 + 7 = 75; \quad \text{Perimeter} = AB + BC + CD + DA = 30 + 50 + 25 + 75 = 180 \;\Rightarrow\; \textbf{(A)}$$

💡 Once every side is known, the Grade 3 definition of perimeter — add the side lengths — finishes the problem.

[1] #1 4.G.A.1 Add auxiliary lines to the picture. Drop a perpendicular from $B$ to $AD$ — call
[2] #1 8.G.B.7 Find $AE$ from the left right triangle. Triangle $ABE$ has hypotenuse $AB = 30$
[3] #1 8.G.B.7 Find $FD$ from the right right triangle. Triangle $CFD$ has hypotenuse $CD = 25$
[4] #1 3.MD.D.8 Assemble the bottom base, then add the four sides. The base splits as $AD = AE +

Review

Reasonableness: Quick sanity pass. The bottom base $AD = 75$ must be longer than the top base $BC = 50$, since the slant sides lean outward in both directions in the picture, and indeed $75 > 50$. The two horizontal overhangs add to $18 + 7 = 25$, which equals $75 - 50$ — the algebra and the picture agree. Both right triangles are standard Pythagorean triples ($3$-$4$-$5$ scaled by $6$, and $7$-$24$-$25$), so no irrational numbers ever appeared, which fits an AMC 8 problem. Finally, the perimeter $30 + 50 + 25 + 75 = 180$ matches the smallest answer choice (A), and the next choice (B) $188$ would require $AD = 83$, which is inconsistent with both Pythagorean computations.

Alternative: Tool #9 (Try an Easier Problem): treat each slant side as the hypotenuse of an unknown right triangle whose vertical leg is the height $24$, and ask only for the horizontal overhang. The left overhang satisfies $x^2 + 24^2 = 30^2$, giving $x = 18$; the right overhang satisfies $y^2 + 24^2 = 25^2$, giving $y = 7$. Adding the top base back in, $AD = 18 + 50 + 7 = 75$ and the perimeter is $30 + 50 + 25 + 75 = 180$. Same answer (A), reached by reducing the trapezoid to two independent right-triangle subproblems.

CCSS standards used (min grade 8)

  • 3.MD.D.8 Solve real world and mathematical problems involving perimeters of polygons (Adding the four side lengths $30 + 50 + 25 + 75 = 180$ to get the perimeter of the trapezoid.)
  • 4.G.A.1 Draw points, lines, line segments, rays, angles, and perpendicular and parallel lines (Drawing the two perpendicular auxiliary lines from $B$ and $C$ to the base $AD$, which decompose the trapezoid into a rectangle and two right triangles.)
  • 8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems (Computing the two horizontal overhangs $AE = \sqrt{30^2 - 24^2} = 18$ and $FD = \sqrt{25^2 - 24^2} = 7$ from the slant sides and the height.)

⭐ Two perpendicular lines turn the trapezoid into a rectangle plus two right triangles. The $3$-$4$-$5$ and $7$-$24$-$25$ triples then hand you the missing base in seconds, and the perimeter is just the sum of the four sides.

⭐ Two perpendicular lines turn the trapezoid into a rectangle plus two right triangles. The $3$-$4$-$5$ and $7$-$24$-$25$ triples then hand you the missing base in seconds, and the perimeter is just the sum of the four sides.

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