AMC 8 · 2003 · #21

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

area-rectanglesarea-trianglespythagorean-theoremperimeter area-differenceidentify-subproblems ↑ Prerequisites: area-trianglesarea-rectanglespythagorean-theoremlinear-equations-one-var

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

The area of trapezoid $ABCD$ is $164\text{ cm}^2$ . The altitude is 8 cm, $AB$ is 10 cm, and $CD$ is 17 cm. What is $BC$ , in centimeters?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Trapezoid $ABCD$ has area $164\text{ cm}^2$, altitude $8\text{ cm}$, leg $AB = 10\text{ cm}$, and leg $CD = 17\text{ cm}$. The parallel sides are the top $BC$ and the bottom $AD$. Find the length of the shorter base $BC$.

Givens: Trapezoid area: $164\text{ cm}^2$; Altitude (perpendicular distance between $BC$ and $AD$): $8\text{ cm}$; Slant side $AB = 10\text{ cm}$, slant side $CD = 17\text{ cm}$; $BC \parallel AD$, with $BC$ on top and $AD$ on the bottom; Answer choices: (A) $9$, (B) $10$, (C) $12$, (D) $15$, (E) $20$

Unknowns: The length of the top base $BC$, in centimeters

Understand

Plan

Primary tool: #7 Break into Subproblems

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

The trapezoid carries three independent facts (area, altitude, two slant sides), so Tool #7 (Break into Subproblems) splits the figure into pieces we already know how to handle: drop perpendiculars from $B$ and $C$ to $AD$, and the trapezoid becomes one rectangle plus two right triangles. Tool #1 (Draw a Diagram) makes the split visible — once the perpendiculars are drawn, the Pythagorean theorem cracks each side triangle. Tool #13 (Convert to Algebra) ties the pieces together: the area formula gives $BC + AD = 41$, and the triangle bases give $AD = BC + 21$. Solve the pair.

Execute — Answer: B

#7 Break into Subproblems 6.G.A.1 Step 1

Subproblem 1 — use the trapezoid area formula.
With parallel bases $BC$ and $AD$ and altitude $8$, the formula $A = \tfrac{1}{2}(b_1 + b_2)h$ becomes a single equation linking the two bases.
Substitute the given area and altitude and simplify.

$$164 = \tfrac{1}{2}(BC + AD) \cdot 8 \;\Rightarrow\; 164 = 4(BC + AD) \;\Rightarrow\; BC + AD = 41$$

💡 Grade 6 finds trapezoid area by decomposing into a rectangle and two triangles, then averaging the two bases. The formula is the algebra version of that average-and-multiply move.

#1 Draw a Diagram 7.G.B.6 Step 2

Subproblem 2 — drop perpendiculars from $B$ and $C$ to $AD$, calling the feet $E$ and $F$.
This carves the trapezoid into three pieces: right triangle $\triangle ABE$ on the left, rectangle $BCFE$ in the middle, and right triangle $\triangle CFD$ on the right.
Both vertical legs $BE$ and $CF$ equal the altitude $8$.

$$AD = AE + EF + FD, \quad EF = BC, \quad BE = CF = 8$$

💡 The diagram exposes the structure: a rectangle hides between the two right triangles, and the rectangle's top is exactly $BC$. The longer base is just the rectangle width plus the two triangle feet.

#7 Break into Subproblems 8.G.B.7 Step 3

Apply the Pythagorean theorem to each side triangle to find its horizontal foot.
In $\triangle ABE$ the hypotenuse is $AB = 10$ and a leg is $BE = 8$; in $\triangle CFD$ the hypotenuse is $CD = 17$ and a leg is $CF = 8$.
Both are familiar Pythagorean triples.

$$AE = \sqrt{10^2 - 8^2} = \sqrt{36} = 6, \quad FD = \sqrt{17^2 - 8^2} = \sqrt{225} = 15$$

💡 $6\text{-}8\text{-}10$ is the $3\text{-}4\text{-}5$ triple doubled, and $8\text{-}15\text{-}17$ is a classic AMC-favorite triple — both pop out without a calculator.

#13 Convert to Algebra 6.EE.A.2 Step 4

Combine the pieces.
The longer base $AD$ is the sum of the two triangle feet plus the rectangle width $EF = BC$.
This gives a second equation linking $BC$ and $AD$.

$$AD = AE + EF + FD = 6 + BC + 15 = BC + 21$$

💡 The rectangle's top side equals its bottom side, so $EF = BC$. The longer base is the shorter base plus a fixed overhang of $6 + 15 = 21$ centimeters.

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

Solve the system $\,BC + AD = 41\,$ and $\,AD = BC + 21\,$ by substituting the second into the first.
The two equations collapse to a single equation in $BC$.

$$BC + (BC + 21) = 41 \;\Rightarrow\; 2BC = 20 \;\Rightarrow\; BC = 10 \;\Rightarrow\; \textbf{(B)}$$

💡 Two linear equations in two unknowns: substitution turns the pair into one equation. The Grade 8 system-of-equations move finishes the geometry problem in one line.

[1] #7 6.G.A.1 Subproblem 1 — use the trapezoid area formula. With parallel bases $BC$ and $AD$

[2] #1 7.G.B.6 Subproblem 2 — drop perpendiculars from $B$ and $C$ to $AD$, calling the feet $E

[3] #7 8.G.B.7 Apply the Pythagorean theorem to each side triangle to find its horizontal foot.

[4] #13 6.EE.A.2 Combine the pieces. The longer base $AD$ is the sum of the two triangle feet plu

[5] #13 8.EE.C.8 Solve the system $\,BC + AD = 41\,$ and $\,AD = BC + 21\,$ by substituting the s

Review

Reasonableness: Plug $BC = 10$ back into the picture. Then $AD = 10 + 21 = 31$, and the trapezoid area is $\tfrac{1}{2}(10 + 31) \cdot 8 = \tfrac{1}{2} \cdot 41 \cdot 8 = 164\text{ cm}^2$, matching the given area exactly. The horizontal feet $6$ and $15$ combine with the altitude $8$ to give hypotenuses $\sqrt{6^2+8^2}=10$ and $\sqrt{15^2+8^2}=17$, matching $AB$ and $CD$. Every input is reproduced, so the answer is consistent. Among the choices, $10$ is the only value that keeps the longer base $AD = 31$ positive and shorter than the trapezoid would need; values like $20$ would force $AD = 41 - 20 = 21$, which is shorter than $BC + 21 = 41$ — a contradiction.

Alternative: Tool #3 (Eliminate Possibilities): for each answer choice, set $BC$ equal to that value and check whether $AD = 41 - BC$ also equals $BC + 21$. Choice (A) $BC = 9$: needs $AD = 32$ and $AD = 30$ — mismatch. (B) $BC = 10$: $AD = 31$ and $AD = 31$ — match. (C) $BC = 12$: $32$ vs $33$ — mismatch. (D) $BC = 15$: $26$ vs $36$ — mismatch. (E) $BC = 20$: $21$ vs $41$ — mismatch. Only (B) satisfies both conditions.

CCSS standards used (min grade 8)

6.G.A.1 Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing and decomposing (Applying the trapezoid area formula $A = \tfrac{1}{2}(b_1 + b_2)h$ to convert area $164$ and altitude $8$ into the equation $BC + AD = 41$.)
7.G.B.6 Solve real-world and mathematical problems involving area, volume, and surface area of two- and three-dimensional objects (Decomposing the trapezoid into a central rectangle and two right triangles after dropping perpendiculars from $B$ and $C$ to $AD$.)
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Finding the horizontal feet $AE = 6$ from the $6\text{-}8\text{-}10$ triangle and $FD = 15$ from the $8\text{-}15\text{-}17$ triangle.)
6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Writing the longer base as $AD = BC + 21$ once the two triangle feet $6$ and $15$ are added to the rectangle width $BC$.)
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations (Solving the system $BC + AD = 41$ and $AD = BC + 21$ by substitution to get $BC = 10$.)

⭐ A trapezoid with two slant sides is really a rectangle hiding between two right triangles. Drop the altitudes, use the Pythagorean theorem on each side triangle, then let the area formula seal the answer.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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