AMC 8 · 2020 · #18

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation

pythagorean-theoremarea-rectanglesarea-circles identify-subproblems ↑ Prerequisites: pythagorean-theoremarea-rectangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$

$\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272$

Pick an answer.

(A)

240

(B)

248

(C)

256

(D)

264

(E)

272

View mode:

Toolkit + CCSS Solution

Understand

Restated: A rectangle $ABCD$ sits inside a semicircle so that side $\overline{DA}$ lies on the diameter $\overline{FE}$, with $D$ and $A$ between $F$ and $E$. The two upper corners $B$ and $C$ touch the curved part of the semicircle. We are told $DA = 16$ and $FD = AE = 9$. Find the area of rectangle $ABCD$.

Givens: $ABCD$ is a rectangle inscribed in a semicircle with diameter $\overline{FE}$; Side $\overline{DA}$ lies on the diameter, $DA = 16$; $FD = AE = 9$ (the two leftover diameter pieces are equal); Corners $B$ and $C$ lie on the semicircular arc; Answer choices: (A) $240$, (B) $248$, (C) $256$, (D) $264$, (E) $272$

Unknowns: The area of rectangle $ABCD$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

The picture already exists, but the key insight is hidden until we *add* one more thing to it (Tool #1): mark the center $O$ of the semicircle and draw the radius $\overline{OC}$. That single extra segment turns the problem into a right triangle $\triangle ODC$ that we can attack directly. Tool #7 (Identify Subproblems) then breaks the area question into three clean pieces: (1) find the radius from the diameter, (2) find the height $CD$ from a right triangle, (3) multiply length $\times$ height to get the area. Each subproblem uses only one idea at a time.

Execute — Answer: A

#1 Draw a Diagram 4.NBT.B.4 Step 1

Add the center $O$ of the semicircle to the diagram.
The diameter is $FE = FD + DA + AE = 9 + 16 + 9 = 34$, so the radius is $r = 34 \div 2 = 17$.
Because $FD = AE = 9$, the figure is symmetric, so $O$ sits exactly at the midpoint of $\overline{DA}$.

$$FE = 9 + 16 + 9 = 34, \quad r = \tfrac{34}{2} = 17$$

💡 Adding $9 + 16 + 9$ and halving the result is just Grade 4 multi-digit addition and a simple division you already know.

#7 Identify Subproblems 3.OA.A.2 Step 2

Find $OD$, the distance from the center to corner $D$.
Since $O$ is the midpoint of $\overline{DA}$ and $DA = 16$, we get $OD = 16 \div 2 = 8$.
This number is the short leg of the right triangle we are about to use.

$$OD = \tfrac{DA}{2} = \tfrac{16}{2} = 8$$

💡 Splitting a length of $16$ in half is a Grade 3 division fact, $16 \div 2 = 8$.

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

Draw the radius $\overline{OC}$.
Since $C$ lies on the semicircle, $OC = r = 17$.
The rectangle's corner at $D$ is a right angle, so $\overline{OD} \perp \overline{DC}$, which makes $\triangle ODC$ a right triangle with legs $OD$ and $DC$ and hypotenuse $OC$.

$$OC = 17, \quad \angle ODC = 90^{\circ}$$

💡 Recognizing the perpendicular sides of a rectangle and labeling the new triangle is the Grade 4 "parallel and perpendicular lines" skill.

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

Apply the Pythagorean theorem to $\triangle ODC$ to find the height $CD$ of the rectangle.
With legs $8$ and $CD$ and hypotenuse $17$, we get $8^{2} + CD^{2} = 17^{2}$, so $CD^{2} = 289 - 64 = 225$ and $CD = \sqrt{225} = 15$.

$$8^{2} + CD^{2} = 17^{2} \;\Rightarrow\; CD^{2} = 289 - 64 = 225 \;\Rightarrow\; CD = 15$$

💡 Using $a^{2} + b^{2} = c^{2}$ on a right triangle to find a missing leg is the Grade 8 Pythagorean-theorem standard exactly.

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

Multiply length by height to get the rectangle's area.
With $DA = 16$ and $CD = 15$, the area is $16 \times 15 = 240$.
That matches choice (A).

$$\text{Area} = DA \times CD = 16 \times 15 = 240 \;\Rightarrow\; \textbf{(A)}$$

💡 Area of a rectangle $=$ length $\times$ width is the Grade 4 rectangle area formula you already use.

[1] #1 4.NBT.B.4 Add the center $O$ of the semicircle to the diagram. The diameter is $FE = FD +

[2] #7 3.OA.A.2 Find $OD$, the distance from the center to corner $D$. Since $O$ is the midpoint

[3] #1 4.G.A.2 Draw the radius $\overline{OC}$. Since $C$ lies on the semicircle, $OC = r = 17$

[4] #7 8.G.B.7 Apply the Pythagorean theorem to $\triangle ODC$ to find the height $CD$ of the

[5] #7 4.MD.A.3 Multiply length by height to get the rectangle's area. With $DA = 16$ and $CD =

Review

Reasonableness: The radius is $17$ and the rectangle's half-width $OD = 8$, so the height must be less than $17$ (it cannot poke out of the semicircle). We got $CD = 15$, which fits — and it lands on the Pythagorean triple $(8, 15, 17)$, a well-known triple. The area $16 \times 15 = 240$ is comfortably the smallest answer choice, which makes sense: the rectangle is wide but not very tall, so its area should be on the smaller end of the listed options.

Alternative: Tool #6 (Guess and Check) on the choices works fast: each candidate area $A$ would force $CD = A \div 16$, giving heights $15, 15.5, 16, 16.5, 17$ for choices (A)-(E). Squaring and adding $8^{2} = 64$ should give $17^{2} = 289$. Only $CD = 15$ works: $15^{2} + 8^{2} = 225 + 64 = 289 = 17^{2}$. The others fail the Pythagorean check, confirming (A).

CCSS standards used (min grade 8)

3.OA.A.2 Interpret whole-number quotients of whole numbers (Halving $DA = 16$ to find $OD = 8$, the short leg of the right triangle.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Adding the three diameter pieces $9 + 16 + 9 = 34$ to find the full diameter $FE$.)
4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines (Recognizing that the rectangle's perpendicular sides make $\triangle ODC$ a right triangle at $D$.)
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems (Computing the rectangle's area as length $\times$ width $= 16 \times 15 = 240$.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Solving $8^{2} + CD^{2} = 17^{2}$ for the height $CD = 15$ of the rectangle.)

⭐ This AMC 8 problem only needs Grade 8 Pythagorean theorem — $a^{2} + b^{2} = c^{2}$ on a right triangle — you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 8)

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