AMC 8 · 2020 · #18
Grade 8 geometry-2dProblem
Rectangle ABCD is inscribed in a semicircle with diameter FE, as shown in the figure. Let DA=16, and let FD=AE=9. What is the area of ABCD?
Pick an answer.
AMC 8 2020 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: 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
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) $272$
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
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 $\overline{DA}$.
💡 Adding $9 + 16 + 9$ and halving the result is just Grade 4 multi-digit addition and a simple division you already know.
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.
💡 Splitting a length of $16$ in half is a Grade 3 division fact, $16 \div 2 = 8$.
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 $OC$.
💡 Recognizing the perpendicular sides of a rectangle and labeling the new triangle is the Grade 4 "parallel and perpendicular lines" skill.
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 $CD = \sqrt{225} = 15$.
💡 Using $a^{2} + b^{2} = c^{2}$ on a right triangle to find a missing leg is the Grade 8 Pythagorean-theorem standard exactly.
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).
💡 Area of a rectangle $=$ length $\times$ width is the Grade 4 rectangle area formula you already use.
4.NBT.B.4 Add the center $O$ of the semicircle to the diagram. The diameter is $FE = FD + 3.OA.A.2 Find $OD$, the distance from the center to corner $D$. Since $O$ is the midpoint 4.G.A.2 Draw the radius $\overline{OC}$. Since $C$ lies on the semicircle, $OC = r = 17$ 8.G.B.7 Apply the Pythagorean theorem to $\triangle ODC$ to find the height $CD$ of the 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.2Interpret whole-number quotients of whole numbers (Halving $DA = 16$ to find $OD = 8$, the short leg of the right triangle.)4.NBT.B.4Fluently 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.2Classify 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.3Apply 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.7Apply 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!
⭐ This AMC 8 problem only needs Grade 8 Pythagorean theorem — $a^{2} + b^{2} = c^{2}$ on a right triangle — you already know!
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