AMC 8 · 2005 · #9
Grade 8 geometry-2dProblem
In quadrilateral ABCD, sides AB and BC both have length 10, sides CD and DA both have length 17, and the measure of angle ADC is 60∘. What is the length of diagonal AC?
Pick an answer.
AMC 8 2005 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: In quadrilateral $ABCD$, the two sides meeting at $B$ have length $\overline{AB} = \overline{BC} = 10$, and the two sides meeting at $D$ have length $\overline{CD} = \overline{DA} = 17$. The angle at $D$ is $\angle ADC = 60^\circ$. Find the length of diagonal $\overline{AC}$.
Givens: $\overline{AB} = \overline{BC} = 10$ (the two sides at vertex $B$); $\overline{CD} = \overline{DA} = 17$ (the two sides at vertex $D$); $\angle ADC = 60^\circ$; Answer choices: (A) $13.5$, (B) $14$, (C) $15.5$, (D) $17$, (E) $18.5$
Unknowns: The length of diagonal $\overline{AC}$
Understand
Restated: In quadrilateral $ABCD$, the two sides meeting at $B$ have length $\overline{AB} = \overline{BC} = 10$, and the two sides meeting at $D$ have length $\overline{CD} = \overline{DA} = 17$. The angle at $D$ is $\angle ADC = 60^\circ$. Find the length of diagonal $\overline{AC}$.
Givens: $\overline{AB} = \overline{BC} = 10$ (the two sides at vertex $B$); $\overline{CD} = \overline{DA} = 17$ (the two sides at vertex $D$); $\angle ADC = 60^\circ$; Answer choices: (A) $13.5$, (B) $14$, (C) $15.5$, (D) $17$, (E) $18.5$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
Tool #1 (Draw a Diagram) shows that the diagonal $\overline{AC}$ slices the quadrilateral into two triangles, $\triangle ABC$ and $\triangle ADC$. Only one of them carries useful given information: $\triangle ADC$ has two known sides ($DA = CD = 17$) and the angle between them ($60^\circ$). The sides $AB$ and $BC$ at vertex $B$ are a distraction. Tool #7 (Identify Subproblems) reduces the original quadrilateral question to a single subproblem about $\triangle ADC$: find its third side $AC$. The isosceles structure plus the $60^\circ$ apex angle forces all three angles to be $60^\circ$, making the triangle equilateral, so $AC = 17$ with no calculation beyond the angle sum.
Execute — Answer: D
4.G.A.2 Step 1 - Draw the quadrilateral and the diagonal $\overline{AC}$.
- The diagonal splits $ABCD$ into $\triangle ABC$ (sides $10$, $10$, and $AC$) and $\triangle ADC$ (sides $17$, $17$, and $AC$).
- The problem only tells us an angle inside $\triangle ADC$, so that triangle is where the work happens.
💡 Picking the triangle that carries the given side lengths and angle is the Grade 4 "classify a figure by its known properties" move.
4.G.A.2 Step 2 - Focus on $\triangle ADC$ as a subproblem.
- Two of its sides are equal ($DA = CD = 17$), so it is isosceles.
- The angles opposite those equal sides — $\angle DCA$ (opposite $DA$) and $\angle DAC$ (opposite $CD$) — must be equal too.
💡 "Two equal sides force two equal opposite angles" is the Grade 4 isosceles fact in its simplest form.
8.G.A.5 Step 3 - Use the triangle angle sum.
- Let $x = \angle DAC = \angle DCA$.
- The three angles of $\triangle ADC$ add to $180^\circ$.
💡 The $180^\circ$ triangle-angle-sum fact is the Grade 8 "informal arguments about triangle angles" standard.
4.G.A.2 Step 4 - All three angles of $\triangle ADC$ are $60^\circ$, so the triangle is equiangular — and therefore equilateral.
- Every side has the same length, so $AC = DA = CD = 17$.
💡 Recognizing the equiangular-equals-equilateral fact is the Grade 4 "classify $2$-D figures" payoff.
4.G.A.2 Draw the quadrilateral and the diagonal $\overline{AC}$. The diagonal splits $AB 4.G.A.2 Focus on $\triangle ADC$ as a subproblem. Two of its sides are equal ($DA = CD = 8.G.A.5 Use the triangle angle sum. Let $x = \angle DAC = \angle DCA$. The three angles 4.G.A.2 All three angles of $\triangle ADC$ are $60^\circ$, so the triangle is equiangul Review
Reasonableness: The answer $AC = 17$ is one of the listed choices and exactly matches $DA = CD = 17$, which fits the equilateral conclusion. It is also between the longest pair of sides ($17$) and the shortest pair ($10$): the diagonal of a convex quadrilateral has to be shorter than the sum of two adjacent sides ($17 + 17 = 34$ and $10 + 10 = 20$) and longer than their difference ($17 - 10 = 7$), which $17$ comfortably satisfies. The fact that the $10$s at vertex $B$ were never used is a hint that the problem really did want only the $\triangle ADC$ subproblem.
Alternative: Tool #6 (Guess and Check) on the choices: the Law of Cosines gives $AC^2 = 17^2 + 17^2 - 2(17)(17)\cos 60^\circ = 289 + 289 - 289 = 289$, so $AC = 17$. The shortcut for kids is the same observation without the law: with two equal sides and a $60^\circ$ included angle, the triangle must be equilateral, so the third side equals the other two — choice (D).
CCSS standards used (min grade 8)
4.G.A.2Classify two-dimensional figures based on properties of their lines and angles (Identifying $\triangle ADC$ as isosceles from $DA = CD$, then concluding it is equilateral once all three angles are $60^\circ$ so $AC = 17$.)8.G.A.5Use informal arguments to establish facts about the angle sum of triangles (Applying the $180^\circ$ triangle angle sum to $x + x + 60 = 180$ and finding the base angles $x = 60^\circ$.)
⭐ The diagonal $\overline{AC}$ lives inside two triangles, but only one of them carries the given $17$, $17$, and $60^\circ$. That isosceles triangle with a $60^\circ$ apex angle has to be equilateral, so the third side just equals $17$.
⭐ The diagonal $\overline{AC}$ lives inside two triangles, but only one of them carries the given $17$, $17$, and $60^\circ$. That isosceles triangle with a $60^\circ$ apex angle has to be equilateral, so the third side just equals $17$.
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