AMC 10 · 2009 · #12
Grade 7 geometry-2dIn quadrilateral ABCD, AB=5, BC=17, CD=5, DA=9, and BD is an integer. What is BD?
Pick an answer.
AMC 10 2009 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 four sides are $AB=5$, $BC=17$, $CD=5$, and $DA=9$. The diagonal $BD$ has a whole-number length. Find $BD$.
Givens: The four side lengths are $AB=5$, $BC=17$, $CD=5$, $DA=9$; $BD$ is the diagonal joining $B$ and $D$; $BD$ is an integer; Answer choices: (A) $11$, (B) $12$, (C) $13$, (D) $14$, (E) $15$
Unknowns: The integer length of the diagonal $BD$
Understand
Restated: In quadrilateral $ABCD$ the four sides are $AB=5$, $BC=17$, $CD=5$, and $DA=9$. The diagonal $BD$ has a whole-number length. Find $BD$.
Givens: The four side lengths are $AB=5$, $BC=17$, $CD=5$, $DA=9$; $BD$ is the diagonal joining $B$ and $D$; $BD$ is an integer; Answer choices: (A) $11$, (B) $12$, (C) $13$, (D) $14$, (E) $15$
Plan
Primary tool: #14 Extreme Principle
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
The diagonal $BD$ belongs to two different triangles at once, so Tool #7 (Identify Subproblems) splits the four-sided figure into $\triangle BCD$ (sides $17$, $5$, $BD$) and $\triangle ABD$ (sides $5$, $9$, $BD$). Inside each triangle the third side cannot be so long or so short that the other two fail to close up — that boundary is exactly the triangle inequality, so Tool #14 (Extreme Principle) pushes each triangle to its flat, zero-area edge to read off how big and how small $BD$ is allowed to be. Those two boundaries trap $BD$ inside a narrow band, and Tool #3 (Eliminate Possibilities) uses the whole-number clue to pick the single integer that survives.
Execute — Answer: C
7.G.A.2 Step 1 Cut the quadrilateral into two triangles
- Draw the diagonal $BD$.
- It splits $ABCD$ into $\triangle BCD$, whose sides are $BC=17$, $CD=5$, and $BD$, and $\triangle ABD$, whose sides are $AB=5$, $DA=9$, and $BD$.
- Both must be genuine triangles, so in each one every side is shorter than the sum of the other two.
- The shared side $BD$ has to obey both triangles at the same time.
💡 One diagonal turns a hard four-sided figure into two triangles that share that diagonal as a common side.
7.EE.B.4 Step 2 Lower bound from the big triangle
- In $\triangle BCD$ the two known sides are $17$ and $5$.
- For the triangle to close, the side $BD$ together with the short side $5$ must out-reach the long side $17$: $BD+5>17$, so $BD>12$.
- At the extreme $BD=12$ the sides $5$ and $12$ line up flat against the $17$ (zero area), so $BD$ must be strictly greater than $12$.
💡 Two sides can only wrap around a long third side if together they are longer than it — otherwise they collapse into a straight line.
6.EE.B.8 Step 3 Upper bound from the small triangle
- In $\triangle ABD$ the two known sides are $5$ and $9$.
- The side $BD$ cannot be longer than those two combined, or they could never meet: $BD<5+9=14$.
- At the extreme $BD=14$ the sides $5$ and $9$ lie flat end to end (zero area), so $BD$ must be strictly less than $14$.
💡 A side can be at most as long as the other two laid end to end — reach that length and the triangle flattens to a segment.
6.NS.C.7 Step 4 Trap the integer
- The two triangles together force $12 < BD < 14$.
- The only whole number strictly between $12$ and $14$ is $13$, and the problem says $BD$ is an integer, so $BD=13$.
- The answer is (C).
💡 Squeeze a value between two consecutive-ish bounds and the whole-number clue leaves exactly one survivor.
7.G.A.2 Draw the diagonal $BD$. It splits $ABCD$ into $\triangle BCD$, whose sides are $ 7.EE.B.4 In $\triangle BCD$ the two known sides are $17$ and $5$. For the triangle to clo 6.EE.B.8 In $\triangle ABD$ the two known sides are $5$ and $9$. The side $BD$ cannot be 6.NS.C.7 The two triangles together force $12 < BD < 14$. The only whole number strictly Review
Reasonableness: Check both triangles with $BD=13$. In $\triangle BCD$ the sides $5,13,17$ satisfy $5+13=18>17$, a valid triangle. In $\triangle ABD$ the sides $5,9,13$ satisfy $5+9=14>13$, also valid. Both hold with a little room to spare, and no other integer works: $12$ fails ($5+12=17$, flat) and $14$ fails ($5+9=14$, flat). So $BD=13$, matching (C).
Alternative: Combine all inequalities at once. From $\triangle BCD$: $|17-5| < BD < 17+5$, i.e. $12 < BD < 22$. From $\triangle ABD$: $|9-5| < BD < 9+5$, i.e. $4 < BD < 14$. Overlapping the two windows gives $12 < BD < 14$, and the only integer inside is $13$.
CCSS standards used (min grade 7)
7.G.A.2Draw geometric shapes with given conditions including triangles (Splitting the quadrilateral along diagonal $BD$ into two triangles and requiring each to satisfy the triangle inequality.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Setting up and solving $BD+5>17$ to get the lower bound $BD>12$ from $\triangle BCD$.)6.EE.B.8Write an inequality of the form x > c or x < c and graph on a number line (Writing the upper-bound condition $BD<14$ from the boundary of $\triangle ABD$.)6.NS.C.7Understand ordering and absolute value of rational numbers (Recognizing that $13$ is the only integer strictly between $12$ and $14$ on the number line.)
⭐ A diagonal cuts a quadrilateral into two triangles; make each triangle just barely close to trap the diagonal between two numbers, then the whole-number clue picks the winner.
⭐ A diagonal cuts a quadrilateral into two triangles; make each triangle just barely close to trap the diagonal between two numbers, then the whole-number clue picks the winner.
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