AMC 10 · 2007 · #15
Grade 8 geometry-2dThe angles of quadrilateral ABCD satisfy ∠A=2∠B=3∠C=4∠D. What is the degree measure of ∠A, rounded to the nearest whole number?
Pick an answer.
AMC 10 2007 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 angles are linked by $\angle A=2\angle B=3\angle C=4\angle D$, meaning $\angle A$ is twice $\angle B$, three times $\angle C$, and four times $\angle D$. Find $\angle A$ in degrees, rounded to the nearest whole number.
Givens: $ABCD$ is a quadrilateral, so its four interior angles add up to $360^\circ$; $\angle A=2\angle B=3\angle C=4\angle D$, a single chain tying all four angles to $\angle A$; Answer choices: (A) 125, (B) 144, (C) 153, (D) 173, (E) 180
Unknowns: The measure of $\angle A$ in degrees, rounded to the nearest whole number
Understand
Restated: In quadrilateral $ABCD$ the four angles are linked by $\angle A=2\angle B=3\angle C=4\angle D$, meaning $\angle A$ is twice $\angle B$, three times $\angle C$, and four times $\angle D$. Find $\angle A$ in degrees, rounded to the nearest whole number.
Givens: $ABCD$ is a quadrilateral, so its four interior angles add up to $360^\circ$; $\angle A=2\angle B=3\angle C=4\angle D$, a single chain tying all four angles to $\angle A$; Answer choices: (A) 125, (B) 144, (C) 153, (D) 173, (E) 180
Plan
Primary tool: #4 Introduce a Variable
Secondary: #13 Convert to Algebra, #3 Eliminate Possibilities
The chain $\angle A=2\angle B=3\angle C=4\angle D$ ties every angle to $\angle A$, so naming $\angle A=x$ (Tool #4) writes all four angles in terms of one unknown. The fact that a quadrilateral's angles sum to $360^\circ$ then turns the picture into a single equation (Tool #13). Solving gives a decimal, and rounding lets us match one of the five listed choices (Tool #3).
Execute — Answer: D
7.EE.B.4 Step 1 Name $\angle A$ and cascade the others
- Let $\angle A=x$.
- The chain says $\angle A=2\angle B$, so $\angle B=\dfrac{x}{2}$.
- It says $\angle A=3\angle C$, so $\angle C=\dfrac{x}{3}$.
- And $\angle A=4\angle D$, so $\angle D=\dfrac{x}{4}$.
- Now all four angles are written using the single unknown $x$.
💡 One equal-value chain lets you pin every angle to a single unknown, so there is really only one number to find.
8.G.A.5 Step 2 Use the quadrilateral angle sum
- The interior angles of any quadrilateral add up to $360^\circ$ (split it along a diagonal into two triangles, each summing to $180^\circ$).
- Add the four expressions and set the total equal to $360$.
💡 A fixed total ($360^\circ$) turns four unknown angles into one equation you can actually solve.
5.NF.A.1 Step 3 Combine the fractions
- Give the terms a common denominator of $12$: $x=\dfrac{12x}{12}$, $\dfrac{x}{2}=\dfrac{6x}{12}$, $\dfrac{x}{3}=\dfrac{4x}{12}$, $\dfrac{x}{4}=\dfrac{3x}{12}$.
- Adding the numerators $12+6+4+3=25$ gives $\dfrac{25x}{12}=360$.
💡 A shared denominator lets the four pieces merge into one clean fraction of $x$.
5.NBT.A.4 Step 4 Solve and round
- Multiply both sides by $12$ and divide by $25$: $x=\dfrac{360\cdot 12}{25}=\dfrac{4320}{25}=172.8$.
- Rounded to the nearest whole number, $\angle A\approx 173^\circ$, which is choice $\textbf{(D)}$.
💡 Undoing the $\tfrac{25}{12}$ coefficient isolates $x$, and $0.8$ rounds up to the next whole degree.
7.EE.B.4 Let $\angle A=x$. The chain says $\angle A=2\angle B$, so $\angle B=\dfrac{x}{2} 8.G.A.5 The interior angles of any quadrilateral add up to $360^\circ$ (split it along a 5.NF.A.1 Give the terms a common denominator of $12$: $x=\dfrac{12x}{12}$, $\dfrac{x}{2}= 5.NBT.A.4 Multiply both sides by $12$ and divide by $25$: $x=\dfrac{360\cdot 12}{25}=\dfra Review
Reasonableness: Recover the other angles from $x=172.8$: $\angle B=86.4$, $\angle C=57.6$, $\angle D=43.2$. Their sum is $172.8+86.4+57.6+43.2=360$ exactly, confirming a valid quadrilateral. Since $\angle A$ is the largest of four angles that average $90^\circ$, it must sit well above $90^\circ$ but below $360^\circ$; $173^\circ$ fits, and it beats the smaller choices 125, 144, 153 while staying under the degenerate $180^\circ$.
Alternative: Work with $\angle D$ instead. Writing $\angle A=4\angle D$, $\angle B=2\angle D$, $\angle C=\tfrac{4}{3}\angle D$ and summing to $360$ gives $\left(4+2+\tfrac43+1\right)\angle D=\tfrac{25}{3}\angle D=360$, so $\angle D=43.2$ and $\angle A=4\angle D=172.8\approx 173^\circ$, the same result.
CCSS standards used (min grade 8)
7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Letting $\angle A=x$ and writing $\angle B,\angle C,\angle D$ as $\tfrac{x}{2},\tfrac{x}{3},\tfrac{x}{4}$.)8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Using that a quadrilateral's interior angles sum to $360^\circ$ to build the equation.)5.NF.A.1Add and subtract fractions with unlike denominators (Combining $x+\tfrac{x}{2}+\tfrac{x}{3}+\tfrac{x}{4}$ into $\tfrac{25x}{12}$ with a common denominator.)5.NBT.A.4Round decimals to any place (Rounding $172.8$ to the nearest whole degree to get $173$.)
⭐ When angles are chained by an equal-value string, name the biggest one, write the rest as fractions of it, and let the fixed $360^\circ$ total do the solving.
⭐ When angles are chained by an equal-value string, name the biggest one, write the rest as fractions of it, and let the fixed $360^\circ$ total do the solving.
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