AMC 8 · 2017 · #6

Grade 8 geometry-2drate-ratio
angle-sum-triangleratio-proportion identify-subproblemsratio-proportion ↑ Prerequisites: ratio-proportionmulti-digit-arithmetic
📏 Short solution 💡 2 insights
📘 View easy version →

Problem

If the degree measures of the angles of a triangle are in the ratio 3:3:43:3:4, what is the degree measure of the largest angle of the triangle?

(A) 18(B) 36(C) 60(D) 72(E) 90\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90

Pick an answer.

(A)
18
(B)
36
(C)
60
(D)
72
(E)
90
View mode:

Toolkit + CCSS Solution

Understand

Restated: A triangle has three interior angles whose measures are in the ratio $3:3:4$. Find the degree measure of the largest of those three angles.

Givens: The three interior angles of a triangle are in the ratio $3:3:4$; The sum of the interior angles of any triangle is $180^\circ$; Answer choices: (A) $18$, (B) $36$, (C) $60$, (D) $72$, (E) $90$ (degrees)

Unknowns: The degree measure of the largest of the three angles

Understand

Restated: A triangle has three interior angles whose measures are in the ratio $3:3:4$. Find the degree measure of the largest of those three angles.

Givens: The three interior angles of a triangle are in the ratio $3:3:4$; The sum of the interior angles of any triangle is $180^\circ$; Answer choices: (A) $18$, (B) $36$, (C) $60$, (D) $72$, (E) $90$ (degrees)

Plan

Primary tool: #7 Identify Subproblems

Secondary: #3 Eliminate Possibilities

Tool #7 (Identify Subproblems) splits the question into two clean pieces: (a) figure out how many degrees one "ratio unit" is worth, then (b) use that unit to size the largest angle. The ratio $3:3:4$ contains $3+3+4=10$ equal units total, and those $10$ units must share the $180^\circ$ of a triangle — so one unit is just $180^\circ \div 10$. Once we know one unit, the largest angle (the $4$-unit part) is immediate. Tool #3 (Eliminate Possibilities) is the back-up check: since this is multiple choice, every candidate answer should be at most $180^\circ$ and should be expressible as $4$ copies of an integer share, which quickly rules out most options.

Execute — Answer: D

#7 Identify Subproblems 3.OA.D.8 Step 1
  • Add the parts of the ratio to see how many equal shares the $180^\circ$ gets split into.
  • The ratio $3:3:4$ has $3+3+4=10$ parts in total.
$$3 + 3 + 4 = 10 \text{ parts}$$

💡 Counting up the pieces of a ratio is a one-step addition word problem from Grade 3.

#7 Identify Subproblems 8.G.A.5 Step 2
  • Use the fact that the three interior angles of any triangle sum to $180^\circ$.
  • So those $10$ equal shares fit exactly inside $180^\circ$, and one share is found by dividing.
$$180^\circ \div 10 = 18^\circ \text{ per share}$$

💡 Knowing that a triangle's interior angles always add to $180^\circ$ is the Grade 8 informal angle-sum fact.

#7 Identify Subproblems 4.OA.A.2 Step 3

The largest angle is the $4$-share part of the ratio, so multiply one share by $4$ to get its measure in degrees.

$$4 \times 18^\circ = 72^\circ \;\Rightarrow\; \textbf{(D)}$$

💡 "$4$ copies of a share" is exactly the multiplicative-comparison move from Grade 4.

#3 Eliminate Possibilities 4.MD.C.7 Step 4
  • Verify by adding the three angles back together, using the ratio: $3$ shares + $3$ shares + $4$ shares should equal $180^\circ$.
  • This rules out every other multiple-choice candidate.
$$54^\circ + 54^\circ + 72^\circ = 180^\circ \;\checkmark$$

💡 Adding angle parts to check that they make the whole is exactly the Grade 4 "angle measure is additive" idea.

[1] #7 3.OA.D.8 Add the parts of the ratio to see how many equal shares the $180^\circ$ gets spl
[2] #7 8.G.A.5 Use the fact that the three interior angles of any triangle sum to $180^\circ$.
[3] #7 4.OA.A.2 The largest angle is the $4$-share part of the ratio, so multiply one share by $
[4] #3 4.MD.C.7 Verify by adding the three angles back together, using the ratio: $3$ shares + $

Review

Reasonableness: The three angles work out to $54^\circ, 54^\circ, 72^\circ$ — all positive, all less than $180^\circ$, and summing to exactly $180^\circ$. The triangle is isosceles (two equal small angles) with one slightly larger angle, which matches the $3:3:4$ ratio (the largest is just a bit bigger than the other two, not dramatically so). The largest angle $72^\circ$ is less than $90^\circ$, so this is an acute triangle — consistent with the modest $3:3:4$ ratio. Answer (D) $72^\circ$ is the right size.

Alternative: Tool #6 (Guess and Check) on the answer choices: the largest angle must be exactly $\tfrac{4}{10} = \tfrac{2}{5}$ of $180^\circ$. Quickly compute $\tfrac{2}{5} \times 180 = 72$ — only choice (D) matches. Alternatively check each option by dividing it by $4$ to find a candidate "share" and seeing whether $10$ shares give $180^\circ$: only $72 \div 4 = 18$ and $10 \times 18 = 180$ works.

CCSS standards used (min grade 8)

  • 3.OA.D.8 Solve two-step word problems using four operations within 100 (Adding the ratio parts $3+3+4=10$ to count the total number of equal shares.)
  • 8.G.A.5 Use informal arguments to establish facts about angle sum and exterior angles (Applying the triangle interior angle-sum fact ($180^\circ$) so the $10$ ratio shares can be matched to a known total.)
  • 4.OA.A.2 Multiply or divide to solve word problems involving multiplicative comparison (Computing the largest angle as $4 \times 18^\circ = 72^\circ$ — "$4$ times as big as one share.")
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems (Verifying $54^\circ + 54^\circ + 72^\circ = 180^\circ$ by adding the three angle parts to the whole.)

⭐ This AMC 8 problem only needs the Grade 8 fact you already know — the three angles of a triangle always add up to $180^\circ$!

⭐ This AMC 8 problem only needs the Grade 8 fact you already know — the three angles of a triangle always add up to $180^\circ$!