Math 1 Quiz: Using Counterexamples
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Using CounterexamplesQuestion 1 of 20

A student claims: "If a quadrilateral has four equal sides, then it must be a square." Which of the following serves as a counterexample to disprove this claim?

A rectangle with sides of length 3, 3, 5, and 5 units
A rhombus with four sides of length 4 units and no right angles
A parallelogram with opposite sides of length 6 and 8 units respectively
A trapezoid with two parallel sides of different lengths but equal diagonals
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Math 1 Quiz

Math 1 Quiz: Using Counterexamples

Practice Using Counterexamples in Math 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Using Counterexamples, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A student claims: "If a quadrilateral has four equal sides, then it must be a square." Which of the following serves as a counterexample to disprove this claim?

  1. A rectangle with sides of length 3, 3, 5, and 5 units
  2. A rhombus with four sides of length 4 units and no right angles (correct answer)
  3. A parallelogram with opposite sides of length 6 and 8 units respectively
  4. A trapezoid with two parallel sides of different lengths but equal diagonals
Explanation: A counterexample must satisfy the hypothesis (four equal sides) but fail the conclusion (being a square). A rhombus has four equal sides but is not a square unless it has right angles. Choice A doesn't have four equal sides, Choice C doesn't have four equal sides, and Choice D doesn't have four equal sides.

Question 2

A student claims: "All prime numbers greater than 2 are odd." Another student wants to use a counterexample to show this is false. What is the fundamental problem with this approach?

  1. The claim is actually true, so no valid counterexample exists for this statement (correct answer)
  2. Counterexamples can only be used for mathematical equations, not for number theory statements
  3. The statement involves infinity, so counterexamples cannot be applied to such claims
  4. Prime numbers are too complex to use as counterexamples in mathematical reasoning
Explanation: The statement "All prime numbers greater than 2 are odd" is actually true. Since 2 is the only even prime number, all other primes must be odd. Therefore, no counterexample exists because the statement is correct. The other choices incorrectly suggest that counterexamples have limitations that don't actually exist in these contexts.

Question 3

Consider the statement: "If f(x)f(x) is continuous on the interval [0,1][0,1], then f(x)f(x) has a maximum value on that interval." A student believes this is false. What should you tell the student about finding a counterexample?

  1. A counterexample exists: f(x)=xf(x) = x is continuous on [0,1][0,1] but has no maximum
  2. A counterexample exists: f(x)=1xf(x) = \frac{1}{x} is continuous and has no maximum on [0,1][0,1]
  3. No counterexample exists because the statement is true by the Extreme Value Theorem (correct answer)
  4. Counterexamples cannot be found for statements involving continuous functions and closed intervals
Explanation: The Extreme Value Theorem guarantees that continuous functions on closed, bounded intervals have both maximum and minimum values. The statement is true, so no counterexample exists. Choice A is wrong because f(x)=xf(x) = x has maximum value 1 at x=1x = 1. Choice B is wrong because f(x)=1xf(x) = \frac{1}{x} is not continuous on [0,1][0,1].

Question 4

A calculus student argues: "If f′(x)>0f'(x) > 0 for all xx in an interval, then f(x)f(x) has no critical points in that interval." Another student wants to find a counterexample. What should the second student conclude?

  1. A counterexample exists: f(x)=x3f(x) = x^3 has f′(x)=3x2>0f'(x) = 3x^2 > 0 everywhere but has a critical point at x=0x = 0
  2. A counterexample exists: f(x)=x2f(x) = x^2 has f′(x)=2x>0f'(x) = 2x > 0 for x>0x > 0 but has critical points
  3. No counterexample exists because critical points occur where f′(x)=0f'(x) = 0, which contradicts f′(x)>0f'(x) > 0 (correct answer)
  4. A counterexample exists: f(x)=sin⁡(x)f(x) = \sin(x) has varying derivative values including positive ones
Explanation: Critical points occur where f′(x)=0f'(x) = 0 or f′(x)f'(x) is undefined. If f′(x)>0f'(x) > 0 throughout an interval, then f′(x)≠0f'(x) \neq 0 in that interval, so no critical points exist there. The statement is true. Choice A is wrong because f′(0)=0f'(0) = 0, not >0> 0. Choice B doesn't specify the interval clearly. Choice D doesn't maintain f′(x)>0f'(x) > 0 throughout any interval.

Question 5

Consider the claim: "If two lines are cut by a transversal and corresponding angles are equal, then the lines must be parallel." A student wants to challenge this with a counterexample. What should you advise?

  1. Find two intersecting lines cut by a transversal where corresponding angles are equal
  2. Find two parallel lines cut by a transversal where corresponding angles are not equal
  3. The statement is a proven theorem, so no valid counterexample can exist (correct answer)
  4. Use two lines that are neither parallel nor intersecting to create the counterexample
Explanation: This statement is the converse of a fundamental theorem in geometry and is itself a proven theorem. If corresponding angles are equal when two lines are cut by a transversal, the lines must be parallel. Since this is mathematically proven, no counterexample exists. Choices A and B describe impossible scenarios, and Choice D is geometrically meaningless.

Question 6

A mathematics teacher states: "If a function passes the vertical line test, then it must also pass the horizontal line test." Which function type provides the clearest counterexample to this statement?

  1. f(x)=x2f(x) = x^2 because it fails both the vertical and horizontal line tests
  2. f(x)=x3f(x) = x^3 because it passes both the vertical and horizontal line tests
  3. f(x)=x2f(x) = x^2 because it passes the vertical line test but fails the horizontal line test (correct answer)
  4. f(x)=±xf(x) = \pm\sqrt{x} because it fails the vertical line test but passes the horizontal line test
Explanation: A counterexample needs to pass the vertical line test (be a function) but fail the horizontal line test (not be one-to-one). Choice C correctly identifies f(x)=x2f(x) = x^2 as passing the vertical line test but failing the horizontal line test. Choice A incorrectly states x2x^2 fails the vertical line test. Choice B supports the claim. Choice D describes a relation that isn't a function.

Question 7

Elena claims: "If a triangle has two equal angles, then all three angles must be equal." To disprove this using a counterexample, which triangle would work?

  1. An equilateral triangle with angles of 60°60°, 60°60°, and 60°60°
  2. A right triangle with angles of 45°45°, 45°45°, and 90°90° (correct answer)
  3. A scalene triangle with angles of 40°40°, 60°60°, and 80°80°
  4. An obtuse triangle with angles of 30°30°, 50°50°, and 100°100°
Explanation: A counterexample needs two equal angles (satisfying the hypothesis) but not all three angles equal (making the conclusion false). The 45°45°-45°45°-90°90° triangle has two equal angles but the third is different. Choice A supports the claim, while Choices C and D have no equal angles, making the hypothesis false.

Question 8

Marcus argues: "For any real numbers xx and yy, if x2=y2x^2 = y^2, then x=yx = y." To show this statement is false, which pair of values would serve as the most direct counterexample?

  1. x=3x = 3 and y=3y = 3, since 32=93^2 = 9 and 32=93^2 = 9
  2. x=4x = 4 and y=−4y = -4, since 42=164^2 = 16 and (−4)2=16(-4)^2 = 16 (correct answer)
  3. x=0x = 0 and y=1y = 1, since 02=00^2 = 0 and 12=11^2 = 1
  4. x=2x = 2 and y=5y = 5, since 22=42^2 = 4 and 52=255^2 = 25
Explanation: A counterexample must make the hypothesis true (x2=y2x^2 = y^2) but the conclusion false (x≠yx \neq y). With x=4x = 4 and y=−4y = -4, we have 42=(−4)2=164^2 = (-4)^2 = 16, so the hypothesis holds, but 4≠−44 \neq -4, so the conclusion fails. Choice A makes both hypothesis and conclusion true, while Choices C and D make the hypothesis false.

Question 9

A researcher states: "If two variables have a correlation coefficient of 0.8, then one variable causes changes in the other." Which scenario best serves as a counterexample?

  1. Height and weight in adults, which have correlation 0.7 and clear causal relationship
  2. Ice cream sales and drowning incidents, which have correlation 0.85 but are both caused by hot weather (correct answer)
  3. Study time and test scores, which have correlation 0.6 and obvious causal connection
  4. Temperature and air conditioner usage, which have correlation 0.9 and direct causal relationship
Explanation: A counterexample needs high correlation (hypothesis true) but no direct causation (conclusion false). Ice cream sales and drowning incidents are highly correlated due to a confounding variable (hot weather) but neither causes the other. The other choices either have lower correlation or demonstrate actual causal relationships.

Question 10

Consider the claim: "If nn is an integer and n2n^2 is even, then nn is divisible by 4." Which of the following provides a valid counterexample?

  1. n=8n = 8, because 82=648^2 = 64 is even and 8 is divisible by 4
  2. n=6n = 6, because 62=366^2 = 36 is even but 6 is not divisible by 4 (correct answer)
  3. n=5n = 5, because 52=255^2 = 25 is odd and 5 is not divisible by 4
  4. n=3n = 3, because 32=93^2 = 9 is odd and 3 is not divisible by 4
Explanation: A counterexample needs the hypothesis (n2n^2 is even) to be true but the conclusion (nn divisible by 4) to be false. For n=6n = 6: 62=366^2 = 36 is even (hypothesis true), but 6 is not divisible by 4 (conclusion false). Choice A supports the claim, while Choices C and D have false hypotheses since the squares are odd.

Question 11

A student argues: "If a polynomial has degree 3, then it must have exactly 3 real roots." Which polynomial serves as the best counterexample?

  1. p(x)=x3+x2+x+1p(x) = x^3 + x^2 + x + 1, which factors as (x+1)(x2+1)(x+1)(x^2+1) (correct answer)
  2. p(x)=x3−6x2+11x−6p(x) = x^3 - 6x^2 + 11x - 6, which factors as (x−1)(x−2)(x−3)(x-1)(x-2)(x-3)
  3. p(x)=x2−4p(x) = x^2 - 4, which factors as (x−2)(x+2)(x-2)(x+2)
  4. p(x)=x4−1p(x) = x^4 - 1, which factors as (x−1)(x+1)(x2+1)(x-1)(x+1)(x^2+1)
Explanation: When you encounter questions about polynomial roots, remember that the degree tells you the maximum number of roots, not the exact number of real roots. A degree-3 polynomial must have exactly 3 roots total (counting multiplicity), but some of these roots might be complex numbers rather than real numbers. To disprove the student's claim, you need a degree-3 polynomial that doesn't have exactly 3 real roots. Looking at choice A, p(x)=x3+x2+x+1=(x+1)(x2+1)p(x) = x^3 + x^2 + x + 1 = (x+1)(x^2+1), you can find the roots by setting each factor to zero. From (x+1)=0(x+1) = 0, you get x=−1x = -1, which is real. From (x2+1)=0(x^2+1) = 0, you get x2=−1x^2 = -1, so x=±ix = ±i, which are complex numbers. This degree-3 polynomial has only 1 real root and 2 complex roots, perfectly contradicting the student's argument. Choice B is wrong because (x−1)(x−2)(x−3)=0(x-1)(x-2)(x-3) = 0 gives three real roots: x=1,2,3x = 1, 2, 3. This actually supports the student's incorrect claim rather than disproving it. Choice C is wrong because x2−4x^2 - 4 has degree 2, not 3, so it's irrelevant to the claim about degree-3 polynomials. Choice D is wrong because x4−1x^4 - 1 has degree 4, not 3, making it unsuitable as a counterexample for degree-3 polynomials. Remember: complex roots always come in conjugate pairs for polynomials with real coefficients. When you see x2+positive number=0x^2 + \text{positive number} = 0, those roots will be complex, reducing the count of real roots.

Question 12

A statistics student claims: "If a dataset has a mean of 50, then at least half the values must be 50 or greater." Which dataset would serve as an effective counterexample?

  1. {40,45,50,55,60}\{40, 45, 50, 55, 60\} with mean 50 and median 50
  2. {45,47,49,51,53}\{45, 47, 49, 51, 53\} with mean 49 and median 49
  3. {50,50,50,50,50}\{50, 50, 50, 50, 50\} with mean 50 and all values equal to 50
  4. {10,20,30,100,90}\{10, 20, 30, 100, 90\} with mean 50 and median 30 (correct answer)
Explanation: When analyzing claims about statistical measures, you need to understand that the mean can be heavily influenced by extreme values, while concepts like the median better represent the "middle" of a dataset. The student's claim confuses the mean with the median. The median is the value where at least half the data points are greater than or equal to it, but the mean has no such guarantee. A few very large values can pull the mean above most of the actual data points. Option D provides the perfect counterexample: {10,20,30,100,90}\{10, 20, 30, 100, 90\} has a mean of 50, but only 2 out of 5 values (40%) are 50 or greater. The two large values (100 and 90) pull the mean up to 50, even though most values fall below this mean. This directly contradicts the student's claim. Option A fails as a counterexample because exactly half the values (50%) are 50 or greater, which doesn't disprove the claim that says "at least half." Option B has a mean of 49, not 50, so it's irrelevant to testing a claim about datasets with mean 50. Option C supports the student's claim rather than contradicting it, since all values equal 50. Remember: the mean can be "pulled" by extreme values away from where most data points actually lie. When you need a counterexample to disprove a statistical claim, look for datasets where outliers create a clear separation between the mean and the bulk of the data.

Question 13

A student states: "If log⁡b(xy)=log⁡b(x)+log⁡b(y)\log_b(xy) = \log_b(x) + \log_b(y), then bb must equal 10." Which choice of bb serves as a counterexample to this claim?

  1. b=10b = 10, since this makes the logarithm property work correctly
  2. b=0b = 0, since log⁡0(xy)\log_0(xy) creates the same addition relationship
  3. b=1b = 1, since log⁡1(xy)\log_1(xy) is undefined but the property still holds
  4. b=eb = e, since the natural logarithm follows the same addition property (correct answer)
Explanation: This question tests your understanding of logarithm properties and what makes a valid counterexample. The student incorrectly believes that the product rule log⁡b(xy)=log⁡b(x)+log⁡b(y)\log_b(xy) = \log_b(x) + \log_b(y) only works when the base b=10b = 10. To disprove this claim, you need to find another valid base where this property still holds. The product rule for logarithms is actually a fundamental property that works for any valid logarithmic base, not just base 10. Choice D is correct because b=eb = e (approximately 2.718) creates a perfectly valid logarithm where log⁡e(xy)=log⁡e(x)+log⁡e(y)\log_e(xy) = \log_e(x) + \log_e(y). Since this property holds for a base other than 10, it serves as a counterexample to the student's claim. Choice A misses the point entirely—using b=10b = 10 supports the student's claim rather than disproving it. Choice B fails because b=0b = 0 creates undefined logarithms; you cannot have a logarithm with base 0. Choice C also involves undefined logarithms since b=1b = 1 is not a valid logarithmic base (log⁡1\log_1 would create the impossible equation 1y=x1^y = x for any x≠1x \neq 1). Remember that logarithm properties like the product rule, quotient rule, and power rule work for any valid base (any positive number except 1). When you see questions about logarithm properties, don't assume they only apply to common bases like 10 or ee—these rules are universal across all valid bases.

Question 14

Jordan claims: "If sin⁡(θ)=12\sin(\theta) = \frac{1}{2}, then θ=30°\theta = 30°." To disprove this claim, which angle measure provides a valid counterexample?

  1. θ=60°\theta = 60°, since sin⁡(60°)=32≠12\sin(60°) = \frac{\sqrt{3}}{2} \neq \frac{1}{2}
  2. θ=90°\theta = 90°, since sin⁡(90°)=1≠12\sin(90°) = 1 \neq \frac{1}{2}
  3. θ=45°\theta = 45°, since sin⁡(45°)=22≠12\sin(45°) = \frac{\sqrt{2}}{2} \neq \frac{1}{2}
  4. θ=150°\theta = 150°, since sin⁡(150°)=12\sin(150°) = \frac{1}{2} but 150°≠30°150° \neq 30° (correct answer)
Explanation: When evaluating claims about trigonometric functions, you need to understand what makes a valid counterexample. Jordan claims that if sin⁡(θ)=12\sin(\theta) = \frac{1}{2}, then θ=30°\theta = 30°. This is claiming a one-to-one relationship that doesn't actually exist. To disprove Jordan's claim, you need to find an angle where sin⁡(θ)=12\sin(\theta) = \frac{1}{2} but θ≠30°\theta \neq 30°. This would prove that multiple angles can have the same sine value, making Jordan's statement false. Option D provides exactly this counterexample: sin⁡(150°)=12\sin(150°) = \frac{1}{2}, but 150°≠30°150° \neq 30°. Since we found a different angle that also produces sin⁡(θ)=12\sin(\theta) = \frac{1}{2}, Jordan's claim is disproven. Options A, B, and C all make the same fundamental error. Choice A shows sin⁡(60°)=32≠12\sin(60°) = \frac{\sqrt{3}}{2} \neq \frac{1}{2}, choice B shows sin⁡(90°)=1≠12\sin(90°) = 1 \neq \frac{1}{2}, and choice C shows sin⁡(45°)=22≠12\sin(45°) = \frac{\sqrt{2}}{2} \neq \frac{1}{2}. While these calculations are correct, they don't disprove Jordan's claim because they don't satisfy the condition sin⁡(θ)=12\sin(\theta) = \frac{1}{2} in the first place. Remember: to disprove a conditional statement "if P, then Q," you need a counterexample where P is true but Q is false. Don't get distracted by examples where P itself is false—those prove nothing about the original claim.

Question 15

Maya claims: "If a number is divisible by 6, then it must be divisible by 12." Her friend wants to disprove this claim. Which number provides the most straightforward counterexample?

  1. 18, because it is divisible by 6 but not by 12 (correct answer)
  2. 24, because it is divisible by both 6 and 12
  3. 15, because it is not divisible by 6 or 12
  4. 8, because it is not divisible by 6 but is divisible by 4
Explanation: When you encounter a claim about divisibility, you need to understand what it means to disprove it. Maya's claim is a conditional statement: "If divisible by 6, then divisible by 12." To disprove this, you need a counterexample—a number that satisfies the first condition (divisible by 6) but fails the second condition (not divisible by 12). Let's think about the relationship between 6 and 12. Since 12=6×212 = 6 \times 2, any number divisible by 12 is automatically divisible by 6. However, the reverse isn't necessarily true. A number can be divisible by 6 without being divisible by 12. Choice A gives us 18. Let's check: 18÷6=318 ÷ 6 = 3 (divisible by 6) and 18÷12=1.518 ÷ 12 = 1.5 (not divisible by 12). This perfectly contradicts Maya's claim, making it an effective counterexample. Choice B uses 24, which is divisible by both 6 and 12 (24÷6=424 ÷ 6 = 4 and 24÷12=224 ÷ 12 = 2). This actually supports Maya's claim rather than disproving it. Choice C presents 15, which isn't divisible by 6 (15÷6=2.515 ÷ 6 = 2.5), so it doesn't even meet the first condition needed for a counterexample. Choice D offers 8, which also fails the first condition since 8÷6=1.33...8 ÷ 6 = 1.33... Remember: to disprove a conditional statement, find an example where the "if" part is true but the "then" part is false. Look for numbers that share some but not all factors with the target number.

Question 16

A probability student claims: "If events A and B are mutually exclusive, then they must also be independent." Which scenario most clearly disproves this statement?

  1. Rolling a die where A = {rolling a 2} and B = {rolling an even number}, since they overlap
  2. Flipping two coins where A = {first coin heads} and B = {second coin tails}, since they're independent
  3. Drawing from a deck where A = {drawing a heart} and B = {drawing a spade}, since P(A∩B)=0P(A \cap B) = 0 but P(A)⋅P(B)≠0P(A) \cdot P(B) \neq 0 (correct answer)
  4. Rolling two dice where A = {sum equals 7} and B = {sum equals 11}, since both events can occur
Explanation: A counterexample needs events that are mutually exclusive but not independent. Choice C gives mutually exclusive events (P(A∩B)=0P(A \cap B) = 0) that are not independent because P(A)⋅P(B)=14⋅14=116≠0P(A) \cdot P(B) = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16} \neq 0. Choice A describes events that aren't mutually exclusive. Choice B describes independent events. Choice D describes events that cannot both occur on a single roll, but doesn't clearly address independence.

Question 17

Maria conjectures: "For any integer nn, if n2n^2 is even, then nn is divisible by 4." To disprove this conjecture, which value of nn serves as the most direct counterexample?

  1. n=3n = 3 because 32=93^2 = 9 is odd, contradicting the hypothesis
  2. n=6n = 6 because 62=366^2 = 36 is even but 6 is not divisible by 4 (correct answer)
  3. n=8n = 8 because 82=648^2 = 64 is even and 8 is divisible by 4
  4. n=5n = 5 because 52=255^2 = 25 is odd and 5 is not divisible by 4
Explanation: A counterexample must make the hypothesis true but the conclusion false. Choice B has n=6n = 6 where n2=36n^2 = 36 is even (hypothesis true) but n=6n = 6 is not divisible by 4 (conclusion false). Choices A and D have odd squares, failing the hypothesis. Choice C supports the conjecture rather than disproving it.

Question 18

A student argues: "If two angles are supplementary, then they must both be acute angles." Which pair of angle measures most effectively disproves this claim?

  1. 45°45° and 45°45° because they are both acute but not supplementary
  2. 60°60° and 30°30° because they are both acute but their sum is 90°90°
  3. 90°90° and 90°90° because they are supplementary but both are right angles
  4. 30°30° and 150°150° because they are supplementary but one is obtuse (correct answer)
Explanation: When you encounter questions about disproving mathematical claims, you need to find a counterexample—a specific case that contradicts the statement. The claim here is that supplementary angles (angles that sum to 180°180°) must both be acute (less than 90°90°). To disprove this claim effectively, you need angles that ARE supplementary but do NOT both fit the "acute" requirement. Let's check: 30°+150°=180°30° + 150° = 180°, so these angles are indeed supplementary. However, 150°150° is obtuse (greater than 90°90°), not acute. This directly contradicts the claim that supplementary angles must both be acute, making option D the perfect counterexample. Option A fails because 45°+45°=90°45° + 45° = 90°, not 180°180°, so these angles aren't supplementary at all—they can't disprove a claim about supplementary angles. Option B has the same problem: 60°+30°=90°60° + 30° = 90°, making them complementary, not supplementary. Option C presents 90°+90°=180°90° + 90° = 180°, which are supplementary, but both angles are right angles (exactly 90°90°), not acute angles (less than 90°90°). While this does show supplementary angles that aren't acute, the original claim specifically mentioned "acute angles," making the obtuse angle in option D a clearer contradiction. When disproving mathematical statements, always look for the most direct counterexample that satisfies the given condition while clearly violating the claimed conclusion. One counterexample is enough to disprove any universal claim.

Question 19

Consider the claim: "If a polynomial function has degree 3, then it must have exactly 3 real zeros." Which polynomial serves as a counterexample to this statement?

  1. p(x)=x3−xp(x) = x^3 - x because it has degree 3 and exactly 3 real zeros
  2. p(x)=x4−1p(x) = x^4 - 1 because it has degree 4 but fewer than 4 real zeros
  3. p(x)=x2−4p(x) = x^2 - 4 because it has degree 2 and exactly 2 real zeros
  4. p(x)=x3+1p(x) = x^3 + 1 because it has degree 3 but only 1 real zero (correct answer)
Explanation: When you encounter questions about counterexamples, you need to find a case that shows the original claim is false. The claim here states that degree 3 polynomials must have exactly 3 real zeros. To disprove this, you need a degree 3 polynomial that has a different number of real zeros. The Fundamental Theorem of Algebra tells us that a polynomial of degree n has exactly n complex zeros (counting multiplicities), but some of these zeros might be complex numbers rather than real numbers. So while a cubic polynomial always has 3 total zeros, they don't all have to be real. Answer choice D, p(x)=x3+1p(x) = x^3 + 1, provides the perfect counterexample. This polynomial has degree 3, but when you solve x3+1=0x^3 + 1 = 0, you get x3=−1x^3 = -1, which gives x=−1x = -1 as the only real solution. The other two solutions are complex: x=1±i32x = \frac{1 \pm i\sqrt{3}}{2}. Since this cubic has only 1 real zero instead of 3, it disproves the claim. Choice A is wrong because p(x)=x3−xp(x) = x^3 - x actually supports the claim rather than contradicting it—it has exactly 3 real zeros. Choice B is wrong because it discusses a degree 4 polynomial, which is irrelevant to a claim about degree 3 polynomials. Choice C is wrong because it involves a degree 2 polynomial, again irrelevant to the cubic claim. Remember: to find a counterexample, look for a case that meets the hypothesis but fails the conclusion.

Question 20

A geometry student claims: "If a triangle is isosceles, then the triangle must be acute." Which triangle measurements provide the best counterexample to disprove this claim?

  1. A triangle with sides 5, 5, 8 because it is isosceles but obtuse (correct answer)
  2. A triangle with sides 3, 4, 5 because it is a right triangle but not isosceles
  3. A triangle with sides 6, 6, 6 because it is isosceles and acute
  4. A triangle with sides 2, 3, 4 because it is scalene and acute
Explanation: When you encounter a claim about geometric properties, you need to understand what would disprove it. A counterexample is a single case that shows the claim is false. Here, the claim states that all isosceles triangles are acute, so you need an isosceles triangle that is NOT acute. To disprove this claim, you need a triangle with two equal sides (isosceles) that has an obtuse angle (greater than 90°). Let's check if the triangle with sides 5, 5, 8 works. Using the Law of Cosines to find the largest angle (opposite the longest side): c2=a2+b2−2abcos⁡(C)c^2 = a^2 + b^2 - 2ab\cos(C) Solving for angle C: 82=52+52−2(5)(5)cos⁡(C)8^2 = 5^2 + 5^2 - 2(5)(5)\cos(C) 64=50−50cos⁡(C)64 = 50 - 50\cos(C) cos⁡(C)=−0.28\cos(C) = -0.28 Since cosine is negative, angle C is obtuse (greater than 90°). This triangle is both isosceles AND obtuse, making it a perfect counterexample. Choice B (3, 4, 5) doesn't help because it's not isosceles—you need an isosceles triangle to disprove the claim about isosceles triangles. Choice C (6, 6, 6) actually supports the claim since this equilateral triangle is both isosceles and acute. Choice D (2, 3, 4) is neither isosceles nor relevant to disproving the claim. Strategy tip: When disproving geometric claims, identify exactly what properties your counterexample must have. Here, you needed "isosceles AND not acute"—only choice A satisfies both requirements.