Math 1 Quiz: Triangle Inequality

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Triangle InequalityQuestion 1 of 11

A student incorrectly states: "If three lengths $a \leq b \leq c$ satisfy $a + b > c$ , then they form a triangle." Which of the following represents the most likely reasoning error behind this incorrect statement?

The student forgot that all sides must be positive

The student only checked one of the three triangle inequality conditions

The student confused necessary and sufficient conditions

The student incorrectly assumed that

c

must be the longest side

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Math 1 Quiz

Math 1 Quiz: Triangle Inequality

Practice Triangle Inequality 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 Triangle Inequality, 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.

All questions

Question 1

The student forgot that all sides must be positive
The student only checked one of the three triangle inequality conditions (correct answer)
The student confused necessary and sufficient conditions
The student incorrectly assumed that $c$ must be the longest side

Explanation: The triangle inequality theorem states that for any triangle with sides

a

b

, and

c

, all three conditions must hold:

a + b > c

a + c > b

, and

b + c > a

. When

a \leq b \leq c

, the condition

a + b > c

is indeed the most restrictive (since

c

is the largest side), but the other two conditions

a + c > b

and

b + c > a

are not automatically satisfied just because

a + b > c

holds. The student's error is checking only one of the three required triangle inequality conditions. While this one condition is the most likely to be violated when

c

is the longest side, it's not sufficient by itself to guarantee that the three lengths form a triangle.

Question 2

Three towns A, B, and C are connected by straight roads. The distance from A to B is 15 km, and from B to C is 8 km. If a new road is built directly from A to C, and the three roads form a triangle, what is the range of possible lengths for the new road?

$0 < AC < 23$
$7 ≤ AC ≤ 23$
$8 < AC < 15$
$7 < AC < 23$ (correct answer)

Explanation: When you see three points forming a triangle with two known side lengths, you're dealing with the triangle inequality theorem. This fundamental geometric principle states that the sum of any two sides of a triangle must be greater than the third side, and the difference of any two sides must be less than the third side. With

AB = 15

km and

BC = 8

km, you need to find the possible values for

AC

. The triangle inequality gives you two key constraints: First:

AB + BC > AC

, so

15 + 8 > AC

, which means

AC < 23

Second:

AC + BC > AB

, so

AC + 8 > 15

, which means

AC > 7

Therefore,

7 < AC < 23

. Looking at the wrong answers: Choice A (

0 < AC < 23

) ignores the lower bound completely—you can't have a triangle where one side is extremely small compared to the others. Choice B (

7 ≤ AC ≤ 23

) incorrectly includes the boundary values. If

AC = 7

, then you'd have

7 + 8 = 15

, which creates a straight line, not a triangle. Similarly, if

AC = 23

, you get

15 + 8 = 23

, again forming a straight line. Choice C (

8 < AC < 15

) severely restricts the range and misses many valid triangle configurations. The correct answer is D:

7 < AC < 23

. Study tip: For triangle inequality problems, always check that the sum of any two sides exceeds the third, and remember that equality conditions create degenerate triangles (straight lines), so use strict inequalities for true triangles.

Question 3

A triangle has two sides of lengths 7 and 12. If the third side has an integer length, how many possible triangles can be formed?

17 triangles (correct answer)
18 triangles
19 triangles
20 triangles

Explanation: Let the third side have length

s

. By the triangle inequality: (1)

7 + 12 > s

gives

s < 19

. (2)

7 + s > 12

gives

s > 5

. (3)

12 + s > 7

gives

s > -5

(always satisfied for positive

s

). Combining:

5 < s < 19

. Since

s

must be a positive integer,

s \in \{6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18\}

. Counting these values: there are 17 possible integer lengths.

Question 4

Three towns A, B, and C are located such that the distance from A to B is $15$ km, the distance from B to C is $8$ km, and the distance from A to C is $d$ km. If the three towns do NOT form a triangle (i.e., they are collinear), what are the possible values of $d$ ?

$d = 7$ or $d = 23$ (correct answer)
$d = 8$ or $d = 15$
$7 < d < 23$
$d \leq 7$ or $d \geq 23$

Explanation: For three points to be collinear (not form a triangle), the triangle inequality must become an equality. This happens when one distance equals the sum of the other two. We have three cases: (1)

d = 15 + 8 = 23

(C is beyond B from A), (2)

15 = d + 8

, so

d = 7

(C is between A and B), (3)

8 = 15 + d

, so

d = -7

(impossible since distance is positive). Therefore,

d = 7

d = 23

. Choice C gives the range where they DO form a triangle, choice D gives where they don't form a triangle but includes impossible values, and choice B gives incorrect specific values.

Question 5

A triangle has sides with lengths in the ratio $3:4:x$ . For what values of $x$ can this triangle exist?

$1 < x < 7$ (correct answer)
$0 < x < 7$
$x > 1$
$x > 0$

Explanation: Let the sides have lengths

3k

4k

, and

xk

for some positive constant

k

. Since

k > 0

, we can divide all triangle inequalities by

k

to get inequalities in terms of the ratios

3

4

, and

x

. The triangle inequalities become: (1)

3 + 4 > x

, so

x < 7

. (2)

3 + x > 4

, so

x > 1

. (3)

4 + x > 3

, so

x > -1

. Since

x

represents a ratio of lengths,

x > 0

. Combining these constraints:

x > 0

x > 1

, and

x < 7

, we get

1 < x < 7

. Choice B incorrectly allows

x \leq 1

, choice C misses the upper bound, and choice D misses both the lower bound

x > 1

and upper bound.

Question 6

Triangle ABC has sides of lengths $a$ , $b$ , and $c$ . If $a + b = 15$ and $c = 8$ , what is the range of possible values for side $a$ ?

$0 < a < 15$
$3.5 < a < 11.5$ (correct answer)
$4 < a < 11$
$7 < a < 8$

Explanation: Given

a + b = 15

and

c = 8

, we have

b = 15 - a

. For a valid triangle, three conditions must hold: (1)

a + b > c

gives

a + (15-a) > 8

, which simplifies to

15 > 8

(always true). (2)

a + c > b

gives

a + 8 > 15 - a

, which simplifies to

2a > 7

, so

a > 3.5

. (3)

b + c > a

gives

(15-a) + 8 > a

, which simplifies to

23 > 2a

, so

a < 11.5

. We also need

a > 0

and

b = 15 - a > 0

, giving

0 < a < 15

. Combining all constraints:

3.5 < a < 11.5

Question 7

A triangle has sides with lengths in the ratio $3:4:k$ where $k$ is a positive real number. What is the range of possible values for $k$ ?

$k > 0$
$0 < k < 7$
$1 < k < 7$ (correct answer)
$1 < k < 6$

Explanation: Let the sides have lengths

3t

4t

, and

kt

where

t > 0

. By the triangle inequality: (1)

3t + 4t > kt

gives

7t > kt

, so

k < 7

. (2)

3t + kt > 4t

gives

kt > t

, so

k > 1

. (3)

4t + kt > 3t

gives

kt > -t

, so

k > -1

(automatically satisfied since

k > 0

). Therefore,

1 < k < 7

Question 8

The sides of triangle PQR have lengths $a$ , $a+d$ , and $a+2d$ where $a > 0$ and $d > 0$ . Which condition must be satisfied for these to form a valid triangle?

$d < a$ (correct answer)
$d < 2a$
$d < \frac{a}{2}$
$d < 3a$

Explanation: The sides are

a

a+d

, and

a+2d

with

a, d > 0

. Since

a+2d

is the longest side, the critical triangle inequality is

a + (a+d) > a+2d

, which simplifies to

2a + d > a + 2d

, giving

a > d

, or equivalently

d < a

. The other inequalities:

a + (a+2d) > a+d

gives

2a + 2d > a + d

, so

a + d > 0

(always true), and

(a+d) + (a+2d) > a

gives

2a + 3d > a

, so

a + 3d > 0

(always true). Therefore, the condition is

d < a

Question 9

In a triangle, the longest side is twice the length of the shortest side. If the shortest side has length $s$ and the middle side has length $m$ , what constraints must $m$ satisfy?

$\frac{s}{2} < m < \frac{3s}{2}$
$s \leq m \leq 2s$
$s < m < 3s$
$s < m < 2s$ (correct answer)

Explanation: When you see a triangle problem involving side lengths, you need to apply the triangle inequality theorem, which states that the sum of any two sides must be greater than the third side. Given that the shortest side has length

s

, the middle side has length

m

, and the longest side has length

2s

, you must check all three triangle inequality conditions: First,

s + m > 2s

, which simplifies to

m > s

. Second,

s + 2s > m

, giving us

m < 3s

. Third,

m + 2s > s

, which is automatically satisfied since all lengths are positive. From the first two conditions, we get

s < m < 3s

. However, since we're told that

m

is the middle side length, it must be between the shortest and longest sides:

s < m < 2s

. Looking at the answer choices: Choice A gives

\frac{s}{2} < m < \frac{3s}{2}

, but this allows

m

to be less than

s

, violating our requirement that

m

is the middle length. Choice B includes

m = s

and

m = 2s

, but these would make sides equal, contradicting the distinct shortest, middle, and longest designations. Choice C gives

s < m < 3s

, which satisfies the triangle inequality but doesn't ensure

m

is actually the middle side length—it could be longer than

2s

. Choice D correctly captures both the triangle inequality constraints and ensures

m

is truly the middle side. Study tip: Always check that your constraints match the given relationships between sides, not just the triangle inequality.

Question 10

Three sticks have lengths $a$ , $b$ , and $c$ where $a \leq b \leq c$ . If $a + b = c + 2$ , which of the following best describes when these sticks can form a triangle?

Always, since $a + b > c$ is automatically satisfied by the given condition
Never, since the triangle inequality $a + b > c$ can never be satisfied when $a + b = c + 2$
Only when $a > 2$ , ensuring all triangle inequalities are satisfied
Always when all lengths are positive, since $a + b > c$ and the other inequalities follow from $a \leq b \leq c$ (correct answer)

Explanation: Given

a + b = c + 2

and

a \leq b \leq c

, we need to check all three triangle inequalities. First,

a + b > c

becomes

(c + 2) > c

, which simplifies to

2 > 0

, always true. Second,

a + c > b

: since

c \geq b

, we have

a + c \geq a + b > b

(the last inequality holds when

a > 0

). Third,

b + c > a

: since

b \geq a

and

c > 0

, we have

b + c \geq a + c > a

. Therefore, when all lengths are positive, all triangle inequalities are satisfied. Choice A is incomplete reasoning, B incorrectly concludes never, and C adds an unnecessary condition on

a > 2

Question 11

A triangle has integer side lengths. Two of the sides have lengths $7$ and $12$ . How many possible integer values are there for the length of the third side?

$11$ possible values
$13$ possible values (correct answer)
$17$ possible values
$18$ possible values

Explanation: Let the third side have length

c

. By triangle inequality: (1)

7 + 12 > c

, so

c < 19

. (2)

7 + c > 12

, so

c > 5

. (3)

12 + c > 7

, so

c > -5

(automatically satisfied for positive

c

). Therefore,

5 < c < 19

. Since

c

must be a positive integer, we have

c \in \{6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18\}

. Counting these values gives us

18 - 6 + 1 = 13

possible integer values for the third side.