Triangle Properties and Angle Theorems Practice Questions — ACT — Page 2

Question 21Question →

A triangle has two sides of lengths $5$ centimeters and $11$ centimeters. Which of the following could be the perimeter of the triangle, in centimeters?

16

22

27

32

34

Show answer & explanation

Answer: 27

Answer

27

The correct answer is

27

centimeters. According to the Triangle Inequality Theorem, the length of the third side,

x

, of a triangle must be strictly greater than the difference of the other two sides (

11 - 5 = 6

) and strictly less than their sum (

11 + 5 = 16

). This gives the inequality range

6 < x < 16

. The perimeter is the sum of all three sides, which is

P = 5 + 11 + x = 16 + x

. Applying the bounds of

x

, we find that the perimeter must satisfy

16 + 6 < P < 16 + 16

, which simplifies to

22 < P < 32

. Among the options,

27

is the only value that is strictly within this range.

Step-by-Step Solution

1

Apply the Triangle Inequality Theorem to find the limits for the third side.

Let the third side be

x

. The length of

x

must satisfy:

11 - 5 < x < 11 + 5 \implies 6 < x < 16

.

The length of any side of a triangle must be strictly between the positive difference and the sum of the lengths of the other two sides.

2

Set up the equation for the perimeter of the triangle.

Perimeter

P = 5 + 11 + x = 16 + x

.

The perimeter of a triangle is defined as the sum of the lengths of all three of its sides.

3

Find the range of possible values for the perimeter

P

by applying the inequality bounds of

x

.

Add

16

to all parts of the inequality

6 < x < 16

:

16 + 6 < 16 + x < 16 + 16 \implies 22 < P < 32

.

We must shift the inequality bounds of the third side by the sum of the two known sides to find the range of the perimeter.

4

Identify the option that falls strictly inside the range

22 < P < 32

.

The value

27

is the only option that satisfies

22 < 27 < 32

.

Only a value strictly between

22

and

32

can represent a mathematically valid perimeter for this triangle.

Key Concept

Triangle Inequality Theorem and Perimeter

Estimated Time:1m 0s

Question 22Question →

A triangle has two sides of length $8$ and $15$ . If the length of the third side, $s$ , is a prime number, how many possible values are there for $s$ ?

3

4

5

6

7

Show answer & explanation

Answer: $4$

Answer

There are 4 possible values for the third side length s.

The correct answer is

4

. According to the Triangle Inequality Theorem, the length of the third side

s

of a triangle with sides of

8

and

15

must satisfy

15 - 8 < s < 15 + 8

. This simplifies to the open interval

7 < s < 23

. The prime numbers strictly between

7

and

23

are

11

,

13

,

17

, and

19

. Counting these gives exactly

4

possible prime values for

s

.

Step-by-Step Solution

1

Apply the Triangle Inequality Theorem to determine the bounds for the third side

s

.

15 - 8 < s < 15 + 8

, which simplifies to

7 < s < 23

.

The Triangle Inequality Theorem states that the length of any side of a triangle must be strictly greater than the difference of the other two sides and strictly less than their sum.

2

Identify all prime numbers that lie strictly within the range

(7, 23)

.

The prime numbers in this range are

11

,

13

,

17

, and

19

.

A prime number is an integer greater than 1 that has no positive divisors other than 1 and itself.

3

Count the number of identified prime numbers.

There are

4

prime numbers (

11

,

13

,

17

,

19

).

This counts the total number of possible valid lengths for

s

.

Key Concept

Triangle Inequality Theorem and basic number properties

Question 23Question →

In a triangle, two of the sides have lengths $13$ and $20$ . The third side has a length of $s$ , where $s$ is an integer. If the side of length $20$ is the longest side of the triangle, and the triangle is obtuse, what is the number of possible values for $s$ ?

Show answer & explanation

Answer: 8

Answer

There are 8 possible integer values for

s

.

To find the number of possible integer values for

s

, we combine the Triangle Inequality Theorem (

13 + s > 20 \implies s > 7

) and the condition for an obtuse triangle with

20

as the longest side (

20^2 > 13^2 + s^2 \implies s^2 < 231 \implies s \leq 15

). This limits

s

to integers in the range

[8, 15]

, which contains exactly

8

values.

Step-by-Step Solution

1

Apply the Triangle Inequality Theorem to find the lower bound for

s

.

s > 7

, so the minimum integer value is

8

.

The sum of the two shorter sides of a triangle must be strictly greater than the longest side.

2

Set up the obtuse triangle inequality with

20

as the longest side.

20^2 > 13^2 + s^2

In any obtuse triangle with longest side

c

, the inequality

c^2 > a^2 + b^2

must hold.

3

Solve the inequality

20^2 > 13^2 + s^2

for

s

.

s^2 < 231 \implies s \leq 15

Simplifying the inequality gives

400 > 169 + s^2 \implies s^2 < 231

. The largest integer whose square is less than

231

is

15

.

4

Determine the number of integers in the range

[8, 15]

.

8 possible values

The integers satisfying both conditions are

\{8, 9, 10, 11, 12, 13, 14, 15\}

, which count to

8

.

Key Concept

Triangle Inequality Theorem and obtuse triangle classification using side lengths

Alternative Method

List the perfect squares and verify which ones satisfy both

s^2 < 231

and the Triangle Inequality Theorem

s > 7

.

Estimated Time:2m 0s

Question 24Question →

A non-degenerate triangle has side lengths of $5$ , $12$ , and $x$ . A second non-degenerate triangle has side lengths of $x$ , $10$ , and $y$ . If $x$ and $y$ must be integers, and the perimeter of the second triangle is the minimum possible integer value, what is the sum of all possible values of $y$ ?

1

3

6

10

27

Show answer & explanation

Answer: 6

Answer

6

The correct answer is 6. By applying the Triangle Inequality Theorem, the shared side length

x

of the first triangle must satisfy

12 - 5 < x < 12 + 5

, which simplifies to

7 < x < 17

. Since

x

must be an integer, its possible values are

\{8, 9, 10, 11, 12, 13, 14, 15, 16\}

. For the second triangle with side lengths

x

,

10

, and

y

, the Triangle Inequality Theorem requires

|x - 10| < y < x + 10

. To minimize the perimeter

P = x + 10 + y

, we minimize

x + y

. Checking the possible values of

x

, we find that when

x = 8

, the minimum integer value for

y

is

3

(giving

P = 21

); when

x = 9

, the minimum integer value for

y

is

2

(giving

P = 21

); and when

x = 10

, the minimum integer value for

y

is

1

(giving

P = 21

). For any

x \ge 11

, the minimum value of

y

is

x - 9

, resulting in a perimeter of at least

23

. Therefore, the minimum perimeter of the second triangle is

21

, which is achieved when

y

is

3

,

2

, or

1

. The sum of these values of

y

is

1 + 2 + 3 = 6

.

Step-by-Step Solution

1

Apply the Triangle Inequality Theorem to the first triangle to find the range of the shared side length

x

.

7 < x < 17

The sum of any two side lengths of a triangle must be strictly greater than the third side length. Thus,

12 - 5 < x < 12 + 5

.

2

List the possible integer values of

x

based on the range.

x \in \{8, 9, 10, 11, 12, 13, 14, 15, 16\}

The problem states that

x

must be an integer.

3

Apply the Triangle Inequality Theorem to the second triangle with sides

x

,

10

, and

y

to express the range of

y

in terms of

x

.

|x - 10| < y < x + 10

The third side length

y

must be strictly between the difference and the sum of the other two sides (

x

and

10

).

4

Determine the minimum integer value of the perimeter of the second triangle,

P = x + 10 + y

, by testing the possible values of

x

and finding the minimum integer

y

for each.

The minimum perimeter is

21

, achieved when

(x, y) = (8, 3)

,

(9, 2)

, or

(10, 1)

.

Minimizing the perimeter

P = x + y + 10

is equivalent to minimizing the sum

x + y

for integer values of

y > |x - 10|

.

5

Sum the possible integer values of

y

that yield the minimum perimeter.

3 + 2 + 1 = 6

We need to find the sum of all possible values of

y

that result in the minimum perimeter of

21

.

Key Concept

Triangle Inequality Theorem and Perimeter Optimization

Estimated Time:3m 0s

Question 25Question →

The measures of the interior angles of a triangle are in the ratio $2:3:7$ . If the measure of the largest angle is decreased by $15^\circ$ and the measure of the smallest angle is increased by $15^\circ$ , what is the ratio of the interior angles of the new triangle, ordered from smallest to largest?

1:2:3

1:3:8

1:1:2

2:3:6

5:6:13

Show answer & explanation

Answer: $1:1:2$

Answer

The ratio of the interior angles of the new triangle is

1:1:2

.

The correct answer is the ratio

1:1:2

. The sum of the interior angles in a triangle is always

180^\circ

. Given the ratio

2:3:7

, the sum of the parts is

12

, which means each part represents

180^\circ / 12 = 15^\circ

. The original angles are therefore

30^\circ

,

45^\circ

, and

105^\circ

. Increasing the smallest angle by

15^\circ

gives

45^\circ

, and decreasing the largest by

15^\circ

gives

90^\circ

. The new angle measures are

45^\circ

,

45^\circ

, and

90^\circ

, which simplifies to

1:1:2

.

Step-by-Step Solution

1

Set up the equation for the sum of the interior angles of a triangle.

2x + 3x + 7x = 180^\circ

The sum of the interior angles of any triangle is always

180^\circ

.

2

Solve for the value of

x

.

12x = 180^\circ \implies x = 15^\circ

Combining like terms simplifies the equation to find the value of one ratio unit.

3

Calculate the original measures of the three angles.

Smallest:

30^\circ

, Middle:

45^\circ

, Largest:

105^\circ

Multiply each part of the ratio by

x = 15^\circ

.

4

Apply the modifications to the smallest and largest angles.

New smallest:

30^\circ + 15^\circ = 45^\circ

; New largest:

105^\circ - 15^\circ = 90^\circ

; Middle:

45^\circ

(unchanged).

Perform the operations described in the problem statement.

5

Order the new angle measures from smallest to largest and simplify the ratio.

45^\circ : 45^\circ : 90^\circ \implies 1 : 1 : 2

Divide each term in the ratio by the greatest common divisor, which is

45

.

Key Concept

Angle sum theorem of a triangle and ratio partition applications

Alternative Method

Instead of calculating the actual angle values, note that the sum of the ratio parts is

2+3+7 = 12

, and the sum of the interior angles of a triangle is

180^\circ

. This means

1

ratio unit is equal to

180^\circ / 12 = 15^\circ

. Since the smallest angle is increased by

15^\circ

(exactly

1

ratio unit) and the largest is decreased by

15^\circ

(exactly

1

ratio unit), we can apply these modifications directly to the ratio terms. The new ratio terms are

2+1 = 3

,

3

(unchanged), and

7-1 = 6

. This gives a ratio of

3:3:6

, which simplifies to

1:1:2

.

Estimated Time:1m 30s

Question 26Question →

In $\triangle ABC$ , the side lengths are $AB = 12$ , $BC = 15$ , and $AC = 18$ . A point $P$ lies strictly inside $\triangle ABC$ . If the lengths of the segments $BP$ and $CP$ are both integers, what is the maximum possible value of the sum of these two lengths?

26

27

28

29

30

Show answer & explanation

Answer: 28

Answer

28

The correct answer is 28. According to the properties of triangles, for any point

P

strictly inside

\triangle ABC

, the sum of the interior segments is strictly less than the sum of the other two sides:

BP + CP < AB + AC = 12 + 18 = 30

. Since

BP

and

CP

are integers, we check the boundary where

P

lies on the side

AC

at an integer distance

CP = z

from

C

. Applying Stewart's Theorem, the boundary length

BP = y_{bound}

is

\sqrt{z^2 - 22.5z + 225}

. For the largest possible integer value

z = 17

, the boundary value is approximately

11.47

. Because the point must lie strictly inside the triangle,

BP

must be strictly less than this boundary, so the maximum integer value for

BP

is 11. This yields a maximum sum of

11 + 17 = 28

. Lower integer values of

z

yield smaller maximum sums (for example, if

z = 16

, the boundary is exactly 11, so

BP

can be at most 10, giving a sum of 26).

Step-by-Step Solution

1

Apply the interior point triangle inequality theorem.

For any point

P

strictly inside

\triangle ABC

, the sum of the distances to two vertices is strictly less than the sum of the other two sides:

BP + CP < AB + AC

.

This establishes the theoretical upper bound for the sum of the two segment lengths.

2

Calculate the theoretical upper bound.

Since

AB = 12

and

AC = 18

, we have

BP + CP < 12 + 18 = 30

. Since

BP

and

CP

must be integers, the sum

BP + CP

can be at most 29.

This sets the initial integer limit before evaluating if it is geometrically possible.

3

Analyze the boundary conditions for integer lengths using Stewart's Theorem.

Let

CP = z

and

BP = y

, where

y

and

z

are integers. As

P

approaches the side

AC

, the boundary value

y_{bound}

represents the distance from

B

to a point on

AC

at distance

z

from

C

. Using Stewart's Theorem, this boundary satisfies:

y_{bound}^2 = z^2 - 22.5z + 225

. Since

P

is strictly inside the triangle,

y

must be strictly less than

y_{bound}

.

This provides the mathematical relationship determining whether a point is inside the triangle for any given integer length of one segment.

4

Test the maximum possible integer value for

z

to maximize the sum

y + z

.

Since

P

is strictly inside,

z

must be strictly less than

AC = 18

, so the maximum integer for

z

is 17. For

z = 17

, the boundary value is

y_{bound} = \sqrt{17^2 - 22.5(17) + 225} = \sqrt{131.5} \approx 11.47

. Since

y < y_{bound}

, the maximum integer value for

y

is 11. This yields a maximum sum of

11 + 17 = 28

.

This determines the actual maximum integer sum that can be geometrically realized within the triangle.

Key Concept

Triangle Inequality Theorem and Interior Point Properties

Question 27Question →

In $\triangle ABC$ , the lengths of sides $AB$ and $AC$ are both $13$ . A point $D$ lies on side $BC$ such that $AD$ is an integer. If the perimeter of $\triangle ABD$ is equal to the perimeter of $\triangle ACD$ , what is the sum of all possible integer values for the length of $BC$ ?

Show answer & explanation

Answer: 34

Answer

34

The perimeters of

\triangle ABD

and

\triangle ACD

are equal, and since

AB = AC = 13

, this forces

BD = CD

, making

D

the midpoint of

BC

. In the isosceles triangle

\triangle ABC

, the median

AD

is perpendicular to

BC

, making

\triangle ABD

a right triangle. By the Pythagorean theorem,

BD^2 + AD^2 = 169

. Since

AD

is an integer,

BD

must also be an integer (a half-integer would result in

AD^2

ending in

.25

, which cannot be a perfect square of an integer). The only positive integer solutions for

(BD, AD)

are

(5, 12)

and

(12, 5)

. This results in

BC = 2 \cdot BD

being either

10

or

24

. The sum of these possible values is

10 + 24 = 34

.

Step-by-Step Solution

1

Set the perimeters of

\triangle ABD

and

\triangle ACD

equal to each other.

BD = CD

Since

AB = AC = 13

, equating

AB + BD + AD = AC + CD + AD

simplifies directly to

BD = CD

.

2

Determine the relationship between

AD

and

BC

.

\triangle ABD

is a right triangle with hypotenuse

13

.

In an isosceles triangle, the median to the base is also the altitude, so

AD \perp BC

.

3

Apply the Pythagorean theorem to

\triangle ABD

.

BD^2 + AD^2 = 169

The sum of the squares of the legs in right triangle

\triangle ABD

must equal the square of the hypotenuse

AB = 13

.

4

Analyze the parity and integer constraints of

BD

and

AD

.

BD

must be a positive integer.

If

BD

were a half-integer,

BD^2

would end in

.25

, preventing

AD^2

from being an integer, which contradicts the given condition that

AD

is an integer.

5

Identify the Pythagorean triples with a hypotenuse of

13

.

(BD, AD) \in \{(5, 12), (12, 5)\}

The only positive integer solutions to

x^2 + y^2 = 13^2

are

(5, 12)

and

(12, 5)

.

6

Calculate the possible lengths of

BC

and sum them.

BC \in \{10, 24\}

, and their sum is

34

.

Since

D

is the midpoint of

BC

, the length of

BC

is

2 \cdot BD

, yielding

2 \cdot 5 = 10

and

2 \cdot 12 = 24

. Both satisfy the triangle inequality because

BC < AB + AC = 26

.

Key Concept

Properties of Isosceles Triangles and the Pythagorean Theorem

Estimated Time:3m 0s

Question 28Question →

In $\triangle ABC$ , the measure of $\angle B$ is $80^\circ$ and the measure of $\angle C$ is $40^\circ$ . A point $D$ lies on side $BC$ such that $AD$ bisects $\angle BAC$ , and a point $E$ lies on side $AC$ such that $AD = AE$ . What is the measure, in degrees, of $\angle CDE$ ?

Show answer & explanation

Answer: 35

Answer

35

The correct answer is

35

. By first finding that

\angle BAC = 60^\circ

, we use the angle bisector

AD

to find

\angle CAD = 30^\circ

. In

\triangle ADC

, we find the interior angle

\angle ADC = 110^\circ

. In the isosceles triangle

\triangle ADE

with

AD=AE

, the base angles are

\angle ADE = \angle AED = 75^\circ

. Finally, subtracting

\angle ADE

from

\angle ADC

gives

\angle CDE = 35^\circ

.

Step-by-Step Solution

1

Find the measure of the third angle of the main triangle,

\angle BAC

.

\angle BAC = 60^\circ

The sum of the interior angles of any triangle is

180^\circ

. Therefore,

\angle BAC = 180^\circ - \angle B - \angle C = 180^\circ - 80^\circ - 40^\circ = 60^\circ

.

2

Determine the measure of the bisected angle

\angle CAD

.

\angle CAD = 30^\circ

Since

AD

bisects

\angle BAC

, it divides the angle into two equal parts:

\angle BAD = \angle CAD = \frac{60^\circ}{2} = 30^\circ

.

3

Calculate the interior angle

\angle ADC

in

\triangle ADC

.

\angle ADC = 110^\circ

In

\triangle ADC

, the sum of angles is

180^\circ

. Therefore,

\angle ADC = 180^\circ - \angle CAD - \angle C = 180^\circ - 30^\circ - 40^\circ = 110^\circ

.

4

Find the base angles of the isosceles triangle

ADE

.

\angle ADE = 75^\circ

Since

AD = AE

,

\triangle ADE

is an isosceles triangle with vertex angle

\angle DAE = 30^\circ

. The two base angles,

\angle ADE

and

\angle AED

, are equal. Thus,

\angle ADE = \frac{180^\circ - 30^\circ}{2} = 75^\circ

.

5

Determine the final angle

\angle CDE

by subtraction.

\angle CDE = 35^\circ

Since point

E

lies on side

AC

, ray

DE

lies between rays

DA

and

DC

. Therefore,

\angle ADC = \angle ADE + \angle CDE

. Rearranging gives

\angle CDE = \angle ADC - \angle ADE = 110^\circ - 75^\circ = 35^\circ

.

Key Concept

Applying triangle angle sum theorem, angle bisector properties, and isosceles triangle base angle properties to perform multi-step angle tracing.

Alternative Method

Use the exterior angle theorem on

\triangle ADC

at vertex

D

:

\angle ADB = \angle CAD + \angle C = 30^\circ + 40^\circ = 70^\circ

. Then, since

E

is on

AC

,

A

,

E

, and

C

are collinear. In

\triangle ADE

, the exterior angle at

E

is

\angle DEC = \angle DAE + \angle ADE = 30^\circ + 75^\circ = 105^\circ

. In

\triangle DEC

, the sum of angles is

180^\circ

, so

\angle CDE = 180^\circ - 105^\circ - 40^\circ = 35^\circ

.

Estimated Time:2m 30s

Question 29Question →

In a certain triangle, the ratio of the measure of the first angle to the measure of the second angle is $1:2$ . The measure of the third angle is $20^\circ$ less than the measure of the second angle. What is the measure of the largest angle in the triangle?

40^\circ

64^\circ

80^\circ

100^\circ

120^\circ

Show answer & explanation

Answer: $80^\circ$

Answer

80^\circ

The correct answer is

80^\circ

. By representing the first angle as

x

, the second angle is

2x

, and the third angle is

2x - 20

. The sum of the interior angles in a triangle is always

180^\circ

, which gives the equation

x + 2x + (2x - 20) = 180

. Simplifying this results in

5x - 20 = 180

, which yields

5x = 200

and

x = 40

. Substituting this value back into the expressions for the three angles gives measures of

40^\circ

,

80^\circ

, and

60^\circ

. Comparing these values shows that the largest angle is

80^\circ

.

Step-by-Step Solution

1

Define the measures of the first and second angles using a single variable based on their ratio.

Let the first angle be

x

and the second angle be

2x

.

Since the ratio of the first angle to the second angle is

1:2

, we can represent them as

x

and

2x

respectively.

2

Express the measure of the third angle in terms of the same variable.

The third angle is

2x - 20

.

The problem states the third angle is

20^\circ

less than the second angle, which has a measure of

2x

.

3

Set up an equation using the triangle angle sum theorem and solve for

x

.

x + 2x + (2x - 20) = 180 \implies 5x - 20 = 180 \implies 5x = 200 \implies x = 40

.

The sum of the measures of the interior angles of any triangle is always

180^\circ

.

4

Calculate the measures of all three angles to determine which is the largest.

First angle =

40^\circ

, Second angle =

2(40^\circ) = 80^\circ

, Third angle =

2(40^\circ) - 20^\circ = 60^\circ

. The largest angle is

80^\circ

.

We must substitute

x = 40

back into our expressions for each angle to find their actual degree measures and identify the largest one.

Key Concept

The interior angles of a triangle always sum to

180^\circ

. Ratios and word problems can be modeled algebraically to determine unknown angle measures.

Estimated Time:1m 15s

Question 30Question →

In $\triangle ABC$ , point $D$ lies on side $BC$ . The segment $AD$ divides the interior angle $\angle BAC$ into two angles, $\angle BAD$ and $\angle DAC$ , whose measures are in the ratio $3:2$ , respectively. The measures of the interior angles $\angle B$ and $\angle C$ are in the ratio $5:4$ , respectively. If the measure of $\angle ADC$ is $104^\circ$ , what is the measure of $\angle BAC$ ?

54^\circ

72^\circ

76^\circ

90^\circ

108^\circ

Show answer & explanation

Answer: $90^\circ$

Answer

90^\circ

The correct answer is

90^\circ

. By expressing the angles in terms of variables using their ratios, we set up two independent linear equations:

3x + 5y = 104

(from the exterior angle theorem on

\triangle ABD

) and

2x + 4y = 76

(from the sum of angles in

\triangle ADC

). Solving this system yields

x = 18

. Since

\angle BAC

is composed of

\angle BAD

and

\angle DAC

, its measure is

3x + 2x = 5x = 5(18^\circ) = 90^\circ

.

Step-by-Step Solution

1

Define variables for the partitioned angles and the base angles using the given ratios.

Let the measures of

\angle BAD

and

\angle DAC

be

3x

and

2x

respectively, so that

\angle BAC = 5x

. Let the measures of

\angle B

and

\angle C

be

5y

and

4y

respectively.

Ratios express quantities as multiples of a common variable, which simplifies setting up equations.

2

Apply the exterior angle theorem to

\triangle ABD

at vertex

D

.

The exterior angle

\angle ADC = \angle BAD + \angle B \Rightarrow 104^\circ = 3x + 5y

.

The measure of an exterior angle of a triangle is equal to the sum of the measures of its two remote interior angles.

3

Apply the angle sum theorem to

\triangle ADC

.

\angle DAC + \angle C + \angle ADC = 180^\circ \Rightarrow 2x + 4y + 104^\circ = 180^\circ \Rightarrow 2x + 4y = 76^\circ

.

The sum of the measures of the interior angles of any triangle is always

180^\circ

.

4

Solve the system of linear equations: (1)

3x + 5y = 104

and (2)

2x + 4y = 76

.

Multiply equation (1) by 2 and equation (2) by 3 to align the coefficients of

x

:

6x + 10y = 208

6x + 12y = 228

Subtract the first aligned equation from the second:

2y = 20 \Rightarrow y = 10

.
Substitute

y = 10

back into equation (2):

2x + 4(10) = 76 \Rightarrow 2x = 36 \Rightarrow x = 18

.

Solving the system of linear equations determines the values of the variables

x

and

y

.

5

Calculate the measure of

\angle BAC

.

\angle BAC = 5x = 5(18^\circ) = 90^\circ

.

We defined the total measure of

\angle BAC

as the sum of its two partitioned parts,

3x + 2x = 5x

.

Key Concept

Using triangle angle properties and exterior angle theorems to set up and solve systems of linear equations.

Question 31Question →

In a triangle, the lengths of the sides are $x$ , $y$ , and $z$ , where $x$ , $y$ , and $z$ are integers such that $x < y < z$ . If $x = 7$ and the perimeter of the triangle is $32$ , what is the number of possible integer values for $z$ ?

Show answer & explanation

Answer: 3

Answer

There are exactly 3 possible integer values for the side length z.

The correct answer is 3. By expressing the second side as

y = 25 - z

and applying the ordering constraint

7 < 25 - z < z

, we determine that

12.5 < z < 18

. Applying the Triangle Inequality Theorem (

7 + y > z

) yields the restriction

z < 16

. Combining these conditions restricts the integer values of

z

to

\{13, 14, 15\}

, which counts to exactly 3 possible values.

Step-by-Step Solution

1

Express the side length

y

in terms of

z

.

y = 25 - z

The perimeter of the triangle is the sum of the side lengths:

x + y + z = 32

. Substituting

x = 7

gives

7 + y + z = 32

, which simplifies to

y = 25 - z

.

2

Apply the given inequality constraint

x < y < z

to find initial bounds for

z

.

12.5 < z < 18

Substituting

x = 7

and

y = 25 - z

into

x < y < z

yields

7 < 25 - z < z

. The left inequality

7 < 25 - z

simplifies to

z < 18

. The right inequality

25 - z < z

simplifies to

25 < 2z

, or

z > 12.5

.

3

Apply the Triangle Inequality Theorem to establish the final constraint on

z

.

z < 16

Since

z

is the longest side, the sum of the two shorter sides must be strictly greater than

z

:

x + y > z

. Substituting

x = 7

and

y = 25 - z

gives

7 + 25 - z > z

, which simplifies to

32 > 2z

, or

z < 16

.

4

Combine all constraints and count the valid integer values for

z

.

3 possible values (

13, 14, 15

)

Combining the bounds from the steps gives

12.5 < z < 16

. The integers satisfying this inequality are

13

,

14

, and

15

, which gives a total of 3 possible integer values.

Key Concept

Triangle Inequality Theorem and algebraic constraints on side lengths

Question 32Question →

A triangle has two sides of length 5 and 12. The third side has a length of $x$ , where $x$ is an integer. If the perimeter of the triangle is a multiple of 5, what is the sum of all possible values of $x$ ?

13

25

21

24

39

Show answer & explanation

Answer: 21

Answer

The sum of all possible values of the third side length is 21.

To find the sum of all possible values of

x

, we first apply the Triangle Inequality Theorem. For a triangle with side lengths 5, 12, and

x

, the third side must satisfy

12 - 5 < x < 12 + 5

, which simplifies to

7 < x < 17

. The perimeter

P

of the triangle is given by

P = 5 + 12 + x = 17 + x

. Given that

7 < x < 17

, the perimeter must be between

17 + 7 = 24

and

17 + 17 = 34

. The only multiples of 5 within this range are 25 and 30. Setting the perimeter equal to these values gives

17 + x = 25 \implies x = 8

, and

17 + x = 30 \implies x = 13

. Both values are integers and satisfy the triangle inequality. The sum of these values is

8 + 13 = 21

.

Step-by-Step Solution

1

Apply the Triangle Inequality Theorem to find the range of possible values for the third side,

x

.

12 - 5 < x < 12 + 5

, which simplifies to

7 < x < 17

.

The length of any side of a triangle must be strictly greater than the difference between the other two sides and strictly less than their sum.

2

Determine the expression for the perimeter of the triangle and find the bounds for the perimeter.

Perimeter

P = 5 + 12 + x = 17 + x

. Since

7 < x < 17

, the perimeter must satisfy

17 + 7 < P < 17 + 17

, which means

24 < P < 34

.

The perimeter of a triangle is the sum of its three side lengths.

3

Identify which values of the perimeter in this range are multiples of 5, and find the corresponding values of

x

.

The multiples of 5 between 24 and 34 are 25 and 30. If

P = 25

, then

17 + x = 25 \implies x = 8

. If

P = 30

, then

17 + x = 30 \implies x = 13

. Both

x = 8

and

x = 13

are integers that satisfy the initial inequality.

We must find the integer values of

x

that make the perimeter a multiple of 5.

4

Calculate the sum of all possible values of

x

.

8 + 13 = 21

.

The question asks for the sum of all valid integer values of

x

.

Key Concept

Triangle Inequality Theorem and Perimeter Calculations

Estimated Time:1m 30s

Question 33Question →

An isosceles triangle has two sides of length $5$ and $11$ . A second triangle has side lengths of $12$ , $18$ , and $d$ , where $d$ is an integer. If $d$ is equal to the perimeter of the first triangle, what is the perimeter of the second triangle?

41

46

51

57

63

Show answer & explanation

Answer: 57

Answer

57

To find the perimeter of the second triangle, we must first determine the value of

d

, which is the perimeter of the first triangle. The first triangle is isosceles with two sides of length

5

and

11

. By the Triangle Inequality Theorem, the sum of any two side lengths must exceed the third. A triangle with sides

5, 5, 11

is impossible because

5 + 5 = 10 < 11

. Thus, the sides of the first triangle must be

11, 11, 5

, giving a perimeter of

11 + 11 + 5 = 27

. This means

d = 27

. The second triangle has sides of length

12

,

18

, and

27

. Since

12 + 18 = 30 > 27

, this is a valid triangle. Its perimeter is

12 + 18 + 27 = 57

.

Step-by-Step Solution

1

Determine the possible side lengths of the first isosceles triangle.

The sides must be

11

,

11

, and

5

.

An isosceles triangle has two equal sides. The side lengths must be either

5, 5, 11

or

11, 11, 5

. According to the Triangle Inequality Theorem, the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the third side. If the sides were

5, 5, 11

, then

5 + 5 = 10 < 11

, which violates this theorem. Thus, the only valid side lengths are

11

,

11

, and

5

(since

5 + 11 = 16 > 11

).

2

Calculate the perimeter of the first triangle to find the value of

d

.

d = 27

The perimeter of the first triangle is the sum of its three sides:

11 + 11 + 5 = 27

. Since

d

is equal to this perimeter,

d = 27

.

3

Verify that a triangle with side lengths

12

,

18

, and

27

is valid.

The triangle is valid.

We check the Triangle Inequality Theorem:

12 + 18 = 30 > 27

,

12 + 27 = 39 > 18

, and

18 + 27 = 45 > 12

. Since all inequalities hold, the second triangle is valid.

4

Calculate the perimeter of the second triangle.

Perimeter =

57

The perimeter of the second triangle is the sum of its side lengths:

12 + 18 + 27 = 57

.

Key Concept

Triangle Inequality Theorem and Isosceles Triangle Properties

Estimated Time:2m 0s

Question 34Question →

In $\triangle ABC$ , the measures of the interior angles $\angle A$ , $\angle B$ , and $\angle C$ are in the ratio $3:4:5$ , respectively. What is the measure of the largest exterior angle of $\triangle ABC$ ?

45^\circ

75^\circ

105^\circ

108^\circ

135^\circ

Show answer & explanation

Answer: $135^\circ$

Answer

135 degrees

The interior angles of a triangle sum to

180^\circ

. Since the angles are in the ratio

3:4:5

, they can be represented as

3x

,

4x

, and

5x

. Adding these gives

12x = 180^\circ

, which simplifies to

x = 15^\circ

. The interior angles are therefore

45^\circ

,

60^\circ

, and

75^\circ

. Because an exterior angle is supplementary to its adjacent interior angle, the largest exterior angle is paired with the smallest interior angle:

180^\circ - 45^\circ = 135^\circ

. Alternatively, the exterior angle is equal to the sum of the two non-adjacent interior angles:

60^\circ + 75^\circ = 135^\circ

.

Step-by-Step Solution

1

Represent the interior angles algebraically using the given ratio.

Let the measures of the interior angles be

3x

,

4x

, and

5x

.

The ratio of the angles is

3:4:5

, so their measures must be multiples of these ratio numbers by the same factor

x

.

2

Set up and solve an equation for

x

using the triangle angle sum theorem.

3x + 4x + 5x = 180^\circ \implies 12x = 180^\circ \implies x = 15^\circ

.

The sum of the interior angles of any triangle is always

180^\circ

.

3

Determine the measures of the three interior angles.

The angles measure

3(15^\circ) = 45^\circ

,

4(15^\circ) = 60^\circ

, and

5(15^\circ) = 75^\circ

.

Multiplying the value of

x

by each term of the ratio gives the individual interior angle measures.

4

Find the largest exterior angle of the triangle.

The largest exterior angle is supplementary to the smallest interior angle:

180^\circ - 45^\circ = 135^\circ

.

An exterior angle and its adjacent interior angle form a linear pair and sum to

180^\circ

. The smallest interior angle will yield the largest exterior angle.

Key Concept

Triangle Angle Sum Theorem and Exterior Angle Relationships

Alternative Method

The exterior angle at any vertex of a triangle is equal to the sum of the measures of the two opposite interior angles. The two largest interior angles are

60^\circ

and

75^\circ

. Therefore, the largest exterior angle is the sum of these two angles:

60^\circ + 75^\circ = 135^\circ

.

Estimated Time:1m 0s

Question 35Question →

In $\triangle ABC$ , point $D$ lies on side $BC$ such that $AD = BD$ . If the measure of $\angle ADC$ is $112^\circ$ and the measure of $\angle BAC$ is $85^\circ$ , what is the measure of $\angle C$ , in degrees?

Show answer & explanation

Answer: 39

Answer

The measure of

\angle C

is

39^\circ

.

The measure of

\angle C

is found by first calculating the interior angle

\angle ADB = 180^\circ - 112^\circ = 68^\circ

since

BDC

forms a straight line. Because

AD = BD

,

\triangle ABD

is isosceles with

\angle B = \angle BAD

. Using the angle sum of

\triangle ABD

, we have

2(\angle B) + 68^\circ = 180^\circ

, which yields

\angle B = 56^\circ

. Finally, using the angle sum of

\triangle ABC

, we calculate

\angle C = 180^\circ - (85^\circ + 56^\circ) = 39^\circ

.

Step-by-Step Solution

1

Find the measure of

\angle ADB

using the linear pair relationship with

\angle ADC

.

\angle ADB = 68^\circ

Angles on a straight line add up to

180^\circ

. Since

D

lies on

BC

,

\angle ADB + \angle ADC = 180^\circ

.

2

Calculate the measure of

\angle B

using the properties of the isosceles triangle

ABD

.

\angle B = 56^\circ

Since

AD = BD

, the base angles opposite to these sides are equal:

\angle BAD = \angle B

. The sum of angles in

\triangle ABD

is

180^\circ

, so

2(\angle B) + 68^\circ = 180^\circ

.

3

Find the measure of

\angle C

using the triangle angle sum theorem on the large triangle

ABC

.

\angle C = 39^\circ

The sum of the angles in

\triangle ABC

is

180^\circ

, meaning

\angle BAC + \angle B + \angle C = 180^\circ

. Substituting the known values gives

85^\circ + 56^\circ + \angle C = 180^\circ

.

Key Concept

Using the Isosceles Triangle Theorem, the Triangle Angle Sum Theorem, and linear pairs to trace unknown angles in a geometric figure.

Practice More

Try finding the missing angles when a transversal cuts two parallel lines that form a triangle with a third intersecting line.

Alternative Method

Instead of finding

\angle B

first and then solving for

\angle C

in

\triangle ABC

, we can find the angle

\angle DAC

first. Since

\angle ADC = 112^\circ

is an exterior angle to

\triangle ABD

, we have

\angle ADC = \angle B + \angle BAD

. Since

\angle B = \angle BAD

, we get

2(\angle BAD) = 112^\circ \implies \angle BAD = 56^\circ

. Because

\angle BAC = 85^\circ

, we have

\angle DAC = 85^\circ - 56^\circ = 29^\circ

. Now looking at

\triangle ADC

, we can solve for

\angle C

directly:

\angle C = 180^\circ - (112^\circ + 29^\circ) = 39^\circ

.

Estimated Time:1m 30s

Question 36Question →

The lengths of the three sides of a triangle are in the ratio $3:4:x$ , where $x$ is an integer. If the perimeter of the triangle is $36$ centimeters, how many different possible values can $x$ have?

1

3

2

4

5

Show answer & explanation

Answer: 2

Answer

There are 2 possible integer values for

x

.

The correct answer is the value of 2. By expressing the side lengths as

3k

,

4k

, and

xk

where

k

must be a positive integer, the perimeter equation becomes

k(7 + x) = 36

. Solving for

7 + x

yields factors of

36

greater than

7

, which are

9, 12, 18,

and

36

. These correspond to

x

values of

2, 5, 11,

and

29

. Checking each set of side lengths against the Triangle Inequality Theorem shows that only the sets corresponding to

x = 2

(sides

12, 16, 8

) and

x = 5

(sides

9, 12, 15

) form valid triangles.

Step-by-Step Solution

1

Define the side lengths using a multiplier

k

.

Let the side lengths of the triangle be

3k

,

4k

, and

xk

for some positive multiplier

k

. Since all three side lengths must be integers, the difference between the first two sides,

4k - 3k = k

, must also be an integer. Thus,

k

must be a positive integer.

This establishes that the scaling factor

k

is a positive integer, allowing us to find discrete solutions.

2

Set up the perimeter equation and express

x

in terms of

k

.

The perimeter is the sum of the side lengths:

3k + 4k + xk = 36 \implies k(7 + x) = 36

. Since

k

and

x

are positive integers,

7 + x

must be a factor of

36

that is greater than

7

.

This constrains the possible values of

x

to the factors of

36

that are larger than

7

.

3

Find the potential values of

x

and their corresponding side lengths.

The factors of

36

greater than

7

are

9, 12, 18,

and

36

. This yields four potential cases:
- If

7+x = 9 \implies x = 2

, then

k = 4

, and the sides are

12, 16, 8

.
- If

7+x = 12 \implies x = 5

, then

k = 3

, and the sides are

9, 12, 15

.
- If

7+x = 18 \implies x = 11

, then

k = 2

, and the sides are

6, 8, 22

.
- If

7+x = 36 \implies x = 29

, then

k = 1

, and the sides are

3, 4, 29

.

This identifies all mathematically possible configurations before checking if they can physically form a triangle.

4

Apply the Triangle Inequality Theorem to each case.

The sum of the lengths of any two sides must be strictly greater than the length of the remaining side:
- For sides

12, 16, 8

:

12 + 8 = 20 > 16

(Valid).
- For sides

9, 12, 15

:

9 + 12 = 21 > 15

(Valid).
- For sides

6, 8, 22

:

6 + 8 = 14 < 22

(Invalid).
- For sides

3, 4, 29

:

3 + 4 = 7 < 29

(Invalid).
Only the cases where

x = 2

and

x = 5

form valid triangles.

This filters the candidate values of

x

to only those that can form a valid geometric triangle.

Key Concept

Triangle Inequality Theorem and Integer Ratio Constraints

Estimated Time:1m 30s

Question 37Question →

A triangle has side lengths of $7$ , $x+2$ , and $2x-1$ , where $x$ is an integer. What is the total number of possible values for $x$ ?

Show answer & explanation

Answer: 7

Answer

The total number of possible integer values for

x

is

7

.

According to the Triangle Inequality Theorem, a triangle is formed if and only if the sum of any two side lengths is strictly greater than the third side length. Solving the three inequalities

7 + (x+2) > 2x-1

,

7 + (2x-1) > x+2

, and

(x+2) + (2x-1) > 7

gives

x < 10

,

x > -4

, and

x > 2

. The overlapping interval is

2 < x < 10

. The integers in this range are

\{3, 4, 5, 6, 7, 8, 9\}

, which gives a total of

7

possible integer values.

Step-by-Step Solution

1

Set up the three inequalities required by the Triangle Inequality Theorem.

1)

7 + (x+2) > 2x-1

2)

7 + (2x-1) > x+2

3)

(x+2) + (2x-1) > 7

The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the remaining side.

2

Solve the first inequality:

7 + (x+2) > 2x-1

.

x < 10

Simplifying the left side yields

x + 9 > 2x - 1

. Subtracting

x

and adding

1

to both sides results in

10 > x

, which means

x < 10

.

3

Solve the second inequality:

7 + (2x-1) > x+2

.

x > -4

Simplifying the left side yields

2x + 6 > x + 2

. Subtracting

x

and

6

from both sides results in

x > -4

.

4

Solve the third inequality:

(x+2) + (2x-1) > 7

.

x > 2

Simplifying the left side yields

3x + 1 > 7

. Subtracting

1

and dividing by

3

results in

x > 2

.

5

Determine the combined range for

x

and identify the valid integers.

The combined range is

2 < x < 10

. The valid integers are

3, 4, 5, 6, 7, 8,

and

9

.

To satisfy all three inequalities,

x

must be greater than

2

, greater than

-4

, and less than

10

, which simplifies to

2 < x < 10

.

6

Count the total number of valid integer values for

x

.

There are

7

integer values.

Counting the elements of the set

\{3, 4, 5, 6, 7, 8, 9\}

yields a total of

7

values.

Key Concept

Triangle Inequality Theorem

Estimated Time:1m 30s

Question 38Question →

In $\triangle PQR$ , the measure of $\angle P$ is $50^\circ$ . The angle bisectors of $\angle PQR$ and $\angle PRQ$ intersect at point $I$ inside the triangle. What is the measure of $\angle QIR$ ?

65^\circ

100^\circ

115^\circ

125^\circ

130^\circ

Show answer & explanation

Answer: $115^\circ$

Answer

115 degrees

To find the measure of the angle at the intersection, we first use the Triangle Angle-Sum Theorem on the larger triangle. The sum of the interior angles in the larger triangle is 180 degrees. Since the angle at one vertex is 50 degrees, the sum of the other two angles must be 130 degrees. The angle bisectors divide these two angles in half, meaning the sum of the two half-angles in the smaller triangle is half of 130 degrees, which is 65 degrees. Applying the Triangle Angle-Sum Theorem to the smaller triangle, the sum of its interior angles is also 180 degrees. Subtracting the sum of the two half-angles (65 degrees) from 180 degrees gives 115 degrees for the angle at the intersection.

Step-by-Step Solution

1

Find the sum of the remaining interior angles of the triangle.

The sum of the angles at Q and R is 130 degrees.

The sum of all interior angles in any triangle is always 180 degrees, and the angle at P is given as 50 degrees.

2

Determine the sum of the bisected angles in the smaller triangle.

The sum of the half-angles is 65 degrees.

Since the lines QI and RI are angle bisectors, the sum of the interior angles of the smaller triangle at vertices Q and R is half the sum of the angles at Q and R of the larger triangle.

3

Calculate the measure of the angle at the intersection point.

The angle at the intersection point is 115 degrees.

The sum of the interior angles in the smaller triangle is also 180 degrees, so the unknown angle is found by subtracting the sum of the two half-angles from 180 degrees.

Key Concept

The Triangle Angle-Sum Theorem states that the sum of the measures of the interior angles of a triangle is always 180 degrees. Angle bisectors divide an angle into two equal parts.

Alternative Method

Alternatively, one can assign specific values to the angles that satisfy the given conditions. Since the sum of the angles in the large triangle must be 180 degrees and the angle at vertex P is 50 degrees, the sum of the other two angles must be 130 degrees. If we assume the triangle is isosceles with the two unknown angles being equal, each of those angles is 65 degrees. Their bisectors would each form an angle of 32.5 degrees with the base. In the smaller triangle, the sum of these two bisected angles is 65 degrees, which leaves 115 degrees for the angle at the intersection point.

Estimated Time:1m 15s

Question 39Question →

In $\triangle ABC$ , the measure of exterior angle $\angle ACD$ is $135^\circ$ , where $D$ lies on the extension of side $BC$ past $C$ . If the measure of interior angle $\angle A$ is $25^\circ$ greater than the measure of interior angle $\angle B$ , what is the measure, in degrees, of $\angle B$ ?

Show answer & explanation

Answer: 55

Answer

55

According to the Exterior Angle Theorem, the measure of exterior angle

\angle ACD

is equal to the sum of the measures of its remote interior angles,

\angle A

and

\angle B

. This gives the equation

\text{m}\angle A + \text{m}\angle B = 135^\circ

. Using the information that

\text{m}\angle A = \text{m}\angle B + 25^\circ

, we substitute this expression into the equation to get

(\text{m}\angle B + 25^\circ) + \text{m}\angle B = 135^\circ

. Simplifying this equation gives

2\text{m}\angle B + 25 = 135

, which simplifies to

2\text{m}\angle B = 110

, and dividing by 2 yields

\text{m}\angle B = 55^\circ

.

Step-by-Step Solution

1

Apply the Exterior Angle Theorem to express the relation between the exterior angle and the two remote interior angles.

\text{m}\angle A + \text{m}\angle B = 135^\circ

The measure of an exterior angle of a triangle is equal to the sum of the measures of its two remote interior angles.

2

Substitute the relationship between the interior angles into the equation.

(\text{m}\angle B + 25^\circ) + \text{m}\angle B = 135^\circ

The problem states that the measure of interior angle

\angle A

is

25^\circ

greater than the measure of interior angle

\angle B

.

3

Solve the algebraic equation for the measure of interior angle

\angle B

.

\text{m}\angle B = 55^\circ

Combining like terms gives

2\text{m}\angle B + 25 = 135

. Subtracting 25 from both sides gives

2\text{m}\angle B = 110

. Dividing by 2 yields

\text{m}\angle B = 55^\circ

.

Key Concept

Exterior Angle Theorem and remote interior angles relation

Question 40Question →

A triangle has two sides of length $9$ and $14$ . The length of the third side, $s$ , is a multiple of $4$ . What is the sum of all possible integer values for $s$ ?

48

60

56

76

80

Show answer & explanation

Answer: 56

Answer

56

According to the Triangle Inequality Theorem, the length of the third side,

s

, of a triangle with sides of length 9 and 14 must satisfy the inequality

14 - 9 < s < 14 + 9

, which simplifies to

5 < s < 23

. The integer values within this range that are multiples of 4 are 8, 12, 16, and 20. Adding these values together yields a sum of 56.

Step-by-Step Solution

1

Apply the Triangle Inequality Theorem to determine the range of possible lengths for the third side.

The third side length,

s

, must satisfy the inequality:

14 - 9 < s < 14 + 9

, which simplifies to

5 < s < 23

.

The Triangle Inequality Theorem states that the length of any side of a triangle must be strictly greater than the positive difference of the other two sides and strictly less than the sum of the other two sides.

2

Identify all integer values in the range

(5, 23)

that are multiples of 4.

The multiples of 4 that are strictly greater than 5 and strictly less than 23 are: 8, 12, 16, and 20.

We must find the integers within the bounds that can be divided by 4 with a remainder of 0.

3

Calculate the sum of the identified multiples of 4.

8 + 12 + 16 + 20 = 56

.

The question asks for the sum of all possible integer values for the third side length

s

.

Key Concept

Triangle Inequality Theorem

Estimated Time:1m 0s

Triangle Properties and Angle Theorems

Answer

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