Big Ideas Math Integrated Math 1 Student Journal Solutions Chapter 10 Maintaining Mathematical Proficiency Exercise

Big Ideas Math Integrated Math Chapter 10 Maintaining Mathematical Proficiency

 

Page 274  Exercise 1  Problem 1

Question 1.

Given the following angle measurements in a diagram:

$\mathrm{\angle }𝐴𝐵𝐷=(5𝑥+2{)}^{\circ }$ $\mathrm{\angle }𝐷𝐵𝐶=(2𝑥+3{)}^{°}$

1. Find the values of ∠ABD and ∠DBC.
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2. Show all steps clearly, including the calculation of the unknown variable 𝑥.

Answer:

Given diagram

Given that

$\mathrm{\angle }𝐴𝐵𝐷=(5𝑥+2{)}^{\circ }$ $\mathrm{\angle }𝐷𝐵𝐶=(2𝑥+3{)}^{°}$

Here

∠ABD = (5x + 2)°

∠DBC = (2x + 3)°

We have to find the angle ∠ABD and ∠DBC

The total sum of angle on one side of a line is 1800.

Thus we add both the angles formed by this BD line on the AC line and equate them with 1800 to find both the angles and by extension the unknown variable.

In this question, we have a line BD meeting a point B on the line AC.

As it meets the line, the sum of both the angles must be the angle that is naturally formed on one side of the line which is 1800.

Thus

​∡ABD + ∡DBC = 180

(5x + 2) + (2x + 3) = 180

5x + 2x + 2 + 3 = 180

7x + 5 = 180

7x = 175

x = 25

​Thus we can calculate both angles.

​∡ABD = 5x + 2

∡ABD = (5 × 25) + 2

∡ABD = 125 + 2

∡ABD = 127
​
Now rather than calculating the angle DBC, we can just subtract this found angle from 1800

So that we can find the supplementary angle DBC.

​∡DBC = 180 − ∡ABD

∡DBC = 180 − 127

∡DBC = 53
​
Both the angles for the given diagram can be written as ∡ABD = 1270 and ∡DBC = 530.

 

Page 274  Exercise 2  Problem 2

Question 2. 

Given the following angle measurements in a right-angle shape EFG with a line FH meeting point F:

$\mathrm{\angle }𝐸𝐹𝐻=(11𝑥+1{)}^{\circ }$ $\mathrm{\angle }𝐻𝐹𝐺=(2𝑥-2{)}^{\circ }$

1. Find the values of ∠𝐸𝐹𝐻 and ∠HFG.
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2. Show all steps clearly, including the calculation of the unknown variable x.

Answer:

Given

$\mathrm{\angle }𝐸𝐹𝐻=(11𝑥+1{)}^{\circ }$ $\mathrm{\angle }𝐻𝐹𝐺=(2𝑥-2{)}^{\circ }$

The total angle in a right angle or a perfect L shape is 900.

Thus we add both the angles formed by this FH line on the EFG shape and equate it with 900 to find both the angles and by extension the unknown variable.

In this question, we have a line FH meeting the point F on the shape EFG.

As it meets the shape, the sum of both the angles must be the angle that is naturally formed on this shape which is a right angle or 900.

Thus

​∡EFH + ∡HFG = 90

(11x + 1)+(2x − 2) = 90

11x + 2x + 1 − 2 = 90

13x − 1 = 90

13x = 90 + 1

x = \(\frac{91}{13}\)

x = 7
​
Thus we can calculate both angles.

​∡EFH = 11x + 1

∡EFH = (11 × 7) + 1

∡EFH = 77 + 1

∡EFH = 78
​
Now rather than calculating the angle HFG, we can just subtract this found angle from 900 so that we can find the complimentary angle HFG.

​∡HFG = 90 − ∡EFH

∡HFG = 90 − 78

∡HFG = 12
​
Both the angles for the given diagram can be written as ∡EFH = 780 and ∡HFG = 120.

 

Page 274  Exercise 3  Problem 3

Question 3. 

Given the following angle measurements in a diagram:

$\mathrm{\angle }𝐽𝐾𝑀=(2𝑥+9{)}^{\circ }$ $\mathrm{\angle }𝐿𝐾𝑀=(3𝑥+1{)}^{\circ }$

1. Find the values of ∠JKM and ∠𝐿𝐾𝑀.
2. Show all steps clearly, including the calculation of the unknown variable x.

Answer:

Given diagram

Given that

$\mathrm{\angle }𝐽𝐾𝑀=(2𝑥+9{)}^{\circ }$ $\mathrm{\angle }𝐿𝐾𝑀=(3𝑥+1{)}^{\circ }$

Here

∠JKM = (2x + 9)°

∠LKM = (3x + 1)°

We have to find out the angles ∠JKM and ∠LKM

The total sum of angle on one side of a line is 1800.

Thus we add both the angles formed by this KM line on the JL line and equate them with 1800 to find both the angles and by extension the unknown variable.

In this question, we have a line KM meeting a point K on the line JL.

As it meets the line, the sum of both the angles must be the angle that is naturally formed on one side of the line which is 1800.

Thus

​∡ JKM + ∡LKM = 1800

(2x+9)+(3x+1) = 180

2x + 3x + 9 + 1 = 180

5x + 10 = 180

5x = 180 − 10

5x = 170

x = 34
​
Thus we can calculate both angles.

​∡JKM = 2x + 9

∡JKM =(2 × 34) + 9

∡JKM = 68 + 9

∡JKM = 77
​
Now rather than calculating the angle LKM, we can just subtract this found angle from 1800 so that we can find the supplementary angle LKM.

​∡LKM = 180 − ∡JKM

∡LKM = 180 − 77

∡LKM = 1030

​Both the angles for the given diagram can be written as ∡JKM = 770 and ∡LKM = 1030

 

Page 274  Exercise 4  Problem 4

Question 4. 

Write the equation of a line that passes through the point (2, 7) and has a slope of 5.

Answer:

Given

The point (2, 7) and has a slope of 5.

When writing the equation of a line, all we need is a slope and a point through which the line passes obeying this slope.

Thus, we use the given point and the given slope to write the equation of the line.

We have the slope of the line represented by m which is 5.

We also have a point through which the line passes represented by (x1,y1) which is (2,7).

Thus we write the equation of a line using

​y − y₁ = m(x − x₁)

y − 7 = 5(x −2)

5x − 10 − y + 7 = 0

5x − y − 3 = 0

5x − y = 3
​
The line passing through point (2,7) and having slope  5 is 5x − y = 3.

 

Page 274 Exercise  5 Problem 5

Question 5. 

Write the equation of a line that passes through the point $(5,-9)$ and has a slope of \(\frac{1}{6}\)

Answer:

Given

The point $(5,-9)$ and has a slope of \(\frac{1}{6}\)

When writing the equation of a line, all we need is a slope and a point through which the line passes obeying this slope.

Thus, we use the given point and the given slope to write the equation of the line.

We have the slope of the line represented by m which is \(\frac{1}{6}\).

We also have a point through which the line passes represented by (x1,y1)which is (5,−9).

Thus we write the equation of a line using the formula y−y₁ = m(x−x₁)

As such y−y₁ = m(x−x₁)

y − (−9) = \(\frac{1}{6}\) (x−5)

6(y + 9) = x − 5

6y + 54 = x − 5

54 + 5 = x − 6y

x − 6y = 59
​
The line passing through point (5,−9)and having slope \(\frac{1}{6}\) is x − 6y = 59.

Page 274  Exercise 6  Problem 6

Question 6.

Write the equation of a line that passes through the point $(0,-8)$ and has a slope of \(\frac{3}{5}\)

Answer:

Given:

The slope of the line is m = \(\frac{3}{5}\)

The coordinate of the point is (0,−8)

We have to find an equation of the line that passes through the given point with the given slope.

We have the slope of the line represented by m which is \(\frac{3}{5}\).

We also have a point through which the line passes represented by (x1,y1) which is (0,−8).

Thus we write the equation of line using the formula y−y1= m(x−x1)

As such

​y−y₁ =  m(x−x₁)

y − (−8) = \(\frac{3}{5}\) (x − 0)

5(y + 8) = 3x

5y + 40 = 3x

3x − 5y = 40

​The line passing through point (0,−8) and having slope \(\frac{3}{5}\) is 3x − 5y = 40

 

Page 274  Exercise 7  Problem 7

Question 7. 

Write the equation of a line that passes through the point $(2,-1)$ and has a slope of 5.

Answer:

Given

The point $(2,-1)$ and has a slope of 5.

We are given a point (2,-1) and m = 5.

And we need to find the equation of a line that passes through (2,-1) and has the slope m =5.

As we know, there are different forms of equations of a line.

We will use the point-slope form of the equation of a line to find the required equation.

We are given a point (2,-1) and m = 5.

And we need to find the equation of a line that passes through (2,-1) and has the slope m =5.

As we know, there are different forms of equations of a line.

We will use the point-slope form of the equation of a line to find the required equation.

The equation of the line that passes through the point (2,-1) and has the slope m = 5 is 5x − y = 11.

 

Page 275 Exercise 8 Problem 8

Question 8. 

Explain the following types of lines and provide their definitions:

1. Parallel Lines
2. Intersecting Lines
3. Coincident Lines
4. Skew Lines

Answer:

We have to find out the meaning when two lines are parallel, intersecting, coincident, or skew.

Two lines are said to be parallel if they never intersect each other, regardless of how far they are extended on either side.

Parallel lines are at equal distances from each other and never meet.

Three sets of parallel lines are shown below:

When two lines cross each other in a plane, they are called intersecting lines. The intersecting lines share a common point.

The point where the lines intersect is called the point of intersection.

 A set of intersecting lines is shown below:

Coincide means that it occurs at the same time.

The coincident lines are lines that lie upon each other in such a way that when we look at them, they appear to be a single line.

A set of Coincident lines is shown below:

Any two lines are said to be skew lines if they do not intersect and also not parallel.

Since two lines in a plane must intersect or be parallel, skew lines can exist only in three or more dimensions.

A set of skew lines is shown below:

Parallel lines are at equal distance from each other and never meet.

When two lines cross each other in a plane, they are called intersecting lines.

The coincident lines are lines that lie upon each other in such a way that when we look at them, they appear to be a single line.

Any two lines are said to be skew lines if they do not intersect and also not parallel.

 

Page 275  Exercise 9  Problem 9

Question 9. 

We are given a right rectangular prism. Use the properties of different types of lines to classify each pair of lines as parallel, intersecting, coincident, or skewed. Refer to the provided rectangular prism diagram.

Given Pairs of Lines:

1. \(\overline{AB}\) and \(\overline{BC}\)
2. \(\overline{AD}\) and \(\overline{BC}\)
3. \(\overline{EI}\) and \(\overline{IH}\)
4. \(\overline{BF}\) and \(\overline{EH}\)
5. \(\overline{EF}\) and \(\overline{CG}\)
6. \(\overline{AB}\) and \(\overline{GH}\)

Answer:

We are given a right rectangular prism

We need to classify given pairs of lines as parallel, intersecting, coincident, or skew.

Pair of lines

1.\(\overline{A B}\) and \(\overline{B C}\)

2. \(\overline{A D}\) and \(\overline{B C}\)

3.\(\overline{E I}\) and \(\overline{I H}\)

4.\(\overline{B F}\) and \(\overline{E H}\)

5.\(\overline{E F}\) and \(\overline{C G}\)

6.\(\overline{A B}\) and \(\overline{G H}\)

We will do this by using properties of different types of lines and a table.

 

Page 276   Exercise 10  Problem 10

Question 10.

We are given a right rectangular prism. Use the properties of different types of lines to classify each pair of lines as parallel, intersecting, coincident, or skew. Refer to the provided rectangular prism diagram. In addition to the pairs above, find and classify three more pairs of lines from the right rectangular prism.

Additional Pairs of Lines:

1. \(\overline{AB}\) and \(\overline{BC}\)
2. \(\overline{AD}\) and \(\overline{BC}\)
3. \(\overline{EI}\) and \(\overline{IH}\)
4. \(\overline{BF}\) and \(\overline{EH}\)
5. \(\overline{EF}\) and \(\overline{CG}\)
6. \(\overline{AB}\) and \(\overline{GH}\)
7. \(\overline{CD}\) and \(\overline{DH}\)
8. \(\overline{AE}\) and \(\overline{FG}\)
9. \(\overline{BG}\) and \(\overline{HF}\)

Answer:

We are given a right rectangular prism.

Similar to Page 275 Exercise 9 Problem 9 we have to find three more pairs of lines and we need to classify each of the pairs of lines as parallel, intersecting, coincident, or skew.

 

Page 280  Exercise 11  Problem 11

Question 11.

Refer to the provided right rectangular prism diagram. Using the properties of parallel lines, determine which line contains point B and is parallel to line \(\overline{FG}\).

Answer:

Given: A diagram

To find – Lines that contain point B and parallel to line FG.

Line that contain point B and parallel to \(\overline{F G}\) is \(\overline{B A}\).

The final answer is that the line which contain point B and parallel to\(\overline{F G}\) is \(\overline{B A}\).

 

Page 280  Exercise 12  Problem 12

Question 12.

Refer to the provided right rectangular prism diagram. Using the properties of parallel planes, determine which plane contains point B and is parallel to plane FGH.

Answer:

Given: A diagram

To find – Planes that contain point B and parallel to plane FGH.

The plane that contains point B and parallel to plane FGH is plane ABC.

The final answer is that the plane that contains point B and parallel to plane FGH is ABC.

 

Page 280  Exercise 13  Problem 13

Question 13.

First, observe the provided diagrams. Then, compare the diagrams and identify the corresponding angles.

Diagrams to Consider:

• Corresponding Angles Diagram 1
• Corresponding Angles Diagram 2
• Corresponding Angles Diagram 3

Answer:

First we observe the Diagram after we compare the diagram

With

Here

Corresponding angles in this diagrams are
(∠1,∠5), (∠2,∠6), (∠3,∠8),(∠4,∠7),(∠5,∠10),(∠8,∠9),(∠6,∠12),(∠7,∠11),(∠1,∠9),(∠3,∠11),(∠2,∠10),(∠4,∠12)

​Corresponding angles of this is are,   (∠1,∠5), (∠2,∠6), (∠3,∠8),(∠4,∠7),(∠5,∠10),(∠8,∠9),(∠6,∠12),(∠7,∠11),(∠1,∠9),(∠3,∠11),(∠2,∠10),(∠4,∠12)

 

Page 280  Exercise 14  Problem 14

Question 14.

First, observe the provided diagrams and compare them to identify the alternate interior angles.

Diagrams to Consider:

• Alternate Interior Angle Diagram 1
• Alternate Interior Angle Lines
• Alternate Interior Angle Diagram 2
• Alternate Interior Angle Diagram 3

Answer:

First we will observe the diagram and compare diagram

With

After that write angles.

Alternate interior angles pairs in this diagram are ​(∠5,∠4 ), (∠2,∠7)

Alternate interior angles pair are (​∠2,∠7 ),(∠5,∠4), (∠2,∠7)

 

Page 280  Exercise 15  Problem 15

Question 15.

First, observe the given diagrams and compare them to identify the alternate exterior angles.

Diagrams to Consider:

• Alternate Exterior Angle Diagram 1
• Alternate Exterior Angle Lines
• Alternate Exterior Angle Diagram 2
• Alternate Exterior Angle Diagram 3

Case 1:

• Transversal Line A: Alternate exterior angles in this diagram are (∠3, ∠6), (∠1, ∠8)

Case 2:

• Transversal Line A: Alternate exterior angles in this diagram are (∠3, ∠6), (∠1, ∠8)

Case 3:

• Transversal Line C: Alternate exterior angles in this diagram are (∠1, ∠12), (∠2, ∠11)

Given Diagram:

• Identify all alternate exterior angles.

Answer:

First we will observe the diagram then compare the diagram with

Compare with

And write the angles.

Given diagram is

Case 1:

Here line A is transversal.

Alternate exterior angles in this diagram are (∠3,∠6), (∠1,∠8)

Case 2:

Here line A is transversal.

Alternate exterior angles in this diagram(∠3,∠6),(∠1,∠8)

Case 3:

Here C is transversal

Alternate exterior angles in this diagram are (∠1,∠12),(∠2,∠11)

Alternate exterior angles in this diagram are (​∠6,∠3),(​∠1,∠8),(​∠9,∠6),(​∠11,∠5),(​∠1,∠12),(∠11,∠2)

Alternate exterior angles of this diagram is  (​∠6,∠3),(​∠1,∠8),(​∠9,∠6),(​∠11,∠5),(​∠1,∠12),(∠11,∠2)

 

Page 280  Exercise 16  Problem 16

Question 16.

First, observe the given diagrams and identify the consecutive interior angles.

Diagrams to Consider:

• Consecutive Interior Angle Diagram 1
• Consecutive Interior Angle Diagram 2

Answer:

First we observe the diagram and according to definitions angles presented in one side of transversal line also inside of two lines are consecutive interior angles.

Consecutive interior angles pair of the diagram are ​(∠4,∠7), (∠7,∠10), (∠4,∠10)

Consecutive interior angles of this diagram are ​(∠4,∠7), (∠7,∠10), (∠4,∠10)

Page 281  Exercise 17  Problem 17

Question 17.

When two parallel lines are cut by a transversal, several pairs of angles are formed. Based on the angle relationships, identify and list the pairs of angles that are congruent.

Definitions to Consider:

1. Corresponding Angle Postulate: If two parallel lines are cut by a transversal, pairs of corresponding angles are congruent.
2. Alternate Interior Angle Theorem: If two parallel lines are cut by a transversal, pairs of alternate interior angles are congruent.
3. Alternate Exterior Angle Theorem: If two parallel lines are cut by a transversal, pairs of alternate exterior angles are congruent.

Answer:

When two parallel lines are cut by a transversal, then the resulting pairs of angles are congruent are

The corresponding angle postulate states that if two parallel lines are cut by a transversal, pairs of corresponding angles are congruent.

The alternate interior angle theorem states that if two parallel lines are cut by a transversal, pairs of alternate interior angles are congruent.

The alternate exterior angle theorem states that if two parallel lines are cut by a transversal, pairs of alternate exterior angles are congruent.

Pairs of corresponding angles, pairs of alternate interior angles, and pairs of alternate exterior angles are congruent.

Page 284  Exercise 19  Problem 18

Question 18.

Given that one angle in a transversal setup is 63°, find the values of the angles m∠1 and m∠2. Identify which theorems are applied to determine these values.

Given:

• One angle is 63º. Assume this angle is represented by m∠3 such that m∠3 = 63º.

Answer:

Given: One angle is given and the value is 63°.

Find the value of the angles m∠1 and m∠2.

Determine which theorem is applied to find the values of m∠1 and m∠2.

Since one of the angle is 63°, assume the given angle is represented by m∠3 such that m∠3 = 63°.

To find, apply the theorem of corresponding angles.

We get, m∠3 = m∠1

m∠1 = 63°

To find m∠2, apply the theorem of consecutive interior angles.

m∠3 + m∠2 = 180°

Substitute the value of m∠3 = 63°.

Thus, we have  63° + m∠2 = 180°

m∠2 = 180° − 63°

m∠2 = 117°

The values are m∠1 = 63° and m∠2 = 117°.The theorems applied are corresponding angles and consecutive interior angles.

Page 284  Exercise 20  Problem 20

Question 19.

Given that one angle in a transversal setup is 95°, find the values of the angles m∠1 and m∠2. Identify which theorems are applied to determine these values.

Given:

• One angle is 95°. Assume this angle is represented by m∠3 such that m∠3 = 95º.

Answer:

Given: One angle is given and the value is 95°.

Find the value of the angles  m∠1 and m∠2.

Determine which theorem is applied to find the values of m∠1 and  m∠2.

Since one of the angle is 95° , assume the given angle is represented by m∠3
such that  m∠3 = 95°.

To find, apply the theorem of consecutive interior angles.

We get, m∠3 = m∠2

m∠2 = 95°

The values are m∠1 = 95°  and  m∠2 = 95° .The theorems applied are consecutive interior angles and alternate exterior angles.

Page 285  Exercise 21  Problem 21

Question 20.

Given two angles, 110° and (x + 12)°, find the value of x by applying the appropriate theorem.

Given:

• One angle is 110°. Assume this angle is represented by m∠1.
• The other angle is (x + 12)°. Assume this angle is represented by m∠2.

Answer:

Given: The angles are 110° and (x + 12)

Find the value of x by applying the theorem.

Since one angle is 110° and the other is (x+12)°, assume the given angle are represented by m∠1 and m∠2.

Therefore, we have m∠1 = 110°

m∠2 = (x + 12)°

By the theorem of alternate exterior angles, we have

m∠2  =  m∠1

x + 12 = 110

x = 110 − 12

x = 98

Hence, the value is x = 98.

Substitute the value of  x = 98 into m∠2.

We get, m∠2 = (98 + 12)°

m∠2 = 110°

The value of x is 98.

Page 285  Exercise 22  Problem 22

Question 21.

Given three angles, 4x°, m∠6, and 52º, find the value of x by applying the appropriate theorems.

Given:

• One angle is 4xº. Assume this angle is represented by m∠1.
• Another angle is 52º. Assume this angle is represented by m∠2.

Answer:

Given: The angles are 4x°, m∠6, and 52°.

Find the value of x by applying the theorem.

Since one angle is 4x∘ and the other is 52° , assume the given angle are represented by m∠1 and m∠2.

Therefore, we have

m∠1 = 4x°

m∠2 = 52°

Find the value of m∠6.

Apply the theorem of corresponding angles.

We get, m∠2 = m∠6

m∠6 = 52°

Consider one interior angle to be m∠3 such thatm∠6 and m∠3 are consecutive interior angles.

Apply the theorem of consecutive interior angles.

Thus, we have  m∠6 + m∠3 = 180°

52° + m∠3 = 180°

m∠3 = 180° −52°

m∠3 = 128°

To find the value of x, apply the corresponding angles.

m∠1 = m∠3

4x°= 128x° 

= 32

The value of x is 32.