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Math - Developmental, section A, Fall 2019
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Alexander & Koeberlein - Geometry 7/e (Homework)

James Finch

Math - Developmental, section A, Fall 2019

Instructor: Dr. Friendly

Current Score : 25 / 25

Due : Monday, December 30, 2030 23:59 EST

Last Saved : n/a Saving... ()

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Question

Points

1	2	3	4	5	6
1/1	1/1	1/1	10/10	2/2	10/10

Total

25/25 (100.0%)

Instructions

Building on the success of previous editions, Elementary Geometry for College Students, 7th edition, by Daniel C. Alexander and Geralyn M. Koeberlein and published by Cengage Learning, explores the important principles and real-world applications of plane, coordinate and solid geometry. Strongly influenced by both NCTM and AMATYC standards, the seventh edition includes intuitive, inductive and deductive experiences in its explorations. It aims to help students develop a comprehensive vocabulary of geometry, an intuitive and inductive approach to the development of principles, and strong deductive skills that lead to both verification of geometric theories and the solution of geometry-based, real-world applications.

These questions show the variety of question types available in WebAssign for use with this title.

Question 1 asks student to select the first statement of an indirect proof.

Question 2 is a multiple-choice question where students consider a given figure then select the correct conclusion based on the figure.

Question 3 is a numerical entry question where students consider a given figure then work through algebraic steps to solve for an unknown number.

Question 4 is an expanded problem with numerical entry answer blanks which provide scaffolding support for students as they step through the process of finding the measure of an angle.

Question 5 asks students to numerically enter an answer to a problem.

Question 6 is a two-column proof with drop-down menus in both columns for students to organize the steps of a proof and choose the correct reasons for each step. This demo assignment allows many submissions and allows you to try another version of the same question for practice wherever the problem has randomized values.

The answer key and solutions will display after the first submission for demonstration purposes. Instructors can configure these to display after the due date or after a specified number of submissions.

Assignment Submission

For this assignment, you submit answers by question parts. The number of submissions remaining for each question part only changes if you submit or change the answer.

Assignment Scoring

Your last submission is used for your score.

1. 1/1 points | Previous Answers AlexGeom7 2.2.013. My Notes

Write the first statement of the indirect proof of the given statement.

AC > BC

ΔABC,

then

m∠B ≠ m∠A.

Suppose that m∠B ≠ m∠A. Suppose that AC ≤ BC. correct

Suppose that m∠B = m∠A.

Suppose that AB

is colinear with AB

Suppose that AC > BC.

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2. 1/1 points | Previous Answers AlexGeom7 2.3.004. My Notes

ℓ and m are cut by transversal v. Determine whether ℓ must be parallel to m if

m∠1 = 104

and

m∠4 = 104.

Three lines with a total of two intersections forming eight angles are shown.
The first line ℓ goes from left to right.
The second line m goes from left to right and is below line ℓ.
The third line v goes up and right and intersects lines m and ℓ.
Four angles are formed where ℓ and v intersect. ∠1 is above ℓ to the left of v. ∠2 is above ℓ to the right of v. ∠3 is below ℓ to the left of v. ∠4 is below ℓ to the right of v.
Four angles are formed where m and v intersect. ∠5 is above m to the left of v. ∠6 is above m to the right of v. ∠7 is below m to the left of v. ∠8 is below m to the right of v.

ℓ ∥ m ℓ ∦ m correct

There is not enough information to determine if ℓ ∥ m.

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3. 1/1 points | Previous Answers AlexGeom7 2.3.023. My Notes

Determine the value of x so that line ℓ will be parallel to line m.

m∠1	=	4x
m∠8	=	3(x + 8)

x =

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4. 10/10 points | Previous Answers AlexGeom7 2.4.049.EP. My Notes

Consider the following.

Three line segments, two dashed line segments, and five points are shown.
The points are labeled M, N, P, Q, and R.
The first line segment

RN

begins at R, goes horizontally right, passes through P, and ends at N.
The second line segment

NM

begins at N, goes up and left, and ends at M.
The third line segment

MP

begins at M, goes down and right, and ends at P.
The first dashed line segment

NQ

begins at N, goes up and left, and ends at Q.
The second dashed line segment

PQ

begins at P, goes up and left, and ends at Q.
∠*MNQ* and ∠*QNP* are congruent with measure a.
∠*MPQ* and ∠*QPR* are congruent with measure b.

Given:		NQ bisects ∠MNP
	PQ bisects ∠MPR
	m∠Q = 44°

Find:

m∠M

Because

bisects

∠MNP

into two equal angles of measure a,

m∠MNQ = m∠QNP =

a

$webMathematica generated answer key$ .

Thus,

m∠MNP =

2a

$webMathematica generated answer key$ .

Because

bisects

∠MPR

into two equal angles of measure b,

m∠MPQ = m∠QPR =

b

$webMathematica generated answer key$ .

Thus,

m∠MPR =

2b

$webMathematica generated answer key$ .

It is given that

m∠Q =

°.

Consider ΔMNP and exterior ∠MPR. According to Corollary 2.4.5,

m∠MPR = m∠M + m∠

MNP

Solve this equation for

m∠M

and substitute the measures of the angles in terms of a and b in place of angle names.

m∠M =

2b−2a

$webMathematica generated answer key$

Consider ΔNQP and exterior ∠QPR. According to Corollary 2.4.5,

m∠QPR = m∠Q + m∠

QNP

Substitute the given information into this equation in place of angle names to find an expression for b.

b =

44+a

$webMathematica generated answer key$

Substitute this expression for b into the equation for

m∠M

above and simplify.

Thus,

m∠M =

°.

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5. 2/2 points | Previous Answers AlexGeom7 2.5.018.MI. My Notes

(a)

Find the number of sides for a regular polygon whose measure of each interior angle is 162°.

sides

(b)

Find the number of sides for a regular polygon whose measure of each interior angle is 172°.

sides

Need Help?

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6. 10/10 points | Previous Answers AlexGeom7 4.2.021. My Notes

Complete the proof.

Given: M-Q-T and P-Q-R such that MNPQ and QRST are parallelograms (▱s)

Prove:

∠N

∠S

Statements	Reasons
1. M-Q-T and P-Q-R such that MNPQ and QRST are ▱s	1. Given.
2. ∠N ≅ ∠MQP	2. Opposite angles of a ▱ are congruent.
3. ∠MQP ≅ ∠RQT	3. If two lines intersect, the vertical ∠s formed are congruent.
4. ∠RQT ≅ ∠S	4. Opposite angles of a ▱ are congruent.
5. ∠N ≅ ∠S	5. Transitive property of congruence.

Solution or Explanation

Statements	Reasons
1. M-Q-T and P-Q-R such that MNPQ and QRST are ▱s	1. Given.
2. ∠N ≅ ∠MQP	2. Opposite angles of a ▱ are congruent.
3. ∠MQP ≅ ∠RQT	3. If two lines intersect, the vertical ∠s formed are congruent.
4. ∠RQT ≅ ∠S	4. Opposite angles of a ▱ are congruent.
5. ∠N ≅ ∠S	5. Transitive property of congruence.

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