Download Problem Set Solutions Chapter 6 and 7 Geometry Correcting

Problem Set Solutions
Chapter 6 and 7
Geometry
Correcting problem set:
• Read the solution on the student’s paper first. Note anything that seems odd or incorrect.
(Note on their paper)
• Read the given solution. Note anything missing on the student’s paper.
• Do your best to understand and read the reasoning of the student. Do not just check for
word-for-word correspondence. Try to understand not just read.
Notes:
(Are there any issues that cam up multiple times? Did they do something particularly well? Be
specific!)
Section 6.3
p.286: 9
9.
R
Given: f ||g
RT W is an isosceles 4
with base T W
S
Prove: 4RSV is isosceles.
f
V
T
g
W
1. f ||g (Given)
2. 4RT W is
(Given)
3. SV ||T W (If two parallel planes are cut by a third plane (4RT W ),
the lines of intersection are parallel.)
4. RT ∼
= RW (If a triangle is isosceles then it has two congruent sides.)
5. ∠T ∼
= ∠W (
⇒
)
6. ∠T ∼
= ∠RSV (P → CAC)
7. ∠W ∼
= ∠RV S (P → CAC)
8. ∠RSV ∼
= ∠RV S (Transitive Property)
9. RS ∼
= RV (
⇒
)
10. 4RSV is isosceles (If a triangle has two sides congruent, then it is isosceles.)
Section 7.1 - p.301: 19
19. Prove that the midpoint of the hypotenuse of a right triangle is equidistant from all three
vertices.
A
Given: 4ABC is
D is the midpoint of AC
D
Prove: D is equidistant from A, C, B.
E
B
C
Draw ED through D parallel to BC (parallel postulate). ED intersects AB at E.
1. ED||BC (Given from auxiliary lines)
2. D is the midpoint of AC (Given)
3. E is the midpoint of AB (If a line passes through the midpoint of one side of a triangle
and is parallel to another side, then it passes through the midpoint
of the third side.)
4. AE ∼
= EB (A midpoint divides a segment into two congruent parts)
5. 4ABC is right (Given)
6. ∠ABC is
(The right angle of a triangle is always opposite the hypotenuse)
7. BC ⊥ AB (
⇒⊥)
8. ED ⊥ AB (If one of two parallel lines is perpendicular to a third line, then
the second parallel line is also perpendicular to the third line.)
9. ∠AED and ∠BED are
(⊥⇒
)
10. ∠AED ∼
= ∠BED (All right angles are congruent.)
11. DE ∼
= DE (Reflexive property)
12. 4AED ∼
= 4BED (SAS)
13. AD ∼
= DB (CPCTC)
14. AD ∼
= DC (A midpoint divides a segment into two congruent parts)
15. AD ∼
= DC ∼
= DB (Transitive Property)
16. D is equidistant from A, C, B (If the segments drawn from a point to other points are
congruent, then that point is equidistant from the other points.)
Note: P → CAC could have been used in place of steps 7-9.
Section 7.3 - p.311: 15, 22
15.
C
Find the restrictions on x.
D
(5x)◦
B
A
We know that 4ABC is a right triangle and it shares ∠A with 4ABD. Given the diagram, we
also know that m∠DBA < m∠CBA (you can assume relative position).
Therefore m∠ADB 6= 90, in fact, m∠ADB > 90 because the Si (4) is always the same so the
smaller second angle in 4ADB forces the third angle to be larger than the corresponding angle in
4ACB
We also know that ∠ADB is not a straight angle, so m∠ADB < 180.
Therefore: 90 < 5x < 180 which means 18 < x < 36.
22. We have stated that in this text the word polygon will mean a convex polygon and that
angles greater than 180◦ will not be considered. Ignore those rules for this problem.
(a) Consider the nonconvex polygon EF GHJ, whose interior angle at J is greater than 180◦ . Can
you demonstrate that the sum of the measure of the angles of this nonconvex polygon is 540?
H
E
J
Si = 180
Si = 180
Si = 180
By drawing diagonals from J, you create three
triangles inside polygon EF GHJ.
The Si (4) = 180, so 3 · 180 = 540 which is the
Si (EF GHJ).
G
F
(b) Can you demonstrate that the sum of the measures of the angles of the nonconvex octagon at
the right is 1080?
2
6
5
3
4
1
You can break up the octagon into six triangles.
6 · Si (4) = 6 · 180 = 1080.So the Si of the octagon
is indeed 540.
(c) Is the sum of the measures of the angles of a nonconvex polygon of n sides 180(n − 2)?
The question here is really, can you draw exactly n − 2 triangles in every polygon? We know
that you can break any polygon up into triangles by definition (no curvy bits). So after some
searching (this isn’t an easy answer to find!) it turns out that yes you can break up any polygon
into n − 2 triangles. The proof has to do with a proof method called induction. We will leave
that for now, you likely have tried out a few concave polygons and it has worked so far (even
if a part of you thought ”But this is proving by example!!!”). Because it does work, the Si of
any polygon will always be 180(n − 2). If you are interested in the proof let me know!
(d) Is the sum of the measures of the exterior angles, one per vertex, of a nonconvex polygon equal
to 360?
Nope! Look at the angle J in the first polygon. It isn’t supplementary to anything! The whole
proof of the Se is based on supplementary angles, so this doesn’t work here.
Section 7.4 - p.316: 6, 14
6.
O
In the stop sign show, is 4N T E scalene, isosceles,
equilateral or undetermined?
I
T
G
Q
S
U
E
There are two acceptable ways of doing this. The first is by symmetry and congruent shapes.
Because QU EST ION is a regular octagon, the two trapezoids GOIT and GQU E have all parts
congruent except for GT and GE. But they must be congruent because there is no other option.
Therefore 4GET is isosceles with base ET .
However, this may seem a bit ”hand-wavy” for you and you want a proof, specifically as to
WHY the trapezoids are congruent. So in come our knights in shining armor: Triangles!
1. QU EST ION is a regular octagon (All stop signs are regular octagons)
2. ∠GQU ∼
= ∠QU E ∼
= ∠GOI ∼
= ∠OIT ( All interior angles of a regular polygon are congruent.)
3. All sides of QU EST ION are congruent (All sides of a regular polygon are congruent)
4. Draw OT and QE (Two points define a line)
5. 4OIT ∼
= 4QU E (SAS)
6. ∠U QE ∼
= ∠IOT (CPCTC)
7. OT ∼
= QE (CPCTC)
8. ∠GOT ∼
= ∠GQE (If congruent angles are subtracted from congruent angles, then their differences are congruent.)
9. 4GQE ∼
= 4GOT (SAS)
10. GE ∼
= GT (CPCTC)
11. 4T EG is isosceles (If a triangle has two congruent sides, then it is isosceles.)
14.
A
B
Hexagon ABCDEF is regular. Prove ACDF is
a rectangle.
C
F
E
D
1. ABCDEF is a regular hexagon (Given)
2. All sides of ABCDEF are congruent (All sides of a hexagon are congruent)
3. Si (ABCDEF ) = 180(6 − 2) = 720 (Sum of the interior angles of a polygon)
4. m∠ABC = 120 ( Si /n = Measure of one interior angle of a regular polygon)
5. ∠BAC ∼
= ∠BCA (
⇒
)
6. m∠BCA = m∠BAC = (180 − 120)/2 = 30 (Sum of the interior angles of a triangle is 180)
7. m∠ACD = m∠F AC = 120 − 30 = 90 (subtraction)
8. ∠F AC is right (A right angle has a measure of ninety degrees)
9. F A ⊥ AC, DC ⊥ AC (
⇒⊥)
10. F A||CD (If two lines are perpendicular to the same line, then they are parallel to each other)
11. ACDF is a parallelogram (If a quadrilateral has two sides parallel and congruent it is a
parallelogram)
12. ACDF is a rectangle (If a parallelogram has one right angle, it is a rectangle)