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Obj. 25 Properties of Polygons November 25, 2013

Posted by Ms. Miller in Geometry.
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Practice: Khan Academy Angles of a Polygon and IXL.com Polygon Vocabulary. IXL.com’s Interior and Exterior Angles of Polygons is also useful.

Unit 5 Test Review Solutions (Reg. and PAP) November 21, 2013

Posted by Ms. Miller in Geometry.
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PAP Unit 5 Test Review

Unit 5 Test Review

Typos:
#11: m∠C=(2x+17)°
#13: side should be 4x−6

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Obj. 24 Special Right Triangles November 15, 2013

Posted by Ms. Miller in Geometry.
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I think it can be useful to see that a 45-45-90 triangle is half of as square and a 30-60-90 triangle is half of an equilateral triangle, so while I’m not expecting you to make your notes look like this, I thought you might find this layout helpful:

Practice: Khan Academy Special Right Triangles or IXL.com Special Right Triangles

Obj. 23 Triangle Theorems November 14, 2013

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Practice: IXL Triangle Midsegments, Angle-Side Relationships in Triangles, and Khan Academy Triangle Inequality Theorem

Obj. 22 Triangle Segments November 12, 2013

Posted by Ms. Miller in Geometry.
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Practice: IXL Identify Medians, Altitudes, and Bisectors and Khan Academy Multiplying Fractions

Obj. 21 Perpendicular and Angle Bisectors November 11, 2013

Posted by Ms. Miller in Geometry.
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Practice: IXL Perpendicular Bisector Theorem

Reg. Geometry Unit 4 Test Review Solutions November 5, 2013

Posted by Ms. Miller in Geometry.
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Unit 4 Test Review

Solutions

1. right isosceles 2. equiangular equilateral
3. 4x-18+8x-6+5x=180
17x=204
x=12		 4.\mathrm{m}\angle 1=127-75=52\textdegree
5. There is a typo on this problem. Replace the 7x with 8x.
8x+6=5x-7+2x+21
8x+6=7x+14
x=8		 6. \mathrm{m}\angle BAD=102\textdegree
7. \angle B \cong \angle C 8. 3x+7=4x-13
x=20
base angles:
3(20)+7=67\textdegree
vertex angle:
180-2(67)=46\textdegree
B
9. \angle BAC\cong\angle BCA
\angle 3\cong\angle 4
\angle 1\cong\angle 2		 10. 5x=2x+18
3x=18
x=6
\mathrm{m}\angle B=120\textdegree and \mathrm{m}\angle BAG=150\textdegree
11. A 12. C
13. B 14. \overline{AB}\cong\overline{LF}
15.

\angle J\cong\angle M \overline{JK}\cong\overline{MN}
\angle K\cong\angle N \overline{KL}\cong\overline{NO}
\angle L\cong\angle O \overline{JL}\cong\overline{MO}
16. \overline{XY}
17. Statement 1 should read \angle D\cong\angle C.
Reason 4: Vertical angles
Reason 5: AAS
18. The Given statement should read \overline{HK} and \overline{IL} bisect each other.

Statements Reasons
1. \overline{HK} and \overline{IL} bisect 1. Given
2. \overline{HJ}\cong\overline{KJ} 2. Def. bisector
3. \angle IJH\cong\angle LJK 3. Vertical angles
4. \overline{IJ}\cong\overline{LJ} 4. Def. bisector
5. \triangle HIJ\cong\triangle KLJ 5. SAS
19. Reason 4: ASA
Reason 5: CPCTC
20. If we prove ΔADB and ΔCDB are congruent, then we can use CPCTC to prove the sides congruent.

Statements Reasons
1. D is the midpoint of \overline{AC} 1. Given
2. \overline{AD}\cong\overline{CD} 2. Def. midpoint
3. \angle ADB\cong\angle CDB 3. Given
4. \overline{BD}\cong\overline{BD} 4. Refl. prop. ≅
5. \triangle ADB\cong\triangle CDB 5. SAS
6. \overline{AB}\cong\overline{CB} 6. CPCTC
7. ΔABC is isosceles 7. Def. isosceles
21.

Statements Reasons
1. M is midpt of \overline{KP} and \overline{LN} 1. Given
2. \overline{KM}\cong\overline{PM} 2. Def. midpoint
3. \angle KML\cong\angle PMN 3. Vertical angles
4. \overline{NM}\cong\overline{LM} 4. Def. midpoint
5. \triangle KLM\cong\triangle PNM

5. SAS
6. \angle K\cong\angle P 6. CPCTC
7. \overline{KL}\parallel\overline{NP} 7. Conv. Alt. Int. ∠s
22. To prove ΔABC is right isosceles, prove the sides congruent and the angles perpendicular.
Sides:
AB=\sqrt{(7-1)^2+(0-2)^2}=\sqrt{40}
BC=\sqrt{(3-7)^2+(-2-0)^2}=\sqrt{20}
AC=\sqrt{(3-1)^2+(-2-2)^2}=\sqrt{20}
Slopes:
AB: m=\frac{0-2}{1-7}=-\frac{1}{3}
BC: m=\frac{-2-0}{3-7}=\frac{1}{2}
AC: m=\frac{-2-2}{3-1}=-2

Therefore, AC=BC, and ΔABC is isosceles.

The product of the slopes is -1, so the lines are perpendicular. Therefore, ∠ACB is right.

PAP Unit 4 Test Review Solutions November 5, 2013

Posted by Ms. Miller in Geometry.
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PAP Unit 4 Test Review

Solutions

1. \mathrm{m}\angle B+\mathrm{m}\angle C=90\textdegree 2. \mathrm{m}\angle ABD=90\textdegree
3. 4x-18+8x-6+5x=180
17x=204
x=12		 4.\mathrm{m}\angle 1=127-75=52\textdegree
5. There is a typo on this problem. Replace the 7x with 8x.
8x+6=5x-7+2x+21
8x+6=7x+14
x=8		 6. \mathrm{m}\angle BAD=102\textdegree
7. \angle B \cong \angle C 8. 3x+7=4x-13
x=20
base angles:
3(20)+7=67\textdegree
vertex angle:
180-2(67)=46\textdegree
B
9. \angle BAC\cong\angle BCA
\angle 3\cong\angle 4
\angle 1\cong\angle 2		 10. 5x=2x+18
3x=18
x=6
\mathrm{m}\angle B=120\textdegree and \mathrm{m}\angle BAG=150\textdegree
11. A 12. C
13. B 14. \overline{AB}\cong\overline{LF}
15.

\angle J\cong\angle M \overline{JK}\cong\overline{MN}
\angle K\cong\angle N \overline{KL}\cong\overline{NO}
\angle L\cong\angle O \overline{JL}\cong\overline{MO}
16. \overline{XY}
17. Statement 1 should read \angle D\cong\angle C.
Reason 4: Vertical angles
Reason 5: AAS
18. The Given statement should read \overline{HK} and \overline{IL} bisect each other.

Statements Reasons
1. \overline{HK} and \overline{IL} bisect 1. Given
2. \overline{HJ}\cong\overline{KJ} 2. Def. bisector
3. \angle IJH\cong\angle LJK 3. Vertical angles
4. \overline{IJ}\cong\overline{LJ} 4. Def. bisector
5. \triangle HIJ\cong\triangle KLJ 5. SAS
19. Reason 4: ASA
Reason 5: CPCTC
20. If we prove ΔAEC and ΔDEB are congruent, then we can use CPCTC to prove the sides congruent.

Statements Reasons
1. \angle EBC\cong\angle ECB 1. Given
2. \overline{EB}\cong\overline{EC} 2. Isosceles Δ thm.
3. \overline{AB}\cong\overline{DC} 3. Given
4. AB=DC 4. Def. congruence
5. AB+BC=BC+CD 5. Add. prop. =
6. AC=BD 6. Segment add. post.
7. \overline{AC}\cong\overline{BD} 7. Def. congruence
8. \triangle AEC\cong\triangle DEB 8. SAS
9. \overline{AE}\cong\overline{DE} 9. CPCTC
10. ΔEAD is isosceles 10. Def. isosceles
21.

Statements Reasons
1. M is midpt of \overline{KP} and \overline{LN} 1. Given
2. \overline{KM}\cong\overline{PM} 2. Def. midpoint
3. \angle KML\cong\angle PMN 3. Vertical angles
4. \overline{NM}\cong\overline{LM} 4. Def. midpoint
5. \triangle KLM\cong\triangle PNM

5. SAS
6. \angle K\cong\angle P 6. CPCTC
7. \overline{KL}\parallel\overline{NP} 7. Conv. Alt. Int. ∠s
22. To prove ΔABC is right isosceles, prove the sides congruent and the angles perpendicular.
Sides:
AC=\sqrt{(3-1)^2+(-2-2)^2}=\sqrt{20}
BC=\sqrt{(3-7)^2+(-2-0)^2}=\sqrt{20}
Therefore, AC=BC, and ΔABC is isosceles.
Slopes:
AC: m=\frac{-2-2}{3-1}=-2
BC: m=\frac{-2-0}{3-7}=\frac{1}{2}
The product of the slopes is -1, so the lines are perpendicular. Therefore, ∠ACB is right.

Obj. 20 Coordinate Proof November 1, 2013

Posted by Ms. Miller in Geometry.
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Homework: IXL Distance Formula