february16 february20
DESCRIPTION
Problems and Solutions for the week of Feb 16 - Feb 20TRANSCRIPT
Math Tutorial Questions
For the week of
February 16-20
Identifying Types of Lines and PlanesQuestions (3.1.1) – February 16, 2009
1. Use the diagram below to identify each of the following.
A. A pair of parallel segments
B. A pair of skew segmentsC. A pair of perpendicular
segmentsD. A pair of parallel planes
2. Use the diagram to the right to identify each of the following.
A. A pair of parallel segmentsB. A pair of skew segmentsC. A pair of perpendicular segmentsD. A pair of parallel planes
Identifying Types of Lines and PlanesSolutions (3.1.1) – February 16, 2009
1. Use the diagram to the right to identify each of the following.3 example solutions are listed, there are other possibilitiesA. A pair of parallel segments
AB & CD, CB & GF, EF & HGB. A pair of skew segments
AE & GF, AB & DH, HG & BFC. A pair of perpendicular segments
DH & HG, AD & DC, EF & FGD. A pair of parallel planes
Plane DCG & Plane ABF, Plane ADH & Plane BCG, Plane ABC & Plane EFG
2. Use the diagram to the right to identify each of the following.3 example solutions are listed, there are other possibilities• A pair of parallel segments
KL & NM, LQ & MR, QR & PSB. A pair of skew segments
KL & MR, LQ & PS, PQ & NSC. A pair of perpendicular segments
KL & LQ, LQ & QR, KN & NSD. A pair of parallel planes
Plane KLM & Plane PQR, Plane MLQ & Plane NKP, Plane NMR & Plane KLQ
Classifying Pairs of Angles
1. Use the diagram to the right to identify each of the following.
A. A pair alternate interior angles
B. A pair of corresponding angles
C. A pair of alternate exterior angles
D. A pair of same-side interior angles
Questions (3.1.2) – February 17, 2009
2. Use the diagram to the right to identify each of the following.
A. A pair alternate interior anglesB. A pair of corresponding anglesC. A pair of alternate exterior anglesD. A pair of same-side interior angles
Classifying Pairs of Angles1. Use the diagram to the right to identify each of the following.
2 example solutions are listed, there are other possibilitiesA. A pair alternate interior angles
Angles 3 & 5, Angles 4 & 6B. A pair of corresponding angles
Angles 2 & 6, Angles 4 & 8C. A pair of alternate exterior angles
Angles 2 & 8, Angles 1 & 7D. A pair of same-side interior angles
Angles 4 & 5, Angles 3 & 6
Solutions (3.1.2) – February 17, 2009
2. Use the diagram to the right to identify each of the following.1 example solution is listed, there are other possibilitiesA. A pair alternate interior angles
Angle EHG & Angle HGKB. A pair of corresponding angles
Angle EHG & Angle FGJC. A pair of alternate exterior angles
Angle IHE & Angle JGKD. A pair of same-side interior angles
Angle EHG & Angle FGH
Angles Formed by Parallel Lines & TransversalsQuestions (3.2.3) – February 18, 2009
1. Use the diagram below to find each angle measure.
A. m ECFB. m DCE
2. Find x and y in the diagram below.
Angles Formed by Parallel Lines & TransversalsSolutions (3.2.3) – February 18, 2009
1. Use the diagram below to find each angle measure.Corresponding angles are equal in measure.Angle ECF & Angle EBG are corresponding.Angle DCE & Angle ABE are corresponding.• m ECF
Angle ECF = Angle EBG (Substitute values in)Angle ECF = 70º
B. m DCEAngle DCE = Angle ABE (Substitute values in)5x = 4x + 22 (Subtract 4x from both sides) x = 22
2. Find x and y in the diagram below.Corresponding angles are equal in measure.
5x + 5y = 60 (Subtract 5x from both sides) 5y = 60 – 5x (Divide both sides by 5) y = 12 – x
Alternate Interior angles are equal in measure.5x + 4y = 55 (Substitute y = 12 – x in for y)5x + 4(12 – x) = 55 (Multiply 4 through 12 –
x)5x + 48 – 4x = 55 (Collect x terms together)x + 48 = 55 (Subtract 48 from both sides)x = 7
y = 12 – x (Substitute 7 in for x)y = 12 – 7 (Subtract 12 and 7)y = 5
Slopes of LinesQuestions (3.5.2) – February 19, 2009
1. Use the diagram below and the information above to determine the slope of each line.
A. ABB. ACC. ADD. CD
2. Justin is driving from home to his college dormitory. At 4:00 P.M., he is 260 miles from home. At 7:00 P.M., he is 455 miles from home. Use the graph of the line that represents Justin’s distance from home at a given time. Find and interpret the slope of the line.
Slopes of LinesSolutions (3.5.2) – February 19, 2009
1. Use the diagram to the right to determine the slope of each line.A(-2, 7), B(3, 7), C(4, 2), D(-2, 1)• AB
Line AB is horizontal, therefore the slope is 0.B. AC
C. ADLine AD is vertical, therefore the slope is UNDEFINED.
D. CD
2. Justin is driving from home to his college dormitory. At 4:00 P.M., he is 260 miles from home. At 7:00 P.M., he is 455 miles from home. Use the graph of the line that represents Justin’s distance from home at a given time. Find and interpret the slope of the line.
5
4
5
4
72
22
AC
ACAC xx
yym
6
1
6
1
42
21
CD
CDAC xx
yym
hr
mi
hr
mi
hr
mi
xx
yym 65
3
195
)47(
)260455(
12
12
The slope is 65 mi/hr, which means his average speed while driving home was at a rate of 65 miles per hour.
Using Slopes to Classify Pairs of LinesQuestions (3.5.3) – February 20, 2009
1. Graph each pair of lines. Find their slopes and use them to determine whether the lines are parallel, perpendicular, or neither.
A. UV and XY for U(0, 2), V(-1, -1), X(3, 1), Y(-3, 3)
B. GH and IJ for G(-3, -2), H(1, 2), I(-2, 4), J(2, -4)
C. CD and EF for C(-1, -3), D(1, 1), E(-1, 1), F(0, 3)
Using Slopes to Classify Pairs of LinesSolutions (3.5.3) – February 20, 2009
A. UV and XYU(0, 2)V(-1, -1)X(3, 1)Y(-3, 3)
B. GH and IJG(-3, -2)H(1, 2)I(-2, 4)J(2, -4)
1. Graph each pair of lines. Find their slopes and use them to determine whether the lines are parallel, perpendicular, or neither.
C. CD and EFC(-1, -3)D(1, 1)E(-1, 1)F(0, 3)
31
3
1
3
01
21
UV
UVUV xx
yym
3
1
6
2
33
13
XY
XYXY xx
yym
The slopes are 3 and -⅓, which multiply to equal -1, or are called opposite reciprocals of each other. Therefore the lines are perpendicular lines.
1
4
4
31
22
GH
GHGH xx
yym
24
8
22
44
IJ
IJIJ xx
yym
The slopes are 1 and -2, which do not multiply to equal -1 and they are not the same slope. Therefore the lines are not perpendicular lines, and they are not parallel lines.
2
2
4
11
31
CD
CDCD xx
yym
21
2
10
12
EF
EFEF xx
yym
The slopes are 2 and 2, which means they have the same slope. Therefore the lines are parallel lines.