geometry
chapter 1
worksheet 2
please use the diagram on the right to answer the questions below.
1. what relationship do the points a, b, c, and d have?
2. name 3 lines that intersect at h.
3. what points do plane adeh and ( overline{fe} ) have in common?
4. are points a, d, and c collinear?
5. are points e, f, g, and c coplanar?
please show your work when solving for x.
6. segment rt with point s: ( rs = 3x + 16 ), ( st = 15x + 6 ), ( rt = 2x + 18 )
7. segment gj with points h, i: ( gh = 27 ), ( hi = x + 6 ), ( ij = 3x - 2 ), ( gj = 12x - 57 )
8. d is between e and f
( de = 4x + 1 )
( df = 3x + 2 )
( ef = 31 )
9. k is between j and m
( jk = 3x + 7 )
( km = 7x - 9 )
( jm = 12x - 11 )

Question

Accepted Answer

### Question 1
# Brief Explanations:
Points A, B, C, D are on the same face (top face) of the rectangular prism, so they are coplanar (lie on the same plane) and form a quadrilateral (rectangle), also they are vertices of the top face.
# Answer:
Points A, B, C, and D are coplanar (lie on the same plane, the top face of the rectangular prism) and are vertices of a rectangle (forming the top face).

### Question 2
# Brief Explanations:
Looking at the diagram (rectangular prism), point H is connected to E, A, and G? Wait, no, let's identify lines. In a rectangular prism, at vertex H, lines like EH, AH, and GH (or EH, DH? Wait, the labels: E, H, G, F; A, D, H, E? Wait, the prism has vertices: A, B, C, D (top), E, F, G, H (bottom), with A connected to E, B to F, C to G, D to H. So at H, lines are EH (connecting E to H), DH (connecting D to H), and GH (connecting G to H)? Wait, no, E is connected to H, H to G, H to D? Wait, maybe EH, DH, and GH? Or EH, AH? Wait, A is connected to E? Wait, the diagram: A is top front, B top back, C top right back, D top right front; E bottom front, F bottom back, G bottom right back, H bottom right front. So edges: A-E, A-B, B-C, C-G, G-F, F-E, D-A, D-C, D-H, H-E, H-G. So at H, the lines (edges) are EH (E to H), DH (D to H), and GH (G to H). So three lines intersecting at H: $\overline{EH}$, $\overline{DH}$, $\overline{GH}$.
# Answer:
Three lines intersecting at H are $\boldsymbol{\overline{EH}}$, $\boldsymbol{\overline{DH}}$, and $\boldsymbol{\overline{GH}}$ (or other valid combinations based on the prism's edge connections, e.g., $\overline{EH}$, $\overline{DH}$, $\overline{GH}$).

### Question 3
# Brief Explanations:
Plane ADEH: let's identify the plane. Points A, D, E, H. $\overline{FE}$: the segment from F to E. Plane ADEH includes point E (since E is a vertex of the plane: A, D, E, H). $\overline{FE}$ has endpoints F and E. So the common point is E.
# Answer:
Plane ADEH and $\overline{FE}$ have point $\boldsymbol{E}$ in common.

### Question 4
# Brief Explanations:
Collinear points lie on the same straight line. Points A, D, C: A is top front, D is top right front, C is top right back. A to D is a horizontal edge (top front to top right front), D to C is a vertical edge (top right front to top right back). These two edges are perpendicular (in a rectangular prism, adjacent edges are perpendicular), so they don't lie on a straight line. Thus, A, D, C are not collinear.
# Answer:
No, points A, D, and C are not collinear.

### Question 5
# Brief Explanations:
Coplanar points lie on the same plane. Points E, F, G, C: E (bottom front), F (bottom back), G (bottom right back), C (top right back). Let's see the plane: E, F, G are on the bottom face (EFGH), and C is on the top face (ABCD). But also, the plane containing E, F, G, and C: E-F-G-C? Wait, E to F to G to C: E-F is bottom front-back, F-G is bottom back-right back, G-C is vertical (bottom right back to top right back). So E, F, G, C lie on a plane (the right face? Wait, E is bottom front, F bottom back, G bottom right back, C top right back. So the plane would be the right - side face? Wait, E (bottom front), F (bottom back), G (bottom right back), C (top right back) – connecting E to F to G to C, these points lie on a plane (a rectangular face: E, F, G, C? Wait, E to F is horizontal, F to G is horizontal, G to C is vertical, C to E? No, but in the prism, E is connected to A, F to B, G to C, so E, F, G, C: E (z - low, x - front, y - left), F (z - low, x - back, y - left), G (z - low, x - back, y - right), C (z - high, x - back, y - right). Wait, maybe the plane is the one containing E, F, G, and C: yes, because E, F, G are on the bottom (z - low) and G to C is vertical, F to B to C? No, better: E, F, G, C – E (0,0,0), F (0,1,0), G (1,1,0), C (1,1,1) (coordinates example). The plane equation: let's see, E(0,0,0), F(0,1,0), G(1,1,0), C(1,1,1). The vectors EF=(0,1,0), EG=(1,1,0), EC=(1,1,1). Wait, actually, E, F, G are on z=0, and C is on z=1, y=1, x=1. So the plane containing E, F, G, C: yes, because F, G, C are on x=0 to 1, y=1, z=0 to 1, and E is (0,0,0). Wait, maybe it's a trapezoid - shaped plane, but they are coplanar. Alternatively, E, F, G are on the bottom face (z=0), and C is above G, but the plane can be defined by E, F, G, C. So yes, they are coplanar.
# Answer:
Yes, points E, F, G, and C are coplanar.

### Question 6
# Explanation:
## Step1: Use segment addition postulate (RT = RS + ST)
Given $ RS = 3x + 16 $, $ ST = 15x + 6 $, $ RT = 2x + 18 $. By segment addition, $ RS+ST = RT $, so $ (3x + 16)+(15x + 6)=2x + 18 $.
## Step2: Simplify left - hand side
Combine like terms: $ 3x+15x + 16 + 6=2x + 18 $, which gives $ 18x+22 = 2x + 18 $.
## Step3: Subtract 2x from both sides
$ 18x-2x+22=2x - 2x+18 $, so $ 16x+22 = 18 $.
## Step4: Subtract 22 from both sides
$ 16x+22 - 22=18 - 22 $, so $ 16x=-4 $.
## Step5: Divide by 16
$ x=\frac{-4}{16}=-\frac{1}{4} $.
# Answer:
$ x = -\frac{1}{4} $

### Question 7
# Explanation:
## Step1: Use segment addition postulate (GJ = GH + HI + IJ)
Given $ GH = 27 $, $ HI=x + 6 $, $ IJ = 3x-2 $, $ GJ=12x - 57 $. By segment addition, $ 27+(x + 6)+(3x - 2)=12x - 57 $.
## Step2: Simplify left - hand side
Combine like terms: $ 27+6 - 2+x+3x=12x - 57 $, so $ 31 + 4x=12x - 57 $.
## Step3: Subtract 4x from both sides
$ 31+4x - 4x=12x - 4x-57 $, so $ 31 = 8x-57 $.
## Step4: Add 57 to both sides
$ 31 + 57=8x-57 + 57 $, so $ 88 = 8x $.
## Step5: Divide by 8
$ x=\frac{88}{8}=11 $.
# Answer:
$ x = 11 $

### Question 8
# Explanation:
## Step1: Use segment addition postulate (EF = DE + DF)
Given $ DE = 4x + 1 $, $ DF = 3x + 2 $, $ EF = 31 $. So $ (4x + 1)+(3x + 2)=31 $.
## Step2: Simplify left - hand side
Combine like terms: $ 4x+3x+1 + 2=31 $, so $ 7x+3 = 31 $.
## Step3: Subtract 3 from both sides
$ 7x+3 - 3=31 - 3 $, so $ 7x = 28 $.
## Step4: Divide by 7
$ x=\frac{28}{7}=4 $.
# Answer:
$ x = 4 $

### Question 9
# Explanation:
## Step1: Use segment addition postulate (JM = JK + KM)
Given $ JK = 3x + 7 $, $ KM = 7x-9 $, $ JM = 12x - 11 $. So $ (3x + 7)+(7x - 9)=12x - 11 $.
## Step2: Simplify left - hand side
Combine like terms: $ 3x+7x+7 - 9=12x - 11 $, so $ 10x-2 = 12x - 11 $.
## Step3: Subtract 10x from both sides
$ 10x-10x - 2=12x - 10x-11 $, so $ - 2=2x - 11 $.
## Step4: Add 11 to both sides
$ -2 + 11=2x-11 + 11 $, so $ 9 = 2x $.
## Step5: Divide by 2
$ x=\frac{9}{2}=4.5 $.
# Answer:
$ x = \frac{9}{2} $ (or $ 4.5 $)

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