geometry
cw 1.2
find the length indicated.
1) ( g \bullet , ? , \bullet h \bullet 11 \bullet i ) (total 15)
2) ( c \bullet , ? , \bullet b \bullet 7 \bullet a ) (total 19)
3) ( q \bullet , ? , \bullet r \bullet 5 \bullet s ) (total 16)
4) ( i \bullet , ? , \bullet h \bullet 1 \bullet g ) (total 13)
solve for ( x ).
5) ( a \bullet x+12 \bullet b \bullet x+6 \bullet c ) (total 12)
6) ( c \bullet x+15 \bullet b \bullet x+8 \bullet a ) (total 15)
7) ( j \bullet x-2 \bullet k \bullet 11 \bullet l ) (total ( 3x-3 ))
8) ( g \bullet 6 \bullet h \bullet x+3 \bullet i \bullet 2x-10 \bullet j ) (total 26)
find the length indicated.
9) find ( ef ): ( d \bullet 4 \bullet e \bullet 4x-1 \bullet f ) (total ( 5x ))
10) find ( eg ): ( e \bullet x+5 \bullet f \bullet 9 \bullet g ) (total ( 4x-1 ))

Question

Accepted Answer

### Problem 1: Find the length indicated (G to H)
# Explanation:
## Step1: Use segment addition postulate
We know that $ GH + HI = GI $. Let $ GH = x $, $ HI = 11 $, and $ GI = 15 $. So the equation is $ x + 11 = 15 $.
## Step2: Solve for x
Subtract 11 from both sides: $ x = 15 - 11 = 4 $.
# Answer: 4

### Problem 2: Find the length indicated (C to B)
# Explanation:
## Step1: Use segment addition postulate
We know that $ CB + BA = CA $. Let $ CB = x $, $ BA = 7 $, and $ CA = 19 $. So the equation is $ x + 7 = 19 $.
## Step2: Solve for x
Subtract 7 from both sides: $ x = 19 - 7 = 12 $. Wait, no, wait. Wait, the diagram shows $ CB $ is the first segment, $ BA $ is 7, and $ CA $ is 19? Wait, no, maybe I misread. Wait, the diagram is $ C $---$ B $---$ A $, with $ CB $ is "?", $ BA $ is 7, and $ CA $ is 19. So $ CB + BA = CA $, so $ x + 7 = 19 $, so $ x = 19 - 7 = 12 $? Wait, no, maybe the other way. Wait, no, the problem says "Find the length indicated" for 2). Wait, maybe I made a mistake. Wait, the original problem: 2) $ C $---"?"---$ B $---7---$ A $, and the total $ CA $ is 19. So $ CB + BA = CA $, so $ CB + 7 = 19 $, so $ CB = 19 - 7 = 12 $. Wait, but maybe the "?" is $ CB $, so the answer is 12? Wait, no, maybe I misread. Wait, let's check again. The user's image: 2) $ C $---?---$ B $---7---$ A $, with $ CA = 19 $. So segment addition: $ CB + BA = CA $, so $ CB = CA - BA = 19 - 7 = 12 $. So the length is 12.
# Answer: 12

### Problem 3: Find the length indicated (Q to R)
# Explanation:
## Step1: Use segment addition postulate
We know that $ QR + RS = QS $. Let $ QR = x $, $ RS = 5 $, and $ QS = 16 $. So the equation is $ x + 5 = 16 $.
## Step2: Solve for x
Subtract 5 from both sides: $ x = 16 - 5 = 11 $.
# Answer: 11

### Problem 4: Find the length indicated (I to H)
# Explanation:
## Step1: Use segment addition postulate
We know that $ IH + HG = IG $. Let $ IH = x $, $ HG = 1 $, and $ IG = 13 $. So the equation is $ x + 1 = 13 $.
## Step2: Solve for x
Subtract 1 from both sides: $ x = 13 - 1 = 12 $.
# Answer: 12

### Problem 5: Solve for x
# Explanation:
## Step1: Use segment addition postulate
We know that $ AB + BC = AC $. $ AB = x + 12 $, $ BC = x + 6 $, and $ AC = 12 $. So the equation is $ (x + 12) + (x + 6) = 12 $.
## Step2: Simplify the equation
Combine like terms: $ 2x + 18 = 12 $.
## Step3: Solve for x
Subtract 18 from both sides: $ 2x = 12 - 18 = -6 $. Then divide by 2: $ x = -3 $.
# Answer: $ x = -3 $

### Problem 6: Solve for x
# Explanation:
## Step1: Use segment addition postulate
We know that $ CB + BA = CA $. $ CB = x + 15 $, $ BA = x + 8 $, and $ CA = 15 $. So the equation is $ (x + 15) + (x + 8) = 15 $.
## Step2: Simplify the equation
Combine like terms: $ 2x + 23 = 15 $.
## Step3: Solve for x
Subtract 23 from both sides: $ 2x = 15 - 23 = -8 $. Then divide by 2: $ x = -4 $.
# Answer: $ x = -4 $

### Problem 7: Solve for x
# Explanation:
## Step1: Use segment addition postulate
We know that $ JK + KL = JL $. $ JK = x - 2 $, $ KL = 11 $, and $ JL = 3x - 3 $. So the equation is $ (x - 2) + 11 = 3x - 3 $.
## Step2: Simplify the equation
Simplify left side: $ x + 9 = 3x - 3 $.
## Step3: Solve for x
Subtract $ x $ from both sides: $ 9 = 2x - 3 $. Add 3 to both sides: $ 12 = 2x $. Divide by 2: $ x = 6 $.
# Answer: $ x = 6 $

### Problem 8: Solve for x
# Explanation:
## Step1: Use segment addition postulate
We know that $ GH + HI + IJ = GJ $. $ GH = 6 $, $ HI = x + 3 $, $ IJ = 2x - 10 $, and $ GJ = 26 $. So the equation is $ 6 + (x + 3) + (2x - 10) = 26 $.
## Step2: Simplify the equation
Combine like terms: $ 6 + x + 3 + 2x - 10 = 26 $ → $ 3x - 1 = 26 $.
## Step3: Solve for x
Add 1 to both sides: $ 3x = 27 $. Divide by 3: $ x = 9 $.
# Answer: $ x = 9 $

### Problem 9: Find EF
# Explanation:
## Step1: Use segment addition postulate
We know that $ DE + EF = DF $. $ DE = 4 $, $ EF = 4x - 1 $, and $ DF = 5x $. So the equation is $ 4 + (4x - 1) = 5x $.
## Step2: Simplify the equation
Simplify left side: $ 3 + 4x = 5x $.
## Step3: Solve for x
Subtract $ 4x $ from both sides: $ 3 = x $.
## Step4: Find EF
Substitute $ x = 3 $ into $ EF = 4x - 1 $: $ EF = 4(3) - 1 = 12 - 1 = 11 $.
# Answer: 11

### Problem 10: Find EG
# Explanation:
## Step1: Use segment addition postulate
We know that $ EF + FG = EG $. $ EF = x + 5 $, $ FG = 9 $, and $ EG = 4x - 1 $. So the equation is $ (x + 5) + 9 = 4x - 1 $.
## Step2: Simplify the equation
Simplify left side: $ x + 14 = 4x - 1 $.
## Step3: Solve for x
Subtract $ x $ from both sides: $ 14 = 3x - 1 $. Add 1 to both sides: $ 15 = 3x $. Divide by 3: $ x = 5 $.
## Step4: Find EG
Substitute $ x = 5 $ into $ EG = 4x - 1 $: $ EG = 4(5) - 1 = 20 - 1 = 19 $.
# Answer: 19

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QUESTION IMAGE

Question

Explanation:

Problem 1: Find the length indicated (G to H)

Step1: Use segment addition postulate

Step2: Solve for x

Step1: Use segment addition postulate

Step2: Solve for x

Step1: Use segment addition postulate

Step2: Solve for x

Answer:

Problem 2: Find the length indicated (C to B)