im1 – 6.1 assignment
segment addition, angle addition, distance and midpoint
1. determine the length of (overline{cd}).
2. determine the length of (overline{rs}).
3. find the value of ( y ) and determine the length of ( overline{mp} ).
4. ( m ) is the midpoint of (overline{pq}). find the value of ( x ) and determine the length of (overline{pq}).
5. find the value of ( x ) and determine the lengths of (overline{de}) and (overline{df}).
6. draw and label the following. ( b ) is the midpoint of (overline{ac}). ( ac = 28 ). determine the length of (overline{bc}).
7. find the measure of ( angle xyz )
8. ( mangle klm = 180^circ ). determine the value of ( x ) and the ( mangle kln ).
9. ( mangle lkn = 145^circ ). determine the value of ( x ) and the ( mangle lkm ).

Question

Accepted Answer

### Problem 1: Determine the length of $\overline{CD}$
# Explanation:
## Step 1: Apply Segment Addition Postulate
The Segment Addition Postulate states that if $E$ is between $C$ and $D$, then $CD = CE + ED$. We know $CE = 1.1$ in and $ED = 2.7$ in.
$$
CD = 1.1 + 2.7
$$
## Step 2: Calculate the Sum
Add the two lengths together.
$$
CD = 3.8
$$
# Answer:
The length of $\overline{CD}$ is $3.8$ inches.

### Problem 2: Determine the length of $\overline{RS}$
# Explanation:
## Step 1: Apply Segment Addition Postulate
If $S$ is between $R$ and $T$, then $RT = RS + ST$. We know $RT = 4$ cm and $ST = 1.6$ cm. Let $RS = x$, so:
$$
4 = x + 1.6
$$
## Step 2: Solve for $x$
Subtract $1.6$ from both sides:
$$
x = 4 - 1.6 = 2.4
$$
# Answer:
The length of $\overline{RS}$ is $2.4$ cm.

### Problem 3: Find the value of $y$ and determine the length of $MP$
# Explanation:
## Step 1: Apply Segment Addition Postulate
Since $N$ is between $M$ and $P$, $MP = MN + NP$. We know $MN = 17$, $NP = 3y$, and $MP = 5y + 9$. So:
$$
5y + 9 = 17 + 3y
$$
## Step 2: Solve for $y$
Subtract $3y$ from both sides: $2y + 9 = 17$
Subtract $9$ from both sides: $2y = 8$
Divide by $2$: $y = 4$
## Step 3: Find $MP$
Substitute $y = 4$ into $MP = 5y + 9$:
$$
MP = 5(4) + 9 = 20 + 9 = 29
$$
# Answer:
$y = 4$ and the length of $MP$ is $29$.

### Problem 4: $M$ is the midpoint of $\overline{PQ}$. Find $x$ and $PQ$
# Explanation:
## Step 1: Midpoint Definition
If $M$ is the midpoint, then $PM = MQ$. So $3x + 10 = 8x + 25$ (wait, no—wait, midpoint means $PM = MQ$, but actually, $PM = 3x + 10$ and $MQ = 8x + 25$? Wait, no, maybe I misread. Wait, the diagram: $P$---$M$---$Q$, with $PM = 3x + 10$ and $MQ = 8x + 25$? Wait, no, midpoint means $PM = MQ$, so:
$$
3x + 10 = 8x + 25
$$
Wait, that would give negative $x$, which is wrong. Wait, maybe it's $PM = 3x + 10$ and $MQ = 8x - 25$? Wait, the original problem: "3x + 10" and "8x + 25" with $M$ as midpoint. Wait, maybe a typo, but assuming the midpoint, so $PM = MQ$, so:
$$
3x + 10 = 8x + 25 \implies -5x = 15 \implies x = -3
$$
That can't be. Wait, maybe the segments are $PM = 3x + 10$ and $MQ = 8x - 25$? Let's check again. Wait, the user's diagram: "3x + 10" and "8x + 25" with $M$ as midpoint. Maybe I made a mistake. Wait, no—wait, midpoint means $PQ = 2 \times PM$ (or $2 \times MQ$). So $PQ = PM + MQ = (3x + 10) + (8x + 25) = 11x + 35$, and since $M$ is midpoint, $PM = MQ$, so $3x + 10 = 8x + 25$ is incorrect. Wait, maybe the labels are reversed. Let's assume $PM = 3x + 10$ and $MQ = 8x - 25$ (to get positive $x$). Then:
$$
3x + 10 = 8x - 25 \implies 5x = 35 \implies x = 7
$$
Then $PM = 3(7) + 10 = 31$, $MQ = 8(7) - 25 = 31$, so $PQ = 31 + 31 = 62$. Maybe the original problem had a typo, but assuming $3x + 10 = 8x - 25$, then:
## Step 1: Set $PM = MQ$
$$
3x + 10 = 8x - 25
$$
## Step 2: Solve for $x$
Subtract $3x$: $10 = 5x - 25$
Add $25$: $35 = 5x$
$x = 7$
## Step 3: Find $PQ$
$PQ = 2 \times PM = 2(3(7) + 10) = 2(31) = 62$
# Answer:
Assuming the correct equation, $x = 7$ and $PQ = 62$. (Note: There might be a typo in the problem's segment labels, but this is the logical solution for a midpoint.)

### Problem 5: Find $x$, $DE$, and $DF$
# Explanation:
## Step 1: Segment Addition Postulate
$DF = DE + EF$, so $6x = (3x - 1) + 13$
## Step 2: Solve for $x$
$$
6x = 3x - 1 + 13 \implies 6x = 3x + 12 \implies 3x = 12 \implies x = 4
$$
## Step 3: Find $DE$
$DE = 3x - 1 = 3(4) - 1 = 11$
## Step 4: Find $DF$
$DF = 6x = 6(4) = 24$ (or $DE + EF = 11 + 13 = 24$)
# Answer:
$x = 4$, $DE = 11$, $DF = 24$.

### Problem 6: $B$ is midpoint of $AC$, $AC = 28$. Find $BC$
# Explanation:
## Step 1: Midpoint Definition
If $B$ is the midpoint of $AC$, then $AB = BC$ and $AC = AB + BC = 2BC$.
## Step 2: Solve for $BC$
$$
BC = \frac{AC}{2} = \frac{28}{2} = 14
$$
# Answer:
The length of $\overline{BC}$ is $14$.

### Problem 7: Find $m\angle XYZ$
# Explanation:
## Step 1: Angle Addition (Straight Angle? Wait, the diagram: $X$---$Y$---(angle), $Y$ with $X$ to $Y$ to $W$ (80°), $Y$ to $Z$ (80°), and the angle between $W$ and $Z$ is 80°? Wait, the total around $Y$ for a straight line? Wait, the diagram: $X$---$Y$---(horizontal), $Y$ with a right angle? Wait, the angles: $m\angle XYW = 80°$, $m\angle WYZ = 80°$, and we need $m\angle XYZ$. Wait, $X$---$Y$---$Z$ with $W$ inside? Wait, the sum: $m\angle XYZ = m\angle XYW + m\angle WYZ = 80° + 80° = 160°$? Wait, no, if $X$---$Y$---$Z$ is a straight line (180°), and $\angle XYW = 80°$, $\angle WYZ = 80°$, then $\angle XYZ = 80° + 80° = 160°$, but that's more than 180? No, maybe $X$---$Y$---$Z$ with $W$ such that $\angle XYW = 80°$, $\angle WYZ = 80°$, so $\angle XYZ = 80° + 80° = 160°$.
## Step 1: Add the Angles
$$
m\angle XYZ = 80° + 80° = 160°
$$
# Answer:
$m\angle XYZ = 160°$.

### Problem 8: $m\angle KLM = 180°$. Find $x$ and $m\angle KLN$
# Explanation:
## Step 1: Angle Addition Postulate
$\angle KLM = \angle KLN + \angle NLM$? Wait, the diagram: $K$---$L$---$M$ (straight line, 180°), with $N$ such that $\angle KLN = (10x - 5)°$ and $\angle NLM = (4x + 3)°$. So:
$$
(10x - 5) + (4x + 3) = 180
$$
## Step 2: Solve for $x$
Combine like terms: $14x - 2 = 180$
Add 2: $14x = 182$
Divide by 14: $x = 13$
## Step 3: Find $m\angle KLN$
Substitute $x = 13$ into $10x - 5$:
$$
10(13) - 5 = 130 - 5 = 125°
$$
# Answer:
$x = 13$ and $m\angle KLN = 125°$.

### Problem 9: $m\angle LKN = 145°$. Find $x$ and $m\angle LKM$
# Explanation:
## Step 1: Angle Addition Postulate
$\angle LKN = \angle LKM + \angle MKN$. $\angle LKM = (2x + 10)°$, $\angle MKN = (4x - 3)°$, so:
$$
(2x + 10) + (4x - 3) = 145
$$
## Step 2: Solve for $x$
Combine like terms: $6x + 7 = 145$
Subtract 7: $6x = 138$
Divide by 6: $x = 23$
## Step 3: Find $m\angle LKM$
Substitute $x = 23$ into $2x + 10$:
$$
2(23) + 10 = 46 + 10 = 56°
$$
# Answer:
$x = 23$ and $m\angle LKM = 56°$.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 1: Determine the length of \(\overline{CD}\)

Step 1: Apply Segment Addition Postulate

Step 2: Calculate the Sum

Step 1: Apply Segment Addition Postulate

Step 2: Solve for \(x\)

Step 1: Apply Segment Addition Postulate

Step 2: Solve for \(y\)

Step 3: Find \(MP\)

Answer:

Problem 2: Determine the length of \(\overline{RS}\)