Question

1. if df = 42, find de. topic 2: segment addition postulate in the diagram below, if jl = 10x - 2, jk = 5x - 8, and kl = 7x - 12, find kl. 10. if line y bisects ac, ab = 4 - 5x, and bc = 2x + 25, find ac. 12. if s is the mid - point of rt, rs = 5x + 17, and st = 8x - 31, find rs. 11. if b is the mid - point of ac, ab = 3x + 4, ac = 11x - 17, and ce = 49, find de. 13.

Explanation:

Step1: Identify segment - addition postulate

For problem 9, by the segment - addition postulate, $DF=DE + EF$. So, $(7x + 1)+(4x-3)=42$.

Step2: Simplify the left - hand side

Combine like terms: $7x+4x + 1 - 3=42$, which gives $11x-2 = 42$.

Step3: Solve for $x$

Add 2 to both sides: $11x=42 + 2=44$. Then divide both sides by 11, so $x = 4$.

Step4: Find $DE$

Substitute $x = 4$ into the expression for $DE$: $DE=7x + 1=7\times4+1=29$.

For problem 10, by the segment - addition postulate, $JL=JK + KL$. So, $(5x - 8)+(7x-12)=10x-2$.

Step1: Simplify the left - hand side

Combine like terms: $5x+7x-8 - 12=10x-2$, which gives $12x-20 = 10x-2$.

Step2: Solve for $x$

Subtract $10x$ from both sides: $12x-10x-20=-2$, then $2x-20=-2$. Add 20 to both sides: $2x=18$, so $x = 9$.

Step3: Find $KL$

Substitute $x = 9$ into the expression for $KL$: $KL=7x-12=7\times9-12=63 - 12=51$.

For problem 11, since $S$ is the mid - point of $RT$, then $RS=ST$. So, $5x + 17=8x-31$.

Step1: Solve for $x$

Subtract $5x$ from both sides: $17=8x-5x-31$, then $17 = 3x-31$. Add 31 to both sides: $3x=17 + 31=48$, so $x = 16$.

Step2: Find $RS$

Substitute $x = 16$ into the expression for $RS$: $RS=5x + 17=5\times16+17=80 + 17=97$.

For problem 12, since line $y$ bisects $\overline{AC}$, then $AB = BC$. So, $4-5x=2x + 25$.

Step1: Solve for $x$

Add $5x$ to both sides: $4=2x+5x + 25$, then $4 = 7x+25$. Subtract 25 from both sides: $7x=4 - 25=-21$, so $x=-3$.

Step2: Find $AC$

First find $AB$: $AB=4-5x=4-5\times(-3)=4 + 15=19$. Then $AC=2AB$ (because $AB = BC$), so $AC = 38$.

For problem 13, since $B$ is the mid - point of $\overline{AC}$, then $AB = BC$ and $AC=AB + BC$. Given $AB = 3x + 4$, $AC=11x-17$, and $BC = DE$. Also, $AC = 2AB$. So, $11x-17=2(3x + 4)$.

Step1: Expand the right - hand side

$11x-17=6x + 8$.

Step2: Solve for $x$

Subtract $6x$ from both sides: $11x-6x-17=8$, then $5x-17=8$. Add 17 to both sides: $5x=8 + 17=25$, so $x = 5$.

Step3: Find $DE$ (equal to $BC$)

First find $AB$: $AB=3x + 4=3\times5+4=19$. Since $AC=11x-17=11\times5-17=55 - 17=38$, then $BC=\frac{AC}{2}=19$, so $DE = 19$.

Answer:

1. $DE = 29$
2. $KL = 51$
3. $RS = 97$
4. $AC = 38$
5. $DE = 19$