Question

name: zelma garner unit 1: geometry basics per: date: homework 2: segment addition postulate this is a 2 - page document! use the diagram below to answer questions 1 and 2. 1. if lm = 22 and mn = 15, find ln. 2. if ln = 54 and lm = 31, find mn. 3. if rt = 36, find the value of x. 4. if df = 9x - 39, find ef. 5. if uw = 6x - 35, find uw. 6. if hj = 7x - 27, find the value of x. 7. if bd = 7x - 10, bc = 4x - 29, and cd = 5x - 9, find each value. 8. if bd ≅ bc, bd = 5x - 26, bc = 2x + 1, and ac = 43, find ab.

Explanation:

Step1: Apply segment - addition postulate for question 1

Since $LN=LM + MN$, substituting $LM = 22$ and $MN=15$, we get $LN=22 + 15$.

Step2: Calculate the value of $LN$

$LN=37$

Step3: Apply segment - addition postulate for question 2

Since $LN=LM + MN$, then $MN=LN - LM$. Substituting $LN = 54$ and $LM = 31$, we have $MN=54 - 31$.

Step4: Calculate the value of $MN$

$MN = 23$

Step5: Apply segment - addition postulate for question 3

Since $RT=RS+ST$, and $RS = 6x + 1$, $ST=x + 7$, $RT = 36$, we get the equation $(6x + 1)+(x + 7)=36$.

Step6: Simplify the left - hand side of the equation

$6x+1+x + 7=7x + 8$. So, $7x+8 = 36$.

Step7: Solve for $x$

Subtract 8 from both sides: $7x=36 - 8=28$. Then $x = 4$.

Step8: Apply segment - addition postulate for question 4

Since $DF=DE + EF$, $DE = 47$, $EF=3x + 10$, $DF=9x - 39$, we have $47+(3x + 10)=9x - 39$.

Step9: Simplify the left - hand side of the equation

$47+3x + 10=57+3x$. So, $57+3x=9x - 39$.

Step10: Solve for $x$

Subtract $3x$ from both sides: $57=6x - 39$. Add 39 to both sides: $6x=57 + 39 = 96$. Then $x = 16$. And $EF=3x + 10=3\times16+10=58$.

Step11: Apply segment - addition postulate for question 5

Since $UW=UV+VW$, $UV = 19$, $VW=4x - 20$, $UW=6x - 35$, we have $19+(4x - 20)=6x - 35$.

Step12: Simplify the left - hand side of the equation

$19+4x - 20=4x - 1$. So, $4x-1=6x - 35$.

Step13: Solve for $x$

Subtract $4x$ from both sides: $-1=2x - 35$. Add 35 to both sides: $2x=34$, $x = 17$. And $UW=6x - 35=6\times17-35=67$.

Step14: Apply segment - addition postulate for question 6

Since $HJ=HI+IJ$, $HI = 3x - 5$, $IJ=x - 1$, $HJ=7x - 27$, we have $(3x - 5)+(x - 1)=7x - 27$.

Step15: Simplify the left - hand side of the equation

$3x-5+x - 1=4x - 6$. So, $4x - 6=7x - 27$.

Step16: Solve for $x$

Subtract $4x$ from both sides: $-6=3x - 27$. Add 27 to both sides: $3x=21$, $x = 7$.

Step17: Apply segment - addition postulate for question 7

Since $BD=BC + CD$, $BD = 7x - 10$, $BC=4x - 29$, $CD=5x - 9$, we have $(4x - 29)+(5x - 9)=7x - 10$.

Step18: Simplify the left - hand side of the equation

$4x-29+5x - 9=9x - 38$. So, $9x-38=7x - 10$.

Step19: Solve for $x$

Subtract $7x$ from both sides: $2x-38=-10$. Add 38 to both sides: $2x=28$, $x = 14$. Then $BC=4x - 29=4\times14-29=27$, $CD=5x - 9=5\times14-9=61$, $BD=7x - 10=7\times14-10=88$.

Step20: Apply the congruence and segment - addition postulate for question 8

Since $BD\cong BC$, $BD = 5x - 26$, $BC=2x + 1$, we have $5x - 26=2x + 1$.

Step21: Solve for $x$

Subtract $2x$ from both sides: $3x-26=1$. Add 26 to both sides: $3x=27$, $x = 9$. Then $BC=2x + 1=2\times9+1 = 19$. Since $AC = 43$, and $AC=AB + BC$, $AB=AC - BC=43 - 19=24$.

Answer:

1. $LN = 37$
2. $MN = 23$
3. $x = 4$
4. $EF = 58$
5. $UW = 67$
6. $x = 7$
7. $x = 14$, $BC = 27$, $CD = 61$, $BD = 88$
8. $AB = 24$