Unit 1: Geometry Basics Homework 2: Segment Addition Postulate age document! ** 1. If LM= 22 and MN = 15, find LN. 2. If LN = 54 and LM = 31, find MN. 4. If DF = 9x – 39, find EF. 47 3x + 10
The Correct Answer and Explanation is:
Let’s go through each of the questions using the Segment Addition Postulate.
1. If LM=22LM = 22LM=22 and MN=15MN = 15MN=15, find LNLNLN.
The Segment Addition Postulate states that if LLL, MMM, and NNN are collinear points, then the sum of the lengths of segments LMLMLM and MNMNMN is equal to the length of LNLNLN:LN=LM+MNLN = LM + MNLN=LM+MN
Substitute the given values:LN=22+15=37LN = 22 + 15 = 37LN=22+15=37
So, LN=37LN = 37LN=37.
2. If LN=54LN = 54LN=54 and LM=31LM = 31LM=31, find MNMNMN.
Again, applying the Segment Addition Postulate:LN=LM+MNLN = LM + MNLN=LM+MN
We are given LN=54LN = 54LN=54 and LM=31LM = 31LM=31, so to find MNMNMN, subtract LMLMLM from LNLNLN:MN=LN−LM=54−31=23MN = LN – LM = 54 – 31 = 23MN=LN−LM=54−31=23
Thus, MN=23MN = 23MN=23.
3. If DF=9x−39DF = 9x – 39DF=9x−39, find EFEFEF given EF=47EF = 47EF=47 and DF=3x+10DF = 3x + 10DF=3x+10.
From the Segment Addition Postulate:DF+EF=DEDF + EF = DEDF+EF=DE
Since we are given that DF=9x−39DF = 9x – 39DF=9x−39 and EF=47EF = 47EF=47, the equation becomes:(9x−39)+47=3x+10(9x – 39) + 47 = 3x + 10(9x−39)+47=3x+10
Now, simplify:9x−39+47=3x+109x – 39 + 47 = 3x + 109x−39+47=3x+109x+8=3x+109x + 8 = 3x + 109x+8=3x+10
Now, isolate the xxx-terms on one side and constants on the other:9x−3x=10−89x – 3x = 10 – 89x−3x=10−86x=26x = 26x=2
Now, solve for xxx:x=26=13x = \frac{2}{6} = \frac{1}{3}x=62=31
Substitute x=13x = \frac{1}{3}x=31 into the equation for DFDFDF:DF=9(13)−39=3−39=−36DF = 9\left(\frac{1}{3}\right) – 39 = 3 – 39 = -36DF=9(31)−39=3−39=−36
Now substitute the value of DFDFDF into EFEFEF:EF=47=47EF = 47 = 47EF=47=47
This confirms that EF=47EF = 47EF=47.
Explanation:
The Segment Addition Postulate is used when you know that two segments are parts of a larger segment. By adding or subtracting known lengths, you can find the missing segment length. In problems like these, basic algebra is used to solve for unknown variables and segment lengths. For the last problem, solving for xxx first and substituting it back in helps us to find DFDFDF and confirm the given EFEFEF length.
