Question

name: lapannall, smerth
date:
per:
unit 1: geometry basics
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 ). (handwritten: ( 22 + 15 = 37 ))
2. if ( ln = 54 ) and ( lm = 31 ), find ( mn ). (handwritten: ( 54 - 31 - ))
3. if ( af = 36 ), find the value of ( x ). (diagram: ( a )---( 6x + 1 )---( b )---( x + 7 )---( f ))
4. if ( df = 9x - 39 ), find ( ef ). (diagram: ( d )---( 47 )---( e )---( 3x + 10 )---( f ))
5. if ( uw = 6x - 35 ), find ( uw ). (diagram: ( u )---( 19 )---( v )---( 4x - 20 )---( w ))
6. if ( hj = 7x - 27 ), find the value of ( x ). (diagram: ( h )---( 2x - 5 )---( i )---( x - 1 )---( j ))
7. if ( bd = 7x - 10 ), ( bc = 4x - 29 ), and ( cd = 5x - 9 ), find each value. (diagram: ( b )---( c )---( d ))

( x = ) ______
( bc = ) ______
( cd = ) ______
( bd = ) ______

1. if ( bd cong bc ), ( bd = 5x - 26 ), ( bc = 2x + 1 ), and ( ac = 43 ), find ( ab ). (diagram: ( a )---( b )---( c ) with ( d ) above ( b ); handwritten: ( 5x - 26 = 2x + 1 ), solving steps, ( x = 9 ))

Explanation:

Let's solve question 1 first:

Step1: Apply Segment Addition Postulate

The Segment Addition Postulate states that if we have three points \( L \), \( M \), and \( N \) collinear with \( M \) between \( L \) and \( N \), then \( LN = LM + MN \).
Given \( LM = 22 \) and \( MN = 15 \).

Step2: Calculate \( LN \)

Substitute the given values into the formula: \( LN = 22 + 15 \)
\( LN = 37 \)

Step1: Apply Segment Addition Postulate (Rearranged)

Using the same postulate \( LN = LM + MN \), we can rearrange it to find \( MN \): \( MN = LN - LM \)
Given \( LN = 54 \) and \( LM = 31 \).

Step2: Calculate \( MN \)

Substitute the values: \( MN = 54 - 31 \)
\( MN = 23 \)

Step1: Apply Segment Addition Postulate

Since \( I \) is between \( H \) and \( J \), \( HJ = HI + IJ \).
So, \( 7x - 27=(2x - 5)+(x - 1) \)

Step2: Simplify the right - hand side

Combine like terms: \( 7x - 27 = 2x - 5+x - 1=3x - 6 \)

Step3: Solve for \( x \)

Subtract \( 3x \) from both sides: \( 7x-3x - 27=3x - 3x-6 \)
\( 4x - 27=-6 \)
Add 27 to both sides: \( 4x-27 + 27=-6 + 27 \)
\( 4x = 21 \)
Wait, there is a mistake. Let's re - do step 1. If the diagram is \( H---I---J \), then \( HJ=HI + IJ \). So \( 7x - 27=(2x - 5)+(x - 1) \) is wrong. Wait, maybe the total length \( HJ \) is composed of \( HI \) and \( IJ \), but maybe the given \( HJ = 7x - 27 \) and \( HI=2x - 5 \), \( IJ=x - 1 \), so \( 7x-27=(2x - 5)+(x - 1) \)
\( 7x-27=3x - 6 \)
\( 7x-3x=27 - 6 \)
\( 4x = 21 \)
\( x=\frac{21}{4}=5.25 \). But maybe I misread the problem. Wait, the user's hand - written part for question 6: the segments are \( HI = 2x - 5 \) and \( IJ=x - 1 \), and \( HJ=7x - 27 \). So according to segment addition, \( HI+IJ = HJ \)
So \( (2x - 5)+(x - 1)=7x - 27 \)
\( 3x-6 = 7x - 27 \)
Subtract \( 3x \) from both sides: \( - 6=4x - 27 \)
Add 27 to both sides: \( 21 = 4x \)
\( x=\frac{21}{4}=5.25 \)

For question 8:

Answer:

\( LN = 37 \)

Now question 2: