Question

name london rodgers
geometry essentials & midpoint - yellow group

1. name this line 3 different ways. - video

(diagram with q, v, r, s, t on a line)
you do: name the line in this picture:
(diagram with e, b, f on a line, labeled c*)

1. find the measure of angle abc.

(diagram with protractor at b, rays a, c, x, z)
you do: find m∠zbx.

1. name angle 1.

(diagram with angles 1,2,3,4,5 at a vertex, rays q, d, w, s)
you do: name angle 3.

1. find x if st=34.

(diagram with s---r---t, sr=10+x, rt=x)
you do: find x if df is 29.
(diagram with d---e---f, de=x, ef=3x-3)

1. find m∠qsr if m∠qst=85°.

(diagram with s, q, r, t, angle rst=27°)
you do: find m∠igh if m∠fgh=112°.
(diagram with g, f, i, h, angle igh=54°)

Explanation:

Let's solve each problem one by one:

Problem 1 (Naming the line)

• A line can be named using any two points on it. For the "YOU DO" part with points E, B, F:
• The line can be named as line \( \overleftrightarrow{EB} \), line \( \overleftrightarrow{BF} \), or line \( \overleftrightarrow{EF} \) (since it's a straight line through E, B, F).

Problem 2 (Finding \( m\angle ABC \) and \( m\angle ZBX \))

• For \( \angle ABC \): Looking at the protractor, ray BA is at \( 30^\circ \) and ray BC is at \( 90^\circ \), so \( m\angle ABC = 90^\circ - 30^\circ = 60^\circ \).
• For \( \angle ZBX \): Ray BZ is at \( 0^\circ \) (or \( 180^\circ \)) and ray BX is at \( 60^\circ \) (from the protractor), so \( m\angle ZBX = 60^\circ - 0^\circ = 60^\circ \)? Wait, maybe better to check the protractor marks. Wait, the protractor has BA at \( 30^\circ \), BC at \( 90^\circ \), BX at \( 120^\circ \), BZ at \( 180^\circ \)? Wait, maybe I misread. Let's re - examine: If BA is at \( 30^\circ \) (from the left - hand scale), BC is at \( 90^\circ \), BX is at \( 120^\circ \), BZ is at \( 180^\circ \). Then \( \angle ABC=90 - 30 = 60^\circ \), \( \angle ZBX=180 - 120 = 60^\circ \)? Or maybe using the right - hand scale. Alternatively, if BA is at \( 150^\circ \) (right - hand scale), BC at \( 90^\circ \), then \( \angle ABC = 150 - 90=60^\circ \). For \( \angle ZBX \), if BZ is at \( 0^\circ \) (right - hand scale) and BX is at \( 60^\circ \), then \( \angle ZBX = 60^\circ \).

Problem 3 (Naming angles)

• For angle 1: It's formed by rays Q and D, with the vertex at the common point (let's say O). So it can be named as \( \angle QOD \), \( \angle DOQ \), or using the number \( \angle 1 \).
• For angle 3 (YOU DO): It's formed by rays S and W, with the vertex at the common point. So it can be named as \( \angle SOW \), \( \angle WOS \), or \( \angle 3 \).

Problem 4 (Finding x when \( ST = 34 \))

Step 1: Set up the equation

Since \( ST=SR + RT \), and \( SR = 10 + x \), \( RT=x \), and \( ST = 34 \), we have the equation \( (10 + x)+x=34 \).

Step 2: Simplify the equation

Combine like terms: \( 10+2x = 34 \).

Step 3: Solve for x

Subtract 10 from both sides: \( 2x=34 - 10=24 \). Then divide both sides by 2: \( x=\frac{24}{2}=12 \).

Step 1: Set up the equation

Since \( DF=DE + EF \), and \( DE = x \), \( EF = 3x-3 \), and \( DF = 29 \), we have the equation \( x+(3x - 3)=29 \).

Step 2: Simplify the equation

Combine like terms: \( 4x-3 = 29 \).

Step 3: Solve for x

Add 3 to both sides: \( 4x=29 + 3=32 \). Then divide both sides by 4: \( x=\frac{32}{4}=8 \).

Step 1: Analyze the angle relationship

We know that \( \angle QST=\angle QSR+\angle RST \), and \( \angle RST = 27^\circ \). Let \( m\angle QSR=y \). So \( y + 27^\circ=85^\circ \).

Step 2: Solve for y

Subtract \( 27^\circ \) from both sides: \( y=85^\circ - 27^\circ = 58^\circ \).

Answer:

\( x = 12 \)

"YOU DO" for Problem 4 (Finding x when \( DF = 29 \))