Question

geometry chapter 1 review

1. jorge wrote the following steps for copying an angle. is jorge missing any step? if yes, please identify the missing step.

step 1
draw a segment.
draw an angle such as ∠a, as shown. then draw a segment.
label point d on the segment.
step 2
draw arcs.
draw an arc with center a. using the same radius, draw an arc with center d.
step 3
draw a ray.
draw (overrightarrow{df}). ∠d has the same measure as ∠a.
a. jorges construction is not missing any step.
b. jorges construction is missing a step. the missing step is: draw an arc. label b, c, and e. draw an arc with a radius smaller than bc and center e. label the intersection f.
c. jorges construction is missing a step. the missing step is: draw an arc. label b, c, and e. draw an arc with a radius larger than bc and center e. label the intersection f.
d. jorges construction is missing a step. the missing step is: draw an arc. label b, c, and e. draw an arc with a radius bc and center e. label the intersection f.

1. a student used a compass and a straightedge to bisect ∠abc in this figure.

which statement best describes point s?
a. point s is located such that (mangle abc=mangle sbc).
b. point s is located such that (mangle pbq = mangle pqs).
c. point s is located such that (mangle apb=mangle apq).
d. point s is located such that (mangle abs=mangle sbc).

1. in the figure below lines ac and bd intersect at point e. select all the statements that are true.

a. (x = 35)
b. (y=100)
c. (mangle dec = 50^{circ})
d. (mangle bec = 100^{circ})
e. (mangle aed = 130^{circ})

Explanation:

12.

To copy an angle, after drawing arcs on the original angle and on the new - constructed figure with the initial steps, the next step is to use the distance between the intersection points on the original - angle arc as the radius for an arc on the new figure. In this case, the radius of the arc with center \(E\) should be equal to \(BC\).

When bisecting an angle \(\angle ABC\), the point \(S\) on the angle - bisector is such that it divides \(\angle ABC\) into two equal angles \(\angle ABS\) and \(\angle SBC\).

Step1: Use vertical - angle property

Vertical angles are equal. \(\angle AEB\) and \(\angle DEC\) are vertical angles, so \(m\angle DEC = 50^{\circ}\), so C is true. Also, \(\angle AED\) and \(\angle BEC\) are vertical angles.

Step2: Use the linear - pair property

\(\angle AEB\) and \(\angle BEC\) form a linear pair, so \(m\angle AEB+m\angle BEC = 180^{\circ}\). Since \(m\angle AEB = 50^{\circ}\), then \(m\angle BEC=180 - 50=130^{\circ}\), and \(m\angle AED = 130^{\circ}\), so E is true.
For the equations:
\(\angle AED=(y + 30)^{\circ}\) and \(\angle AED = 130^{\circ}\), then \(y+30 = 130\), so \(y = 100\), so B is true.
\(\angle DEC=(x + 15)^{\circ}\) and \(\angle DEC = 50^{\circ}\), then \(x+15 = 50\), so \(x = 35\), so A is true.

Answer:

D. Jorge's construction is missing a step. The missing step is: Draw an arc. Label \(B\), \(C\), and \(E\). Draw an arc with a radius \(BC\) and center \(E\). Label the intersection \(F\).

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