Question

geometry quiz 1 unit 2
namemt madder

1. give another name for line jk.
2. name a line containing point j.
3. name a point coplanar to point k.
4. name the intersection of plane r and line jl.
5. name three non - collinear points.
6. give another name for line s.
7. name the intersection of lines r and s.
8. give another name for plane n.
9. name a segment containing point c.
10. name three collinear points.

use the figure below for questions 9 - 12.

1. if df = 45 and ef = 18, find de.
2. if de = 4x - 1, ef = 9, and df = 9x - 12, find the value of x.
3. if df = 64, de = 5x - 9, and ef = 2x + 10, find ef.
4. if de = 4x + 10, ef = 2x - 1, and df = 9x - 24, find df.

answer the following questions. draw a diagram if necessary.

1. m is the midpoint of the segment. find x.
2. m is the midpoint of pq. pm = 6x + 13 and mq = 9x - 8, find pq.
3. c is the midpoint of df. dc = 9x + 3 and cf = 11x - 11, find dc.
4. de bisects ab at c. if ac = 8x - 18 and cb = 3x + 22, find x.

Explanation:

Step1: Recall line - naming convention

A line can be named using any two points on it. For line JK, another name is line KJ.

Step2: Identify line containing point J

From the figure, line JL contains point J.

Step3: Find coplanar point to K

Points in the same plane are coplanar. Point J is in the same plane R as point K.

Step4: Determine intersection of plane and line

The intersection of plane R and line JL is point K.

Step5: Select non - collinear points

Non - collinear points are not on the same line. Points J, K, and M are non - collinear.

Step6: Rename line s

Line s can also be named line DC.

Step7: Find intersection of lines

The intersection of lines r and s is point B.

Step8: Rename plane N

Plane N can also be named plane ABCD.

Step9: Identify segment with point C

Segment BC contains point C.

Step10: Find collinear points

Points B, C, and D are collinear.

Step11: Use segment addition postulate

Given DF = 45 and EF = 18, by DF=DE + EF, we have DE=DF - EF. So DE = 45 - 18=27.

Step12: Apply segment addition postulate

Since DE + EF = DF, then (4x - 1)+9 = 9x - 12. Simplify: 4x+8 = 9x - 12. Move x terms to one side: 12 + 8=9x - 4x, 20 = 5x, x = 4.

Step13: Use segment addition postulate

Given DF = 64, DE = 5x - 9 and EF = 2x + 10. Then (5x - 9)+(2x + 10)=64. Simplify: 7x+1 = 64, 7x = 63, x = 9. So EF = 2x + 10=2×9 + 10 = 28.

Step14: Apply segment addition postulate

Since DE + EF = DF, (4x + 10)+(2x - 1)=9x - 24. Simplify: 6x + 9 = 9x - 24. Move x terms to one side: 24 + 9=9x - 6x, 33 = 3x, x = 11. Then DF = 9x - 24=9×11 - 24 = 75.

Step15: Use mid - point property

If M is the mid - point of segment JK, then JM = MK. So 6x + 11=9x - 13. Move x terms to one side: 11 + 13=9x - 6x, 24 = 3x, x = 8.

Step16: Use mid - point property

If M is the mid - point of PQ, then PM = MQ. So 6x + 13=9x - 8. Move x terms to one side: 13 + 8=9x - 6x, 21 = 3x, x = 7. Then PM = 6x + 13=6×7+13 = 55, and PQ = 2PM = 110.

Step17: Use mid - point property

If C is the mid - point of DF, then DC = CF. So 9x + 3=11x - 11. Move x terms to one side: 3 + 11=11x - 9x, 14 = 2x, x = 7. Then DC = 9x + 3=9×7 + 3 = 66.

Step18: Use bisection property

If DE bisects AB at C, then AC = CB. So 8x - 18=3x + 22. Move x terms to one side: 8x - 3x=22 + 18, 5x = 40, x = 8.

Answer:

1. Line KJ
2. Line JL
3. Point J
4. Point K
5. Points J, K, M
6. Line DC
7. Point B
8. Plane ABCD
9. Segment BC
10. Points B, C, D
11. 27
12. 4
13. 28
14. 75
15. 8
16. 110
17. 66
18. 8