PDF) Lesson 1.1 @BULLET Building Blocks of Geometry | Connor ... - Free Printable

Problem Analysis and Solution

The provided image contains a set of geometry exercises from "Lesson 1.1: Building Blocks of Geometry." Below, I will solve each problem step by step.

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#### Exercise 1–7: Completing Statements

We are given that \( \overline{PS} = 3 \) cm. Let's analyze the diagram and solve each part:

1. The midpoint of \( \overline{PQ} \) is ______.
- From the diagram, the midpoint of \( \overline{PQ} \) is labeled as \( S \).
- Answer: \( S \)

2. \( NQ = \) ______.
- The segment \( \overline{NQ} \) spans from \( N \) to \( Q \). Since \( P \) and \( S \) are midpoints, we can infer:
- \( NP = PS = SQ = 3 \) cm (each segment is equal because \( P \) and \( S \) divide \( \overline{NQ} \) into three equal parts).
- Therefore, \( NQ = NP + PS + SQ = 3 + 3 + 3 = 9 \) cm.
- Answer: \( 9 \) cm

3. Another name for \( \overline{NS} \) is ______.
- The segment \( \overline{NS} \) can also be named as \( \overline{SN} \) since the order of points in a line segment does not matter.
- Answer: \( \overline{SN} \)

4. \( S \) is the ______ of \( \overline{SQ} \).
- From the diagram, \( S \) is the starting point of \( \overline{SQ} \), so it is the endpoint of \( \overline{SQ} \).
- Answer: endpoint

5. \( P \) is the midpoint of ______.
- From the diagram, \( P \) is the midpoint of \( \overline{NS} \).
- Answer: \( \overline{NS} \)

6. \( \overline{NS} \cong \) ______.
- The segment \( \overline{NS} \) is congruent to \( \overline{SQ} \) because both are equal in length (each is 3 cm).
- Answer: \( \overline{SQ} \)

7. Another name for \( \overline{SN} \) is ______.
- Similar to question 3, \( \overline{SN} \) can also be named as \( \overline{NS} \).
- Answer: \( \overline{NS} \)

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#### Exercise 8: Congruent Segments in \( KLMN \)

We are asked to name all pairs of congruent segments in the quadrilateral \( KLMN \).

- From the diagram:
- \( KN = LM = 8 \) cm (given).
- \( KL = NM \) (opposite sides of a parallelogram are congruent).
- \( KO = OM \) and \( NO = OL \) (diagonals bisect each other).

Thus, the pairs of congruent segments are:
- \( \overline{KN} \cong \overline{LM} \)
- \( \overline{KL} \cong \overline{NM} \)
- \( \overline{KO} \cong \overline{OM} \)
- \( \overline{NO} \cong \overline{OL} \)

Answer:
\[
\overline{KN} \cong \overline{LM}, \quad \overline{KL} \cong \overline{NM}, \quad \overline{KO} \cong \overline{OM}, \quad \overline{NO} \cong \overline{OL}
\]

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#### Exercise 9: Midpoint Formula

Given:
- \( M(-4, 8) \) is the midpoint of \( \overline{DE} \).
- Coordinates of \( D \) are \( (6, 1) \).
- Find the coordinates of \( E \).

The midpoint formula is:
\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]
where \( M(x_m, y_m) \), \( D(x_1, y_1) \), and \( E(x_2, y_2) \).

Substitute the known values:
\[
(-4, 8) = \left( \frac{6 + x_2}{2}, \frac{1 + y_2}{2} \right)
\]

Solve for \( x_2 \) and \( y_2 \):
1. For \( x \)-coordinate:
\[
-4 = \frac{6 + x_2}{2}
\]
Multiply both sides by 2:
\[
-8 = 6 + x_2
\]
Subtract 6:
\[
x_2 = -14
\]

2. For \( y \)-coordinate:
\[
8 = \frac{1 + y_2}{2}
\]
Multiply both sides by 2:
\[
16 = 1 + y_2
\]
Subtract 1:
\[
y_2 = 15
\]

Thus, the coordinates of \( E \) are \( (-14, 15) \).

Answer: \( (-14, 15) \)

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#### Exercises 10 and 11: Drawing Figures

These exercises require drawing figures using a ruler and marking congruent parts. Since this is a text-based solution, I will describe the steps instead of drawing:

10. Draw \( \overline{AB} \) and \( \overline{CD} \) with \( M \) as the midpoint of both.
- Draw \( \overline{AB} \) with length 6.4 cm.
- Draw \( \overline{CD} \) with length 4.0 cm.
- Mark \( M \) as the midpoint of both segments.
- Label the congruent parts.

11. Draw \( \overline{AB} \) and \( \overline{CD} \) with \( C \) as the midpoint of \( \overline{AB} \) and \( D \) as the midpoint of \( \overline{AE} \).
- Draw \( \overline{AB} \) with \( AC = 1.5 \) cm and \( CB = 1.5 \) cm (since \( C \) is the midpoint).
- Extend \( \overline{AB} \) to \( E \) such that \( AD = 2BC \).
- Mark \( D \) as the midpoint of \( \overline{AE} \).
- Label the congruent parts.

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#### Exercise 12: Sketching Points and Lines

Sketch six points \( A, B, C, D, E, F \) such that no three are collinear. Name the lines defined by these points.

- When six points are placed such that no three are collinear, the number of lines formed is given by the combination formula \( \binom{n}{2} \), where \( n \) is the number of points.
- Here, \( n = 6 \):
\[
\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15
\]

Thus, there are 15 lines.

Answer: 15 lines

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#### Exercise 13: Sets of Coplanar Points

In the figure (a cube), we are given that \( \{B, C, H, E\} \) is a set of four coplanar points. We need to find two other sets of four coplanar points.

From the cube diagram:
- One face of the cube is \( \{B, C, G, F\} \).
- Another face of the cube is \( \{C, D, H, G\} \).

Thus, two other sets of four coplanar points are:
1. \( \{B, C, G, F\} \)
2. \( \{C, D, H, G\} \)

Answer: \( \{B, C, G, F\} \) and \( \{C, D, H, G\} \)

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Final Answers:

1. \( S \)
2. \( 9 \) cm
3. \( \overline{SN} \)
4. endpoint
5. \( \overline{NS} \)
6. \( \overline{SQ} \)
7. \( \overline{NS} \)
8. \( \overline{KN} \cong \overline{LM}, \overline{KL} \cong \overline{NM}, \overline{KO} \cong \overline{OM}, \overline{NO} \cong \overline{OL} \)
9. \( (-14, 15) \)
10. Draw as described.
11. Draw as described.
12. 15 lines
13. \( \{B, C, G, F\} \) and \( \{C, D, H, G\} \)

\boxed{S, 9 \text{ cm}, \overline{SN}, \text{endpoint}, \overline{NS}, \overline{SQ}, \overline{NS}, \overline{KN} \cong \overline{LM}, \overline{KL} \cong \overline{NM}, \overline{KO} \cong \overline{OM}, \overline{NO} \cong \overline{OL}, (-14, 15), 15, \{B, C, G, F\}, \{C, D, H, G\}} \)