Your Turn
10.1
1.
\overrightarrow {BD} is the ray that starts at point B and extends infinitely in the direction of point D.
3.
\overleftarrow {BA} represents the ray that starts at point A and extends infinitely in the direction of point B.
4.
\overleftrightarrow {AD} represents a line that contains the points A and D. Notice the arrowheads on both ends of the line above AD, which means that the line continues infinitely in both directions.
10.2
1.
Answers will vary. One way {\overline {AE}} ,\,{\overline {EI}} ,\,{\overline {IE}} ,\,{\overline {EB}} ,\,{\overline {BC}} ,\,{\overline {CD}} ,\,{\overline {DG}} \,; Second way {\overline {AH}} ,\,{\overline {IE}} ,\,{\overline {EB}} ,\,{\overline {BC}} ,{\overline {CF}} ,\,{\overline {FG}} .
10.3
1.
\overleftrightarrow {RS}\parallel \overleftrightarrow {UV}
\overleftrightarrow {RS} \bot \overleftrightarrow {XY}
\overleftrightarrow {UV} \bot \overleftrightarrow {XY}
\overleftrightarrow {RS} \bot \overleftrightarrow {XY}
\overleftrightarrow {UV} \bot \overleftrightarrow {XY}
10.4
10.5
1.
Point A is located at ( - 5, - 2); Point B is located at (- 3,4); Point C is located at (3, - 6); Point D is located at (5, - 2); Point E is located at (3,2); Point F is located at (5,5).
5.
Yes, this represents a plane. One reason is that the figure contains four points that are not on the line \overleftrightarrow {AB}.
10.6
10.7
1.
Acute Angles | Obtuse Angles | Right Angles | Straight Angles |
---|---|---|---|
\angle AOB \angle AOC \angle BOC \angle BOD \angle COD \angle DOE \angle FOE |
\angle AOE \angle BOF \angle COF |
\angle AOD \angle BOE \angle DOF |
\angle AOF |
10.8
10.9
1.
m\measuredangle (6x) = {24^ \circ}, m\measuredangle (9x) = {36^ \circ}, m\measuredangle (7x + 2) = {30^ \circ}
10.10
1.
m\measuredangle 4 = {67^ \circ }, m\measuredangle 1 = {113^ \circ }, m\measuredangle 3 = {113^ \circ }
10.11
10.12
1.
m\measuredangle 5 = {120^ \circ}, m\measuredangle 4 = {120^ \circ}, m\measuredangle 8 = {120^ \circ}
10.13
10.14
1.
m\measuredangle 5 = {56^ \circ }
m\measuredangle 6 = {62^ \circ }
m\measuredangle 7 = {118^ \circ }
m\measuredangle 8 = {62^ \circ }
m\measuredangle 9 = {118^ \circ }
m\measuredangle 6 = {62^ \circ }
m\measuredangle 7 = {118^ \circ }
m\measuredangle 8 = {62^ \circ }
m\measuredangle 9 = {118^ \circ }
10.15
1.
m\measuredangle x = {66^ \circ }, m\measuredangle (x + 1) = {67^ \circ }, m\measuredangle (x - 19) = {47^ \circ }
10.16
1.
m\measuredangle 1 = {35^ \circ }; m\measuredangle 2 = 2 = {85^ \circ }; m\measuredangle 3 = {60^ \circ }
10.17
10.18
10.19
10.20
10.21
10.22
10.23
10.25
1.
Shapes 1, 2, 4, and 6 are triangles; shape 3 is a pentagon; shape 5 is a parallelogram; and shape 7 is a rectangle.
10.26
10.27
10.28
10.29
1.
We have the sum of interior angles is {360^ \circ }. Then, x = - 9. The other angles measure {54^ \circ }, {111^ \circ }, {44^ \circ }.
10.30
10.31
10.32
10.33
10.34
10.35
10.36
1.
This tessellation could be produced with a reflection of the triangle vertically, then each triangle is rotated {180^ \circ } and translated to the right.
10.37
1.
From the first square on the left, rotate the square {30^ \circ } to the right, or A1 \to A2. Then, reflect the square over the horizontal, or A2 \to A3. Next, reflect all three squares over the vertical line. The lavender triangles comprise another pattern that tessellates, fits in with the squares, and fills the gaps.
10.38
10.39
1.
Not by themselves, but by adding an equilateral triangle, the two regular polygons do tessellate the plane without gaps.
10.40
1.
We made a tessellation with a regular decagon (10 sides) and an irregular hexagon. We see that the regular decagon will not fill the plane by itself. The gap is filled, however, with an irregular hexagon. These two shapes together will fill the plane.

10.41
10.42
10.43
10.44
10.45
10.46
10.47
10.48
10.49
10.50
10.51
10.52
10.53
10.54
10.55
10.56
10.57
10.58
10.59
10.60
1.
\begin{array}{rcl}
{SA}&{ = }&{527.78\text{ cm}{^2}}\\
{V}&{ = }&{769.69\text{ cm}{^3}}\\ \end{array}
10.61
10.62
10.63
1.
1.26\,{\text{ft}}\,{\text{wide}}\, \times \,1.26\,{\text{ft}}\,{\text{long}}\, \times \,1.26\,{\text{ft}}\,{\text{high}}, $95.26
10.64
10.65
10.66
10.67
10.68
10.69
10.70
10.71
10.72
10.73
10.74
10.75
Check Your Understanding
1.
The line containing point D and point A is a line segment from point D to point A, \overline {DA} , or from point A to point D, \overline {AD} .
2.
The line containing points C and B is a straight line that extends infinitely in both directions and contains points C and B.
3.
This is a ray that begins at point E, although it does not contain point E, and extends in the direction of point F.
4.
{\overline {AB}} \cup {\overline {BD}} = \overline {AD}. The union of line segment \overline {AB} and the line segment {\overline {BD}} contains all points in each line segment combined.
5.
\overrightarrow {BD} \cap \overline {BC} = \overline {BC} . The intersection of the ray \overrightarrow {BD} and the line segment {\overline {BC}} contains only the points common to each set, \overline {BC} .
6.
\overleftarrow {BA} \cup \overrightarrow {BD} = \overleftrightarrow {AD}. The union of the ray starting at point B and extending infinitely in the direction of A and the ray starting at point B and extending infinitely in the direction of D is the straight line extending infinitely in both directions containing points A, B, C, and D.
7.
Two lines are parallel if the distance between the lines is constant implying that the lines cannot intersect.
20.
These are similar triangles, so we can solve using proportions.
Then, x = \frac{{14}}{3} and y = \frac{{16}}{3}.
\begin{array}{*{20}{rcl}}{\frac{3}{2}}&{ = }&{\frac{{(3 + 4)}}{x}} \\ {3x}&{ = }&{14} \\ {x}&{ = }&{\frac{{14}}{3}} \\ \end{array} | \begin{array}{*{20}{rcl}}{\frac{4}{2}}&{ = }&{\frac{{4 + y}}{{\frac{{14}}{3}}}} \\{\frac{{14}}{3}(4)}&{ = }&{2(4 + y)} \\{\frac{{56}}{3}}&{ = }&{8 + 2y} \\{56}&{ = }&{3(8 + 2y)} \\{56}&{ = }&{24 + 6y} \\{32}&{ = }&{6y} \\{\frac{{16}}{3}}&{ = }&{y}\end{array} |
21.
Set up the proportions.
\begin{array}{*{20}{rcl}}{\frac{6}{a}}&{ = }&{\frac{{12}}{{14}}} \\{6(14)}&{ = }&{12a} \\{84}&{ = }&{12a} \\{7}&{ = }&{a}\end{array}
Thus, t = 20 and a = 7.
\begin{array}{*{20}{rcl}}{\frac{6}{a}}&{ = }&{\frac{{12}}{{14}}} \\{6(14)}&{ = }&{12a} \\{84}&{ = }&{12a} \\{7}&{ = }&{a}\end{array}
Thus, t = 20 and a = 7.
28.
\begin{array}{*{20}{c}}{360}&{=}&{152 + 9x + (5x + 1) + (x + 12)}\\{}&{=}&{15x + 165}\\{195}&{=}&{15x}\\{13}&{=}&{x}\\{x + 12}&{=}&{25^ \circ },\,9x = {117^ \circ },\,5x + 1 = {66^ \circ }\end{array}
30.
The patterns are repeated shapes that can be transformed in such a way as to fill the plane with no gaps or overlaps.
31.
Starting with the triangle with the point labeled , the triangle is translated point by point 3 units to the right and 3 units up to point A'. Then, the triangle labeled A' is translated 3 units to the right and 3 units up to point A''.
33.
The dark triangle is reflected about the vertical line showing the light back, and then reflected about the horizontal line. The pattern is repeated leaving a white diamond between the shapes.