Download Fill in the blank in each sentence with the vocabulary term that best

5. Name the intersection of lines a and c.
SOLUTION: The lines a and c intersect at the point P.
Study Guide and Review
Fill in the blank in each sentence with the vocabulary term that
best completes the sentence.
1. A _____ is a flat surface made up of points that extends infinitely in all
directions.
SOLUTION: A plane is a flat surface made up of points that extends infinitely in all
directions. Therefore, the correct answer is plane.
6. Give another name for line b.
SOLUTION: There are two point S and T marked on the line b. So, the line b can also
be called 7. Name a point that is not contained in any of the three lines a, b, or c.
SOLUTION: Here, W is a point on the plane R and is not on any of the lines on the
plane.
2. A set of points that all lie on the same line are said to be _____.
SOLUTION: A set of points that all lie on the same line are said to be collinear points.
Therefore, the correct answer is collinear.
8. Give another name for plane WPX.
SOLUTION: The plane WPX is also named as plane R.
3. If two lines intersect to form four right angles, the lines are called _____.
Name the geometric term that is best modeled by each item.
SOLUTION: If two lines intersect to form four right angles, the lines are called
perpendicular.
4. If the sum of the measures of two angles is 180, then the angles are
called _____ angles.
9. SOLUTION: If the sum of the measures of two angles is 180, then the angles are
called supplementary angles.
SOLUTION: In the figure, two flat surfaces intersect each other. The two flat
surfaces model two planes. Since the intersection of two planes form a
line, the item models a line.
Use the figure to complete each of the following.
5. Name the intersection of lines a and c.
SOLUTION: The lines a and c intersect at the point P.
6. Give another name for line b.
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SOLUTION: There are two point S and T marked on the line b. So, the line b can also
10. SOLUTION: Each bead denotes a location. So, it models a point.
Find the value of the variable and XP, if X is between P and Q.
Page 1
11. XQ = 13, XP = 5x – 3, PQ = 40
SOLUTION: 6k – 10k = –2 – 16
–4k = –18
k = 4.5
10. SOLUTION: Study
Guide and Review
Each bead denotes a location. So, it models a point.
Find the value of the variable and XP, if X is between P and Q.
11. XQ = 13, XP = 5x – 3, PQ = 40
So, XP = 7(4.5) – 2 = 29.5.
Determine whether each pair of segments is congruent.
13. SOLUTION: Here X is between P and Q. So, PQ = XP + XQ.
SOLUTION: Segments that have the same measure are called congruent segments.
Here, the lengths of the segments AB, BC, CD, and AD are given to be
equal. So, they are congruent.
We have XQ = 13, XP = 5x – 3, and PQ = 40.
40 = 13 + 5x – 3
30 = 5x
6=x
14. So, XP = 5(6) – 3 = 27.
12. XQ = 3k, XP = 7k – 2, PQ = 6k + 16
SOLUTION: Here X is between P and Q. So, PQ = XP + XQ.
We have XQ = 3k, XP = 7k – 2, PQ = 6k + 16.
6k + 16 = 7k – 2 + 3k
6k – 10k = –2 – 16
–4k = –18
k = 4.5
So, XP = 7(4.5) – 2 = 29.5.
Determine whether each pair of segments is congruent.
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SOLUTION: Segments that have the same measure are called congruent segments.
Substitute for a to find the lengths.
YZ = 3a + 2 = 3(4) + 2 = 14
YX = 4a = 4(4) = 16
The lengths of the segments XY and YZ are not equal. So, they are not
congruent.
15. DISTANCE The distance from Salvador’s job to his house is 3 times
greater than the distance from his house to school. If his house is
between his job and school and the distance from his job to school is 6
miles, how far is it from Salvador’s house to school?
SOLUTION: Salvador’s house is between his job and school. Let x be the distance
from his house to school. Then the distance from his job to his house is
3x. The distance from his job to school is 6 miles. Page 2
That is,
6 = 3x + x.
YZ = 3a + 2 = 3(4) + 2 = 14
YX = 4a = 4(4) = 16
The
lengths
ofReview
the segments XY and YZ are not equal. So, they are not
Study
Guide
and
congruent.
15. DISTANCE The distance from Salvador’s job to his house is 3 times
greater than the distance from his house to school. If his house is
between his job and school and the distance from his job to school is 6
miles, how far is it from Salvador’s house to school?
SOLUTION: Salvador’s house is between his job and school. Let x be the distance
from his house to school. Then the distance from his job to his house is
3x. The distance from his job to school is 6 miles. That is,
6 = 3x + x.
6 = 4x
1.5 = x
The distance between A and B is
or about 15.6 units.
17. P(2, –1), Q(10, –7)
SOLUTION: Use the Distance Formula.
Substitute.
Therefore, the house is 1.5 miles away from the school.
The distance between P and Q is 10 units.
Find the distance between each pair of points.
16. A(–3, 1), B(7, 13)
SOLUTION: Use the Distance Formula.
Find the coordinates of the midpoint of a segment with the given
endpoints.
18. L(–3, 16), M (17, 4)
SOLUTION: Use the Midpoint Formula
.
Substitute.
Substitute.
The midpoint of
The distance between A and B is
17. P(2, –1), Q(10, –7)
SOLUTION: Use Manual
the Distance
Formula.
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or about 15.6 units.
19. C(32, –1), D(0, –12)
SOLUTION: Use the Midpoint Formula
. is (7, 10).
Substitute.
Page 3
Study
Guide
andof
Reviewis (7, 10).
The
midpoint
19. C(32, –1), D(0, –12)
The midpoint of
is (16, –6.5).
Find the coordinates of the missing endpoint if M is the midpoint
of
20. X(–11, –6), M (15, 4)
SOLUTION: Use the Midpoint Formula
SOLUTION: Let the coordinates of Y be (x, y).
. Substitute.
Then by the Midpoint Formula,
.
The midpoint of
is (16, –6.5).
Find the coordinates of the missing endpoint if M is the midpoint
of
20. X(–11, –6), M (15, 4)
SOLUTION: Let the coordinates of Y be (x, y).
Write two equations to find the coordinates of Y.
Then by the Midpoint Formula,
.
The coordinates of Y are (41, 14).
Write two equations to find the coordinates of Y.
21. M (–4, 8), Y(19, 0)
SOLUTION: Let the coordinates of X be (x, y).
Then by the Midpoint Formula,
.
Write two equations to find the coordinates of X.
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coordinates
of Y
(41,
21. M (–4, 8), Y(19, 0)
14).
Page 4
Study
Guide
and Review
The
coordinates
of Y are (41, 14).
The coordinates of X are (–27, 16).
21. M (–4, 8), Y(19, 0)
22. HIKING Carol and Marita are hiking in a state park and decide to take
separate trails. The map of the park is set up on a coordinate grid.
Carol’s location is at the point (7, 13) and Marita is at (3, 5).
a. Find the distance between them.
b. Find the coordinates of the point midway between their locations.
SOLUTION: Let the coordinates of X be (x, y).
Then by the Midpoint Formula,
SOLUTION: a. Use the Distance Formula.
.
Write two equations to find the coordinates of X.
Substitute.
The distance between Carol and Marita is
b. Use the Midpoint Formula.
The coordinates of X are (–27, 16).
22. HIKING Carol and Marita are hiking in a state park and decide to take
separate trails. The map of the park is set up on a coordinate grid.
Carol’s location is at the point (7, 13) and Marita is at (3, 5).
a. Find the distance between them.
b. Find the coordinates of the point midway between their locations.
SOLUTION: a. Use the Distance Formula.
or about 8.9 units
Substitute.
The midpoint of the locations of Carol and Marita is (5, 9).
For Exercises 23–26, refer to the figure below.
Substitute.
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The distance between Carol and Marita is
Page 5
or about 8.9 units
23. Name the vertex of
7.
26. Name a pair of opposite rays.
Guide and Review
Study
The midpoint of the locations of Carol and Marita is (5, 9).
For Exercises 23–26, refer to the figure below.
23. Name the vertex of
7.
SOLUTION: Here, 7 is same as the angle CGJ. So, the vertex of the angle is G.
24. Write another name for
SOLUTION: Here, 4 is same as the
25. Name the sides of
4.
SOLUTION: Any point on a line determines exactly two opposite rays. Here,
are two opposite rays.
27. SIGNS A sign at West High School has the shape shown. Measure each
of the angles and classify them as right, acute, or obtuse.
SOLUTION: The angles A and B measure 90°. So, they are right angles. The angles E
and C measure more than 90°. So, they are obtuse angles.
The angle D measures less than 90°. So, it is an acute angle.
For Exercises 28–30, refer to the figure below.
CDG.
2.
SOLUTION: Here,
2 is same as the
ACH. Therefore, its sides are
26. Name a pair of opposite rays.
SOLUTION: Any point on a line determines exactly two opposite rays. Here,
are two opposite rays.
27. SIGNS A sign at West High School has the shape shown. Measure each
of the angles and classify them as right, acute, or obtuse.
28. Name an angle supplementary to
TVY.
SOLUTION: Supplementary angles are two angles with measures that have a sum of
180°. Here, TVR is supplementary to TVY.
29. Name a pair of vertical angles with vertex W.
SOLUTION: Vertical angles are two nonadjacent angles formed by two intersecting
lines. Here, QWP and XWV are a pair of vertically opposite angles.
30. If m SXW = 5x – 16, find the value of x so that
SOLUTION: Since
SOLUTION: The angles A and B measure 90°. So, they are right angles. The angles E
and C measure more than 90°. So, they are obtuse angles.
The angle D measures less than 90°. So, it is an acute angle.
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That is, 5x – 16 = 90.
Add 16 to both sides.
5x = 106
Page 6
29. Name a pair of vertical angles with vertex W.
SOLUTION: The angle between the initial position of the arm and the vertical position
will be 90°. So, once it is moved 24°, to reach 90° it has to be moved (90 – 24) or 66° more.
SOLUTION: Vertical
are two nonadjacent angles formed by two intersecting
Study
Guideangles
and Review
lines. Here, QWP and XWV are a pair of vertically opposite angles.
Name each polygon by its number of sides. Then classify it as
convex or concave and regular or irregular.
30. If m SXW = 5x – 16, find the value of x so that
SOLUTION: Since
That is, 5x – 16 = 90.
Add 16 to both sides.
5x = 106
Divide both sides by 5.
x = 21.2
32. SOLUTION: The polygon has 3 sides. So, it is a triangle. No line containing any of the
sides will pass through the interior of the triangle, so it is convex. All of
the sides are congruent, so it is equilateral. All of the angles are
congruent, so it is equiangular. Since the polygon is convex, equilateral,
and equiangular, it is regular. So this is a regular triangle.
31. PARKING The parking arm shown below rests in a horizontal position
and opens to a vertical position. After the arm has moved 24°, how many
more degrees does it have to move so that it is vertical?
33. SOLUTION: The angle between the initial position of the arm and the vertical position
will be 90°. So, once it is moved 24°, to reach 90° it has to be moved (90 – 24) or 66° more.
SOLUTION: The polygon has 12 sides. So, it is a dodecahedron.
Four of the lines containing the sides of the polygon will pass through the
interior of the dodecahedron, so it is concave.
Name each polygon by its number of sides. Then classify it as
convex or concave and regular or irregular.
32. SOLUTION: The polygon has 3 sides. So, it is a triangle. No line containing any of the
sides will pass through the interior of the triangle, so it is convex. All of
the sides are congruent, so it is equilateral. All of the angles are
congruent, so it is equiangular. Since the polygon is convex, equilateral,
and equiangular, it is regular. So this is a regular triangle.
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Only convex polygons can be regular, so this is an irregular
dodecahedron.
5), 7
34. Find the perimeter of quadrilateral ABCD with vertices A(–3, 5), B(0,Page
C(2, 0), and D(–5, 0).
SOLUTION: Study
Guide
andpolygons
Review can be regular, so this is an irregular
Only
convex
dodecahedron.
34. Find the perimeter of quadrilateral ABCD with vertices A(–3, 5), B(0, 5),
C(2, 0), and D(–5, 0).
SOLUTION: Use the Distance Formula
of each side of the quadrilateral.
to find the length
or about 20.8 Therefore, the perimeter of the quadrilateral is
units.
35. PARKS Westside Park received 440 feet of chain-link fencing as a
donation to build an enclosed play area for dogs. The park administrators
need to decide what shape the area should have. They have three
options: (1) a rectangle with length of 100 feet and width of 120 feet, (2)
a square with sides of length 110 feet, or (3) a circle with radius of
approximately 70 feet. Find the areas of all three enclosures and
determine which would provide the largest area for the dogs.
SOLUTION: The area of a rectangle of length l and width w is l × w. So, the area of a
2
rectangle of length 100 feet and width 120 feet is 100(120) = 12,000 ft .
2
The area of a square of side length s is s .
2
2
So, the area of a square of side length 110 feet is (110) = 12,100 ft .
2
The area of a circle of radius r is πr .
So, the area of a square of radius 70 feet is
Therefore, the third option provides the largest area.
Identify each solid. Name the bases, faces, edges, and vertices.
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Page 8
SOLUTION: The base of the pyramid is a square. So, it is a square pyramid. The other
faces are triangles. The base is
The other faces are So, the area of a square of radius 70 feet is
The edges are
Study Guide and Review
Therefore, the third option provides the largest area.
The vertices are A, B, C, D, E, F, G, H, J, K, L, and M .
Identify each solid. Name the bases, faces, edges, and vertices.
Find the surface area and volume of each solid.
38. 36. SOLUTION: The surface area of a cylinder of radius r and height h is given by the
formula,
SOLUTION: The base of the pyramid is a square. So, it is a square pyramid. The other
faces are triangles. The base is
The other faces are Here, the radius of the cylinder is 6 cm and the height is 10 cm. So, the
area is
The edges are
The vertices are A, B, C, D, and X.
2
Therefore, the area of the cylinder is about 603.2 cm .
The volume of a cylinder of radius r and height h is given by the formula,
37. SOLUTION: The bases of the prism are hexagons. So, it is a hexagonal prism. The
other faces are rectangles. The bases are ABCDEF and GHJKLM. The
other faces are
The edges are
Therefore, the volume of the cylinder is about 1131.0 cm .
Here, the radius of the cylinder is 6 cm and the height is 10 cm. So, the
volume is
3
The vertices are A, B, C, D, E, F, G, H, J, K, L, and M .
Find the surface area and volume of each solid.
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38. SOLUTION: 39. Page 9
SOLUTION: The surface area of a pyramid of regular base with an area B, base
2
Here, the base area of the pyramid is 144 in and the height is 8 in. So,
the volume is
3
Study
Guide and
Therefore,
the Review
volume of the cylinder is about 1131.0 cm .
3
Therefore, the volume of the pyramid is 384 in .
39. SOLUTION: The surface area of a pyramid of regular base with an area B, base
perimeter P, and slant height is given by the formula,
40. SOLUTION: The surface area of a cylinder of radius r, slant height , and height h is
given by the formula,
Here, the height of the pyramid is 8 in. and each side of the base is 12 in.
So, the slant height can be found using the Pythagorean Theorem.
Here, the radius of the cylinder is 3 ft, the slant height is 5 ft, and the
height is 4 ft. So, the area is
So, the slant height of the pyramid is 10 in. Then, the area is
2
Therefore, the area of the cone is about 75.4 in .
The volume of a cone of radius r and height h is given by the formula,
2
Therefore, the area of the pyramid is 384 in .
Here, the radius of the cylinder is 3 ft. and the height is 4 ft. So, the
volume is
The volume of a pyramid of base area B and height h is given by the
formula,
2
Here, the base area of the pyramid is 144 in and the height is 8 in. So,
the volume is
3
Therefore, the volume of the cone is about 37.7 ft .
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3
Therefore, the volume of the pyramid is 384 in .
Page 10
Guide and Review
Study
3
Therefore, the volume of the cone is about 37.7 ft .
3
Therefore, the volume of the prism is 36 m .
42. BUILDING Chris is building a trunk like the one shown below. His
design is a square prism. What is the volume of the trunk?
41. SOLUTION: The triangular prism has isosceles triangles as bases and rectangular
lateral faces. The total surface area of the prism is the sum of the areas
of the bases and the lateral faces. The altitude of each base is 4 m and
the length of the base is 6 m.
SOLUTION: The volume of a prism of base area B and height h is given by the
formula,
So, the area of each base is
2
Here, the area of each base of the prism is 3(3) = 9 ft and the height is 2
ft. So, the volume is
Each of the lateral faces has a width of 3 m and the lengths 6 m, 5 m,
and 5 m respectively.
So, the area of the lateral surfaces is
3
Therefore, the volume of the trunk is 18 ft .
Therefore, the total surface area is
The volume of a prism of base area B and height h is given by the
formula,
2
Here, the area of each base of the prism is 12 m and the height is 3 m.
So, the volume is
43. HOCKEY A regulation hockey puck is a cylinder made of vulcanized
rubber 1 inch thick and 3 inches in diameter. Find the surface area and
volume of a hockey puck.
SOLUTION: The surface area of a cylinder of radius r and height h is given by the
formula,
Here, the radius of the hockey puck is 1.5 in. and the height is 1 in. So,
the area is
3
Therefore, the volume of the prism is 36 m .
42. BUILDING Chris is building a trunk like the one shown below. His
design is a square prism. What is the volume of the trunk?
2
Therefore, the surface area of the hockey puck is about 23.6 in .
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The volume of a cylinder of radius r and height h is given by the formula,
Page 11
Guide and Review
Study
3
Therefore, the volume of the trunk is 18 ft .
43. HOCKEY A regulation hockey puck is a cylinder made of vulcanized
rubber 1 inch thick and 3 inches in diameter. Find the surface area and
volume of a hockey puck.
SOLUTION: The surface area of a cylinder of radius r and height h is given by the
formula,
Here, the radius of the hockey puck is 1.5 in. and the height is 1 in. So,
the area is
2
Therefore, the surface area of the hockey puck is about 23.6 in .
The volume of a cylinder of radius r and height h is given by the formula,
Here, the volume is
3
Therefore, the volume of the hockey puck is about 7.1 in .
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