Download 2013-14 Semester 2 Practice Final

GeometrySemester2InstructionalMaterials
‐
2013‐2014
Geometry Semester 2
Instructional Materials for the WCSD Math Common Finals
The Instructional Materials are for student and teacher use and are aligned
to the Math Common Final blueprint for this course. When used as test
practice, success on the Instructional Materials does not guarantee success
on the district math common final.
Students can use these Instructional Materials to become familiar with the
format and language used on the district common finals. Familiarity with
standards vocabulary and interaction with the types of problems included
in the Instructional Materials can result in less anxiety on the part of the
students.
Teachers can use the Instructional Materials in conjunction with the course
guides to ensure that instruction and content is aligned with what will be
assessed. The Instructional Materials are not representative of the depth
or full range of learning that should occur in the classroom
Note from Earl. Many of the solutions in this document use techniques presented in
the Geometry Handbook, which is available on the www.mathguy.us website. If you
have trouble following any of the techniques used, try looking in the handbook for
pages that deal with the issue you are struggling with.
I solve the problems in this sample test using the quickest method available in most
cases. Occasionally, I also make comments about some of the math involved in an
effort to enhance your understanding of what is going on in the problem. You may
have learned different techniques in your classroom. Use whichever techniques work
for you.
Finally, if there is a conflict between the content of this document and what you have
learned in class, your teacher should be the primary source for how any problem should
be solved. 1 | P a g e GeometrySemester2InstructionalMaterials
‐
MultipleChoice:Identifythechoicethatbestcompletesthestatementoranswersthe
question.Figuresarenotnecessarilydrawntoscale.
1. The ratios of the areas of square A to square B is . If the area of square B is 100
what is the length of a side of square A?
A. 4
C. 10
B. 8
,
D. 64
For this kind of problem, we set up a proportion, with the ratio of the areas on the other. side and the actual areas 16
25
on one 100
Cross multiply the two fractions to get: 16 ∙ 100
25 . Then: 64. We are not yet finished because the problem asks for the length of the side of Square . For this, we use the area formula, where represents the length of a side: 64, which we solve to obtain . AnswerB
2. What is the scale factor for the dilation of ∆
to image ∆ ′ ′ ′ ?
2
A.
B. 1
C. 2
D. 3
To determine the scale factor of a dilation from the origin, pick corresponding points on the two figures and divide the ‐ and ‐values of the image by those of the pre‐image. If the two values that result are the same, you have most likely determined the scale factor without error. I have selected Points 1, 4 and ′
3, 12 for this purpose. 3
1
12
4
2 | P a g e GeometrySemester2InstructionalMaterials
‐
Since the calculated scale factors are both , I can conclude with confidence that AnswerD is correct. 3. Apply the dilation : , → 4 , 4 to the polygon with the given vertices. Name the
coordinates of the image points.
A.
8, 4 ,
16, 4 ,
16, 12
8, 4 ,
16, 4 ,
16, 12
B.
C.
0.5, 0.25 ,
1, 0.25 ,
1, 0.75
0.5, 0.25 ,
1, 0.25 ,
1, 0.75
D.
Determine the coordinates of the points; then apply the dilation, as follows: 2, 1 4 ∙ 2, 4 ∙ 1
8, 4 4, 1 4 ∙ 4, 4 ∙ 1
16, 4 4, 3 4 ∙ 4, 4 ∙
3
16,
12 AnswerA 4. The pair of triangles is similar. What is the value for
A.
18
B.
16
C.
13
D.
3
?
To help with this problem, let’s assign some letters to the vertices of the two triangles. ~∆
by following the angle congruences from angle to angle Then, we note that ∆
(∠ ≅ ∠ ; ∠ ≅ ∠ ; ∠ ≅ ∠ ). is the measure of side sides, we have: 9
6
12
⇒ 108
, so we want it in our proportion. Then, using corresponding 6 ⇒ AnswerA The following page contains information about the similarity theorems from the Geometry Handbook available at www.mathguy.us/BySubject/Geometry.php. 3 | P a g e GeometrySemester2InstructionalMaterials
‐
Geometry Similar Triangles The following theorems present conditions under which triangles are similar. Side‐Angle‐Side (SAS) Similarity SAS similarity requires the proportionality of two sides and the congruence of the angle between those sides. Note that there is no such thing as SSA similarity; the congruent angle must be between the two proportional sides. Side‐Side‐Side (SSS) Similarity SSS similarity requires the proportionality of all three sides. If all of the sides are proportional, then all of the angles must be congruent. AA similarity requires the congruence of two angles and the side between those angles. Angle‐‐Angle (AA) Similarity Similar Triangle Parts In similar triangles,  Corresponding sides are proportional.  Corresponding angles are congruent. Establishing the proper names for similar triangles is crucial to line up corresponding vertices. In the picture above, we can say: ∆
∆
~∆
~∆
or∆
or∆
~∆
~∆
or∆
or∆
~∆
~∆
or All of these are correct because they match corresponding parts in the naming. Each of these similarities implies the following relationships between parts of the two triangles: ∠ ≅ ∠ and∠ ≅ ∠ and∠ ≅ ∠ 4 | P a g e GeometrySemester2InstructionalMaterials
‐
5. In the figure, ∆
A.
6
B.
7
C.
16
D.
20
~∆
. What is the length of side
?
Note the angle congruences that result from the parallel lines: ∠ ≅ ∠ and ∠ ≅ ∠
Then, we note that ∆
~∆
from the AA Similarity Theorem. Then, using corresponding sides, we have the following proportion: 16
4
8
4
⇒ 80
5
16
Finally, we want to determine: AB
32 ⇒ 112
4
8
4∙ 7
8
16 ⇒ 7 . AnswerD 6. Two triangles are similar and the ratio of each pair of corresponding sides is 2: 1 . Which
statement regarding the two triangles is not true?
A. Their areas have a ratio of 4: 1
B. The scale factor is a ratio of 2: 1
C. Their perimeters have a ratio of 2: 1
D. Their corresponding angles have a ratio of 2: 1
Some thoughts: If the ratio of side lengths is : in a pair of similar figures. Then:  The measures of the angles in the two figures are the same.  The ratio of any one‐dimensional measurements is : .  The ratio of any two‐dimensional measurements (e.g., areas) is :  The ratio of any three‐dimensional measurements (e.g., volumes) is : Relating these rules to this problem, we note that the scale factor is one‐dimensional, so A, B, and C are all true. D is false because the angle measures in the triangles are the same. AnswerD 5 | P a g e GeometrySemester2InstructionalMaterials
‐
7. What is the value of
A.
39
B.
54
C.
63
D.
90
?
To help with this problem, let’s assign some letters to the vertices of the two triangles. Then, we note that ∆
~∆
. is the measure of side , so we want it in our proportion. Using corresponding sides, we have: ⇒ 9
21
9
27
⇒ 30 ∙ 27
9 ⇒ AnswerD 8. In ∆
, is the midpoint of
, is the midpoint of
Given the following, what is the perimeter of ∆
?

8.2

4.3

6
A. 11.4
B. 15.5
, and R is the midpoint of
.
.
.
C. 18.4
D. 18.5
The Perimeter of ∆
following: 1
∙
2
. Since points , and are midpoints, we have the is: 1
∙6
2
3 4.3 (given) 8.2 3
The perimeter then is: 4.3
8.2
.
AnswerB 6 | P a g e GeometrySemester2InstructionalMaterials
‐
9. In the figure, ∆
is a right triangle and ∆
A.
20
B.
40
C.
2√10
D.
10√20
~∆
. What is the length of
?
There are special rules for these “three‐triangle problems,” which may be learned as words instead of formulas. I list all three here because the problem on the final may involve different parts of the three triangles. The height squared The left side squared The right side squared = the product of: = the product of: = the product of: the two parts of the base the part of the base below it and the entire base the part of the base below it and the entire base ∙
∙
Using the first column of this table, we have: ∙
∙
5∙8
or √
Then, taking the square root of both sides of the equation, we get: 40 √
. AnswerC
10. In the right triangle, and
What is the length of side
A.
2
B.
4
C.
2√3
D.
3√2
In a ˚
˚
represent unknown side lengths.
?
˚ Triangle, the proportions of the sides are: : √ : for short side : long side : hypotenuse respectively. In this problem, we are given the short side and asked to calculate the hypotenuse. The length of the hypotenuse is two times the length of the short side. So, we have: 2∙2
. AnswerB 7 | P a g e GeometrySemester2InstructionalMaterials
‐
11. In the figure, what is the distance a ball travels when thrown from second base to home
plate?
A. 90
B. 180
C. 90√2
D. 2√90
In a ˚
˚
˚ Triangle, the proportions of the sides are: : : √ for side : side : hypotenuse respectively. In this problem, we are given the two sides and asked to calculate the hypotenuse. The length of the hypotenuse is √2 times the size of the length of a side. So, we have: √2 ∙ 90
√ feet. AnswerC
12. What is cos ° in the triangle?
A. 40
41
C.
41
9
9
41
D.
41
40
B.
SOH‐CAH‐TOA sin
In this problem: cos °
cos
tan
AnswerA 8 | P a g e GeometrySemester2InstructionalMaterials
‐
13. What is the value of
A.
16 cos 35°
B.
cos 35°
16
C.
16 sin 35°
D.
16
sin 35°
in the triangle?
Using SOH‐CAH‐TOA, we first note that relative to the angle of 35˚, we are given the opposite side and the hypotenuse. The trig function that uses these two sides is the sine function. So, sin 35°
⇒ sin 35°
16
⇒ x
16
sin 35°
AnswerD
14. What is the measure of angle A in the triangle, rounded to the nearest degree?
A.
∠
35°
B.
∠
44°
C.
∠
46°
D.
∠
72°
Using SOH‐CAH‐TOA, we first note that relative to angle , we are given the adjacent side and the hypotenuse. The trig function that uses these two sides is the cosine function. So, cos
⇒ cos
1 7
~
10
° (using a calculator) AnswerC
9 | P a g e GeometrySemester2InstructionalMaterials
‐
15. A person is standing at ground level with the base of the Empire State Building in New
York City. The angle formed by the ground and a line segment from his position to the top
of the building is 48.4°. The height of the Empire State Building is 1472 feet. Find the
distance that he is standing from the base of the Empire State Building to the nearest foot.
A. 8
C. 1968
B. 1307
D. 2217
We need a picture for this one, so I drew the one to the right. Using SOH‐CAH‐TOA, we first note that relative to the 48.4° angle, we are given the adjacent side and the opposite side. The trig function that uses these two sides is the tangent function. So, tan 48.4°
,
⇒ ,
. °
~ ,
AnswerB
16. In the figure, ∆
is a right triangle with the hypotenuse
. Given the segments
lengths of
4,
2, and
1, what is cos ?
√5,
A. √5
B.
12√5
25
C.
√5
5
D.
5√5
12
I have added the given dimensions to the diagram, as You should on any problem that gives lengths or angles. Next, notice that calculating the cosine of angle B requires us to use only ∆
√
at the right. Using SOH‐CAH‐TOA, cos
cos
√
1
√5
1
∙
√5
√5 √5
, but we do not want to leave a radical in the denominator, so, √5
5
AnswerC
10 | P a g e GeometrySemester2InstructionalMaterials
‐
17. Using a string a student decided to determine the diameter of a large trash can. If a string
60
long will wrap around the trash can, what is the approximate diameter of the
trash can?
A. 25
C. 9.55
B. 19.1
D. 9
Drawing a picture of the situation helps – see the one at right. Next, we need the formula for circumference in terms of diameter: 60, so we get: In this problem, 60
⇒ ~
. inches AnswerB
18. A rectangle is inscribed in a circle as shown below. Find the exact circumference of the
circle.
A. 13 B. 17 C. √119 D. 169 Draw the diameter in the picture to the right. 5
Calculate using the Pythagorean Theorem: Taking the square root of both sides gives: 12
25
144
169. 13. Next, we need the formula for circumference in terms of diameter: In this problem, ∙ 13
13, so we get: mm AnswerA
11 | P a g e GeometrySemester2InstructionalMaterials
‐
19. What is the area of a regular hexagon with an apothem of 10
side length of
√
and a
?
A. 200√3
B. 300√3
C. 600√3
D. 600
We need the formula for area of a regular polygon based on the apothem and side length: , where: In this problem, length of apothem, 10 and 10 40√3
6
perimeter of the regular polygon. √
∙
2
√ cm 40√3, so we get: AnswerA
20. Given that the side of the regular pentagon is 8 and that the apothem is approximately
5.51 , what is the approximate area of the shaded triangle?
A. 20
B. 22
C. 64
D. 110
We want the area of the shaded region, which is a triangle. So, we will use the formula for area of a triangle. , where: length of the base, height of the triangle. Note that the base of the triangle is the length of a side of the polygon (8 ft.). Also, the height of the triangle is the length of the apothem (5.51 ft.). In this problem, 8 and 8 5.51 ~
5.51, so we get: ft2 AnswerB
12 | P a g e GeometrySemester2InstructionalMaterials
‐
21. If two pieces of ice have the same volume, the one with the greater surface area will melt
faster because more of its surface area is exposed to the air, which is warmer than the ice.
Two pieces of ice labeled and B have the same volume. Each piece of ice is shaped like a
rectangular prism. Which piece of ice melts the fastest?
A. Piece
melts the fastest.
B. Piece
melts the fastest.
C. They take the same amount of time to melt.
D. A relationship cannot be determined
First check that the volumes are equal: ∙
∙
288 ft3 32 ∙ 3 ∙ 3
18 ∙ 8 ∙ 2
288 ft3 Since the volumes are equal, we can check the surface areas: 2
2
2 18 8
2
2 32 3
2 18 2
2 32 3
2 3 3
402 ft2 392 ft2 2 8 2
The one that melts fastest is the one with the greatest surface area, which is Piece . AnswerA 22. What best describes the cross section shown on the cube?
A. square
C. trapezoid
B. triangle
D.
rectangle
The cross section shown is a quadrilateral. It appears to have two parallel sides (the top and bottom) and two non‐parallel sides (the left and right). This describes a trapezoid. AnswerC The following page contains information about the various kinds of quadrilaterals from the Geometry Handbook available at www.mathguy.us/BySubject/Geometry.php. 13 | P a g e GeometrySemester2InstructionalMaterials
‐
Geometry Figures of Quadrilaterals Kite Trapezoid
Isosceles Trapezoid
 2 consecutive pairs of congruent sides  1 pair of congruent opposite angles  Diagonals perpendicular  1 pair of parallel sides (called “bases”)  Angles on the same “side” of the bases are supplementary  1 pair of parallel sides  Congruent legs  2 pair of congruent base angles  Diagonals congruent Parallelogram Rectangle





 Parallelogram with all angles congruent (i.e., right angles)  Diagonals congruent Both pairs of opposite sides parallel Both pairs of opposite sides congruent Both pairs of opposite angles congruent Consecutive angles supplementary Diagonals bisect each other Rhombus Square
 Parallelogram with all sides congruent  Diagonals perpendicular  Each diagonal bisects a pair of opposite angles  Both a Rhombus and a Rectangle  All angles congruent (i.e., right angles)  All sides congruent 14 | P a g e GeometrySemester2InstructionalMaterials
‐
23. A layered cake is a solid of revolution. Which of the following is the drawing of a twodimensional shape and an axis of rotation that could form the cake?
A.
C.
B.
D.
Rotating a straight edge creates a circular shape in three dimensions. We want half of the shape that we see looking at a cross‐section of the cake in 2 dimensions. AnswerD 24. What is the volume of the cylinder in terms of x ?
A.
15
12
B.
15
12
C.
45
36
D.
75
120
489
We need the formula for volume of a cylinder based on the radius of a base and the height: , where: 3 and In this problem, 3
radius of a base, 5
5
4
4, so we get: 9
height of the cylinder. 5
4
cm3 AnswerC
The following page contains formulas for the volumes and surface areas for various solids from the Geometry Handbook available at www.mathguy.us/BySubject/Geometry.php. 15 | P a g e GeometrySemester2InstructionalMaterials
‐
Geometry Summary of Surface Area and Volume Formulas – 3D Shapes Shape Figure Surface Area Volume Sphere Right Cylinder Cone Square Pyramid ∙
Rectangular Prism Cube General Right Prism 16 | P a g e GeometrySemester2InstructionalMaterials
‐
25. What is the height of a square pyramid that has a side length of 13
volume of 1521
?
A.
3
C.
27
B.
9
D.
39
and a
We need the formula for volume of a square pyramid based on the length of a side of the base and the height: , where: length of a side of the base, 13 and In this problem, 1,521, so we get: ⇒ 1,521
,
Solving for , we get: height of the pyramid. 13
∙
⇒ 1,521
ft. AnswerC
26. A food manufacturer sells yogurt in cone shaped cups with the dimensions shown. To the
nearest tenth, how many fluid ounces of yogurt does the cup hold?
(Hint: 1
0.034 )
A. 0.6 B. 5.7 C. 17.1 D. 22.8 We need the formula for volume of a cone based on the radius of the base and the height: , where: In this problem, radius of a base, 8
2
4
10
4 and 160
height of the cone. 10, so we get: ~167.55 cm3 In fluid ounces, this is ~167.55 0.034 ~ . fl. oz. AnswerB
17 | P a g e GeometrySemester2InstructionalMaterials
‐
27.
What is the volume of the sphere in terms of
A. 36
B. 48
C. 288
D. 864
?
We need the formula for volume of a sphere based on the radius of the sphere: , where: In this problem, radius of the sphere. 6, so we get: 6
216
ft3 AnswerC
28. You want to design a cylindrical container for oatmeal that has a volume of 77
. You
also want the height of the container to be 2 times the radius. To the nearest tenth, what
should the radius of the container be?
A.
2.3
B.
2.9
C.
3.0
D.
3.1
We need the formula for volume of a cylinder based on the radius of a base and the height: , where: In this problem, 77, 2
77
77
radius of a base, 2
and 2
height of the cylinder. 2 , so we get: , and solving for gives: ~ . inches AnswerA
18 | P a g e GeometrySemester2InstructionalMaterials
‐
29. Find the volume of the composite figure. Round your answer to the nearest tenth.
A.
245
B.
441
C.
539
D.
735
The volume of the composite figure is the sum of the volumes of the square pyramid on top, and the rectangular prism on the bottom. For the square pyramid: 7
For the rectangular Prism: 7 7 9
Total volume is: 98
441
6
m3 98 m3 441 m3 AnswerC 30. What is the ratio of the volumes of the two cubes?
The cubes have edges of lengths 3 inches and 12 inches.
A.
1: 4
B. 1: 16
C. 1: 64
D. 1: 256
Some thoughts: If the ratio of side lengths is : in a pair of similar figures. Then:  The measures of the angles in the two figures are the same.  The ratio of any one‐dimensional measurements is : .  The ratio of any two‐dimensional measurements (e.g., areas) is :  The ratio of any three‐dimensional measurements (e.g., volumes) is : In this problem, the ratios of the lengths is 3: 12, which simplifies to 1: 4. Then, the ratio of the volumes is: 1 : 4
:
Note that the ratio of the surface areas is: 1 : 4
this on the real final. AnswerC
1: 16. Watch out for a question like 19 | P a g e GeometrySemester2InstructionalMaterials
‐
31. Given mAC = mBC and ∠
A.
18, mBC
136°
B.
9.2, mBC
88°
C.
18, mBC
88°
D.
9.2, mBC
136°
is a central angle, what is the value of
and mBC ?
Let’s add a couple of indicators (orange dashes) for measures that are equal. Now, let’s work with the arc and angle shown. Recall that the measure of an inscribed angle is half the measure of the arc it subtends. This tells us that: ∠
⇒ 44
5
Solving for x, we get: To determine Since 88°
Solving for , note that: 5
2
88°) 360° (the whole circle) 360° or 88°
°. 2 , we can convert the above equation to: gives us 5
(Note also that 88° and 2 ⇒ 88
2∙
360° AnswerA 32. What is the measure of angle x ?
A. 50°
B. 35°
C. 25°
D. 5°
An angle with a vertex outside the circle is half the difference of its subtended arcs. 60
10
° AnswerC The following pages contain information about circles, angles and their subtended arcs from the Geometry Handbook available at www.mathguy.us/BySubject/Geometry.php.
20 | P a g e GeometrySemester2InstructionalMaterials
‐
Geometry Parts of Circles Center – the middle of the circle. All points on the circle are the same distance from the center. Radius – a line segment with one endpoint at the center and the other endpoint on the circle. The term “radius” is also used to refer to the distance from the center to the points on the circle. Diameter – a line segment with endpoints on the circle that passes through the center. Arc – a path along a circle. Minor Arc – a path along the circle that is less than 180⁰. Major Arc – a path along the circle that is greater than 180⁰. Semicircle – a path along a circle that equals 180⁰. Sector – a region inside a circle that is bounded by two radii and an arc. Secant Line – a line that intersects the circle in exactly two points. Tangent Line– a line that intersects the circle in exactly one point. Chord – a line segment with endpoints on the circle that does not pass through the center. 21 | P a g e GeometrySemester2InstructionalMaterials
‐
Geometry Angles and Circles Central Angle Inscribed Angle Vertex inside the circle Vertex outside the circle ∠
∠
∠
∠
Tangent on one side Tangents on two sides ∠
∠
22 | P a g e GeometrySemester2InstructionalMaterials
‐
33. What is the measure of RS ?
A. 54°
B. 38°
C. 32°
D. 27°
An angle with a vertex inside the circle is half the sum of its subtended arcs. 54°
76°
⇒ 108°
76°
⇒ °
AnswerC 34. What is the measure of the inscribed angle
, if the ray is tangent to the circle?
A. 140°
B. 110°
C. 70°
Arc measure is: °
°
D. 55°
°
This is a special case of the measure of an inscribed angle. Recall that the measure of an inscribed angle is half the measure of the arc it subtends. The arc we care about is the one for which the measure is not given in the problem. We must calculate it, as show above in magenta, to be °. Then, 220°
° AnswerB 23 | P a g e GeometrySemester2InstructionalMaterials
‐
35. What is the length of the minor arc AB in the circle with a radius of 36
?
A. 9 B. 6 36
C. 3 D. 1.5 First, let’s get the circumference of the whole circle: 2
36
72 Next, let’s find out what part of the whole circle is represented by the arc. The arc is 30° out of a total of 360° in a complete circle. This is of the whole circle. Multiply these two values together to get the length of minor arc ∙ 72
cm. AnswerB . 36. What is the area of a circular pool that has a circumference of 100
A. 10 C. 100
B. 50 D. 2500
?
First, find the radius of the pool from the circumference. ⇒ 100
2
⇒ 50 ft. Next, calculate the area of the circle from the formula: ft2 50
AnswerD The following page contains information about arc length and sector area from the Geometry Handbook available at www.mathguy.us/BySubject/Geometry.php. 24 | P a g e GeometrySemester2InstructionalMaterials
‐
Geometry Circle Lengths and Areas Circumference and Area ∙ is the circumference (i.e., the perimeter) of the circle. is the area of the circle. where: is the radius of the circle. Length of an Arc on a Circle A common problem in the geometry of circles is to measure the length of an arc on a circle. Definition: An arc is a segment along the circumference of a circle. ∙ where: ∠AB is the measure (in degrees) of the arc. Note that this is also the measure of the central angle ∠
. is the circumference of the circle. Area of a Sector of a Circle Another common problem in the geometry of circles is to measure the area of a sector a circle. Definition: A sector is a region in a circle that is bounded by two radii and an arc of the circle. ∙ where: ∠AB is the measure (in degrees) of the arc. Note that this is also the measure of the central angle ∠
. is the area of the circle. 25 | P a g e GeometrySemester2InstructionalMaterials
‐
37. The diameter of a circular pizza pan is 18
. Two-thirds of the pizza is eaten by your
friends. What is the approximate area of the pizza pan that is covered by the remaining
pizza? (Assume that the diameter of the pan and the diameter of the pizza are the same.)
A. 170
C. 54
B. 85
D. 27
If your friends eat of a pizza, there is left for you. 9 inches. Note also that the radius of the pizza pan is The area of the whole pizza is: 9
81 in2 The portion of the pizza left for you, then, is: ∙ 81
in2 27 ~
AnswerB A sector of a circle has an area of 75
radius of the circle?
38.
and an arc measure of 120°. What is the
A. 10.6
C. 5
B. 21.2
D. 15
First, let’s find out what part of the whole circle is represented by the arc. The arc is 120° out of a total of 360° in a complete circle. This is of the whole circle. So, the Area of the circle must be three times the size of the sector: 3 ∙ 75
225 . The radius of the circle can then be determined from the area formula: ⇒ 225
⇒ 225
⇒ cm AnswerD 26 | P a g e GeometrySemester2InstructionalMaterials
‐
39. Given the circle inscribed in the square with side length 12. What is the probability that the
point lies inside the circle, if a point is chosen at random inside the square?
A.
B.
C.
1
D.
4
4
The probability is the ratio of the shaded area to the total area. 6 Note that the radius of the circle is half the length of the side of the square: AnswerD 40. Find the probability that a point chosen at random in the trapezoid shown lies in either of
the shaded regions. Round your answer to the nearest hundredth.
A. 0.61
B. 0.56
C. 0.39
D. 0.35
The probability is the ratio of the shaded area to the total area. I have added a couple of measurements to the above diagram to make the calculations easier. ~ .
AnswerC The following page contains information regarding perimeters and areas of 2‐dimensional shapes from the Geometry Handbook available at www.mathguy.us/BySubject/Geometry.php. One interesting thing to keep in mind is that a square is also a kite. So if you are given the diagonal of a square and asked to find the area, use the formula 27 | P a g e GeometrySemester2InstructionalMaterials
‐
Geometry Summary of Perimeter and Area Formulas – 2D Shapes Shape Figure Perimeter Area Kite ,
,
Trapezoid ,
,
b ,b
bases h height Parallelogram ,
Rectangle ,
Rhombus ,
Regular Polygon Square ,
∙ Circle 28 | P a g e GeometrySemester2InstructionalMaterials
‐
A grab bag contains 7 football cards and 3 basketball cards. An experiment consists of
taking one card out of the bag, then selecting another card. What is the probability of
selecting a football card, replacing it, and then selecting a basketball card?
41.
A. 0.23
C. 0.21
B. 0.09
D. 0.49
Let’s look at the probabilities for a draw. Notice that the probabilities don NOT change after the first draw because you replace the card chosen. Cards Probability Football Cards 7 7/10
0.7 Basketball Cards 3 3/10
0.3 Total Cards 10 Notice that the two draws are independent of each other. Two events are independent when neither one affects the other. When events are independent, we can multiply the probabilities of each event to get the overall probability. .
0.7 ∙ 0.3
.
AnswerC 42. A bag contains hair ribbons for a spirit rally. The bag contains 5 black
ribbons and 7 green ribbons. Lila selects a ribbon at random, then Jessica
selects a ribbon at random from the remaining ribbons. Find the probability
that both events and occur.
:
:
A.
7
33
B.
.
35
144
C.
.
35
132
D.
5
22
Let’s look at the probabilities for each draw. Notice that the probabilities change after the first draw because you do not replace the ribbon chosen. Lila Draw Jessica Draw Black Ribbons 5 4 Green Ribbons 7 7 Total Ribbons 12 11 Notice that the probabilities change between the two draws because they are not independent. We still multiply the two resulting probabilities. ∙
∙
AnswerC 29 | P a g e GeometrySemester2InstructionalMaterials
‐
43. The table shows the distribution of male and female students and left- and right-handed
students in the math club. Find the probability that a female student selected at random is
left-handed. Which is the correct answer as a fraction in simplest form?
Left-handed
Right-handed
Male
2
35
Female
6
36
3
4
A.
1
7
B.
Total
42
1
6
C.
D.
6
79
We are given that the student is female, so we confine ourselves to that row. I added the total column so we can do our calculation: |
6
42
AnswerB
44.The table shows the distribution of the labor force in a city in the year 2000. Suppose that a
worker is selected at random. Find the probability of randomly selecting a worker in the
Industry field given that the worker is female. Which is the correct answer as a decimal
rounded to the nearest thousandth?
Agriculture Industry Services Total Male 3,132 25,056 50,112 Female 667 8,004 57,362 66,033 A.
0.141
0.242
B.
C.
0.121
D.
0.312
We are given that the worker is female, so we confine ourselves to that row. I added the total column so we can do our calculation: |
8,004
~ .
66,033
AnswerC 30 | P a g e GeometrySemester2InstructionalMaterials
‐
45. Events
and are independent. Find the missing probability.
_______ (call it )
0.3
0.06
A. 0.7
C. 0.2
B. 0.24
D. 0.3
If two events are independent, then ∙
.
For this problem, ∙
⇒ 0.06
.
Solving for , we get . .
∙ 0.3 AnswerC |
0.43 and
46. If
0.89, find
.
A. 0.51
C. 0.11
B. 0.48
D. 0.38
|
The key formula to use here is: |
⇒ 0.89
0.43
⇒ 0.89 0.43 ~ .
AnswerD 47. The sections on a spinner are numbered from 1 through 8. If the probability of landing on a
given section is the same for all the sections, what is the probability of spinning a number
less than 4 or greater than 7 in a single spin?
A. 1
2
C.
3
8
1
8
D.
5
8
B.
The successes are values 4 or 7. So, 1, 2, 3, 8 . There are 4 possible successes out of a set of 8 total possibilities when spinning. The resulting probability of success is: AnswerA 31 | P a g e GeometrySemester2InstructionalMaterials
‐
48. Given ∆
, a student constructed point as shown. Next the student will draw a circle
. Which statement is true and why?
with center and radius
A. Circle will be inscribed in ∆
bisectors of ∆
.
because point
is the intersection of two angle
B. Circle will be circumscribed about ∆
angle bisectors of ∆
.
because point
is the intersection of two
C. Circle will be circumscribed about ∆
perpendicular bisectors of sides of ∆
.
because point
is the intersection of two
D. Circle will be inscribed in ∆
because point
perpendicular bisectors of sides of ∆
.
is the intersection of two
The center of the circle (O) in this problem is the intersection of angle bisectors of the triangle. This point of intersection is called the incenter of the triangle. The incenter is also the center of the circle inscribed in the triangle. The key words in the paragraph above are “angle bisectors” and “inscribed.” These are the words in AnswerA.
The following page contains information regarding other centers of triangles with which the student should be familiar. The page is from the Geometry Handbook available at: www.mathguy.us/BySubject/Geometry.php. 32 | P a g e GeometrySemester2InstructionalMaterials
‐
Geometry Centers of Triangles The following are all points which can be considered the center of a triangle. Centroid (Medians) The centroid is the intersection of the three medians of a triangle. A median is a line segment drawn from a vertex to the midpoint of the line opposite the vertex. 

The centroid is located 2/3 of the way from a vertex to the opposite side. That is, the distance from a vertex to the centroid is double the length from the centroid to the midpoint of the opposite line. The medians of a triangle create 6 inner triangles of equal area. Orthocenter (Altitudes) The orthocenter is the intersection of the three altitudes of a triangle. An altitude is a line segment drawn from a vertex to a point on the opposite side (extended, if necessary) that is perpendicular to that side. 


In an acute triangle, the orthocenter is inside the triangle. In a right triangle, the orthocenter is the right angle vertex. In an obtuse triangle, the orthocenter is outside the triangle. Circumcenter (Perpendicular Bisectors) The circumcenter is the intersection of the perpendicular bisectors of the three sides of the triangle. A perpendicular bisector is a line which both bisects the side and is perpendicular to the side. The circumcenter is also the center of the circle circumscribed about the triangle. 


In an acute triangle, the circumcenter is inside the triangle. In a right triangle, the circumcenter is the midpoint of the hypotenuse. In an obtuse triangle, the circumcenter is outside the triangle. Euler Line: Interestingly, the centroid, orthocenter and circumcenter of a triangle are collinear (i.e., lie on the same line, which is called the Euler Line). Incenter (Angle Bisectors) The incenter is the intersection of the angle bisectors of the three angles of the triangle. An angle bisector cuts an angle into two congruent angles, each of which is half the measure of the original angle. The incenter is also the center of the circle inscribed in the triangle. 33 | P a g e GeometrySemester2InstructionalMaterials
‐
49. Find the angle measures of
.
A.
∠
∠
∠
∠
34°
54°
146°
126°
C.
∠
∠
∠
∠
35°
134°
145°
54°
B.
∠
∠
∠
∠
71°
54°
109°
126°
D.
∠
∠
∠
∠
71°
50°
109°
130°
Opposite angles of a quadrilateral inscribed in a circle add to 180°. So, we have: 2
3
3
7
180 ⇒ 5
10
180. 34. Solving for , we get: 34 into the measures of each angle to get our solution: Then, substitute ∠
2
3 °
2 34
3°
71° ∠
3
7 °
3 34
7°
109° ∠
4
10 °
4 34
10 °
126° 126°
54° These two angles add to 180°. Finally, ∠
180°
∠
180°
AnswerB 34 | P a g e GeometrySemester2InstructionalMaterials
‐
50. Which circle is inscribed in the triangle?
A. Circle
B. Circle
C. Circle
D. All of the above
Inscribed means inside the triangle. So that would be Circle . AnswerA
Geometry Semester 2 Instructional Material Answers
1.
B
11.
C
21.
A
31. A
41.
C
2.
D
12.
A
22.
C
32. C
42.
C
3.
A
13.
D
23.
D
33. C
43.
B
4.
A
14.
C
24.
C
34. B
44.
C
5.
D
15.
B
25.
C
35. B
45.
C
6.
D
16.
C
26.
B
36. D
46.
D
7.
D
17.
B
27.
C
37. B
47.
A
8.
B
18.
A
28.
A
38. D
48.
A
9.
C
19.
A
29.
C
39. D
49.
B
10. B
20.
B
30.
C
40. C
50.
A
35 | P a g e