LESSON 2: CIRCLES AND SECTORS ANSWERSmsdoughtymath.weebly.com/uploads/3/9/0/8/39087729/g-6... ·...

High School: Circles and Parabolas

Copyright © 2015 Pearson Education, Inc. 48

ANSWERSLESSON 2: CIRCLES AND SECTORS

ANSWERS

G.C.1 1. B similar

G.C.1 2. Similar figures have the same shape and proportional size differences. This is true of circles in which the radius is used to scale the figure larger or smaller.

G.C.5 3. 66.99 units2

G.C.5 4. 9.77 units2

G.C.5 5. 153.86 cm2

G.C.5 6. 76.93 cm2

G.C.5 7. 51.29 cm2

Challenge Problem

G.C.5 8. a. The surface area of the cone without the base is equal to the area of the remaining 2

3 of the circle.

2

36 75 362 2π ≈ . in

b. The circumference of the base of the cone will equal 2

3 of the circle circumference.

2

312 25 12( • ) .π ≈ in.

You can use the circumference to determine the cone base diameter and radius.

25 12

8.

π≈ in.

So, the diameter is 8 in. and the radius is 4 in.

c. Since you have the cone base radius, you can calculate the cone base area. π42 ≈ 50.24 in2



ANSWERSLESSON 3: DEGREES AND RADIANS

ANSWERS

G.C.5 1. C 2π radians

G.C.5 2. B

G.C.5 3. D 180°

G.C.5 4. or 0.79 radians

G.C.5 5. or 2.36 radians

G.C.5 6. 30°

G.C.5 7. 120°

G.C.5 8. 4.19 radians

Challenge Problem

G.C.5 9.

2radians

π

π4

3π4

θ =

=

≈

sr307

4 286. radians



ANSWERSLESSON 4: AREA OF A SECTOR

ANSWERS

G.C.5 1. C 14.13 m2

G.C.5 2. B 6.28 m2

G.C.5 3. 150.72 cm2

G.C.5 4. 58.88 in2

G.C.5 5. 133.97 cm2

G.C.5 6. 2.36 m2

G.C.5 7. The total pizza area is π72 ≈ 153.96 in2.

One piece is a sector with a central angle of 72°.

The surface area of each piece is 153.93 • 72

360 ≈ 30.79 in2.

Challenge Problem

G.C.5 8. The total pizza area is π(8 in.)2 = 200.96 in2.

The larger piece has a central angle of 65°, so the surface area is

200 9665

36036 28 2. • .≈ in .

One of the smaller pieces has a surface area of 200 9659360

32 94 2. • .≈ in .

The larger piece has about 3.34 in2 more surface area than one of the smaller pieces.



ANSWERSLESSON 5: ANGLES IN CIRCLES

ANSWERS

G.C.2 1. D 230°

G.C.2 2. C 90°

G.C.2 3. A 76°

G.C.2 4. B 40°

G.C.2 5. A 25°

G.C.2 6. B 50°

G.C.2 7. D 120°

G.C.2 8. C 50°

Challenge Problem

G.C.2 9. Here are the steps I followed to find the measure of ∠CFE. I know that ∠DEF is a right angle since DF is the diameter. Therefore, based on ∆DEF, you can find the measure of ∠CFE = 43°.

Here is another method to determine the measure of ∠CFE .

∠ABD forms a straight angle with ∠DBE, so the two angles are supplementary. Thus, the measure of ∠DBE = 104°.

You know the measures of two of the three angles for ∆DBE, so you can determine the measure of the third angle: the measure of ∠BED = 29°.

You know the measure of ∠DEF = 90°, since it spans a diameter. Thus, ∠BED and ∠BEF are complementary, so you can compute the measure of ∠BEF = 61°.

∠FBE is vertical with ∠ABD, so the two angles are equal in measure: the measure of ∠FBE = 76°.

Now you know the measures of two of the three angles of ∆BFE, so you can compute the measure of ∠BFE = 43°.

∠BFE and ∠CFE are the same angle since points C and B are both on the same diameter. So, the measure of ∠CFE = 43º.

A

DE

F

B76° 76°

43°

104°

29°61°47°

C



ANSWERSLESSON 6: CHORDS AND TANGENTS

ANSWERS

G.C.2 1. ∠CAD = 90º

G.C.2 2. ∠ACD = 60º

G.C.2 3. B 35°

G.C.2 4. D 13 units

G.C.2 5. A 17 units

G.C.2 6. ∠BCA = 70º

G.C.2 7. FG is 10 cm long.

G.C.2 8. ∠ADC = 130º

Challenge Problem

G.C.2 9. Here are the steps I followed to find the measure of ∠FEC.

As I found the measures of new angles, I recorded them in this diagram.

∠DBC makes a straight angle with the given 140° angle, so the measure of ∠DBC = 40°.

∠CDB is a right angle since DB is tangent at point D. I now have two of the three angle measures for ∆CBD. Therefore, the measure of ∠DCB = 50°.

∠DCB is vertical with ∠ECF, so the angle measures are equal: ∠ECF = 50°.

CE is congruent to CF since both are radii of the same circle, so ∆CEF is isosceles.

Thus, the measures of ∠FEC and ∠EFC must be equal. Since I know ∠ECF = 50°, then the measures of ∠FEC and ∠EFC must each be half of 130°, or 65°.

Therefore, ∠FEC = 65°.

140°

65°

65°50° 50°

40°

G

D

EB

C

F



ANSWERSLESSON 7: CIRCLES AND QUADRILATERALS

ANSWERS

G.C.3 1. B cyclic

G.C.3 2. A supplementary (add up to 180°)

G.C.3 3. ∠ADC = 76°

G.C.3 4. ∠BDA = 30°

G.C.3 5. To find the measure of ∠BDA, I started by finding the measure of the third angle in ∆ABC.

180 = 110 + 40 + ∠BCA

∠BCA = 30º

∠BCA ≅ ∠BDA because of the theorem stating that if two inscribed angles of a circle intercept the same arc or arcs of equal measure, then the inscribed angles have equal measure. Therefore, ∠BDA = 30º.

G.C.3 6. ∠DAP = 50º

G.C.3 7. D isosceles

G.C.3 8. B 29°

G.C.3 9. ∠BPD = 150°

Challenge Problem

G.C.3 10. The diagonals intersect and make vertical angles, so you can fill in all of the angle measures around that intersection.

With this information, you can then fill in all of the other angle measures shown, using the fact that the sum of the angle measures of a triangle add up to 180°.

In order to show that the quadrilateral is cyclic, you need to show that opposite angles add up to 180°.

∠CAE = 68° and ∠CDE = 112°: 68 + 112 = 180

∠ACD = 81° and ∠AED = 99°: 81 + 99 = 180

Both pairs of opposite angle measures of the quadrilateral add up to 180°. Therefore, the quadrilateral is cyclic.

40°28°

110°

110°

70° 70°

70°29°

71° 41°

42°39°

D

C

E

B

A



ANSWERSLESSON 8: DESCRIBING CIRCLES

ANSWERS

G.GPE.1 1. B (–3, 2)

G.GPE.1 2. B 5 units

G.GMD.4 3. C Parallel to the base of the cone

G.GPE.1 4. A (x + 3)2 + (y – 8)2 = 9

G.GPE.1 5. The center is at (–2, 4).

G.GPE.1 6. The radius = 4 units

G.GPE.1 7.

G.GPE.1 8. The center is at (3, 7).


G.GPE.1 10.

2

4

6

8

10

–2

–4

4 62–2–4–6–8–10 x

y

(–2, 4) 4

2

4

6

8

10

–24 6 8 102–2–4–6 x

y

12

14

(3, 7)

7



ANSWERSLESSON 8: DESCRIBING CIRCLES

G.GPE.1 11. The center is at (–4, 2).


G.GPE.1 13.

2

4

6

8

–2

–4

42–2–4–6–8–10 x

y

(–4, 2)

5

Challenge Problem

G.GPE.1 14. a. This equation is different from the equation of a circle because the two denominators are different. Instead of indicating the same radius in both horizontal and vertical directions, this equation indicates different radii horizontally and vertically.

b. This equation describes an ellipse with a center at (3, –2).

c.

2

4

–2

–4

–6

4 6 8 102–2–4–6 x

y



ANSWERSLESSON 9: CIRCLE: AS AN EQUATION

ANSWERS

G.GPE.1 1. Center: (3, –5) Radius: 4 units

G.GPE.1 2. B x2 + 12x + y2 – 8y + 16 = 0

E

–12 –8 –4 –2–14 –10 –6 2 x

y

–4

–2

4

8

12

2

6

10

G.GPE.1 3. (x + 4)2 + (y + 2)2 = 25

G.GPE.1 4. D x2 – 10x + y2 + 6y – 2 = 0

G.GPE.1 5. (x + 1)2 + (y – 8)2 = 16 x2 + 2x + 1 + y2 – 16y + 64 = 16 x2 + 2x + y2 – 16y + 49 = 0

G.GPE.1 6. x2 + y2 + 6x – 12y + 41 = 0 x2 + 6x + y2 – 12y = –41 x2 + 6x + 9 + y2 – 12y + 36 = –41 + 9 + 36 (x + 3)2 + (y – 6)2 = 22

G.GPE.1 7. x2 + y2 – 14x + 4y + 28 = 0 x2 – 14x + y2 + 4y = –28 x2 – 14x + 49 + y2 + 4y + 4 = –28 + 49 + 4 (x – 7)2 + (y + 2)2 = 52

G.GPE.1 8. Center: (2, –6) Radius: 4 units



ANSWERSLESSON 9: CIRCLE: AS AN EQUATION

G.GPE.1 9.2

–2

–4

–6

–8

–10

4 6 8 102–2–4–6 x

y

(2, –6)

4

G.GPE.1 10. Center: (–6, 9) Radius: 8 units

G.GPE.1 11.

4

6

8

10

–4

2

12

14

16

18

20–2–6–8–10–12–14 x

y

(–6, 9)8

Challenge Problem

G.GPE.1 12. This equation does not represent a circle. You can tell just from the equation, because it has the 3xy term. Because of this term, the equation will not simplify into an equation of a circle after completing the square. Here is the graph of this equation. It makes a rotated ellipse.

4x2 + 3xy + y2 + 14x + 17y = 16

10 20

10

–10

–10–20

–20

–30

x

y



ANSWERSLESSON 10: DESCRIBING PARABOLAS

ANSWERS

G.GMD.4 1. A Parabola

G.GPE.2 2. B directrix, focus

G.GPE.2 3. C point C

G.GPE.2 4. The directrix is y = –5, or line l.

G.GPE.2 5. D FO and DO are the same length.

G.GPE.2 6. The coordinates of the focus are (0, 0).

G.GPE.2 7. The focus is at (0, 2) and the directrix is y = –2.

2

4

6

–2

–4

–6

4 6 8 102–2–4–6–8–10 x

y

8

10

Directrix:y = –2

Focus:(0, 2)

18

y = x2



ANSWERSLESSON 10: DESCRIBING PARABOLAS

G.GPE.2 8. The focus is at (0, –5.5) and the directrix is y = –6.5.

21 4 5 6 73

2

1

–2

–3

–5

–7

–8

–1–2 –1–3–4–5–6–7

–4

–6

x

y

Directrix: y = –6.5

Focus: (0, –5.5)

12

y = x2 – 6

Challenge Problem

G.GPE.2 9. The focus is at (0, ) and the directrix is .−12732

y = –129

32



ANSWERSLESSON 11: PARABOLA: AS AN EQUATION

ANSWERS

G.GPE.2 1. A

G.GPE.2 2.

G.GPE.2 3.

G.GPE.2 4.

G.GPE.2 5. B Focus at (0, 4) and directrix of y = –4

G.GPE.2 6. The focus is at (3, 0). The directrix is x = 1.

G.GPE.2 7. The focus is at . The directrix is .

Challenge Problem

G.GPE.2 8. This equation describes a sideways parabola with vertex at (4, 2).

04 8 12 x

y

2 6 10

–8

–4

4

8

12

–2

–6

2

6

10

y x= +18

22

y x= 18

2

x y= +112

32

y x= +16

3 52 .

04712

, –

y = − 4912



ANSWERSLESSON 12: CIRCLES AND PARABOLAS

ANSWERS

G.GMD.4 G.MG.1

1. C = 24π cm A = 144π cm2

G.GMD.4 G.MG.1

2. C = 16π cm A = 64π cm2

G.GMD.4 G.MG.1

3. C = 8π cm A = 16π cm2

G.GMD.4 G.MG.1

4. The circumference decreases proportionally with how far up the circle the cross section is. In this case, the circumference of the base is 75.4 cm, the circumference of the middle is 50.26 cm, and the circumference of the top is 25.13 cm (or 24 π cm,

16 π cm, and 8π cm), which follows the 1, ,23

13

pattern of distances from the top of the cone.

The area changes from 144π cm2 to 64π cm2 to 16π cm2. These areas do not follow

a proportional pattern: 144

642 25

64

164

ππ

butππ

= =.

The pattern is derived from the ratios squared. The middle area is 23

49

2

= of the

base area and the top area is 13

19

2

= of the base area.

G.GPE.1 5. This sketch is of a cross section of the scenario. Since the parabola must be parallel with the side of the cone, you can determine that these triangles are similar.

4

6

8

10

4 6 8 10 12 14 16

2

0

12

14

16

20 x

y

A

B



ANSWERS LESSON 12: CIRCLES AND PARABOLAS

G.GPE.1 6. Because the triangles are similar, the length of AB is proportional to the large triangle created from the cone. The slant height of the cone can be computed using the Pythagorean Theorem.

7 5 15 56 25 225 281 25 16 772 2. . . .+ = + = ≈

The length of AB is 23

16 77 11 18• . .≈ cm.

G.GPE.1 7. x2 + (y – 7.5)2 = 56.25

Challenge Problem

G.GMD.4 8. The different conic sections are determined by the angle at which the plane intersects the double cone. The circle only occurs when the plane is parallel to the base of the double cone. The parabola only occurs if the plane is parallel to the side of the cone. And the hyperbola occurs when the plane is at a steeper angle than parallel to the side of the cone (closer to perpendicular to the base of the cone).



LESSON 13: PUTTING IT TOGETHER ANSWERS

ANSWERS

G.C.2 2.Word of Phrase Definition Example

Central angle

An angle with its vertex at the center of the circle and two radii as its sides

You can measure central angles with degrees or radians.

s

Arc length s = radius r

r

= 1 radian

Circle Conic Section Definition: The intersection of a right circular cone with a plane such that the plane is perpendicular to the altitude of the cone

Points Definition: A collection of points on a plane that have a fixed distance r (the radius) from a fixed point C (the center) in the plane

Sectors A region bounded by two radii and an arc of a circle

s

r

Arc Length The distance along an arc for a particular angle or portion of the circumference of a circle

The arc length of the above sector is s = r • θ (in radians).

(continues)




G.C.2 2. (continued)

Word of Phrase Definition Example

Chord A line segment with a circle that has its end points on the circle

MN is a chord.

O

T

QN

M

P

S

Diameter The longest chord of a circle

x

A

B

Tangent A line that intersects a circle at just one point

PS is a tangent.

O

T

QN

M

P

S

(continues)





Word of Phrase Definition Example

Secant A line that intersects a circle at two points

PQ is a secant.

O

T

QN

M

P

S



ANSWERSLESSON 13: PUTTING IT TOGETHER

G.C.2 3.Theorem

(Lesson Number) DefinitionExplain the Theorem

Using Words or Diagrams

Arc Length-Radius Theorem (Lesson 3)

In sectors with a common central angle θ, the arc length of the sector is proportional to the radius.

The arc length of this sector is determined by both the radius and the central angle. s = r • θ (in radians)

s

r

Area of a Sector of a Circle (Lesson 4)

The area of a sector with central angle θ in degrees is

A r= π 2

360•

θ.

The area of a sector with central angle θ in radians is

Ar=

2

2θ

.

The radius is squared and then multiplied by pi and the portion of the circle that the central angle sweeps. If the sector is 210º, or ,

and the radius is 3, then the area of

the sector is 976

2 5 25• .π π units2÷ = .

When calculated with the degree

formula, π π units2• • .9210360

5 25= .

Inscribed Angle Theorem (Lesson 5)

An inscribed angle of a circle has a measure exactly half of the angle measure of the intercepted arc.

A

a

b

m

m

B

C

n

nx y

In this diagram, ∠a = 2∠b because ∠b is an inscribed angle that intercepts arc AB while ∠a is the central angle.

(continues)

76π



EXERCISESLESSON 13: PUTTING IT TOGETHER


Theorem (Lesson Number) Definition

Explain the Theorem Using Words or Diagrams

Tangents From an External Point Theorem (Lesson 6)

Tangent segments to a circle from the same external point are congruent.

O

T

P

S

PT is congruent to PS.

Circumscribed Angle Theorem (Lesson 6)

When a circumscribed angle of a circle shares a common chord with the central angle, the two angles are supplementary.

In the above tangents diagram, ∠TPS is supplementary to ∠T0S because they share the common chord TS.

Cyclic Quadrilateral Angles Theorem (Lesson 7)

A quadrilateral is cyclic if and only if its opposite angles are supplementary.

B

F

A

C

D

∠ABC + ∠ADC = 180º and ∠BAD + ∠BCD = 180º




G.GPE.1 G.GPE.2 G.GMD.4

4.Circle

x00

1

2

3

4

5

6

1 2 3 4 5

y

(2, 3)

2

Parabola

–12 –8 –4–10 –6 –2 4 8 122 6 10

–4

4

8

y

x

–6

–2

2

6

Focus: (0, 4)

Directrix: y = –4

y x= 116

2

General Formula for a Circle

With a center at (h, k) and a radius of r, the circle has the formula (x – h)2 + (y – k)2 = r2.

In polynomial form, it is x2 + Dx + y2 + Ey + F = 0, which can be converted to center-radius form by completing the square.

General Formula for a Parabola

With a focus at (0, p) and a directrix of y = –p and the parabola opening up or down,

the parabola has the formula yp

x= 14

2 .

If the parabola opens left or right, then x and y are switched such that the y-term is squared and the x-term is not squared.

Circle as a Conic Section

The intersection of a right circular cone with a plane such that the plane is perpendicular to the altitude of the cone.

Parabola as a Conic Section

The intersection of a right circular cone with a plane such that the plane is parallel to the side of the cone.

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