Copyright © by Holt, Rinehart and Winston



Reteach

Developing Formulas for Circles and Regular Polygons

Find the circumference of circle S in which A ( 81( cm2.

Step 1 Use the given area to solve for r.

A ’ πr 2 Area of a circle

81π cm2 ’ πr 2 Substitute 81π for A.

81 cm2 ’ r 2 Divide both sides by π.

9 cm ’ r Take the square root of both sides.

Step 2 Use the value of r to find the circumference.

C ’ 2πr Circumference of a circle

C ’ 2π(9 cm) ’ 18π cm Substitute 9 cm for r and simplify.

Find each measurement.

1. the circumference of circle B 2. the area of circle R in terms of π

[pic] [pic]

3. the area of circle Z in terms of π 4. the circumference of circle T in terms of π

[pic] [pic]

5. the circumference of circle X in 6. the radius of circle Y in which C ’ 18π cm

which A ’ 49π in2

Reteach

Developing Formulas for Circles and Regular Polygons continued

Find the area of a regular hexagon with side length 10 cm.

Step 1 Draw a figure and find the measure of a central angle. Each central

angle measure of a regular n-gon is [pic].

Step 2 Use the tangent ratio to find the apothem. You could also use the

30°-60°-90° ( Thm. in this case.

[pic] Write a tangent ratio.

[pic] Substitute the known values.

[pic] Solve for a.

Step 3 Use the formula to find the area.

[pic]

[pic] [pic] , P ’ 6 ( 10 or 60 cm

A ≈ 259.8 cm2 Simplify.

Find the area of each regular polygon. Round to the nearest tenth.

7. [pic] 8. [pic]

9. a regular hexagon with an apothem of 3 m 10. a regular decagon with a perimeter of 70 ft

4. [pic]

5. m∠K ’ tan−1[pic] 6. 35.5°

Problem Solving

1. 120 ft 2. 154 m

3. 57 ft 4. A

5. H 6. C

7. J

Reading Strategies

1. elevation 2. 21 m

3. 35 ft

Answers for Unit 3

Developing Formulas for Circles and Regular

Polygons

PRACTICE A

1. [pic]aP 2. πd

3. πr2 4. A ’ 25π ft2

5. A ’ 100π in2 6. C ’ 18π cm

7. C ’ 26π mi 8. 12 m

9. 2 km 10. 256π yd2

11. 314.2 in2; 530.9 in2; 572.6 in2

12. A ’ 27.6 ft2 13. A ’ 1086 mm2

14. s ’ 4 in.

Practice B

1. A ’ 625π m2 2. A ’ 4a2π in2

3. C ’ (2x + 2y)π yd 4. C ’ 1200π mi

5. r ( ( cm 6. d ( (2x ( 2) km

7. 353.0 mm2; 248.8 mm2; 471.4 mm2

8. 0.01 cent/mm2; 0.04 cent/mm2; 0.05 cent/mm2

9. the nickel

10. 2833 nickels; $141.65

11. A ≈ 1122.4 in2 12. A ≈ 85.6 m2

Practice C

1. Possible answer: By the definition of the center of a regular polygon,[pic], so (ABC is isosceles. By the Isosceles Triangle Theorem, ∠ACB ” ∠ABC. By the definition of apothem, ∠ADC and ∠ADB are right angles so ∠ADC ” ∠ADB by the Right Angle Congruence Theorem. [pic] by the reflexive property, so (ADC ” (ADB by AAS. By CPCTC, ∠CAD ” ∠BAD and [pic]. Thus, by the definition of angle bisector and segment bisector, [pic] bisects ∠BAC

and [pic].

2. A ≈ 2.828 units2; P ≈ 6.123 units

3. A ≈ 3.037 units2; P ≈ 6.231 units

4. A ≈ 3.139 units2; P ≈ 6.282 units

5. A ≈ 3.142 units2; P ≈ 6.283 units

6. The ratio of the perimeter to the area approaches 2.

7. Possible answer: As the number of sides in a regular polygon increases, the polygon gets closer to the shape of a circle, the distance from the center to a vertex gets closer to a radius, and the perimeter gets closer to a circumference. In a circle with a radius of 1 unit, the area is π units2 and the circumference is 2π units. So the ratio of the circumference to the area is 2.

8. 5 or 7

9. 3 10. 10

Reteach

1. C ’ 6 cm 2. A ’ 25π m2

3. A ’ 121π ft2 4. C ’ 20π in.

5. C ’ 14π in. 6. r ’ 9 cm

7. A ’ 695.3 cm2 8. A ’ 58.1 in2

9. A ’ 31.2 m2 10. A ’ 377.0 ft2

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|Circumference and Area of Circles |

|A circle with diameter d and radius r has |

|circumference C ’ πd or C ’ 2πr. |

|A circle with radius r has area A ’ πr2. |

[pic]

|Area of Regular Polygons |

|The area of a regular |

|polygon with apothem |

|a and perimeter P |

|is [pic] |

The apothem is

the distance from

the center to a side.

The center is equidistant from the vertices.

[pic]

A central angle has its vertex at the center. This central angle measure is

[pic]

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