Name ________________________________________ Date __________________ Class__________________
LESSON
9-2
Reteach
Developing Formulas for Circles and Regular Polygons
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.
Find the circumference of circle S in which A = 81π cm2.
Step 1
Use the given area to solve for r.
A = πr 2
2
81π cm = πr
2
81 cm = r
Area of a circle
2
Substitute 81π for A.
2
Divide both sides by π.
9 cm = r
Step 2
Take the square root of both sides.
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.
2. the area of circle R in terms of π
1. the circumference of circle B
_________________________________________
3. the area of circle Z in terms of π
________________________________________
4. the circumference of circle T in terms of π
_________________________________________
________________________________________
6. the radius of circle Y in which C = 18π cm
5. the circumference of circle X in
which A = 49π in2
_________________________________________
________________________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
9-14
Holt Geometry
Name ________________________________________ Date __________________ Class__________________
LESSON
9-2
Reteach
Developing Formulas for Circles and Regular Polygons continued
Area of Regular Polygons
The area of a regular
polygon with apothem
a and perimeter P
1
is A = aP.
2
The apothem is
the distance from
the center to a side.
The center is
equidistant from
the vertices.
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
360°
angle measure of a regular n-gon is
.
n
A central angle has its
vertex at the center. This
central angle measure is
360°
= 60°.
n
Step 2
Use the tangent ratio to find the apothem. You could also use the
30°-60°-90° U Thm. in this case.
leg opposite 30° angle
tan 30° =
Write a tangent ratio.
leg adjacent to 30° angle
5 cm
a
5 cm
a=
tan 30°
tan 30° =
Step 3
Substitute the known values.
Solve for a.
Use the formula to find the area.
1
A = aP
2
1⎛ 5 ⎞
60
A= ⎜
2 ⎝ tan 30° ⎟⎠
a=
A ≈ 259.8 cm2
5
, P = 6 × 10 or 60 cm
tan 30°
Simplify.
Find the area of each regular polygon. Round to the nearest tenth.
8.
7.
_________________________________________
9. a regular hexagon with an apothem of 3 m
________________________________________
10. a regular decagon with a perimeter of 70 ft
_________________________________________
________________________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
9-15
Holt Geometry
5. JL; JL; Two-Transversal Proportionality
Theorem
6.
Practice C
1. Possible answer: By the definition of the
center of a regular polygon, AC ≅ AB , so
UABC 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.
AD ≅ AD by the reflexive property, so
UADC ≅ UADB by AAS. By CPCTC,
∠CAD ≅ ∠BAD and CD ≅ DB . Thus, by
the definition of angle bisector and
segment bisector, AD bisects ∠BAC and
BC .
PK
MN
18
=
=
PK + KL TS
21
7. 18 cm
1
1
(PK)(MN) = (18 cm)(18 cm) =
2
2
2
162 cm
8. A =
Problem Solving
1. 21.8 mm
2. about $263.84
3. 12 ft by 22 ft
4. 26.3 cm2
5. C
6. F
7. B
8. G
2. A ≈ 2.828 units2; P ≈ 6.123 units
3. A ≈ 3.037 units2; P ≈ 6.231 units
Reading Strategies
1. A = 108 cm2
2. A = 35 yd2
4. A ≈ 3.139 units2; P ≈ 6.282 units
3. b = 10 ft
4. A = 150 mm2
5. A ≈ 3.142 units2; P ≈ 6.283 units
5. h = 12 in.
6. b = 4 cm
6. The ratio of the perimeter to the area
approaches 2.
LESSON 9-2
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.
Practice A
1.
1
aP
2
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
8. 5 or 7
12. A = 27.6 ft2
9. 3
13. A = 1086 mm2
14. s = 4 in.
Reteach
Practice B
1. A = 625π m2
2. A = 4a2π in2
3. C = (2x + 2y)π yd
4. C = 1200π mi
5. r = π cm
6. d = (2x + 2) km
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
7. 353.0 mm2; 248.8 mm2; 471.4 mm2
8. 0.01 cent/mm2; 0.04 cent/mm2; 0.05
cent/mm2
10. A = 377.0 ft2
Challenge
1. A0 =
9. the nickel
10. 2833 nickels; $141.65
11. A ≈ 1122.4 in2
10. 10
3 2
s
4
12. A ≈ 85.6 m2
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
A17
Holt Geometry