Areas of Parallelograms and Triangles
GEOMETRY LESSON 10-1
(For help, go to Lesson 1-9.)
Find the area of each figure.
1. a square with 5 cm sides
2. a rectangle with base 4 in.
and height 7 in.
3. a 4.6 m-by-2.5 m rectangle
4. a rectangle with length 3 ft
and width 1 ft
2
Each rectangle is divided into two congruent triangles.
Find the area of each triangle.
5.
6.
7.
Check Skills You’ll Need
10-1
Areas of Parallelograms and Triangles
GEOMETRY LESSON 10-1
Solutions
1. A = s2 = 52 = 25 cm2
2. A = bh = (4)(7) = 28 in.2
3. A = bh = (4.6)(2.5) = 11.5 m2
4. A = bh = (3)( 1 ) = 3 ft2
2
2
5. The rectangle consists of 4 rows of 3 squares in each row, or 12 square units.
By Postulate 1-9, congruent figures have congruent areas, so the area of
each triangle is half the area of the rectangle, or 6 units2.
6. The rectangle consists of 2 rows of 2 squares in each row, or 4 square units.
By Postulate 1-9, congruent figures have congruent areas, so the area of
each triangle is half the area of the rectangle, or 2 units2.
7. The rectangle consists of 4 rows of 4 squares in each row, or 16 square units.
By Postulate 1-9, congruent figures have congruent areas, so the area of
each triangle is half the area of the rectangle, or 8 units2.
10-1
Areas of Parallelograms and Triangles
GEOMETRY LESSON 10-1
Find the area of the parallelogram.
You are given two sides with lengths 12 m and 10.5 m and an altitude
that measures 8 m to the side that measures 12 m. Choose the side
with a corresponding height to use as a base.
A = bh
Area of a parallelogram
A = 12(8)
Substitute 12 for b and 8 for h.
A = 96
Simplify.
The area of the parallelogram is 96 m2.
Quick Check
10-1
Areas of Parallelograms and Triangles
GEOMETRY LESSON 10-1
A parallelogram has 9-in. and 18-in. sides. The height
corresponding to the 9-in. base is 15 in. Find the height corresponding
to the 18-in. base.
Find the area of the parallelogram using the 9-in. base and
its corresponding 15-in. height.
A = bh
Area of a parallelogram
A = 9(15)
Substitute 9 for b and 15 for h.
A = 135
Simplify.
The area of the parallelogram is 135 in.2
10-1
Areas of Parallelograms and Triangles
GEOMETRY LESSON 10-1
(continued)
Use the area 135 in.2 to find the height to the 18-in. base.
A = bh
Area of a parallelogram
135 = 18h
Substitute 135 for A and 18 for b.
135 = h
18
Divide each side by 18.
7.5 = h
Simplify.
The height corresponding to the 18-in. base is 7.5 in.
Quick Check
10-1
Areas of Parallelograms and Triangles
GEOMETRY LESSON 10-1
Find the area of
1
bh
2
1
A = (30)(13)
2
A=
XYZ.
Area of a triangle
A = 195
Substitute 30 for b and 13 for h.
Simplify.
XYZ has area 195 cm2.
Quick Check
10-1
Areas of Parallelograms and Triangles
GEOMETRY LESSON 10-1
The front of a garage is a square 15 ft on each side with a
triangular roof above the square. The height of the triangular roof is
10.6 ft. To the nearest hundred, how much force is exerted by an
80 mi/h wind blowing directly against the front of the garage? Use the
formula F = 0.004Av2.
Draw the front of the garage, and then
use the area formulas for rectangles and
triangles to find the area of the front of the garage.
Area of the square: bh = 152 = 225 ft2
Area of the triangular roof: 1 bh = 1 (15)(10.6)
= 79.5
ft2
2
2
The total area of the front of the garage is
225 + 79.5 = 304.5 ft2.
10-1
Areas of Parallelograms and Triangles
GEOMETRY LESSON 10-1
(continued)
Find the force of the wind against the front of the garage.
F = 0.004Av2
Use the formula for force.
F = 0.004(304.5)(80)2
Substitute 304.5 for A and 80 for v.
A = 7795.2
Simplify.
A
Round to the nearest hundred.
7800
An 80 mi/h wind exerts a force of about 7800 lb against the front of
the garage.
Quick Check
10-1
Areas of Parallelograms and Triangles
GEOMETRY LESSON 10-1
1. Find the area of the parallelogram.
4. Find the area of
150 ft2
RST.
15 m2
2. Find the area of
XYZW with
vertices X(–5, –3), Y(–2, 3),
Z(2, 3) and W(–1, –3).
24 square units
3. A parallelogram has 6-cm and 8-cm sides.
The height corresponding to the 8-cm base
is 4.5 cm. Find the height corresponding
to the 6-cm base.
6 cm
10-1
5. A rectangular flag is divided into
four regions by its diagonals.
Two of the regions are shaded.
Find the total area of the
shaded regions.
187 in.2
Areas of Trapezoids, Rhombuses, and Kites
GEOMETRY LESSON 10-2
(For help, go to Lesson 10-1.)
Write the formula for the area of each type of figure.
1. rectangle
2. a triangle
Find the area of each trapezoid by using the formulas for area
of a rectangle and area of a triangle.
3.
4.
5.
Check Skills You’ll Need
10-2
Areas of Trapezoids, Rhombuses, and Kites
GEOMETRY LESSON 10-2
Solutions
1. A = bh
2. A = 1 bh
2
3. Draw a segment from S perpendicular to UT. This forms a rectangle and a triangle.
The area A of the triangle is 1 bh = 1 (1)(2) = 1. The area A of the rectangle is
2
2
bh = (4)(2) = 8. By Theorem 1-10, the area of a region is the sum of
the area of the nonoverlapping parts. So, add the two areas: 1 + 8 = 9 units2.
10-2
Areas of Trapezoids, Rhombuses, and Kites
GEOMETRY LESSON 10-2
Solutions (continued)
4. Draw two segments, one from M perpendicular to CB and the other from K
perpendicular to CB. This forms two triangles and a rectangle between them.
The area A of the triangle on the left is 1 bh = 1 (1)(2) = 1. The area A of
2
2
1
1
the triangle on the right is bh = (2)(2) = 2. The area A of the rectangle is
2
2
bh = (2)(2) = 4. By Theorem 1–10, the area of a region is the sum of the area
of the nonoverlapping parts. So, add the three areas: 1 + 2 + 4 = 7 units2.
5. Draw two segments, one from A perpendicular to CD and the other from B
perpendicular to CD. This forms two triangles and a rectangle between them.
The area A of the triangle on the left is
1
1
bh = (2)(3) = 3. The area A of the
2
2
triangle on the right is 1 bh = 1 (3)(3) = 4.5. The area A of the rectangle is
2
2
bh = (2)(3) = 6. By Theorem 1–10, the area of a region is the sum of the area
of the nonoverlapping parts. So, add the three areas: 3 + 4.5 + 6 = 13.5 units2.
10-2
Areas of Trapezoids, Rhombuses, and Kites
GEOMETRY LESSON 10-2
A car window is shaped like the trapezoid shown.
Find the area of the window.
A=
1
h(b1 + b2)
2
1
Area of a trapezoid
A = 2 (18)(20 + 36)
Substitute 18 for h, 20 for b1, and 36 for b2.
A = 504
Simplify.
The area of the car window is 504 in.2
10-2
Quick Check
Areas of Trapezoids, Rhombuses, and Kites
GEOMETRY LESSON 10-2
Find the area of trapezoid ABCD.
Draw an altitude from vertex B to DC that divides trapezoid ABCD
into a rectangle and a right triangle.
Because opposite sides of rectangle ABXD
are congruent, DX = 11 ft and
XC = 16 ft – 11 ft = 5 ft.
10-2
Areas of Trapezoids, Rhombuses, and Kites
GEOMETRY LESSON 10-2
(continued)
By the Pythagorean Theorem, BX 2 + XC2 = BC2, so BX 2 = 132 – 52 = 144.
Taking the square root, BX = 12 ft. You may remember
that 5, 12, 13 is a Pythagorean triple.
1
A = 2 h(b1 + b2)
A=
1
(12)(11 + 16)
2
A = 162
Use the trapezoid area formula.
Substitute 12 for h, 11 for b1, and 16 for b2.
Simplify.
The area of trapezoid ABCD is 162 ft2.
Quick Check
10-2
Areas of Trapezoids, Rhombuses, and Kites
GEOMETRY LESSON 10-2
Find the area of kite XYZW.
Find the lengths of the diagonals of kite XYZW.
XZ = d1 = 3 + 3 = 6 and YW = d2 = 1 + 4 = 5
A = 1 d1d2
2
A = 1 (6)(5)
2
A = 15
Use the formula for the area of a kite.
Substitute 6 for d1 and 5 for d2.
Simplify.
Quick Check
The area of kite XYZW is 15 cm2.
10-2
Areas of Trapezoids, Rhombuses, and Kites
GEOMETRY LESSON 10-2
Find the area of rhombus RSTU.
To find the area, you need to know the lengths of both diagonals.
Draw diagonal SU, and label the intersection of the diagonals point X.
10-2
Areas of Trapezoids, Rhombuses, and Kites
GEOMETRY LESSON 10-2
(continued)
SXT is a right triangle because the diagonals of
a rhombus are perpendicular.
The diagonals of a rhombus bisect each other, so TX = 12 ft.
You can use the Pythagorean triple 5, 12, 13 or the Pythagorean Theorem
to conclude that SX = 5 ft.
SU = 10 ft because the diagonals of a rhombus bisect each other.
1
A = d1d2
Area of a rhombus
2
1
A = (24)(10)
Substitute 24 for d1 and 10 for d2.
2
A = 120
Simplify.
The area of rhombus RSTU is 120 ft2.
10-2
Quick Check
Areas of Trapezoids, Rhombuses, and Kites
GEOMETRY LESSON 10-2
1. Find the area of a trapezoid with bases 3 cm and 19 cm
and height 9 cm. 99 cm2
2. Find the area of a trapezoid in a coordinate plane with
26 square units
vertices at (1, 1), (1, 6), (5, 9), and (5, 1).
Find the area of each figure in Exercises 3–5.
Leave your answers in simplest radical form.
3. trapezoid ABCD
4. kite with diagonals 20 m and 10
94.5
2 m long
5. rhombus MNOP
3 in.2
100
840 mm2
10-2
2 m2
Areas of Regular Polygons
GEOMETRY LESSON 10-3
(For help, go to Lesson 8-2.)
Find the area of each regular polygon. If your answer involves
a radical, leave it in simplest radical form.
1.
2.
3.
Find the perimeter of the regular polygon.
4. a hexagon with sides of 4 in.
5. an octagon with sides of 2
3 cm
Check Skills You’ll Need
10-3
Areas of Regular Polygons
GEOMETRY LESSON 10-3
Solutions
1. The triangle is equilateral and equiangular, so each of its angles is 60°.
The altitude divides the triangle into two 30°-60°-90° triangles. Since the short
leg of the 30°-60°-90° triangle is 5cm, the long leg, which is the altitude of the
equilateral triangle, is 5 3 cm. The base is 10 cm and the height is 5
The area A = 1 bh = 1 (10)(5 3) = 25 3 cm2.
2
2
3 cm.
2. The diagonal is 10 cm and divides the square into two 45°-45°-90° triangles.
The legs are each 10 , or 5 2 ft. The base is 5 2 and the height is 5 2.
2
The area A = bh = (5 2)(5 2) = (25)(2) = 50 ft2.
10-3
Areas of Regular Polygons
GEOMETRY LESSON 10-3
Solutions (continued)
3. The triangle is equilateral and equiangular, so each of its angles is 60°.
The altitude divides the triangle into two 30°-60°-90° triangles. The altitude is
10 m. Since the long leg of the 30°-60°-90° triangle is 10m the short leg is
10
m and the hypotenuse is 20 m. Thus the base of the equilateral triangle
3
3
is 20 m. The area A = 1 bh = 1 ( 20 )(10) = 100 , or 100 3 m2.
3
3
3
2
2
3
4. The perimeter of a polygon is the sum of the lengths of its sides. A regular
hexagon has six sides of the same length. Since each side has length 4 in.,
the perimeter is = (6)(4) = 24 in.
5. The perimeter of a polygon is the sum of the lengths of its sides. A regular
octagon has eight sides of the same length. Since each side has length
2
3 cm, the perimeter is (8)(2
3) = 16
10-3
3 cm.
Areas of Regular Polygons
GEOMETRY LESSON 10-3
A portion of a regular hexagon has an apothem
and radii drawn. Find the measure of each numbered angle.
m
1 = 360 = 60
m
2=2 m
m
2 = 1 (60) = 30
Substitute 60 for m
m
3 = 180 – (90 + 30) = 60
The sum of the measures of the angles
of a triangle is 180.
m
1 = 60, m
6
1
1
Divide 360 by the number of sides.
The apothem bisects the vertex angle of
the isosceles triangle formed by the radii.
2
2 = 30, and m
3 = 60.
10-3
1.
Quick Check
Areas of Regular Polygons
GEOMETRY LESSON 10-3
Find the area of a regular polygon with twenty 12-in. sides
and a 37.9-in. apothem.
p = ns
Find the perimeter.
p = (20)(12) = 240
Substitute 20 for n and 12 for s.
A=
1
ap
2
Area of a regular polygon
A=
1
(37.9)(240)
2
Substitute 37.9 for a and 240 for p.
A = 4548
Simplify.
The area of the polygon is 4548 in.2
10-3
Quick Check
Areas of Regular Polygons
GEOMETRY LESSON 10-3
A library is in the shape of a regular octagon. Each side is
18.0 ft. The radius of the octagon is 23.5 ft. Find the area of the library
to the nearest 10 ft2.
Consecutive radii form an isosceles triangle, as shown below,
so an apothem bisects the side of the octagon.
1
To apply the area formula A = 2 ap,
you need to find a and p.
10-3
Areas of Regular Polygons
GEOMETRY LESSON 10-3
(continued)
Step 1: Find the apothem a.
a2 + (9.0)2 = (23.5)2
a2 + 81 = 552.25
a2 = 471.25
a 21.7
Step 2: Find the perimeter p.
p = ns
p = (8)(18.0) = 144
Pythagorean Theorem
Solve for a.
Find the perimeter.
Substitute 8 for n and 18.0 for s,
and simplify.
10-3
Areas of Regular Polygons
GEOMETRY LESSON 10-3
(continued)
Step 3: Find the area A.
A = 1 ap
2
A
A
1 (21.7)(144)
2
1562.4
Area of a regular polygon
Substitute 21.7 for a and 144 for p.
Simplify.
To the nearest 10 ft2, the area is 1560 ft2.
Quick Check
10-3
Areas of Regular Polygons
GEOMETRY LESSON 10-3
Use the portion of the regular decagon for Exercises 1–3.
1. Find m 1.
36
2. Find m 2.
18
3. Find m 3.
72
4. Find the area of a regular 9-sided figure with a 9.6-cm
apothem and 7-cm side.
302.4 cm2
For Exercises 5 and 6, find the area of each regular polygon.
Leave your answer in simplest radical form.
5.
6.
48
3 m2
6
10-3
3 in.2
Perimeters and Areas of Similar Figures
GEOMETRY LESSON 10-4
(For help, go to Lesson 1-9.)
Find the perimeter and area of each figure.
1.
2.
3.
Find the perimeter and area of each rectangle with the given base
and height.
4. b = 1 cm, h = 3 cm
5. b = 2 cm, h = 6 cm
6. b = 3 cm, h = 9 cm
Check Skills You’ll Need
10-4
Perimeters and Areas of Similar Figures
GEOMETRY LESSON 10-4
Solutions
1. All sides are congruent, so each side measures 7 in. The perimeter is
the sum of all the sides. 4  7 = 28 in.; the area is the square of a side
s. A = s2 = 72 = 49 in.2
2. Two sides measure 4 m and two sides measure 8 m. The perimeter is
the sum of all sides: 2  4 + 2  8 = 2(4 + 8) = 2(12) = 24 m; the area is
the product of the base b and the height h: A = bh = (8)(4) = 32 m2
3. Use the Pythagorean Theorem: a2 + b2 = c2
(6)2 + (8)2 = c2 36 +
64 = c2 c2 = 100 c = 10 cm; the perimeter is the sum of all the
sides: 10 + 8 + 6 = 24 cm; the area is half the product of the base b
and the height h; A = 1 bh = 1 (8)(6) = 24 cm2
2
2
10-4
Perimeters and Areas of Similar Figures
GEOMETRY LESSON 10-4
Solutions (continued)
4. The perimeter is the sum of the sides: 1 + 3 + 1 + 3 = 8 cm; the area is
the product of the base b and the height h: A = bh = (1)(3) = 3 cm2
5. The perimeter is the sum of the sides: 2 + 6 + 2 + 6 = 16 cm; the area
is the product of the base b and height h: A = bh = (2)(6) = 12 cm2
6. The perimeter is the sum of the sides: 3 + 9 + 3 + 9 = 24 cm; the area
is the product of the base b and height h: A = bh = (3)(9) = 27 cm2
10-4
Perimeters and Areas of Similar Figures
GEOMETRY LESSON 10-4
The triangles below are similar. Find the ratio
(larger to smaller) of their perimeters and of their areas.
The shortest side of the triangle to the left has length 4, and the shortest side
of the triangle to the right has length 5.
5
From larger to smaller, the similarity ratio is 4.
By the Perimeters and Areas of Similar Figures Theorem, the ratio of the
2
perimeters is also 5, and the ratio of the areas is 52, or 25.
4
4
16
Quick Check
10-4
Perimeters and Areas of Similar Figures
GEOMETRY LESSON 10-4
The ratio of the lengths of the corresponding sides of two
8
regular octagons is 3. The area of the larger octagon is 320 ft2. Find
the area of the smaller octagon.
All regular octagons are similar.
Because the ratio of the lengths of the corresponding sides of the regular
2
octagons is 8 , the ratio of their areas is 82 , or 64 .
3
64
320
=
9
A
64A = 2880
A = 45
3
9
Write a proportion.
Use the Cross-Product Property.
Divide each side by 64.
The area of the smaller octagon is 45 ft2.
Quick Check
10-4
Perimeters and Areas of Similar Figures
GEOMETRY LESSON 10-4
Benita plants the same crop in two rectangular fields. Each
dimension of the larger field is 3 12 times the dimension of the smaller
field. Seeding the smaller field costs $8. How much money does
seeding the larger field cost?
The similarity ratio of the fields is 3.5 : 1, so the ratio of the areas of the
fields is (3.5)2 : (1)2, or 12.25 : 1.
Because seeding the smaller field costs $8, seeding 12.25 times as much
land costs 12.25($8).
Seeding the larger field costs $98.
Quick Check
10-4
Perimeters and Areas of Similar Figures
GEOMETRY LESSON 10-4
The areas of two similar pentagons are 32 in.2 and 72 in.2
What is their similarity ratio? What is the ratio of their perimeters?
Find the similarity ratio a : b.
a2
32
b2 = 72
The ratio of the areas is a2 : b2.
a2
16
=
2
b
36
Simplify.
a
4
2
=
=
b
6
3
Take the square root.
The similarity ratio is 2 : 3. By the Perimeters and Areas of Similar
Figures Theorem, the ratio of the perimeters is also 2 : 3.
Quick Check
10-4
Perimeters and Areas of Similar Figures
GEOMETRY LESSON 10-4
1. For the similar rectangles, give the ratios (smaller
to larger) of the perimeters and of the areas.
4
16
perimeters: 9 ; areas: 81
2. The triangles are similar. The area of the larger
triangle is 48 ft2. Find the area of the smaller triangle.
27 ft2
3. The similarity ratio of two regular octagons is 5 : 9. The area of the
smaller octagon is 100 in.2 Find the area of the larger octagon. 324 in.2
4. The areas of two equilateral triangles are 27 yd2 and 75 yd2. Find their
similarity ratio and the ratio of their perimeters. 3 : 5; 3 : 5
5. Mulch to cover an 8-ft by 16-ft rectangular garden costs $48. At the
same rate, what would be the cost of mulch to cover a 12-ft by 24-ft
rectangular garden? $108
10-4
Trigonometry and Area
GEOMETRY LESSON 10-5
(For help, go to Lesson 10-3.)
Find the area of each regular polygon.
1.
2.
3.
Check Skills You’ll Need
10-5
Trigonometry and Area
GEOMETRY LESSON 10-5
Solutions
1. The perimeter is 4(6) = 24 m. The area A of a regular polygon is half the
apothem a times the perimeter p: A = 1 ap = 1 (3)(24) = 36 m2
2
2
2. The perimeter is 6(42) = 252 in. The area A of a regular polygon is half
the apothem a times the perimeter p: A = 1 ap = 1 (36)(252) = 4536 in.2
2
2
3. The perimeter is 6(8) = 48 ft. The area A of a regular polygon is half the
apothem a times the perimeter p: A = 1 ap = 1 (7)(48) = 168 ft2
2
10-5
2
Trigonometry and Area
GEOMETRY LESSON 10-5
Find the area of a regular polygon with 10 sides and side
length 12 cm.
Find the perimeter p and apothem a, and then find
the area using the formula A = 1 ap.
2
Because the polygon has 10 sides and each side
is 12 cm long, p = 10 • 12 = 120 cm.
Use trigonometry to find a.
Because the polygon has 10 sides, m


ACB = 360 = 36.
CA and CB are radii, so CA = CB. Therefore,
1
10
ACM
1
BCM by the HL
Theorem, so m ACM = 2 m ACB = 18 and AM = 2 AB = 6.
10-5
Trigonometry and Area
GEOMETRY LESSON 10-5
(continued)
tan 18° =
6
a
Use the tangent ratio.
6
Solve for a.
a = tan 18°
Now substitute into the area formula.
A = 1 ap
2
6 .
1
A = 2 • tan 18° • 120
360
360
18
Substitute for a and p.
A = tan 18°
Simplify.
1107.966073
Use a calculator.
The area is about 1108 cm2.
Quick Check
10-5
Trigonometry and Area
GEOMETRY LESSON 10-5
The radius of a garden in the shape of a regular pentagon is
18 feet. Find the area of the garden.
Find the perimeter p and apothem a, and then find
the area using the formula A = 1 ap.
2
Because the pentagon has 5 sides, m ACB = 360 = 72.
5
CA and CB are radii, so CA = CB. Therefore,
Theorem, so m ACM = 1 m ACB = 36
2
10-5
ACM
BCM by the HL
Trigonometry and Area
GEOMETRY LESSON 10-5
(continued)
Use the cosine ratio
to find a.
cos 36° =
a
18
a = 18(cos 36°)
Use the sine ratio
to find AM.
Use the ratio.
Solve.
Use AM to find p. Because ACM
pentagon is regular, p = 5 • AB.
sin 36° =
AM
18
AM = 18(sin 36°)
BCM, AB = 2 • AM. Because the
So p = 5 • (2 • AM) = 10 • AM = 10 • 18(sin 36°) = 180(sin 36°).
10-5
Trigonometry and Area
GEOMETRY LESSON 10-5
(continued)
Finally, substitute into the area formula A = 1 ap.
2
A=
1
• 18(cos 36°) • 180(sin 36°)
2
Substitute for a and p.
A = 1620(cos 36°) • (sin 36°)
Simplify.
A
Use a calculator.
770.355778
The area of the garden is about 770 ft2.
Quick Check
10-5
Trigonometry and Area
GEOMETRY LESSON 10-5
A triangular park has two sides that measure 200 ft and 300 ft
and form a 65° angle. Find the area of the park to the nearest hundred
square feet.
Use Theorem 9-1: The area of a triangle is one
half the product of the lengths of two sides and the
sine of the included angle.
1
Area = • side length • side length
2
• sine of included angle
Theorem 9-1
Area = 1 • 200 • 300 • sin 65°
Substitute.
Area = 30,000 sin 65°
Simplify.
2
Use a calculator
27189.23361
Quick Check
The area of the park is approximately 27,200 ft2.
10-5
Trigonometry and Area
GEOMETRY LESSON 10-5
Find the area of each figure. Give answers to the nearest unit.
1. regular hexagon with perimeter 90 ft
585 ft2
2. regular pentagon with radius 12 m
342 m2
3. regular polygon with 12 sides of length 1 in.
11 in2
4.
5.
490 mm2
70 yd2
10-5
Circles and Arcs
GEOMETRY LESSON 10-6
(For help, go to Lesson 1-9 and Skills Handbook, page 761.)
Find the diameter or radius of each circle.
1. r = 7 cm, d =
2. r = 1.6 m, d =
3. d = 10 ft, r =
4. d = 5 in., r =
Round to the nearest whole number.
5. 9% of 360
6. 38% of 360
7. 50% of 360
8. 21% of 360
Check Skills You’ll Need
10-6
Circles and Arcs
GEOMETRY LESSON 10-6
Solutions
1. The diameter is twice the radius: (7)(2) = 14 cm
2. The diameter is twice the radius: (1.6)(2) = 3.2 m
3. The radius is half the diameter: 10 ÷ 2 = 5 ft
4. The radius is half the diameter: 5 ÷ 2 = 2.5 in.
5. 9% = 0.09. Read of as times, so 9% of 360 is 0.09 times 360.
(0.09)(360) = 32.4. The nearest whole number to 32.4 is 32.
6. 38% = 0.38. Read of as times, so 38% of 360 is 0.38 times 360.
(0.38)(360) = 136.8. The nearest whole number to 136.8 is 137.
7. 50% = 0.50. Read of as times, so 50% of 360 is 0.50 times 360.
(0.50)(360) = 180.
8. 21% = 0.21. Read of as times, so 21% of 360 is 0.21 times 360.
(0.21)(360) = 75.6. The nearest whole number to 75.6 is 76.
10-6
Circles and Arcs
GEOMETRY LESSON 10-6
A researcher surveyed 2000 members of a club to
find their ages. The graph shows the survey results. Find
the measure of each central angle in the circle graph.
Because there are 360° in a circle,
multiply each percent by 360 to
find the measure of each central angle.
65+ : 25% of 360 = 0.25 • 360 = 90
45–64: 40% of 260 = 0.4 • 360 = 144
25–44: 27% of 360 = 0.27 • 360 = 97.2
Under 25: 8% of 360 = 0.08 • 360 = 28.8
10-6
Quick Check
Circles and Arcs
GEOMETRY LESSON 10-6
Identify the minor arcs, major arcs, and semicircles in . P with
point A as an endpoint.
Minor arcs are smaller than semicircles.
Two minor arcs in the diagram have point A
as an endpoint, AD and AE.
Major arcs are larger than semicircles.
Two major arcs in the diagram have point A
as an endpoint, ADE and AED.
Two semicircles in the diagram have
point A as an endpoint, ADB and AEB.
Quick Check
10-6
Circles and Arcs
GEOMETRY LESSON 10-6
Find mXY and mDXM in . C.
mXY = mXD + mDY
Arc Addition Postulate
mXY = m
The measure of a minor arc is the
measure of its corresponding
central angle.
XCD + mDY
mXY = 56 + 40
Substitute.
mXY = 96
Simplify.
mDXM = mDX + mXWM
Arc Addition Postulate
mDXM = 56 + 180
Substitute.
mDXM = 236
Simplify.
10-6
Quick Check
Circles and Arcs
GEOMETRY LESSON 10-6
A circular swimming pool with a 16-ft diameter will be
enclosed in a circular fence 4 ft from the pool. What length of fencing
material is needed? Round your answer to the next whole number.
Draw a diagram of the situation.
The pool and the fence are concentric circles.
The diameter of the pool is 16 ft, so the
diameter of the fence is 16 + 4 + 4 = 24 ft.
Use the formula for the circumference of a
circle to find the length of fencing material needed.
C= d
Formula for the circumference of a circle
C = (24)
Substitute.
C 3.14(24)
Use 3.14 to approximate .
C 75.36
Simplify.
About 76 ft of fencing material is needed.
10-6
Quick Check
Circles and Arcs
GEOMETRY LESSON 10-6
Find the length of ADB in . M in terms of
.
Because mAB = 150,
mADB = 360 – 150 = 210.
length of ADB =
mADB
•2
360
length of ADB =
210
•2
360
r
Arc Addition Postulate
Arc Length Formula
(18)
Substitute.
length of ADB = 21
The length of ADB is 21
cm.
Quick Check
10-6
Circles and Arcs
GEOMETRY LESSON 10-6
1. A circle graph has a section marked “Potatoes: 28%.”
What is the measure of the central angle of this section?
100.8
2. Explain how a major arc differs from a minor arc.
A major arc is greater than a semicircle. A minor arc is smaller
than a semicircle.
Use . O for Exercises 3–6.
3. Find mYW.
30
4. Find mWXS.
270
5. Suppose that . P has a diameter 2 in. greater than the diameter of
. O. How much greater is its circumference? Leave your answer in terms of
2
6. Find the length of XY. Leave your answer in terms of .
9
10-6
.
Areas of Circles and Sectors
GEOMETRY LESSON 10-7
(For help, go to Lesson 10-6.)
1. What is the radius of a circle with diameter 9 cm?
2. What is the diameter of a circle with radius 8 ft?
3. Find the circumference of a circle with diameter 12 in.
4. Find the circumference of a circle with radius 3 m.
Check Skills You’ll Need
10-7
Areas of Circles and Sectors
GEOMETRY LESSON 10-7
Solutions
1. The radius is half the diameter: 9 ÷ 2 = 4.5 cm
2. The diameter is twice the radius: (8)(2) = 16 ft
3. C =
d=
(12) = 12
4. C = 2 r = 2
(3) = 6
, or about 37.7 in.
, or about 18.8 m
10-7
Areas of Circles and Sectors
GEOMETRY LESSON 10-7
A circular archery target has a 2-ft diameter. It is
yellow except for a red bull’s-eye at the center with a 6-in.
diameter. Find the area of the yellow region. Round your
answer to the nearest whole number.
Find the areas of the archery target and the bull’s-eye.
The radius of the archery target is
2
= 1 ft.
2
Because the diameters are in different units, convert 1 ft to 12 in.
The radius of the archery target is 1 ft = 12 in.
The area of the archery target is
r2 =
10-7
(12)2 = 144
in.2
Areas of Circles and Sectors
GEOMETRY LESSON 10-7
(continued)
The radius of the red region is
The area of the red region is
6
= 3 in.
2
r2 =
(3)2 = 9
in.2
area of archery target – area of red region = area of yellow region
144
Use a calculator. 135
–
9
=
135
424.11501
The area of the yellow region is about 424 in.2
10-7
Quick Check
Areas of Circles and Sectors
GEOMETRY LESSON 10-7
.
. of .
Find
the area of sector ACB. Leave your answer in terms
mAB
area of sector ACB = 360 •
=
100
•
360
r2
(6)2
= 5 • 36
18
= 10
The area of sector ACB is 10
m2.
Quick Check
10-7
Areas of Circles and Sectors
GEOMETRY LESSON 10-7
Find the area of the shaded segment. Round your answer to
the nearest tenth.
Step 1: Find the area of sector AOB.
area of sector AOB = mAB •
360
r2
Use the formula for
area of a sector.
120
= 360 •
= 1 • 576
3
10-7
(24)2
= 192
Substitute.
Simplify.
Areas of Circles and Sectors
GEOMETRY LESSON 10-7
(continued)
Step 2: Find the area of AOB.
You can use a 30°-60°-90° triangle to find the height h of
24 = 2h
hypotenuse = 2 • shorter leg
12 = h
AB = 3 • 12 = 12
2
AB = 24 3
Divide each side by 2.
AOB has base 12
1
A = bh
2
A = 1 (24 3 )(12)
2
A = 144 3
3
longer leg =
AOB and AB.
3• shorter leg
Multiply each side by 2.
3 ft + 12
3 ft, or 24
3 ft and height 12 ft.
Area of a triangle
Substitute 24 for b and 12 for h.
Simplify.
10-7
Areas of Circles and Sectors
GEOMETRY LESSON 10-7
(continued)
Step 3: Subtract the area of AOB from the area of sector AOB
to find the area of the segment of the circle.
area of segment = 192
– 144
353.77047
3
Use a calculator.
To the nearest tenth, the area of the shaded segment is 353.8 ft2.
Quick Check
10-7
Areas of Circles and Sectors
GEOMETRY LESSON 10-7
1. A park contains two circular playgrounds. One has a diameter of 60 m,
and the other has a diameter of 40 m. How much greater is the area of
the larger playground? Round to the nearest whole number.
1571 m2
2. A circle has an 8-in. radius. Find the area of a sector whose
arc measures 135. Leave your answer in terms of .
24 in.2
For Exercises 3 and 4, find the area of the shaded segment.
Round to the nearest whole unit.
3.
4.
15 cm2
138 in.2
10-7
Geometric Probability
GEOMETRY LESSON 10-8
(For help, go to the Skills Handbook, pages 756 and 762.)
Find and simplify each ratio.
1. BD
AE
2. CE
3. AB
AF
BC
4. Two circles have radii 1 m and 2 m, respectively. What is
the simplest form of the fraction with numerator equal to the
area of the smaller circle and denominator equal to the area
of the larger circle?
You roll a number cube. Find the probability of rolling each
of the following.
5. 4
6. an odd number
7. 2 or 5
8. a prime number
10-8
Check Skills You’ll Need
Geometric Probability
GEOMETRY LESSON 10-8
Solutions
1. BD = 5 – 2 = 3; AE = 9 – 0 = 10; BD = 3 = 1
AE
9
3
2. CE = 9 – 4 = 5; AF = 10 – 0 = 10; CE = 5 = 1
AF
3. AB = 2 – 0 = 2; BC = 4 – 2 = 2;
4. The area of the smaller circle is
circle is
r2 =
(2)2 = 4
10
2
AB
= 2 =1
BC
2
r2 =
(1)2 =
m; smaller : larger =
m; The area of the larger
:4
= 1 : 4, or 1 .
4
5. 4 is one of six numbers on the number cube. So, the probability is 1 out
of 6 chances or 1 .
6
10-8
Geometric Probability
GEOMETRY LESSON 10-8
Solutions (continued)
6. The numbers on a number cube are 1, 2, 3, 4, 5, and 6. Three numbers, 1, 3,
and 5, are odd. 3 out of 6 numbers are odd, so the probability is 3 out of 6
or 3 = 1 .
6
2
7. There are 6 numbers on a number cube. The numbers 2 and 5 are two of
them. So, 2 out of 6 numbers are desired. The probability is 2 out of 6 or 2 = 1 .
6
8. The numbers on a number cube are 1, 2, 3, 4, 5, and 6. The numbers 2, 3,
and 5 are prime. So 3 out of 6 numbers are prime. The probability is 3 out of
6 or 3 = 1 .
6
2
10-8
3
Geometric Probability
GEOMETRY LESSON 10-8
A gnat lands at random on the edge of the ruler
below. Find the probability that the gnat lands on a point
between 2 and 10.
The length of the segment between 2 and 10 is 10 – 2 = 8.
The length of the ruler is 12.
P(landing between 2 and 10) =
length of favorable segment
8
2
=
, or
length of entire segment
12
3
Quick Check
10-8
Geometric Probability
GEOMETRY LESSON 10-8
Quick Check
A museum offers a tour every hour. If Benny arrives at the
tour site at a random time, what is the probability that he will have to
wait at least 15 minutes?
Because the favorable time is given in minutes, write 1 hour as 60 minutes.
Benny may have to wait anywhere between 0 minutes and 60 minutes.
Represent this using a segment.
Starting at 60 minutes, go back 15 minutes. The segment of length 45
represents Benny’s waiting more than 15 minutes.
P(waiting more than 15 minutes) =
45
, or 3
60
4
The probability that Benny will have to wait at least 15 minutes is 3 , or 75%.
4
10-8
Geometric Probability
GEOMETRY LESSON 10-8
A circle is inscribed in a square target with 20-cm sides. Find
the probability that a dart landing randomly within the square does not
land within the circle.
Find the area of the square.
A = s2 = 202 = 400 cm2
Find the area of the circle. Because the square has sides of length
20 cm, the circle’s diameter is 20 cm, so its radius is 10 cm.
A = r 2 = (10)2 = 100 cm2
Find the area of the region between the square and the circle.
A = (400 – 100 ) cm2
10-8
Geometric Probability
GEOMETRY LESSON 10-8
(continued)
Use areas to calculate the probability that a dart landing randomly in
the square does not land within the circle. Use a calculator. Round to
the nearest thousandth.
P (between square and circle)
=
=
area between square and circle
area of square
400 – 100
400
=1–
4
0.215
The probability that a dart landing randomly in the square does not land
within the circle is about 21.5%.
Quick Check
10-8
Geometric Probability
GEOMETRY LESSON 10-8
To win a prize, you must toss a quarter so that it lands entirely
within the outer region of the circle below. Find the probability that this
happens with a quarter of radius 15 in. Assume that the quarter is
32
equally likely to land anywhere completely inside the large circle.
The center of a quarter with a radius of 15 in. must land
32
15
at least
in. beyond the boundary of the inner circle in
32
order to lie entirely outside the inner circle. Because
the inner circle has a radius of 9 in., the quarter must
land outside the circle whose radius
is 9 in. + 15 in., or 9 15 in.
32
10-8
32
Geometric Probability
GEOMETRY LESSON 10-8
(continued)
Find the area of the circle with a radius of 9
A=
r2 =
(9 15 )2
32
15
in.
32
281.66648 in.2
Similarly, the center of a quarter with a radius of
15
in. must land at least
32
15 in. within the outer circle. Because the outer circle has a radius of 12 in.,
32
the quarter must land inside the circle whose radius is 12 in. – 15 in., or
32
17
11 in.
32
Find the area of the circle with a radius of 11
A=
r2 =
(11 17 )2
32
417.73672 in.2
10-8
17
in.
32
Geometric Probability
GEOMETRY LESSON 10-8
(continued)
Use the area of the outer region to find the probability that the quarter
lands entirely within the outer region of the circle.
area of outer region
P (outer region) =
area of large circle
136.07024
417.73672
417.73672 – 281.66648
=
417.73672
0.32573
The probability that the quarter lands entirely within the outer region
of the circle is about 0.326, or 32.6%.
Quick Check
10-8
Geometric Probability
GEOMETRY LESSON 10-8
1. A point on AF is chosen at random. What is the probability
that it is a point on BE?
3
5
2. Express elevators to the top of a tall building leave the
ground floor every 40 seconds. What is the probability
that a person would have to wait more than 30 seconds
for an express elevator? 1
4
A dart you throw is equally likely to land at any point on each
board shown. For Exercises 3–5, find the probability of its landing
in the shaded area.
3. regular octagon
4. square
3
, or 37.5%
8
5. circle
1
, or 50%
2
10-8
8
, or 32%
25