Areas of Parallelograms and Triangles : 10-1 (For help, go to Lesson 1-9.) GEOMETRY LESSON 10-1 Areas of Parallelograms and Triangles Find the area of each figure. Each rectangle is divided into two congruent triangles.
Find the area of each triangle. Check Skills You’ll Need
Areas of Parallelograms and Triangles : Solutions GEOMETRY LESSON 10-1 Areas of Parallelograms and Triangles 10-1
Areas of Parallelograms and Triangles : 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. 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. GEOMETRY LESSON 10-1 Areas of Parallelograms and Triangles Quick Check 10-1
Areas of Parallelograms and Triangles : 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 GEOMETRY LESSON 10-1 Areas of Parallelograms and Triangles 10-1
Areas of Parallelograms and Triangles : Use the area 135 in.2 to find the height to the 18-in. base. The height corresponding to the 18-in. base is 7.5 in. GEOMETRY LESSON 10-1 (continued) Areas of Parallelograms and Triangles Quick Check 10-1
Areas of Parallelograms and Triangles : A = 195 Simplify. GEOMETRY LESSON 10-1 Areas of Parallelograms and Triangles Quick Check 10-1
Areas of Parallelograms and Triangles : 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. The total area of the front of the garage is
225 + 79.5 = 304.5 ft2. GEOMETRY LESSON 10-1 Areas of Parallelograms and Triangles 10-1
Areas of Parallelograms and Triangles : (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. An 80 mi/h wind exerts a force of about 7800 lb against the front of
the garage. GEOMETRY LESSON 10-1 Areas of Parallelograms and Triangles Quick Check 10-1
Areas of Parallelograms and Triangles : 187 in.2 150 ft2 24 square units 6 cm 15 m2 GEOMETRY LESSON 10-1 Areas of Parallelograms and Triangles 10-1
Areas of Trapezoids, Rhombuses, and Kites : 10-2 (For help, go to Lesson 10-1.) GEOMETRY LESSON 10-2 Areas of Trapezoids, Rhombuses, and Kites Write the formula for the area of each type of figure. Find the area of each trapezoid by using the formulas for area
of a rectangle and area of a triangle. Check Skills You’ll Need
Areas of Trapezoids, Rhombuses, and Kites : Solutions GEOMETRY LESSON 10-2 Areas of Trapezoids, Rhombuses, and Kites 10-2
Areas of Trapezoids, Rhombuses, and Kites : 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 bh = (1)(2) = 1. The area A of
the triangle on the right is bh = (2)(2) = 2. The area A of the rectangle is
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 bh = (2)(3) = 3. The area A of the
triangle on the right is bh = (3)(3) = 4.5. The area A of the rectangle is
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. 1
2 1
2 1
2 1
2 1
2 1
2 1
2 1
2 Solutions (continued) GEOMETRY LESSON 10-2 Areas of Trapezoids, Rhombuses, and Kites 10-2
Areas of Trapezoids, Rhombuses, and Kites : A = 504 Simplify. The area of the car window is 504 in.2 GEOMETRY LESSON 10-2 Areas of Trapezoids, Rhombuses, and Kites Quick Check 10-2
Areas of Trapezoids, Rhombuses, and Kites : Because opposite sides of rectangle ABXD
are congruent, DX = 11 ft and
XC = 16 ft – 11 ft = 5 ft. GEOMETRY LESSON 10-2 Areas of Trapezoids, Rhombuses, and Kites 10-2
Areas of Trapezoids, Rhombuses, and Kites : A = 162 Simplify. The area of trapezoid ABCD is 162 ft2. GEOMETRY LESSON 10-2 Areas of Trapezoids, Rhombuses, and Kites Quick Check 10-2
Areas of Trapezoids, Rhombuses, and Kites : Find the lengths of the diagonals of kite XYZW.
XZ = d1 = 3 + 3 = 6 and YW = d2 = 1 + 4 = 5 A = 15 Simplify. The area of kite XYZW is 15 cm2. Find the area of kite XYZW. GEOMETRY LESSON 10-2 Areas of Trapezoids, Rhombuses, and Kites Quick Check 10-2
Areas of Trapezoids, Rhombuses, and Kites : Find the area of rhombus RSTU. GEOMETRY LESSON 10-2 Areas of Trapezoids, Rhombuses, and Kites 10-2
Areas of Trapezoids, Rhombuses, and Kites : 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. A = 120 Simplify. The area of rhombus RSTU is 120 ft2. GEOMETRY LESSON 10-2 Areas of Trapezoids, Rhombuses, and Kites Quick Check 10-2
Areas of Trapezoids, Rhombuses, and Kites : 1. Find the area of a trapezoid with bases 3 cm and 19 cm
and height 9 cm.
2. Find the area of a trapezoid in a coordinate plane with
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 2 m long
5. rhombus MNOP 99 cm2 26 square units 840 mm2 GEOMETRY LESSON 10-2 Areas of Trapezoids, Rhombuses, and Kites 10-2
Areas of Regular Polygons : 1. 2. 3.
4. a hexagon with sides of 4 in.
5. an octagon with sides of 2 3 cm (For help, go to Lesson 8-2.) GEOMETRY LESSON 10-3 Areas of Regular Polygons Find the area of each regular polygon. If your answer involves
a radical, leave it in simplest radical form. Find the perimeter of the regular polygon. 10-3 Check Skills You’ll Need
Areas of Regular Polygons : GEOMETRY LESSON 10-3 Areas of Regular Polygons 10-3
Areas of Regular Polygons : GEOMETRY LESSON 10-3 Areas of Regular Polygons 10-3
Areas of Regular Polygons : A portion of a regular hexagon has an apothem and radii drawn. Find the measure of each numbered angle. GEOMETRY LESSON 10-3 Areas of Regular Polygons Quick Check 10-3
Areas of Regular Polygons : p = ns Find the perimeter. p = (20)(12) = 240 Substitute 20 for n and 12 for s. A = 4548 Simplify. The area of the polygon is 4548 in.2 GEOMETRY LESSON 10-3 Areas of Regular Polygons Quick Check 10-3
Areas of Regular Polygons : GEOMETRY LESSON 10-3 Areas of Regular Polygons 10-3
Areas of Regular Polygons : Step 2: Find the perimeter p.
p = ns Find the perimeter.
p = (8)(18.0) = 144 Substitute 8 for n and 18.0 for s,
and simplify. (continued) GEOMETRY LESSON 10-3 Areas of Regular Polygons 10-3
Areas of Regular Polygons : To the nearest 10 ft2, the area is 1560 ft2. (continued) GEOMETRY LESSON 10-3 Areas of Regular Polygons Quick Check 10-3
Areas of Regular Polygons : 1. Find m 1.
2. Find m 2.
3. Find m 3.
4. Find the area of a regular 9-sided figure with a 9.6-cm
apothem and 7-cm side.
For Exercises 5 and 6, find the area of each regular polygon.
Leave your answer in simplest radical form.
5. 6. 36 18 72 302.4 cm2 GEOMETRY LESSON 10-3 Areas of Regular Polygons Use the portion of the regular decagon for Exercises 1–3. 10-3
Perimeters and Areas of Similar Figures : 10-4 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 (For help, go to Lesson 1-9.) GEOMETRY LESSON 10-4 Perimeters and Areas of Similar Figures Check Skills You’ll Need
Perimeters and Areas of Similar Figures : GEOMETRY LESSON 10-4 Perimeters and Areas of Similar Figures 10-4
Perimeters and Areas of Similar Figures : 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 GEOMETRY LESSON 10-4 Perimeters and Areas of Similar Figures Solutions (continued) 10-4
Perimeters and Areas of Similar Figures : 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. GEOMETRY LESSON 10-4 Perimeters and Areas of Similar Figures Quick Check 10-4
Perimeters and Areas of Similar Figures : The ratio of the lengths of the corresponding sides of two regular octagons is . The area of the larger octagon is 320 ft2. Find the area of the smaller octagon. 8
3 All regular octagons are similar. The area of the smaller octagon is 45 ft2. GEOMETRY LESSON 10-4 Perimeters and Areas of Similar Figures Quick Check 10-4
Perimeters and Areas of Similar Figures : 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. GEOMETRY LESSON 10-4 Perimeters and Areas of Similar Figures Quick Check 10-4
Perimeters and Areas of Similar Figures : 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? The similarity ratio is 2 : 3. By the Perimeters and Areas of Similar Figures Theorem, the ratio of the perimeters is also 2 : 3. GEOMETRY LESSON 10-4 Perimeters and Areas of Similar Figures Quick Check 10-4
Perimeters and Areas of Similar Figures : 1. For the similar rectangles, give the ratios (smaller to larger) of the perimeters and of the areas.
2. The triangles are similar. The area of the largertriangle is 48 ft2. Find the area of the smaller triangle.
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.
4. The areas of two equilateral triangles are 27 yd2 and 75 yd2. Find their similarity ratio and the ratio of their perimeters.
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? 27 ft2 324 in.2 3 : 5; 3 : 5 $108 GEOMETRY LESSON 10-4 Perimeters and Areas of Similar Figures 10-4
Trigonometry and Area : (For help, go to Lesson 10-3.) Find the area of each regular polygon. 1. 3. 2. GEOMETRY LESSON 10-5 Trigonometry and Area 10-5 Check Skills You’ll Need
Trigonometry and Area : Solutions GEOMETRY LESSON 10-5 Trigonometry and Area 10-5
Trigonometry and Area : Find the area of a regular polygon with 10 sides and side length 12 cm. Because the polygon has 10 sides and each side is 12 cm long, p = 10 • 12 = 120 cm. Use trigonometry to find a. GEOMETRY LESSON 10-5 Trigonometry and Area 10-5
Trigonometry and Area : (continued) Now substitute into the area formula. The area is about 1108 cm2. GEOMETRY LESSON 10-5 Trigonometry and Area Quick Check 10-5
Trigonometry and Area : The radius of a garden in the shape of a regular pentagon is 18 feet. Find the area of the garden. GEOMETRY LESSON 10-5 Trigonometry and Area 10-5
Trigonometry and Area : (continued) So p = 5 • (2 • AM) = 10 • AM = 10 • 18(sin 36°) = 180(sin 36°). GEOMETRY LESSON 10-5 Trigonometry and Area 10-5
Trigonometry and Area : (continued) Finally, substitute into the area formula A = ap. 1
2 The area of the garden is about 770 ft2. GEOMETRY LESSON 10-5 Trigonometry and Area Quick Check 10-5
Trigonometry and Area : 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. The area of the park is approximately 27,200 ft2. GEOMETRY LESSON 10-5 Trigonometry and Area Quick Check 10-5
Trigonometry and Area : Find the area of each figure. Give answers to the nearest unit. 1. regular hexagon with perimeter 90 ft
2. regular pentagon with radius 12 m
3. regular polygon with 12 sides of length 1 in. 4. 5. 585 ft2 342 m2 11 in2 490 mm2 70 yd2 GEOMETRY LESSON 10-5 Trigonometry and Area 10-5
Circles and Arcs : 10-6 (For help, go to Lesson 1-9 and Skills Handbook, page 761.) 1. r = 7 cm, d = 2. r = 1.6 m, d =
3. d = 10 ft, r = 4. d = 5 in., r =
5. 9% of 360 6. 38% of 360
7. 50% of 360 8. 21% of 360 GEOMETRY LESSON 10-6 Circles and Arcs Find the diameter or radius of each circle. Round to the nearest whole number. Check Skills You’ll Need
Circles and Arcs : 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. Solutions GEOMETRY LESSON 10-6 Circles and Arcs 10-6
Circles and Arcs : 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 GEOMETRY LESSON 10-6 Circles and Arcs Quick Check 10-6
Circles and Arcs : GEOMETRY LESSON 10-6 Circles and Arcs Quick Check 10-6
Circles and Arcs : GEOMETRY LESSON 10-6 Circles and Arcs Quick Check 10-6
Circles and Arcs : 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. About 76 ft of fencing material is needed. GEOMETRY LESSON 10-6 Circles and Arcs Quick Check 10-6
Circles and Arcs : GEOMETRY LESSON 10-6 Circles and Arcs Quick Check 10-6
Circles and Arcs : A major arc is greater than a semicircle. A minor arc is smaller
than a semicircle. 100.8 270 30 GEOMETRY LESSON 10-6 Circles and Arcs 10-6
Areas of Circles and Sectors : (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. GEOMETRY LESSON 10-7 Areas of Circles and Sectors 10-7 Check Skills You’ll Need
Areas of Circles and Sectors : 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 , or about 37.7 in.
4. C = 2 r = 2 (3) = 6 , or about 18.8 m Solutions GEOMETRY LESSON 10-7 Areas of Circles and Sectors 10-7
Areas of Circles and Sectors : Find the areas of the archery target and the bull’s-eye. GEOMETRY LESSON 10-7 Areas of Circles and Sectors 10-7
Areas of Circles and Sectors : The area of the yellow region is about 424 in.2 GEOMETRY LESSON 10-7 Areas of Circles and Sectors Quick Check 10-7
Areas of Circles and Sectors : . . GEOMETRY LESSON 10-7 Areas of Circles and Sectors Quick Check 10-7
Areas of Circles and Sectors : GEOMETRY LESSON 10-7 Areas of Circles and Sectors Step 1: Find the area of sector AOB. 10-7
Areas of Circles and Sectors : GEOMETRY LESSON 10-7 Areas of Circles and Sectors 10-7
Areas of Circles and Sectors : To the nearest tenth, the area of the shaded segment is 353.8 ft2. GEOMETRY LESSON 10-7 Areas of Circles and Sectors Quick Check 10-7
Areas of Circles and Sectors : 1571 m2 15 cm2 138 in.2 GEOMETRY LESSON 10-7 Areas of Circles and Sectors 10-7
Geometric Probability : 10-8 GEOMETRY LESSON 10-8 Geometric Probability (For help, go to the Skills Handbook, pages 756 and 762.) Check Skills You’ll Need
Geometric Probability : 1. BD = 5 – 2 = 3; AE = 9 – 0 = 10; = =
2. CE = 9 – 4 = 5; AF = 10 – 0 = 10; = =
3. AB = 2 – 0 = 2; BC = 4 – 2 = 2; = = 1
4. The area of the smaller circle is r2 = (1)2 = m; The area of the larger
circle is r2 = (2)2 = 4 m; smaller : larger = : 4 = 1 : 4, or .
5. 4 is one of six numbers on the number cube. So, the probability is 1 out
of 6 chances or . BD
AE CE
AF AB
BC 3
9 1
3 5
10 1
2 2
2 1
6 1
4 Solutions GEOMETRY LESSON 10-8 Geometric Probability 10-8
Geometric Probability : 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 = .
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 = .
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
6 1
2 2
6 1
3 3
6 1
2 GEOMETRY LESSON 10-8 Solutions (continued) Geometric Probability 10-8
Geometric Probability : The length of the segment between 2 and 10 is 10 – 2 = 8. The length of the ruler is 12. GEOMETRY LESSON 10-8 Geometric Probability Quick Check 10-8
Geometric Probability : 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. Starting at 60 minutes, go back 15 minutes. The segment of length 45
represents Benny’s waiting more than 15 minutes. GEOMETRY LESSON 10-8 Geometric Probability Quick Check 10-8
Geometric Probability : Find the area of the square.
A = s2 = 202 = 400 cm2 GEOMETRY LESSON 10-8 Geometric Probability 10-8
Geometric Probability : The probability that a dart landing randomly in the square does not land
within the circle is about 21.5%. GEOMETRY LESSON 10-8 Geometric Probability Quick Check 10-8
Geometric Probability : 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 in. Assume that the quarter is equally likely to land anywhere completely inside the large circle. 15
32 GEOMETRY LESSON 10-8 Geometric Probability 10-8
Geometric Probability : (continued) GEOMETRY LESSON 10-8 Geometric Probability Find the area of the circle with a radius of 9 in.
A = r2 = (9 )2 281.66648 in.2 15
32 15
32 10-8
Geometric Probability : The probability that the quarter lands entirely within the outer region
of the circle is about 0.326, or 32.6%. GEOMETRY LESSON 10-8 Geometric Probability Quick Check 10-8
Geometric Probability : 1. A point on AF is chosen at random. What is the probability
that it is a point on BE?
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?
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 5. circle 3
5 1
4 GEOMETRY LESSON 10-8 Geometric Probability 10-8