GEOMETRY AND MEASUREMENT - Cengage · 2008-06-27 · Change 4 yards to feet 2. Change 8 1/2 ft to yards 3. Convert 9.7 kilometers to meters 4. Convert 90 centimeters to hectometers

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GEOMETRY AND MEASUREMENT

CLAST MATHEMATICS COMPETENCIES IB1: Round measurements to the nearest given unit of the measuring device used IB2a: Calculate distances IB2b: Calculate areas IB2c: Calculate volumes IIB1: Identify relationships between angle measures IIB2: Classify simple plane figures by recognizing their properties IIB3: Recognize similar triangles and their properties IIB4: Identify appropriate units of measurement for geometric objects IIIBl: Infer formulas for measuring geometric figures IIIB2: Identify applicable formulas for measuring geometric figures IVB1: Solve real-world problems involving perimeters, areas, volumes of geometric figures IVB2: Solve real-world problems involving the Pythagorean theorem

132

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3.1 MEASURING DISTANCE, AREA AND VOLUME

In this Chapter we shall discuss the measurement of distances, areas, volumes and their applications. We shall then learn about the measurement of angles, the relationship between different types of angles, the classification of plane figures and similar triangles.

A. Rounding Measurements Objective IB1 CLAST SAMPLE PROBLEMS

1. Round 43.846 to the nearest hundredth of a centimeter 2. Round 19,283 tons to the nearest ten thousand tons 3. Round 9 pounds, 6 ounces to the nearest tenth of a pound 4. Round the measurement of the screw to the nearest 1/4 inch

In Section 1.3 we learned how to round numbers. We are now ready to round measurements in such units as feet, yards, centimeters, meters, pounds, and liters.

1 ROUNDING MEASUREMENTS RULE

1. Underline the place to which you are rounding. 2. If the first number to the right of the

underlined place is 5 or more, add one to the underlined number. Otherwise, do not change the underlined number.

3. Change all the numbers to the right of the underlined number to zeros.

EXAMPLES Round 38.67 centimeters to the nearest centimeter. Since we want to round to the nearest centimeter, we underline the units place. 1. Underline the 8. 38.67 2. The first number to the right of 8 is 6, so

we add one to 8. (8 + 1 = 9) 3. Change all the numbers to the right of 8 to

zeros obtaining 39.00 or 39. Note: We are asking to round 3138 pounds to the nearest hundred pounds. The idea and procedure would be the same if you were asked to round 3138 grams to the nearest hundred grams or 3138 liters to the nearest hundred liters.

Round 3138 pounds to the nearest hundred pounds. 1. Underline the hundreds place. 3138 2. The first number to the right of 1 is 3. Do

not change the underlined number. 3. Change all the numbers to the right of 1 to

0's obtaining 3100. Thus, 3138 pounds rounded to the nearest hundred pounds becomes 3100 pounds.

ANSWERS 1. 43.85 cm 2. 20,000 tons 3. 9.4 pounds 4. 3/4 inch

SECTION 3.1 Measuring Distance, Area and Volume 133


CLAST EXAMPLE

Example 1. Round the measurement of the length of the

paper clip to the nearest 14 inch.

A. 1 in. B. 1

14 in. C. 1

34 in. D. 2 in.

Solution Note: The abbreviation for inch is in.

If we are measuring (or counting) by 14 in., the

measures would be: 14 in.,

24 in.,

34 in.,

44 in.,

54 in., and so on. The end of the paper

clip is closest to the 54 in. mark.

Since 54 in. = 1

14 in., the correct answer is B.

B. Calculating Distances, Areas and Volumes

Objectives IB2a,IB2b,IB2c CLAST SAMPLE PROBLEMS 1. Change 4 yards to feet 2. Change 8 1/2 ft to yards 3. Convert 9.7 kilometers to meters 4. Convert 90 centimeters to hectometers 5. Change 18 inches to yards 6. Change 2.2 miles to feet 7. Find the perimeter of a rectangle that has a length of 4 yards and a width of 4 feet 8. Find the distance around the polygon shown in kilometers 9. Find the distance around a regular pentagon with one side having a

length of 6 meters 10. What is the distance around a circular walk with a diameter of 6

ft.

300 m

400 m500 m

600 m

11. What is the area in square centimeters of the triangle shown? 12. What is the area of a circular region with a 6 yard diameter? 13. Find the area of the 3 meter wide rectangle shown 14. What is the surface area of a rectangular solid that is 6 feet long, 1

yard wide and 4 feet high? 15. Find the volume of a right circular cone 3 meters high and with a

radius of 20 centimeters 16. What is the volume of a rectangular solid 5 feet long, 4 feet

wide and 9 inches high?

13 cm

30 cm 40 mm

3 m5 m

17. Find the volume of a sphere with a diameter of 2 yards 18. What is the volume in liters of a 850-milliliter bottle? 19. What is the volume in cubic centimeters of a 25.8-liter container? 20. What is the volume in cubic meters of a 675 -liter container?

ANSWERS 1. 12 ft 2. 2 5/6 yd 3. 9700 meters 4. 0.009 hm 5. 1/2 yd 6. 11,616 ft 7. 32 ft 8. 1.8 km 9. 30 m 10. 6π ft 11. 60 cm2 12. 9p yd2 13. 12 m2 14. 108 ft2 15. 40,000π cm3 16. 15 ft3

17. 43 π yd3 18. 0.85 L 19. 25,800 cm3 20. 0.675 m3

134 CHAPTER 3 Geometry and Measurement


When we measure a distance we measure length and the result is a linear unit. In the U. S. system we use inches, feet, yards and miles. In the metric system we use millimeters, centimeters, meters, and kilometers. Here are some relationships you should know. T TERMINOLOGY -- LINEAR MEASURES

U. S. UNITS 12 inches (in.) = 1 foot (ft)

1 (in.) = 112 ft

3 ft = 1 yard (yd)

1 ft = 13 yd

5280 ft = 1 mile (mi)

1 ft = 1

5280 mi

EXAMPLES

52 inches = 52 ft121 =

5212 ft = 4

13 ft.

8 ft = 8 yd31 =

83 yd = 2

23 yd.

7920 ft = 7920 mi5280

1 = 79205280 mi

= 1 12 mi.

METRIC UNITS kilometer hectometer dekameter meter (km ) (hm) (dam) m 1000 m 100m 10 m meter decimeter centimeter millimeter m (dm) (cm) (mm)

110 m

1100 m

11000 m

Chart: km hm dam m dm cm mm

To convert from one metric unit to another is a matter of moving the decimal point. For example, to convert from km to meters, we move three places right in the chart. Thus, to convert 73 km to meters, we move the decimal point in 73. three places right obtaining 73. km = 73,000 m. To convert 3847 mm to meters, we have to move the decimal point in 3847. three places left. Thus, 3847. mm = 3.847 m.

You can recall the order in the metric chart if you remember that: King Henry Died Monday Drinking Chocolate Milk!

2 FINDING PERIMETERS RULE

A polygon is a closed plane figure having three or more sides. (See figure at right.) The perimeter of (distance around) a polygon is the sum of the lengths of its sides.

EXAMPLES

5 in.

15 in

10 in9 in

The perimeter of the polygon, in inches, is (10 + 9 + 5 + 15) in. = 39 in. The perimeter

in feet is 3912 = 3

312 = 3

14 ft.



3 FINDING CIRCUMFERENCES The distance around a circle is called the circumference C and given by the formula

C = π d = 2π r

Note that the diameter d of the circle is twice

the radius, that is, d = 2r or r = d2 .

Note: The CLAST uses π as a factor in its answers, you need not approximate π

diameter d

radius r

If the radius of the circle is 7 ft, (r = 7) the circumference C = 2π r = 2π (7) = 14π ft.

CLAST EXAMPLE

Example 2. What is the distance around the polygon, in

meters?

75 cm

80 cm95 cm

78 cm

A. 328 m B. 32.8 m C. 3.28 m D. 0.328 m

Solution The distance around the polygon is: 95 + 75 + 80 + 78 = 328 cm Since 100 cm is 1 m, 328 cm = 3.28 m. (If you look in the chart, to go from cm to m we have to go two places left, so we move the decimal point in 328. two places left obtaining 3.28 m.) The answer is C. Hint: When adding the length of the sides, we added 95 + 75 by writing 75 as 5 + 70 and then adding 95 + 5 + 70 = 170. Then add 80 to 170, obtaining 250 and finally, add 78 to get 328. Be careful with the arithmetic!

Sometimes we have to use more than one formula to get the perimeter (distance around) of a figure. For example, the next problem involves a rectangle and two semicircles (half circles). In order to find the distance around the figure we need to know the formulas for the perimeter of a rectangle and the perimeter (circumference) of a circle.



CLAST EXAMPLES

Example

3. The figure below shows a running track in the shape of a rectangle with semi-circles at each end.

2y

3x What is the distance around the track? A. 6x + 4y + 2π y B. 2π y + 6x C. π y2 + 3x D. π y2 + 6x Note that the straight distance around the track is 3x + 3x = 6x. Thus, the answer must involve 6x. Eliminate C.

Solution

The total distance around the track is: Straight + Distance around the Distance two semicircles

The straight distance is 3x + 3x or 6x and for convenience, you can think of the two semicircles as a complete circle of diameter d = 2y as shown.

2y

The circumference C of this circle is C = π d = π •2y = 2π y Thus, the total distance around the track is: Straight + Distance around the Distance two semicircles

6x + 2π y Since 6x + 2π y = 2π y + 6x, the answer is B.

To find the area of a plane figure such as a circle, square, rectangle, parallelogram or triangle, we need the formulas that follow. Note: When we measure area or surface area the result is in square units (abbreviated as sq. units or units2.)

4 FINDING AREAS RULE

The area of a circle of radius r is: 2rA π=

The area of a square of side s is: A = s2 The area of a rectangle of base b and height h is: A = bh The area of a triangle of base b and height h is:

A = 12 bh

EXAMPLES The area of a circle of radius 5 in. is:

A = π •52 sq. in. = 25π sq. in.

The area of a square of side 3 m is: A = 32 sq. m = 9 sq. m

The area of a rectangle of base 4 mm and height 8 mm is:

A = 4•8 sq. mm = 32 sq. mm

The area of a triangle of base 2 yd and height 4 yd is

A = 12 •2•4 = 4 sq. yd



CLAST EXAMPLE

Example 4. What is the area of a circular region whose

diameter is 6 centimeters? A. 36π sq. cm B. 6π sq. cm C. 12π sq. cm D. 9π sq. cm

Solution The formula to find the area of a circle is A = π r2. Since the diameter is 6 cm, the radius is half of that, or 3 cm. Thus, the area is: A = π •(3)2 sq. cm = 9π sq. cm. The answer is D. (Some books use cm2 but the CLAST uses sq. cm.)

In case the answer to the problem requires conversions (cm to meters or in. to ft), do the conversions before you use the correct area formula, as shown in the next example. CLAST EXAMPLE

Example

5. What is the area of a triangle with a base of

24 inches and a height of 12 inches? A. 12 sq. ft B. 1 sq. ft C. 24 sq. ft D. 144 sq. ft Note that the answers are given in square feet.

Solution

First, convert the original dimensions to feet. 24 in. = 2 ft, 12 in. = 1 ft. The area of the

triangle is A = 12 bh, where b = 2 and h = 1.

Thus, A = 12 •2•1 = 1 sq. ft.

The answer is B.

5 FINDING SURFACE AREAS The surface area of a rectangular solid of length L, width W, and height H is: A = 2HW + 2LH + 2WL

H

W L

Note that to obtain the surface area of the solid, we first need the area of the front and the back (2HW), the two sides (2LH) and the top and bottom (2WL). The surface area of a rectangular solid 4 ft long, 3 ft wide and 2 ft high is: A = 2•(2•3) + 2•(4•2) + 2•(3•4) = 12 + 16 + 24 = 52 sq. ft



CLAST EXAMPLE Example

6. What is the surface area of a rectangular solid that is 12 inches long, 5 inches wide and 6 inches high?

A. 360 cubic inches B. 324 square inches C. 324 cubic inches D. 360 square inches Note: Draw a picture whenever possible!

6

5 12

Solution

Since we are looking for a surface area the answer must be in square units (eliminate choices A and C). The area of the front is 6•5 = 30 sq. in. The area of the back is 6•5 = 30 sq. in. The area of the right side is 12•6 = 72 sq. in. The area of the left side is 12•6 = 72 sq. in. The area of the top is 5•12 = 60 sq. in. The area of the bottom is 5•12 = 60 sq. in. The total surface area is: 324 sq. in. The answer is B.

When we measure volume we measure the amount of space in a solid three dimensional object and the result is in cubic units, sometimes written as units3. In the U.S. system we have cubic inches, cubic feet and cubic yards. The metric system uses cubic centimeters, cubic meters and so on. The volume of liquids in the metric system is measured in liters.

T TERMINOLOGY -- METRIC CUBIC MEASURES METRIC UNITS

kiloliter hectoliter dekaliter liter (kL) (hL) (daL) L 1000 L 100L 10 L liter deciliter centiliter milliliter L (dL) (cL) (mL)

110 L

1100 L

11000 L

Chart: kL hL daL L dL cL mL

King Hendy Died Late Drinking Chocolate Milk

To convert from one metric unit to another is a matter of moving the decimal point. For example, to convert from kL to liters, we move three places right in the chart. Thus, to convert 37 kL to liters, we move the decimal point in 37. three places right obtaining 37. kL = 37,000 L. To convert 2647 mL to liters, we have to move the decimal point in 2647. three places left. Thus, 2647. mL = 2.647 L.

CLAST EXAMPLE

Example

7. Find the volume in centiliters (cl) of a 2.95 liter bottle.

A. 29.5 cl B. 295 cl C. 0.295 cl D. 2950 cl

Solution Since centi means 100, one liter = 100 cl. Thus, 2.95 liter s= 2.95•100 cl = 295 cl. The answer is B. You do not need any formulas but remember your prefixes like milli, centi, deci and kilo.



To find the volume of a rectangular solid (a box), right circular cylinder, circular cone or sphere we need the formulas that follow.

6 FINDING VOLUMES

H

W L

RULE The volume of a rectangular solid of length L, width W and height H is:

V = LWH

EXAMPLES The volume of a rectangular solid 10 ft long, 9 ft wide and 8 ft high is:

V = 10•9•8 cubic ft = 720 cubic ft

r

h

The volume of a right circular cylinder of radius r and height h is: hrV 2π=

The volume of a right circular cylinder of radius 5 cm and height 7 cm is:

hrV 2π= = π •52•7 cubic cm

= 175π cubic cm

r

h

The volume of a right circular cone of radius r and height h is:

hrV 2

31 π=

The volume of a right circular cone of radius 6 in and height 12 in is:

V = 13 π r2h =

13 π •62•12 cubic in.

= 144π cubic in.

r

The volume of a sphere of radius r is:

3

34 rV π=

The volume of a sphere of radius 3 m is

V = 43 π r3 =

43 π •33 cubic m

= 43 π •27 cubic m

= 36π cubic m

CLAST EXAMPLE

Example 8. What is the volume of a sphere with a 12 in.

diameter? A. 48π cubic in. B. 48π sq. in.

C. 288π cubic in. D. 288π sq. in. Caution: To use the formula for the volume of a sphere you need the radius r.

Solution Since the volume must be in cubic in., eliminate B and D. Since the diameter is 12 in, the radius is 6 in.

Substituting 6 for r in V = 43 π r3 we

obtain:

V = 43 π •216 cubic in = 288π cubic in.

The answer is C.



C. Identifying the Appropriate Unit of Measurement

Objective IIB4 CLAST SAMPLE PROBLEMS 1. Which unit of measurement would be used to give the amount of wall surface to be covered by

the contents of a can of paint? A. gallons B. liters C. square feet D. cubic feet 2. Which unit of measurement would be used to find the distance around a building? A. liters B. cubic feet C. square meters D. meters 3. What unit of measurement would be used to find the amount of paper needed for the label on the

can? A. inches B. square inches C. cubic inches D. fluid ounces

7 PROPER MEASUREMENT UNITS RULE

When measuring length use linear units (kilometers, miles, meters, yards, centimeters, inches, or millimeters) When measuring area use square units. (Square yards, square meters, square feet, square centimeters) When measuring volumes use cubic units. (Cubic centimeters, cubic inches, cubic feet, cubic meters, gallons, liters)

EXAMPLES

The diameter of a coin may be measured in inches or centimeters. The distance around a circular track may be measured in meters or yards. The amount of paper needed to wrap a gift may be in square inches. The amount of wall surface covered by a gallon of paint may be in square meters. The amount of gasoline in a fuel storage tank may be measured in cubic feet, cubic meters, gallons or liters.

CLAST EXAMPLE

Example 9. Which of the following would not be used to

measure the amount of water needed to fill a swimming pool?

A. Cubic feet B. Liters C. Gallons D. Meters

Solution We are to measure the volume of the water. Since volume is measured in cubic units, meters is not an appropriate unit of measure. The answer is D.

ANSWERS 1. C 2. D 3. B



Section 3.1 Exercises

WARM-UPS A

1. Round the measurement of the length of the key to the nearest inch.

2. Round the measurement of the length of the key to the nearest 12 inch.

CLAST PRACTICE A PRACTICE PROBLEMS: Chapter 3, # 1-2

3. Round the measurement of the length of the key to the nearest 14 inch.

A. 2 in. B. 214 in. C. 2

12 in. D. 2

34 in.

WARM-UPS B 4. Find the distance around the polygon in meters. 5. Find the distance around the polygon in kilometers.

150 m

250 m

300 m

200 m

6. Find the distance around the polygon in feet. 7. Find the distance around the polygon in yards.

30 ft

20 ft

35 ft

40 ft

8. Find the circumference of a circle of radius 8 cm. 9 Find the circumference of a circle whose diameter is 10 in. 10. Find the area of a circle of radius 10 m.



11. Find the area of a circle whose diameter is 12 ft. 12. Find the area of a rectangle of base 8 mm and height 10 mm. 13. Find the area of a rectangle of base 2 yards and height 4 yards. 14. Find the area of a triangle of base is 3 km and height 4 km. 15. Find the area of a triangle of base 2 ft and height 7 ft. 16. Find the surface area of a rectangular solid of length 8 cm, width 10 cm and height 7 cm. 17. Find the surface area of a shoe box that is 12 inches long, 8 inches wide and 4 inches high.

18. Find the volume of the shoe box in Problem 17. 19. Find the volume of a rectangular solid of length 30 mm, width 20 mm and height 10 mm. 20. Find the volume of a right circular cylinder of radius 8 in and height 9 in. 21. Find the volume of a right circular cylinder whose diameter is 10 cm and whose height is 12

cm. 22. Find the volume in cubic feet of a right circular cylinder of radius 6 feet and height 30 inches. 23. Find the volume in cubic feet of a sphere whose diameter is 24 inches. 24. Find the volume in cubic yards of a sphere whose radius is 6 feet. 25. Find the volume in centiliters (cl) of a 1.95 liter bottle. 26. Find the volume in centiliters (cl) of a 3.90 liter bottle.



CLAST PRACTICE B PRACTICE PROBLEMS: Chapter 3, # 3-8 27. Find the distance around the polygon in kilometers

A. 2300 km B. 23 km C. 2.3 km D. 230 km

600 m

400 m

600 m400 m

300 m

28. Find the circumference of a circle of radius 12 in. A. 24π sq. in. B. 144π in. C. 24π in. D. 22π in. 29. What is the area of a circular region of diameter 10 centimeters? A. 20π sq. cm B. 25π sq. cm, C, 10π sq. cm. D. 100π sq. cm. 30. What is the area of a circular region of radius 6 ft? A. 12π sq. ft B. 24π sq. ft. C. 36π sq. ft D. 144π sq. ft. 31. What is the area of a square that has sides 6 centimeters in length? A. 24 sq. cm B. 12 sq. cm C. 25 sq. cm. D. 36 sq. cm 32. Find the area of a rectangle that is 6 inches long and 4 inches wide. A. 20 sq. in. B. 24 sq. in. C. 10 sq. in. D. 30 sq. in. 33. Find the area of a triangle with base 8 in. and height 10 in. A. 40 sq. in. B. 80 sq. in. C. 40 in. D. 80 in. 34. Find the surface area of a rectangular solid that is 14 inches long, 5 inches wide and 4 inches high. A. 292 sq. in. B. 650 in. C. 900 sq. in. D. 650 sq. in.



35. Find the volume of a rectangular solid that is 7 cm long, 5 cm wide and 4 cm high. A. 140 sq. cm B. 168 cubic cm C. 140 cubic cm D. 166 cm 36. Find the volume of a right circular cylinder 7 inches high and with a 5 in. radius. A. 35π cubic in. B. 42π cubic in. C. 175π sq. in. D. 175π cubic in. 37. Find the volume of a right circular cone of radius 9 ft and height 10 ft. A. 270π sq. ft B. 810π cubic ft C. 270π cubic ft D. 870π sq. ft 38. Find the volume of a sphere with a radius of 9 mm. A. 108π sq. mm B. 972π cubic mm C. 18π mm D. 729π cubic mm WARM-UPS C 39. What type of measure is needed to express the circumference of a circle? 40. What type of measure is needed to express the diameter of a coin? 41. What type of measure is needed to express the area of a rectangle? 42. What type of measure is needed to express the area of a triangle? 43. What type of measure is needed to express the amount of water in a pool? 44. What type of measure is needed to express the amount of gas in your gas tank? CLAST PRACTICE C PRACTICE PROBLEMS: Chapter 3, # 9-11 45. What type of measure is needed to express the length of the line

segment AC of the given rectangle?

A. Square B. Linear

C. Cubic D. Equilateral

A B

C D

46. What type of measure is needed to express the volume of the rectangular solid?

A. Square B. Linear


H

W L



47. What type of measure is needed to express the surface area of the rectangular solid?

A. Square B. Linear


H

W L

48. What type of measure is needed to express the thickness of a coin? A. Square B. Linear C. Cubic D. Equilateral 49. What type of measure is needed to express the amount of paint in a can? A. Square B. Linear C. Cubic D. Equilateral 50. What type of measure is needed to express the amount of sand in a box? A. Square B. Linear C. Cubic D. Equilateral EXTRA CLAST PRACTICE 51. What is the area of the figure in square meters? A. 1400 square meters B. 14 square meters C. 14 square meters D. 24 square meters

10m 140 cm

52. What is the area of a triangle whose base is 36 feet and whose height is 40 inches? A. 60 square feet B. 180 square feet C. 360 square feet D. 720 square feet



3.2 APPLICATIONS AND PROBLEM SOLVING The information we have learned in Section 3.1 will now be used to solve problems involving perimeters, areas, and volumes. In addition, we shall learn how to select and create new formulas to measure other geometric figures. If you have forgotten about the RSTUV procedure for solving problems, review it before you go on. A. Applications and the Pythagorean Theorem

Objectives IVB1, IVB2 CLAST SAMPLE PROBLEMS 1. A new street connecting A Street and B Avenue is to be

constructed. Construction costs are about $100 per linear foot. What is the estimated cost for constructing the new street?

2. A football player runs 8 yards to the right as shown. He

then turns left, runs down field and catches a 10 yard pass thrown from the spot from where he started. How far down field did he run?

5 miles

12 milesNew

A Street

B Avenue

10 yd

8 yd

d = ?

CLAST EXAMPLES

Example

1. What will be the cost of tiling a room

measuring 12 feet by 15 feet if square tiles costing $2 each and measuring

12 inches on each side are used? A. $180 B. $4320 C. $360 D. $3600 Note: The measurements of the tile and of the room are not in the same units. It is easier to change the measurement of the tile to feet. Since 12 inches = 1 ft, the tile measures 1 ft on each side. If the tiles were 8 in. on each side, you would need to convert 8 in. to ft.

Solution 1. Read the problem. You are given the dimensions of the room and the cost of each tile. 2. Select the unknown. We want to find the cost of tiling the room. 3. Think of a plan. Find how many tiles are needed and multiply by the cost of each tile ($2). 4. Use the fact that we need to cover a rectangle measuring 12 tiles by 15 tiles. Thus, we need 12 × 15 = 180 tiles. Total cost = Cost of each tile × 180 = $2 ×180 or $360 5. Verify your multiplication. The answer is C.

ANSWERS 1. $6,864,000 2. 6 yards

SECTION 3.2 Applications and Problem Solving 147


Example 2. Find the cost of building a rectangular

driveway measuring 12 feet by 30 feet if concrete costs $12.50 per square yard.

A. $4500 B. $400 C. $1500 D. $500

Solution Since the cost is per sq. yd, we change 12 ft to 4 yd and 30 ft to 10 yd. In square yards, the area of the driveway is: 4 yd × 10 yd = 40 sq. yd. Total Cost = Cost per sq. yd × sq. yd needed = $12.50 × 40 = $500 The answer is D.

Some applications in the CLAST require the use of the Pythagorean theorem. 1 PYTHAGOREAN THEOREM

RULE For any right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. In symbols, c2 = a2 + b2 Note: In a right triangle (a triangle with a 90 degree angle) the hypotenuse c is the side opposite the 90 degree angle.

a

bc

EXAMPLES A jogger runs 4 miles east and then 3 miles north. How far is the jogger from his starting point? If we let a = 4 and b = 3 in the diagram, we have to find the distance c. By the Pythagorean theorem, c2 = a2 + b2 Thus, c2 = 32 + 42 c2 = 9 + 16 c2 = 25 c = 5

Thus, he is 5 miles from the starting point.

CLAST EXAMPLE

Example 3. A television antenna 12 ft high is to be

anchored by three wires each attached to the top of the antenna and to points on the roof 5 ft from the base of the antenna. If wire costs $0.75 per foot, what will be the cost of the wire needed to anchor the antenna?

A. $27 B. $29.25 C. $9.75 D. $38.25 Hint: After you read the problem, see if you can draw a picture of the situation.

12 ft

5 ft

c

Solution 1. Read the problem. The antenna is 12 ft high and each wire is 5 ft from the base. The cost of each foot of wire is $0.75. 2. Select the unknown. We need to find the length of each wire, multiply by 3 (we need 3 wires) and then by $0.75 (each foot of wire is $0.75) 3. Think of a plan. First, find the length of each wire. Draw a picture like the one at left. 4. Use the Pythagorean theorem c2 = 52 + 122 c2 = 25 + 144 c2 = 169 c = 13 Thus, each wire is 13 ft long. Since we need three wires, we need 3 ×13 ft = 39 ft. The total cost would be $0.75 ×39 = $29.25. The answer is B.



B. Inferring and Selecting Geometric Formulas

Objectives IIIB1, IIIB2 CLAST SAMPLE PROBLEMS 1. In the figure, S represents the sum of the measures of the interior angles. Study the figure and

calculate the sum S of the measures of the interior angles of an eight-sided convex polygon.

3 sides 4 sides 6 sides 1 triangle 2 triangles 4 triangles S = 180o S = 360o S = 720o 2. The figure consists of a rectangle and a semi-circle of

radius r. Determine the formula for finding the area of the shaded region.

Many geometric measurements can be generalized to be used in more complicated situations. Look at the three spheres with radii 1, 2 and 3 units and surface areas (SA) 4π , 16π and 36π , respectively. The CLAST asks us to generalize this information and find the surface area of a sphere with a radius of 4 units. To solve this problem set up a table and see if there is a pattern. Radius Surface Area 1 4π = 1•4π 2 16π = 2•8π 3 36π = 3•12π 4 = 4•?

r = 1

r = 2

r = 3 SA = 4π SA = 16π SA = 36π

The underlined numbers are 4π , 8π and 12π . The next number in the pattern is 16π so the surface area of a sphere with radius 4 should be 4•16π or 64π . (You can also reason that since the surface area is measured in square units, the pattern should involve r2, the square of the radius, that is 12•? = 4π , 22•? = 16π , 32•? = 36π , where we multiply the square of the radius by 4π . For a radius of 4 units, the answer should be 42 • 4π = 64π .)

ANSWERS 1. S = 1080o 2. 2r2 -

2

2rπ



CLAST EXAMPLE

Example

4. In each of the figures, S represents the shaded area when a triangle is removed from a square

Side = 4 S = 8

Side = 6 S = 18 Side = 8

S = 32 Find S if the side of the square is 10 A. 100 B. 75 C. 50 D. 40

Solution

First find the areas of each of the squares. The area of the first square is 42 = 16 The area of the second square is 62 = 36 The area of the third square is 82 = 64 Now, from the diagram, S = 8, 18 and 32. What is the relationship between the area of the squares and the area S? Area of squares 16 36 64 Area of S 8 18 32 In each case, S is half of the area of the corresponding square. If the side of a square is 10, its area is 100 and S is half of that or 50. The answer is C.

CLAST EXAMPLE

Example

5. Study the information given with the regular hexagons. Then calculate the height h of the corresponding triangle in a regular hexagon with each side 4 units long.

h

b h

b h

b s = 1 unit s = 2 units s = 3 units

h = 3

2 units h = 3 units h = 3 3

2 units

A. 2 3 units B. 3 3 units

C. 4 3 units D. 5 3

2 units

Solution Look at the pattern involving the length s of the side of the hexagon and the height h. Length of side Height

1 1• 3

2

2 2• 3

2 = 3

3 3• 3

2 = 3 3

2

In each case multiply the length s of the side by 3

2 . For a hexagon whose side measures 4

units, the height is 4•3

2 = 2 3 . The answer is

A. Note: We were looking for a linear pattern, so we used a linear factor.



Some of the patterns appearing in the CLAST involve the measure of an angle which is in degrees. You should know that the sum of the measures of the angles of any triangle is 180o. (Read "180 degrees".) CLAST EXAMPLE

Example

6. Study each figure below and then calculate

S, the sum of the measures of the interior angles of a 9-sided convex polygon.

4 sides 5 sides 6 sides 2 triangles 3 triangles 4 triangles S = 360o S = 540o S = 720o A. 14400 B. 1260o C. 1080o D. 1000o

Solution

We want to discover a pattern involving the number of sides, the number of triangles and S, the sum of the measures of the angles. No. of sides No. of triangles S 4 2 2•180o = 360o

5 3 3•180o = 540o 6 4 4•180o = 720o 9 ? ?•180o

Note that the number of triangles is 2 less than the number of sides. If the number of sides is 9, the number of triangles is 9 - 2, or 7, and S = 7•180o = 1260o. The answer is B.

So far we have inferred new formulas for different geometric figures. A different CLAST skill requires you to calculate the measure of a geometric figure composed of several figures for which the formulas are known. Here is the rule you need.

2 FORMULAS FOR AREAS AND VOLUMES RULE

1. Write the formulas necessary to find the area or volume of the individual figures. 2. Use the figure to find the variables to be used in the formula. 3. Simplify the area or volume formula. 4. Find the sum of the areas or volumes. 5. If the area (or volume) to be found is shaded, you may be able to find it by calculating the difference between the areas or volumes. (See Problems 16, 30 and 31.)

EXAMPLES Study the figure of a right circular cylinder with a hemisphere (half a sphere) mounted on top. Find the volume.

h

r

h

r

r

The volume of the sphere is

43 π r3. One half

of this is 12 ×

43 π r3 =

23 π r3 The volume of

the cylinder is π r2h. Thus the total volume is

the indicated sum 23 π r3 + π r2h.



CLAST EXAMPLE

Example 7. The figure shows a regular hexagon. Select

the formula for computing the total area of the hexagon.

h

b A. Area = 3h + b B. 6(h + b) C. 6hb D. 3hb

Solution The total area of the hexagon is composed of 6 identical triangles as shown. The area

h

b of each triangle is

12 bh. Since there are 6 of

them, the total area is 6•12 bh = 3bh.

Since 3bh = 3hb, the answer is D. Section 3.2 Exercises

WARM-UPS A 1. Find the cost of tiling a room measuring 12 feet by 18 feet if square tiles costing $2 each and

measuring 8 in. by 8 in. are used.

2. A gardener wants to prepare a flower bed measuring 10 feet by 9 feet and 6 inches deep. How many cubic yards of soil are needed to do this?

3. Roofing material come in 100 sq. ft. bundles costing $90 each. How much would the material

cost to repair a roof measuring 15 ft by 20 ft? 4. How many square yards of wallpaper costing $10 per roll are needed to cover an area 10 ft

long and 8 ft high? 5. A baking pan in the shape of a rectangular solid is 20 cm long, 10 cm wide and 4 cm deep.

How many liters will it hold? Hint: 1 liter = 1000 cc. 6. What will be the cost of carpeting an office measuring 12 ft by 18 ft if carpeting costs $15 per

square yard? 7. How much would it cost to fence a 30 ft by 60 ft yard if fencing material costs $12 per linear

foot? 8. A paper cup is in the shape of a right circular cone 10 cm high and 6 cm in diameter. How

many cubic centimeters does the cup hold?



9. A radio station is going to construct a 16 foot antenna anchored by three cables, each attached to the top of the antenna and to points on the roof of the building that are 12 feet from the base of the antenna. What is the total length of the three cables?

10. An airplane leaves the airport and flies 40 miles north. It then flies 30 miles west. How far

from the airport is the plane? 11. The door in a warehouse is 9 ft tall and 48 inches wide. If a sheet of paneling is 10 ft long,

what is the widest it can be and still fit through the door? 12. A contractor wants to put a sidewalk in front of the

triangular garden shown. If it costs $50 per linear foot to build the sidewalk, how much will the job cost?

10 yd 8 yd

Front

CLAST PRACTICE A PRACTICE PROBLEMS: Chapter 3, # 12-15 13. A farmer wants to build a fence to enclose a rectangular area 20 ft long by 12 ft wide. How

many feet of fencing will be needed? A. 30 ft B. 23 ft C. 64 ft D. 112 ft 14. Find the cost of carpeting an office that measures 21 ft by 24 ft if carpet costs $13 per square

yard. A. $6552 B. $2184 C. $1092 D. $728 15. The outside dimensions of a picture frame are 4 ft by 30 inches. If its inside dimensions are

234 ft by 21 inches, find the area of the frame.

A. 759 sq. in. B. 747 sq. in. C. 67.50 sq. ft D. 182.25 sq. ft 16. A gardener uses a six-inch mulch border around a garden

measuring 6 ft in diameter. What is the area of the border? A. 3.2 sq. in. B. 5π sq. ft C. 2.75π sq. ft D. 2.75 sq. ft

6 feet

6 inches

17. The city commission wants to construct a new street that

connects Main Street and North Boulevard as shown in the diagram. If the construction cost is $90 per linear foot, find the estimated cost for constructing the street. Hint: 1 mile = 5280 ft.

A. $33,808 B. $3690 C. $2,376,000 D. $475,200

ES

NNorth Blvd

Main St

New3 miles

4 mi



WARM-UPS B

18. Study the information given with the right triangles then calculate the area of a right triangle with hypotenuse 10 and base 5 3 .

2

3

4

2 3

6

3 3

8

4 3

A = 12 • 3 •1 A =

12 •2 3 •2 A =

12 •3 3 •3 A =

12 •4 3 •

19. Study the given figures then find the area of a the regular 6- sided polygon.

h

b h

b h

b 4 sides 5 sides 6 sides

A = 2 bh A = 52 bh A = ?

20. Study the figure formed from a semicircle and a triangle.

Then give the formula for calculating the total area of the figure.

h

r

21. 22.

Study the figure showing a square with a semicircle attached to each side. Find the formula for computing the perimeter of the figure. Find the formula for the total area of the figure in Problem 21.

s



CLAST PRACTICE B PRACTICE PROBLEMS: Chapter 3, #16-20

23. Study the information given with the regular hexagons, then calculate the area A of a regular hexagon with a side equal to 8. A. 96 3 B. 72 3 C. 48 3 D. 36 3

h

b h

b h

b

Side 2 Side 4 Side 6 A = 6 3 A = 24 3 A = 54 3

24

Study the information given for the 3, 4 and 6 sided figures. If S represents the sum of the measures of the interior angles, find S for a 10 sided polygon. A. 1080o B. 1800o C. 1440o D. 1260o

3 sides 4 sides 6 sides 1 triangle 2 triangles 4 triangles S = 180o S = 360o S = 720o

25 The given figure is formed by joining a triangle and a square as shown. Study the figure then select the formula for computing its total area. A. A = s4 B. A = s3

C. A = 3s2

4 + s2 D. A = 5 s2

4

s s

s

s 26.

A cylindrical container can be formed from two tops and a rectangle as shown. Based on the figure, what is the formula for calculating the surface area SA of the right circular cylinder? A. SA = 2(2π r) + rh B. SA = π r2h C. SA = 2π r2 + 2π rh D. SA = 2π r2 + 2rh

h

r

r

r

h



27.

The figure shows a rectangular box without a top formed by cutting out squares x units on each side from a rectangular piece of length L and width W. Select the formula for calculating the surface area SA of the box. A. SA = LW - 4x2 B. SA = LW - x2 C. SA = (L - 2x)W - 2x) - 4x2 D. SA = (L - 2x)(W - 2x)

28. Study the figure showing a regular pentagon. Then select the formula for computing the total area A of the pentagon.

A. A = 52 hb B. A = 5(h + b)

C. A = 5h + b D. A = 5hb

h

b 29. Study the figure showing a trapezoid,

then select the formula for computing the total area A of the trapezoid, if B = x + b + y. A. A = b(B - x - y) + xh2 B. A = h(B + b) C. A = b(B - x - y)h

D. A = 12 h(B + b)

h h

bx y

30. Study the figure showing a square inscribed in a circle. Select the formula for computing the area of the shaded region. A. A = π r2 - s2

B. A = π s2

2 - s2

C. A = s2

2

D. A = πs2

2

s

x

x

W

L



31.

Study the figure showing a triangle inscribed in a circle. Select the formula for computing the area of the shaded region. A. A = π r2 - 2r2 B. A = π - 2 C. A = π D. A = π r2 - r2

r

EXTRA CLAST PRACTICE

32. Study the information given in the

figures and then calculate A, the exterior angle in the last figure. A. 65o B. 135o C. 110o D. 115o

60

70

130

90

25

115 o

o

o

o

o o

o

o o 10020

120

65

45

A o

o

33. The base of a tool shed is to be a slab of concrete 15 feet long by 12 feet wide by 6 inches thick. If one cubic yard of concrete costs $39, how much will the concrete for the base of the tool shed cost? A. $3510 B. $1560 C. $130 D. $65

34. A rectangular flower bed measures 212 feet by 50 inches. The outside dimensions of a path

around the bed are 4 ft by 60 inches. What is the area of the path? A. 115 square feet B. 200.50 square feet C. 1200 square inches D. 1380 square inches 35. The diameter of a tree trunk is 2.4 meters. What is its circumference? A. 0.72π square meters B. 1.2π meters C. 2.4π meters D. 2.88π square meters

SECTION 3.3 Lines, Angles and Triangles 157


3.3 LINES, ANGLES AND TRIANGLES Before we proceed to discuss the different types of lines, angles and triangles we need to learn some of the basic terminology. T TERMINOLOGY--TYPES OF ANGLES

PLANE ANGLES A plane angle is a figure formed by two rays with a common endpoint called the vertex. The rays AB and AC are the sides of the angle and the angle can be named ∠ 1 (read "angle 1"), ∠ CAB , ∠ BAC or ∠ A

EXAMPLES

A

vertex

B

C

1

The vertex of angle BAC is A.

Angles are measured by the amount of rotation needed to turn one side of an angle so that it coincides with (falls exactly on top of) the other side. Note that ∠ YXZ. is "greater than" ∠ BAC.

AB

C

X Y

Z

The rotation of an angle is measured in degrees and is denoted by m∠ BAC. One complete revolution is 360o (read "360

degrees") and 1

360 of a complete revolution is

1o. The CLAST sometimes states that the measure of ∠ A, for example, is represented by "x".

A complete revolution is 360o

xA

B

C

30°A

B

C

m∠ BAC = 30o x = 30o

One-half of a complete revolution is 180o and results in an angle called a straight angle.

180°

One-quarter of a complete revolution is 90o and results in an angle called a right angle (denoted by using the little square ).

90°

An acute angle is an angle whose measure is less than 90o.

An obtuse angle is an angle whose measure is greater than 90o and less than 180o.



We now give several relationships between angles. T TERMINOLOGY--COMPLEMENTARY, SUPPLEMENTARY, VERTICAL ANGLES

COMPLEMENTARY ANGLES Two angles whose measures add to 90o are called complementary angles. Note that m∠ A + m∠ B = 90o or m∠ A = 90o - m ∠ B

EXAMPLES

A

B

SUPPLEMENTARY ANGLES Two angles whose measures add to 180o are supplementary angles. Note that m∠ A + m∠ B = 180o or m ∠ A =180o - m∠ B

A

B

VERTICAL ANGLES When two lines intersect, the opposite angles are called vertical angles and they have the same measure. If x, y, u and v are the measures then x = y and u = v.

x y uv

We are now ready to study the properties existing between different plane figures such as angles and triangles. Before we do that, let us see what type of questions we will encounter.

A. Properties of Plane Figures

Objective IIB2 CLAST SAMPLE PROBLEMS 1. What type of triangle is Δ ABC below?

45

65o

o

A B

C In the figure to the right:

2. Find two right angles 3. Find two pairs of complementary angles 4. Find two pairs of vertical angles that are not right angles 5. Which geometric figure has all of the following

characteristics? i. quadrilateral ii. opposite sides are parallel iii. diagonals are not equal

123

45

6

ANSWERS 1. Scalene 2. 1 and 4 3. 5 and 6; 2 and 3 4. 3 and 6; 2 and 5 5. Rhombus



CLAST EXAMPLE

Example 1.

1

234

56

Which of the following pairs of angles are complementary? A. 1 and 2 B. 5 and 2 C. 3 and 5 D. 4 and 1

Solution Complementary angles are angles whose sum is 90o, so 1 can not be complementary to any other angle. (Eliminate answers A and D.) Now, 5 and 6 are complementary, but this is not one of the choices. However, angles 6 and 3 are vertical angles, so they must have the same measure. Since angles 5 and 6 are complementary and angles 3 and 6 have equal measures, ∠ 5 and ∠ 3 are complementary. The answer is C. If you are not convinced, here is the proof: m∠ 5 + m∠ 6 = 90o m∠ 6 = m∠ 3 m∠ 5 + m ∠ 3 = 90o, so ∠ 3 and ∠ 5 are complementary.

Triangles can be classified according to their angles (right, acute or obtuse) or according to the number of equal sides (scalene, isosceles, or equilateral).

T TERMINOLOGY -- CLASSIFYING TRIANGLES BY ANGLES AND SIDES CLASSIFICATION

A right triangle is a triangle containing a right angle.

EXAMPLES

An acute triangle is a triangle in which all the angles are acute.

An obtuse triangle is a triangle containing an obtuse angle.

A scalene triangle is a triangle with no equal sides. Note that the sides are labeled |, || and ||| to indicate that the lengths of the sides are different.



T TERMINOLOGY--CLASSIFYING TRIANGLES BY ANGLES AND SIDES (CONT.) An isosceles triangle is a triangle with two equal sides and two equal base angles. Note that the equal angles are indicated by arcs, the equal sides by slashes.

An equilateral triangle is a triangle with three equal sides and three 60o angles. As before the equal angles are indicated by arcs and the equal sides by slashes.

Next, we give the rules used to name specific triangles, and a very important rule that applies to the measures of the sum of the angles of any type of triangle. Do you remember what that is?

T TERMINOLOGY--NAMING TRIANGLES NAMING TRIANGLES

Triangles are named using the letters at the vertices and the symbol Δ . The given triangle is Δ ABC (read "triangle ABC").

EXAMPLES

A

BC

1 SUM OF THE ANGLES OF A TRIANGLE RULE

The sum of the measures of the angles of a triangle is 180o. You can use this fact to find the measure of the third angle when the measures of the other two angles are given.

60 35o o

The measure of the other angle, call it x, is such that: 60o + 35o + x = 180o or 95o + x = 180o x = 85o

CLAST EXAMPLE

Example 2. What type of triangle is Δ ABC?

oo55 70A B

C A. Isosceles B. Right C. Equilateral D. Scalene

Solution First, we have to determine the measure "x" of the missing angle. Since the sum of the three angles in a triangle is 180o, we have: 55o + 70o + x = 180o 125o + x = 180o x = 180o - 125o x = 55o Since two of the angles are of equal measure, the triangle is an isosceles triangle. Thus, A.



In the preceding sections we have mentioned geometric figures such as squares, circles, rectangles and polygons. These figures have definite properties that should be familiar to you. We shall give the definitions and examples for several polygons and quadrilaterals.

T TERMINOLOGY--TYPES OF POLYGONS TYPES OF POLYGONS

A polygon is a closed figure whose sides are line segments that do not cross each other.

EXAMPLES

A regular polygon is a polygon with all sides of equal length and all angles of equal measure. They are usually named by using the number of sides. For example: Pentagon is a 5-sided polygon. Hexagon is a 6-sided polygon. Octagon is an 8-sided polygon.

A

B

CD

E

A

B

CD

E

F

A quadrilateral is a four-sided polygon (Squares and rectangles are quadrilaterals).

A trapezoid is a quadrilateral with one pair of parallel sides.

A parallelogram is a quadrilateral in which the opposite sides are parallel.

A rhombus is a parallelogram with all sides equal.

A rectangle is a parallelogram with a right angle.

A square is a rectangle with all sides equal.

A diagonal is a line segment joining two non-consecutive vertices of a polygon.



The CLAST emphasizes the characteristics of these polygons as shown in the example.

CLAST EXAMPLE

Example 3. Select the geometric figure that possesses all

of the following characteristics: i. polygon ii. quadrilateral iii. two and only two sides are parallel. A. Parallelogram B. Rectangle C. Rhombus D. Trapezoid

Solution First note that all of the responses give figures that are polygons and quadrilaterals, so we have to concentrate on the figure having two and only two parallel sides. The parallelogram, the rectangle and the rhombus have more than two parallel sides. The only quadrilateral with exactly one pair of parallel sides is the trapezoid. The answer is D.

B. Relationships Between Angle Measures

Objective IIB1 CLAST SAMPLE PROBLEMS Referring to the diagram on the right m ∠ g = 130o: 1. Which angles are supplements of g? 2. Find m∠ f, m∠ h and m∠ j 3. Find three angles with the same measure as m ∠ c 4. Find two pairs of supplementary angles that are not right

angles.

b c f gd e h j

a

The material we have studied can be used to find additional relations between angles. One of these relations is congruence as detailed next.

T TERMINOLOGY--CONGRUENT ANGLES AND SIDES CONGRUENT ANGLES

Angles ∠ 1 and ∠ 2 are congruent, denoted by ∠ 1 ≅ ∠ 2, means that m ∠ 1 = m ∠ 2. The congruence symbol should remind you of "=".

EXAMPLES

1

2

3

B

CA The angles of an equilateral triangle are congruent, that is, ∠ 1 ≅ ∠ 2 ≅ ∠ 3

CONGRUENT SIDES Two sides of a triangle are congruent, denoted

by AB___

≅ AC___

means that side AB___

is the

same length as side AC___

.

In the equilateral triangle above

AB___

≅ B C___

≅ AC___

ANSWERS 1. f and j 2. 50o, 130o, 50o 3. b, d, e 4. f and g, g and j, j and h, f and h



CLAST EXAMPLE

Example

4. In Δ ABC, AB___

≅ CB___

. Which of the following statements is true for the figure shown? (The measures of the angles are represented by x, y, z as indicated.)

100°

A

yC

B

x

z

w

A. z = 80o B. z = x C. y = 80o D. z = 10o

Solution

AB___

≅ CB___

means that the line segments AB___

and CB___

are the same length. This makes the triangle an isosceles triangle, so y = x. Now, x + 100o = 180o and x = 80o

Since y = x = 80o, y = 80o The correct answer is C.

The CLAST combines the idea of perpendicular and parallel lines to establish other relationships between the measures of angles. Here are the concepts we need. T TERMINOLOGY--PARALLEL AND PERPENDICULAR LINES

PARALLEL LINES

Lines AB___

and CD___

are parallel, denoted by

AB___

|| CD___

, if the lines AB___

and CD___

are in the same plane and do not intersect..

EXAMPLES A B

C D

E

F

G

H

These pairs of parallel lines are denoted by

AB___

|| CD___

and EF___

|| GH___

.

PERPENDICULAR LINES

Lines AB___

and CD___

are perpendicular,

denoted by AB___

⊥ CD___

, if they intersect at right angles.

A BC

D

Lines AB___

and CD___

are perpendicular.



CLAST EXAMPLE

Example

5. Given that CD___

|| BE___

and AE___

⊥ CD___

which of the following statements is true for the figure at the right? (The measure of angle DCB is represented by "v".)

A. w = x B. y = x C. ∠ EDC and ∠ DEB are vertical angles D. ∠ DCB and ∠ DEB are supplementary angles

Solution

x

o

A

B

C D

Eu

v w

135y z

Since y and the 135o angle are supplementary, y = 45o. Now, z = 90o and x + y + z = 180o. For y = 45o, z = 90o, x + 45o + 90o = 180o x + 135o = 180o x = 45o

Since both x and y are 45o, y = x and the correct answer is B.

There is one more important fact concerning parallel lines and the measures of related angles. Here it is: T TERMINOLOGY--CORRESPONDING, ALTERNATE AND VERTICAL ANGLES

ANGLE RELATIONSHIPS If L1 and L2 are parallel lines crossed by a transversal as shown in the Example, the following relationships hold: Corresponding angles are congruent. ∠ 1 ≅ ∠ 5, ∠ 3 ≅ ∠ 7, ∠ 2 ≅ ∠ 6 and ∠ 4 ≅ ∠ 8 These are the pairs of angles that lie on the same side of the transversal. Alternate interior angles are congruent. ∠ 3 ≅ ∠ 6, and ∠ 4 ≅ ∠ 5. These are the pairs of angles between the parallel lines and on opposite sides of the transversal Vertical angles are congruent. ∠ 1 ≅ ∠ 4, ∠ 2 ≅ ∠ 3, ∠ 5 ≅ ∠ 8, and ∠ 6 ≅ ∠ 7

EXAMPLES

1 2

3 4

5 6

7 8

L1

L 2

Here is a quick way of remembering all these facts: Look at angles 1 and 2. Angle 1 is acute and 2 is obtuse. Think of ∠ 1 as “small" and ∠ 2 as "big". All acute ("small" ) angles are congruent. Thus, ∠ 1, ∠ 4, ∠ 5, and ∠ 8 are congruent. All obtuse ("big") angles are congruent Thus, ∠ 2, ∠ 3, ∠ 6 and ∠ 7 are congruent.



CLAST EXAMPLES

Example 6. Which statement is true for the figure shown

at right given that L1 and L2 are parallel lines?

A. Since m∠ T = 750, m∠ S = 60o B. Since m∠ T = 750, m∠ S = 105o C. m∠ V = m∠ R D. None of the statements is true.

Solution

P R75°

Q

W XY 45°

L1

L 2S T

U V

Look at the transversal on the left m∠ T = 75o

(Remember, all the "small" angles are congruent). Now, ∠ S and ∠ T are suppplementary, so m∠ S = 180o - m ∠ T = 180o - 75o = 105o.This is answer B.

Example

7. If L1 || L2 and L3 || L4, which of the following statements is true?

40 o L

L

1 2

3

4

x y

z r s t

LL

A. z = 40o B. r = 140o C. z = 140o D. s = t

Solution x = 40o because x and the 40o angle are corresponding angles. y = 180o - 40o = 140o, since x and y are supplementary. z = y = 140o, since z and y are alternate interior angles (remember "big" angles are congruent) The correct answer is C.



C. Similar Triangles Objective IIB3 CLAST SAMPLE PROBLEMS In the figure to the right, the measure of angle A is x

1. x = ? . 2. CBAB =

? AD 3.

BABD =

? ?

4. If the measure of angle A is z, what is AC ?

E

D

A

B

C

30

30

o

ow

xy

z

5

4

7.5

AC

BD E120o 120o

x

yz

Geometric figures with exactly the same shape, but not necessarily the same size, are called similar figures. If you know that two triangles are similar, there are several relationships you should know. We list them next. T TERMINOLOGY--SIMILAR TRIANGLES DEFINITION OF SIMILAR TRIANGLES

When two triangles are similar, denoted by Δ ABC ~ Δ DEF

(1) Corresponding angles are equal and (2) Corresponding sides are proportional. To remember the information in (1) note:

Δ ABC ~ Δ DEF

∠A ≅ ∠D, ∠B ≅ ∠E, ∠C ≅ ∠ ∠ F. Match first and first (A and D), second and second (B and E), third and third (C and F). To remember the information in (2), note:

Δ ABC ~ Δ DEF match AB___

and DE___

.

Δ ABC ~ Δ DEF match BC___

and EF___

.

Δ ABC ~ Δ DEF match AC___

and DF___

.

EXAMPLES

2

3

4

A C

B

6

8

F

E

4

D The two triangles above are similar. (1) ∠ A ≅ ∠ D, ∠ B ≅ ∠ E, ∠ C ≅ ∠ F,

(2) AB___

DE___

=

BC___

EF___ =

CA___

FD___ =

24 =

48 =

36

Δ ABC ~ Δ ADE

A

B C

D E

In the figure at right, if BC is parallel to DE

ANSWERS 1. y 2. CE 3.

BCBE 4. AC = 6



The CLAST asks two types of questions about similar triangles: 1. From a group of triangles, select the ones that are similar 2. Find specific measures for angles or sides of one of two similar triangles. ÜCLAST EXAMPLES

Example

8. Which triangles are similar: A.

5 5

555° 55°

Solution

To find the correct answer, we have to check each of the responses. The first triangle in A is an equilateral triangle. For the second triangle to be similar, it has to be equilateral and the third angle must also be 55o. But 55o + 55o + 55o = 165o and not 180o, so the second triangle is not equilateral. A is not the answer.

B.

5

7

9

630°

C.

A

C

BED

25o

60o

D.

A

CB55°

80°

S

Q

R

45°

80°

If the triangles are similar, the sides must be

proportional. Thus, 69 =

57 . But this is not

true, so the triangles are not similar. The two triangles in C are obviously not similar since corresponding angles are not equal. The correct answer must be D. Let us see why. The measure of the angles in the first triangle are 55o and 80o, so the measure of the third angle must be 180o - 80o - 55o = 45o. But the measures of the angle in the second triangle are 80o and 45o, so the third angle must be 55o, making the corresponding angles equal and the triangles similar.



Example 9. Which statements are true for the pictured

triangles?

A

B

C

D

E12

3

5

i. m ∠ 1 = m ∠ 2 ii. AB___

ED___

=

AC___

EC___

iii. AB___

is parallel to ED___

A. i only B. ii only C. i and ii only D. i, ii, and iii

Solution Since ∠ 1 and ∠ 2 are vertical angles, they have the same measure, so i is true.

Triangle BAC and triangle DEC have equal corresponding angles (the 90o angle and angles 1 and 2, which make the third angles equal). Thus, Δ BAC ~ Δ DEC which makes statement ii true.

Finally, AB___

is parallel to ED___

because

AB___

⊥ BD___

and DE___

⊥ BD___

. This makes iii a true statement and the correct answer is D.

Example Solution 10. Which statements are true for the pictured

triangles?

B

A

C

7.5

5 4

E

D40°

40° GH

i. m ∠ A = m∠ E ii. AC___

= 6

iii. CE___

CA___

=

CB___

CD___

A. i only B. ii only C. i and ii only D. i, ii, and iii

Since m ∠ G = m ∠ H (vertical angles) and m ∠ D = m ∠ B = 40o, the third angles must be equal. Thus, m ∠ A = m ∠ E, and i is true. Since Δ CDE ~ Δ ABC, corresponding sides are proportional. Thus,

45 =

AC___

7.5 or AC

___ =

4•7.55 =

305 = 6

hence ii is true. Since the two triangles are similar,

CE___

CA___

=

CD___

CB___ , so iii is not true.

The correct answer is C, i and ii are the only true ones.



Section 3.3 Exercises

WARM-UPS A 1. Name two pairs of complementary angles in the figure. 2. Name a pair of supplementary angles in the figure. 3. Name two pairs of vertical angles in the figure. 4. Name two right angles in the figure. 5. Name two acute angles in the figure. 6. Is there an obtuse angle in the figure?

1

23

4

56

7. Find m ∠ ABC 8. Find m ∠ 3 9. Find x 10. Find y and z

A

B

C 3

x

60 o 60 o 50 o 30o100 o

60o o 75

y

z

Triangle 1 Triangle 2 Triangle 3 Triangle 4 In Exercises 11-14, classify the triangles according to their angles and sides. 11. What type of triangle is triangle 1? 12. What type of triangle is triangle 2? 13. What type of triangle is triangle 3? 14. What type of triangle is triangle 4? 15. Name the geometric figure that is a parallelogram in which the diagonals are perpendicular and

of the same length. 16. Name a quadrilateral in which only two sides are parallel.



CLAST PRACTICE A PRACTICE PROBLEMS: Chapter 3, # 21-22 17. Which of the following pairs of angles

are supplementary, given that L1 and L2 are parallel lines?

A. ∠ P and ∠ S B. ∠ Y and ∠ Q D. ∠ R and ∠ Q D. ∠ R and ∠ X

L1

L 2

P Q

X Y

R S

18. What type of triangle is Δ ABC? A. Isosceles B. Equilateral C. Right D. Obtuse

40 50o oA B

C 19. Which of the angles is an obtuse angle?

A. B. C. D. 20. Select the geometric figure that possesses all of the following characteristics:

i. Triangle ii. All acute angles iii. All sides are equal A. Scalene triangle B. Equilateral triangle C. Acute triangle D. Isosceles triangle 21. The words "right," "acute," "obtuse," "scalene," "isosceles," and "equilateral" are used

individually to describe triangles. Which of the following combinations of words are impossible to describe a triangle?

A. Right; scalene B. Isosceles; obtuse C. Right; obtuse D. Isosceles; right



WARM-UPS B 22. Find m ∠ 1. 23. Find m ∠ 4. 24. Find m ∠ 2.

1 2

3 45 115 o

L

L

1

2

L and L are parallel1 2

25. In D ABC, AB___

≅ CB___

. Find m ∠ y.

26. In D ABC, AB___

≅ CB___

. Find m ∠ x.

27. In D ABC, AB___

≅ CB___

. Find m ∠ z.

A

yC

B

x

z

w

o110

Given that CD___

|| BE___

and AD___

⊥ CD___

. 28. Find m ∠ x. 29. Find m ∠ y. 30. Find m ∠ u. 31. Find m ∠ z.

x

y z

o

A

B

C D

Euv

w45



Given that L1 and L2 are parallel 32. Find m ∠ Q. 33. Find m ∠ T. 34. Find m ∠ W.

P R75°

Q

W X

Y

L1

L 2S T

U V

30o

CLAST PRACTICE B PRACTICE PROBLEMS: Chapter 3, # 23-24 35. Which statement is true for the figure? A. m ∠ B = 100o B. m∠ E = 20o C. m ∠ C ≠ m ∠ H D. ∠ A and ∠ B are complementary

100°A B

C D

E

F

G H

8 8

4

36. Which statement is true for the figure? A. m ∠ E π m ∠ C B. m∠ I = 120o C. m ∠ B = m ∠ G D. ∠ A and ∠ I are supplementary

A BC D

E

FG H

6 6

I6

37. Which statements are true for the figure if M1 and M2 are parallel?

i. m ∠ 3 = 50o ii. m ∠ 4 = 70o iii. ∠ 1 and ∠ 2 are supplementary A. i only B. ii only C. iii only D. i and iii only

1 23 4

M 1

M2

70°

50°

38. Which of the statements is true for the

figure shown, given that L1 and L2 are parallel lines?

A. m ∠ P = m ∠ Y B. m∠ P = 100o C. ∠ X and ∠ P are supplementary D. ∠ Q and ∠ X are complementary

L1

L 2

P Q

X Y

80°



39. The diagram to the right shows lines in the same plane. If line L1 is a horizontal line, which statement is true?

i. L2 and L3 are the only parallel lines ii. L2 and L4 are perpendicular lines iii. Lines L2 and L3 are vertical lines. A. i only B. ii only C. iii only D. i and iii only

L3

L 4

L1

L2

L5

60°60°

40. Given that L1 || L2, which of the following

statements is true?

A. u + v = 180o B. AC___

⊥ BC___

C. u = 40o D. y = 50o

L

LAB

Cu v

yzx

40

130

o

o

1

2

WARM-UPS C 41. The two triangles to the right are similar.

Find the lengths marked x and y.

3

45

4

xy

42. The parallelograms to the right are similar.

Find the length of the diagonal PR__

.

A B

CD

8

10

P Q

RS

10

10



In the figure to the right, the line PQ__

is

parallel to the line AB In each problem, find the missing lengths.

AP__

PC__

BQ___

BC__

43. 3 4 6 ? 44. 5 4 ? 6 45. 2 ? 3 5 46. ? 4 4 8

A B

C

PQ

CLAST PRACTICE C PRACTICE PROBLEMS: Chapter 3, # 25-27 47. Study figures A, B, C, D, then select the figure in which all triangles are similar.

A. B.

80°

70°

80°

75°C. D..

48. Study figures A, B, C, D, then select the figure in which all triangles are similar.

60°60°

5 5

5

58°58°

30°

96

75

A. B.

C. D.



49. Which statement is true for the triangles to the right?

A. m ∠ A = m∠ E B. m∠ x = 30o

C. AC___

= 3.5 D. CE___

CA___

=

CB___

CD___

B

A

C

5

4 3

E

D30°

30° x

y

z

50. Which statement is true for the triangles

to the right? A. m ∠ X = m∠ Z B. m∠ X = m∠ Y

C. CE___

AD___

=

AB___

CB___ D. None is true

A

D E

C

B

X

Y

Z130° 130°

EXTRA CLAST PRACTICE 51. Given that AB

___ || CD

___ and

BD___

⊥ CD___

which of the following statements is true for the figure shown. (The measure of angle ACD is represented by x) A. w = v B. ∠ EAB and ∠ EBA are supplementary C. ∠ AEB and ∠ ECD are complementary angles D. z = x

A u z B

v

C x y Dw

E

52.

Select the geometric figure that possesses all of the following characteristics i. four sided ii. right angle iii. adjacent sides equal and opposite sides parallel A. rectangle B. square C. trapezoid D. rhombus

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GEOMETRY AND MEASUREMENT - Cengage · 2008-06-27 · Change 4 yards to feet 2. Change 8 1/2 ft to yards 3. Convert 9.7 kilometers to meters 4. Convert 90 centimeters to hectometers