Unit 6 – Area, and Volume Lesson 1 – Area, Parallelogram, Triangle 1 Black – Area, Parallelogram, Triangle 1. Each of the squares here actually measures 4 cm by 4 cm. A B C D E a. b. Which pattern has the smallest shaded area? Which pattern has the largest shaded area? 2. Calculate the area of these sectors: a. b. 3. A cell phone tower provides a clear signal for up to 6.25 km in any direction. Accurately construct a circle using the scale 1 km = 1 cm and find the area that gets a clear signal. 4. Coach Dickinson is teaching his track and field athletes about the distance they can throw the discus. They know that if they throw the discus 18 metres, they have a good chance of being in the top 8 at the IASIS competition. Coach Dickinson tells them that the circular area that would be created by a throw of 18 metres is about 113 m 2 . Coach Lee disagrees and tells the athletes that the circular area created by a throw of 18 m is about 1017 m². Who is correct? Explain both coaches’ calculations. 5. In the figure shown, ABC is a right-angled triangle and ADB and BDC are semi-circles of radii 18 cm. Find the area of the shaded parts. (Take π = 3.14)
Transcript
Unit 6 – Area, and Volume Lesson 1 – Area, Parallelogram, Triangle
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Black – Area, Parallelogram, Triangle 1. Each of the squares here actually measures 4 cm by 4 cm.
A
B
C
D
E
a.
b.
Which pattern has the smallest shaded area?
Which pattern has the largest shaded area?
2. Calculate the area of these sectors:
a.
b.
3. A cell phone tower provides a clear signal for up to 6.25 km in any direction. Accurately construct a circle using the scale 1 km = 1 cm and find the area that gets a clear signal.
4. Coach Dickinson is teaching his track and field athletes about the distance they can throw the discus. They know that if they throw the discus 18 metres, they have a good chance of being in the top 8 at the IASIS competition. Coach Dickinson tells them that the circular area that would be created by a throw of 18 metres is about 113 m 2. Coach Lee disagrees and tells the athletes that the circular area created by a throw of 18 m is about 1017 m². Who is correct? Explain both coaches’ calculations.
5.
In the figure shown, ABC is a right-angled triangle and ADB and BDC are semi-circles of radii 18 cm. Find the area of the shaded parts. (Take π = 3.14)
Unit 6 – Area, and Volume Lesson 1 – Area, Parallelogram, Triangle
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6.
The radius of circle O is 8 cm. How many centimeters are in the length of the diagonal of the rectangle?
OO
7. 4’
6’3’
4’
6’3’
John ties the leash of his dog to the corner of the doghouse. The dimensions of the dog house are 3’ x 4’, and the leash is 6’ long. Over how many square feet can John’s dog wander? Express your answer as a decimal to the nearest hundredth.
8. In a race, athletes run three laps around an oval race track formed by a rectangle and two semicircles as shown. The length of a radius of each semicircle is 12 meters, and the length of the rectangle is twice the length of the diameter of the semicircle. What is the number of meters in the length of the race? Express your answer in terms ofπ .
9. A square dartboard has an inscribed circle. A thrown dart hits the square target. What is the probability that the dart lands outside of the inscribed circle? Express your answer as a common fraction in terms of π .
10.
A wheel on a bike turns 1320 times during a 1-mile trip. What is the diameter of the wheel?
11. Lawn-Mowing Chore
At my house, summer vacation means added chores for my three sons. Every Saturday, the lawn must be mowed. Mark starts the mower and completes 1/3 of the lawn. Sam takes over and mows exactly 1/4 of the grass. Josh then finishes off the last 700 square feet of the yard.
What is the area of my yard?
Bonus: The boys devised the plan for dividing the work and consider it to be fair based on their ages. My youngest son is nine. Can you match the names and ages of my three sons?
Don't forget to explain how you found your answer, and say whether you think the plan is fair.
Unit 6 – Area, and Volume Lesson 1 – Area, Parallelogram, Triangle
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12. More Mowing Madness Matt does such a good job with his mowing business that word-of-mouth has brought him more work than he can complete by himself. His friend Georgette has joined the business, and they have agreed to split the work evenly on each yard they mow. Since they will use only Matt's mower, which cuts a path 2 feet wide, one will mow while the other works on edging, shrub clipping, and any other yard care requested.
Georgette is also a big math fan, so she would like to continue to use the "Matt-mowing" technique, starting on the outside portion of the lawn and mowing a path completely around the outside edge, then continuing with a second path around the remaining portion of the lawn, and so on. (Visiting the highlights of the previous problem will help you visualize the pattern.)
Where should Matt and Georgette change places if each is to mow exactly half of a 72-ft. by 96-ft. yard? In other words, will Matt and Georgette switch places at the end of a complete "Matt-mowed" rectangle? If so, after which one? If not, how many more feet into the next rectangle should the first person mow before the switch?
13. Stained Glass Window I have fallen in love with a particular shape I want made into a stained glass window for my front door. The square grid in the image shown is a six inch grid - that is, every grid line is six inches away from the next line.
The stained glass is quite expensive, so I'd like to know exactly what its area is. Also, to insert the stained glass into the clear glass, I'll have a line of lead between the two colors of glass. I need to know how long this perimeter is.
You probably don't know a formula for the area or perimeter of the shape I'm making out of stained glass, but the curve is made up of circular pieces and it all fits into one 24-inch by 24-inch square.
Make sure you describe how you found the area and perimeter in your answer.
Bonus: Give these answers in exact form.
Unit 6 – Area, and Volume Lesson 1 – Area, Parallelogram, Triangle
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Solutions
1. They are all the same Shaded area = 16 - 4�
2. a.
b.
135360
x �(25)²
= 0.375 x � x 625 = 234.375� cm²
71360
x �(10)²
= 19.722π cm²
3. �r² �(6.25)²= 39.0625� km²
4. Coach Dickinson or Coach Lee? Radius of circle = 18 m à �(18)² à 1017.8 m² Coach Dickinson: 133 m² Coach Lee: 1017 m² �(18)(2) He didn’t square it! Coach Lee is correct He multiplied it by 20
Coach Lee is correct. He squared the radius. Coach Dickinson multiplied the radius by 2.
5.
Area: π(18)²= 1017.87 cm²(½)= 508.935 cm²
Area: π(18)²= 1017.87 cm²(½)= 508.935 cm²
18 cm
cm²1017.872
cm²1017.872
Area of triangle ½ x (18 x 2) x (18 x 2) = 648 cm² 648 : 2 324 cm²
Unit 6 – Area, and Volume Lesson 1 – Area, Parallelogram, Triangle
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??
508.935 – 324 = 184.935 cm² 184.935
2x 4 = 369.36 cm²
6. 8 cm
7.
3
46’
3
46’
33
quarter ¾ x 6² x �
½ x (3)² x � = 27� = 2¼�
¼ x (4 – 6)² x � 27 + 2¼ + 1 = ¼ x 4 x � = 30¼� ft² = �
30.25� ≈ 94.99 ft²
8.
24
24 x 2 = 48
1224
24 x 2 = 48
12
3 laps Diameter à �d 24� 24� + (48 x 2) = 24� + 96 = (24� + 96)3 = 72� + 288 = 514.08 meters
Unit 6 – Area, and Volume Lesson 1 – Area, Parallelogram, Triangle
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9.
Area of = x² Area of = �(½�)² n² - �(½n)² n² - �¼n² n(1 - �¼)/ n²
= 14π
−
72� + 288
9. 1 - ¼� ≈ 21.5%
10. Perimeter = 11320
-- 1 mile = 5280 feet
52801320
= 4 ft
4 ft4 ft
�d = 4
d = 4π
= 1.27 feet
The wheel turns 1320 times during a 5280 foot trip, so each turn is 5280 + 1320 = 4 feet. A wheel with a circumference of 4 feet has a diameter of: 4 ÷ 3.14 = 1.27 feet
11. M = 1/3 of L S = ¼ of L J = 1 – (1/3 + ¼) 512
= 700 ft²
Area: 512
= 700
1212
= ?
Unit 6 – Area, and Volume Lesson 1 – Area, Parallelogram, Triangle
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=
12 70012
512
×
= 700512
= 84005
= 1680 ft²
Youngest son = 9 years old
Hence least amt of work 312
= 9 years
Sam = 9 years old
Using proportions. 312
L = 9 yrs
412
L = ? yrs
4 912312
× = 12
Mark = 12 years old
312
L = 9 yrs
512
L = ? yrs
5 912312
× = 15
Josh = 15 years old
Your yard is 1,680 sq. ft. Bonus: Sam is 9; Mark is 12; and Josh is 15. Mark mows 1/3 or 4/12 of the lawn. (1/3) x (4/4) = 4/12 Sam mows 1/4 or 3/12 of the lawn. (1/4) x (3/3) = 3/12 Josh mows 5/12 of the lawn. 12/12 - 4/12 - 3/12 = 5/12 Josh mows 700 sq. ft. so 5/12 of the lawn is 700 sq. ft. 1/12 of the lawn is 700/5 = 140 sq. ft. The whole lawn is 12 x 140 sq. ft. = 1,680 sq. ft.
Bonus: Sam mows 3/12 of the lawn, and he is 9 years old. For each 1/12 of the lawn, Sam has 3 years (3 x 3 = 9). Mark mows 4/12 of the lawn, so he is 12 years old (4 x 3 = 12). Josh is 15 years old because he mows 5/12 of the lawn (5 x 3 = 15). I think it is fair because the youngest does the least and the oldest does the most, and each additional 3 years adds the same amount of work.
12. I found that Matt mows exactly 6 paths of the yard to mow half. To solve this problem, I started out by drawing a rectangle with a length of 96 feet and a width of 72 feet. Next I calculated the area of the rectangle by multiplying 96 feet x 72
Unit 6 – Area, and Volume Lesson 1 – Area, Parallelogram, Triangle
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feet = 6912 square feet. One half of 6912 = 3456 square feet. Now I know that Matt should mow 3456 square feet.
To figure out how many rectangles Matt has to mow, I first made a path just inside the border that went all the way around the rectangle and represented a width of 2 feet. On the long side of the rectangle, I wrote 96 for 96 feet. On the short side I wrote 72 for 72 feet. The length from end point to endpoint on the 96-foot side, is 96 feet long, but the length from end point to endpoint on the 72 foot side is 68 feet long. This is because at each end of the 68- foot side there would be 2 feet that would already have been mowed when the long side was mowed. Therefore, the perimeter would be 68 + 68 + 96 + 96, which equals 328 feet x 2 feet = 656 square feet.. I multiplied by 2 because the lawnmower cuts a 2 foot path. Next I subtract 656 from 3456 = 2800. Next I calculate the second rectangle, which will have a length of 92 feet and a width of 64 because I have to subtract out the area already mowed. Once again I compute the perimeter by adding 92 + 92 + 64 + 64 = 312. Then I multiply 312 by 2 = 624. I subtracted 624 from 2800 = 2176. After the 2nd rectangle, Matt would have 2176 more square feet to mow. Next I calculate the third rectangle, which will have a length of 88 and a width of 60. Once again I compute the perimeter by adding 88 + 88 + 60 + 60 = 296 square feet. Then I multiply 296 square feet by 2 feet = 592 square feet. I subtract 592 from 2176 = 1584 square feet left to go.
At this point, I saw a pattern. I saw that every time I made a new path, the new path is 16 feet shorter. So for the fourth rectangle, I just subtract 16 from 296 and I get 280. I multiply 280 x 2 = 560 square feet, which I subtract from 1584 to get 1024. For the fifth, rectangle I subtract 16 from 280 and I get 264 which I multiply by 2 to get 528. I subtracted 528 from 1024 to get 496. For the sixth rectangle I subtracted 16 from 264 to get 248 and multiplied by 2 to get 496. When I subtract 496 from 496 I see that after mowing the sixth rectangle Matt is done with his turn.
13. My answers are: the perimeter of the stained glass is 75.36 in. The area of the stained glass is 288 sq. in.
Method 1: Grid length = 6 in. In the following, the (2,3) represents the 2nd row and the 3rd column grids; etc.
Perimeter Area Description Grid No. in. sq. in. (1,1) 0 0 Non-stained glass (1,2) 9.42 7.74 Non-stained glass is 1/4 of a circle (1,3) 9.42 7.74 Non-stained glass is 1/4 of a circle (1,4) 0 0 Non-stained glass (2,1) 9.42 28.26 Stained glass is 1/4 of a circle (2,2) 0 36 Stained glass (2,3) 0 36 Stained glass (2,4) 9.42 28.26 Stained glass is 1/4 of a circle
Unit 6 – Area, and Volume Lesson 1 – Area, Parallelogram, Triangle
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(3,1) 9.42 7.74 Non-stained glass is 1/4 of a circle (3,2) 0 36 Stained glass (3,3) 0 36 Stained glass (3,4) 9.42 7.74 Non-stained glass is 1/4 of a circle (4,1) 0 0 Non-stained glass (4,2) 9.42 28.26 Stained glass is 1/4 of a circle (4,3) 9.42 28.26 Stained glass is 1/4 of a circle (4,4) 0 0 Non-stained glass Total 75.36 288.0
Square grid area = 6 x 6 sq. in (side length x side length) A circle's area = 6 x 6 x 3.14 (r x r x pi) or 12 x 12 x 3.14 / 4 (d x d x pi / 4)
unit: sq. in. A circle's perimeter = 6 x 2 x 3.14 ( d x pi or 2 x r x pi ) unit: in.
Method 2: For perimeter calculation, I rearranged the stained pieces in grids (1,2), (1,3), (2,1), (2,4), as one circle and the pieces in grids (3,1), (3,4), (4,2), (4,3) as another circle. The perimeter of the stained glass = the circumferences of 2 circles = 2 x (d x pi) = 2 x 12 in. x 3.14 = 75.36 in. where d = 12 in., which is the diameter of the circle, and pi = 3.14
For area calculation, I rearranged the stained pieces in grids (1,2), (1,3), (2,1), (2,4), (3,1), (3,4), (4,2), (4,3) as 4 square grids. The area of the stained glass = the 8 square grids = 8 x a x a = 8 x 6 x 6 = 288 sq. in. where a = 6 in., the length of a grid.
Bonus: A circle's perimeter = d x pi The perimeter of the stained glass = the circumferences of 2 circles = 2 x (d x pi) = 2 x 12 in. x 3.14 = 75.36 in. where d = 12 in., which is the diameter of the circle, and pi = 3.14 (only take 2 decimals from 3.1459...)
The area of the stained glass = the 8 square grids = 8 x a x a = 8 x 6 x 6 = 288 sq. in. where a = 6 in., the length of a grid.
Unit 6 – Area, and Volume Lesson 1 – Area, Parallelogram, Triangle
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Bibliography Information Teachers attempted to cite the sources for the problems included in this problem set. In some cases, sources may not have been known.
Problems Bibliography Information
11 - 13 The Math Forum @ Drexel (http://mathforum.org/)
6 - 9 Math Counts (http://mathcounts.org)
10 Zaccaro, Edward. Challenge Math (Second Edition): Hickory Grove Press, 2005.
