ch. 11 putting showdown hosted by parker and robert
TRANSCRIPT
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Ch. 11 Putting Showdown
Hosted by Parker and Robert
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How to Play• Split up into groups of 4 and everybody
get one whiteboard and marker.
• When instructed, solve the problem given on the power point. If finished, DO NOT put your whiteboard up right away, just raise your hand.
• Everybody must have their own answer and work on their whiteboard.
• Parker and Robert will check your answer.
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How to Play (Con.)• The group that everybody had the right
answer in and raised their hand the earliest wins and gets one point.
• Each team member of the winning team will get to take one putt. Each putt equals one point if made.
• At the end of the game, there will be a bonus question in which your group will bet any amount of your group’s points before you know the question.
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How to Play (Con.)
• The bonus question will be presented on the power point.
• Within two minutes, as long as each member in your group has the right answer, your group will select one person to take a single putt.
• If made, your bet is added to your score. If not, it is subtracted from your score.
• The group with the most point at the end of the game wins.
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Question #1
Find the area of the shaded region.
34 inches
21 inches 29 inches
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Answer to Question #1
A = bh Hypotenuse=21 in
= (21)(34) 292=212+x2 X=base
A = 714 in2 841=441+x2
x2= 400 x=20
A =1/2(b)(h)
= ½(20)(21)
A = 210 in2
A - A = Area of shaded region=x
714-210=X
X=504
Area of shaded region= 504 in2
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Question #2Find the area of the regular octagon when AB is the radius of circle A. Also find the area of the circle in terms of Pi. Carry all decimals over and round to the nearest tenth for your final answer.
A
B
10 feet
C
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Answer to Question #2M<BAC=360 = 22.5°
16Sin(22.5=x Sin(22.5(10)=X 10 0.382683432(10)=x
x=3.82683424side=2x Side=7.653668647 feet
Cos) 22.5=x Cos)22.5(10)=x 10 0.923879533(10)=x X=9.238795325 feet
x= apothemA = A(s)(n)
A= apothem =(9.238795325)(7.653668647)(8)S=side =565.6854249n= # of sides A =565.7 feet2
A =(pi)r2
=(pi)102
=100piA =100pi feet2
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Question #3Find the area of the shaded region when the hypotenuse of the right triangle is 15 inches.
12 inches
7 inches
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Answer to Question #3
A =s2 x2+ 122= 152 b1=6+9
= 62 x2+ 144=225 =15 in
A= 36 in2 x2=81 b2=6+7
x=9 b2=13
h=12
A =1/2(b1+b2)h
=½(15+13)12
=6(28)
= 168 in2 Area of shaded=
168-36
Area of shaded=132 in2
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Question #4Find the area of the shaded regular pentagon and the ratio of the area (shaded to non-shaded).
10 inches
7 inches
Area = 175 in.2
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Answer To Question #4
Ratio of sides- 10:7
Ratio of area- a2: b2
=102: 7
=100:49
(x=a of non shaded)
100= 175 100x=8,575
49 x x=85.75
A= 85.75 in2
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Impossible is nothing. Will
you be a hero?
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Now for the Bonus
Question
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Bonus Question
If a golf ball with a diameter of 4.5cm rotates 54 times before getting in the hole on an exactly straight putt, then how far is the putt? Round your answer to the nearest foot in terms of pi. P.S. 1ft=30cm
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Answer to Bonus Question
4.5/2= r r=2.25 cm
C=2(pi)r c=4.5pi cm
4.5 pi(45 rotations) =243 pi cm
243 pi/30= distance in feet
distance= 8.1 pi feet
distance =8 pi feet