Download 12-21 a. The slant height of the cone is ≈ 9.22 m, LA(cone) ≈ 6π

March 17, 2017
12-21 a. The slant height of the cone is ≈ 9.22 m, LA(cone) ≈ 6π(9.22)
≈173.78 m2, and LA(cylinder) = 12π(11) = 132π ≈ 414.69 m2, so total
surface area is ≈ 173.78 + 414.69 = 588.47 m2
b. V(cylinder) = 36π(11) = 396π ≈ 1244.07 m3 and V(cone)
= 1/3(36π)(7) = 84π ≈ 263.89m3, so total the volume is ≈ 1244.07 +
263.89 = 1507.96 m3
12-22 a. 36º
b. b = c = 108º, d = 72º
12-23 a. E(1, 3) and F(7, 3); AB = 9, DC = 3, EF = 6; EF seems to be the
average of AB and CD.
b. Yes; EF = 4, while AB = 6 and CD = 2
c. Sample response: The midsegment of a trapezoid is parallel to the
bases and has a length that is the average of the lengths of the bases.
12-24 a. (2, –3), r = 5
b. (–1, –3), r = 4
12-25 a. sin27º = , x ≈ 8.17 b. x ≈ 6.32
c. tanx = , x ≈ 56.31º
12-26 a. Vertical angles are equal, 2x + 9º = 4x – 2º, x = 5.5
b. The sum of the angles of a quadrilateral is 360°,
116º + (3x + 8) + 32º + (2x – 1) = 360º, x = 41°
c. When lines are parallel, same-side exterior angles are supplementary,
so 7x – 3º + 4x + 12º = 180º and x = 15.55°.
12-27 D
March 17, 2017
Warmup in your spiral: Find the volume and total
surface area of each solid below. Show all work.
March 17, 2017
Warmup in your spiral: Find the volume and total
surface area of each solid below. Show all work.
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Quiz Time: 3/17/17
Do only your own work. Show all work.
Use only your own INB, pencil, eraser,
scientific calculator; NO SHARING.
No phones out when quizzes are out
Turn quiz over when finished for me to
collect; then, preview Lesson 12.2.4.
If you want to access accommodations
for extended time/alternative location,
please write where to route your quiz,
and/or Room Number & date/time where
& when you will complete the quiz, such
as, "➔ Rebecca ➔ Jordan," or, "Finish in
Rm 704, 9/15 @ 3:35."
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P(Favorable)=
Favorable
Total
CW: 12-98 to 12-100
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CW: 12-98 to 12-100
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CW: 12-98 to 12-100
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53
0°
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53
0°
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CW: 12-98 to 12-100
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58
0°
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SISTER
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12‑100 The solutions are as follows:
a. The area of the grazing region is ¾(100π) = 75π ≈ 236 ft2.
Therefore, the probability is 236/3500 ≈ 7%.
b. When the circle of radius 30 feet intersects with the top fence
(point D in the diagram at right), a right triangle is formed
(ΔBED). Since ED = √(302-202) ≈ 22.36 ft, then the area
2
of ΔBED = ½(20)(22.26) ≈ 223.61ft . Since m∠EBD =
–1 20
cos ( /30) ≈ 48.19°, then m∠DBA ≈ 90° – 48.19° ≈ 41.81°.
Then the area of sector DBA is 41.81°/360°(302π) ≈ 328.37 ft2.
The total area is approximately 223.61 + 328.37 ≈ 552 ft2.
Thus, the probability is 552/3500 ≈ 16%.
c.
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Since tan x = 20/10 = 2, then x ≈ 63.43° and
y = 360° − 90° − 63.43° ≈ 206.57°. The radius of Sector #4
is r = 30 - √(500) ≈ 7.64 ft. The central angle for Sector #4
equals x because the barn sides are parallel and their
corresponding angles are equal.
Area of Sector #1:
¼(400π) ≈ 314.15 ft2
206.57°
Area of Sector #2:
/360°(900π) ≈ 1622.40 ft2
Area of Triangle #3: ½(20)(10) = 100 ft2
63.43°
Area of Sector #4:
/360°(7.642π) ≈ 32.30 ft2
2069
∴ the probability is /3500 ≈ 59%.
March 17, 2017
Homework:
• 12-100c
• 12-50 to 12-56 (skip 12-55)
• MN Conic Sections (CPM 749/INB 138)
**Extra-Credit Quiz on Monday:
• Area of a Polygon
• Volume of a Solid
• Area- & Volume- Scale Factors
• Chords, Secants, Tangets, Arcs & Sectors
March 17, 2017
HW Answers; 12-100c on next slide.
12-56
B
12-54 base radius = 14 in;
V = ⅓(196π)(18) ≈ 3695 in3;
SA - πrl = π(14)(√520) ≈ 1003 in2
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100c
Mom
parallel lines & corresponding
angle with 𝛳
P=
%
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Challenge:
Calculate the radius of a cylinder that has a surface
area of 200pi square units and a height of 21 units.