Mod3and4QuestReviewKey
3.1
1. 3p » 9.42; its height
2. Nets should show 8 equilateral triangles arranged so that there is no overlap when the nets
are folded. Possible answer:
3.
4. point, circle
5. trapezoid
6. ellipse (oval)
7. triangle
8. 1 in. ´
2 in. (rectangle)
9.
10.
3.2
.Surface area = lateral surface area + area of bases
Lateral surface area = x2p
x 2p
Area of bases = 2p r 2 =
2
3 2
Surface area = x p
2
2. 7.4. (Answers may vary slightly due to rounding.)
Height of triangle = 2.0sin39° » 1.3
Area of both triangles = 2.2 ´ 1.3 » 2.9
Area of rectangles: 2.2 ´ 0.8 = 1.8;
2.0 ´ 0.8 = 1.6; 1.4 ´ 0.8 » 1.1
Surface area » 2.9 + 1.8 + 1.6 + 1.1 = 7.4
3. Surface area = 4(12)(3.2) + 2(3.2)(3.2) = 153.6 + 20.48 »174 cm2
3
1
1
1
3
1
7
)(1 ) + 2(1 )(1 ) = 23 + 3 = 26 = 26.875 in2.
4
4
4
4
4
8
8
2
2
2
2
26.875 in ´ 6.45 cm /in » 173 cm
4. Surface area = 4(4
5. 22; 22; 22
6. Surface area = cube surface area + cylinder lateral area - cylinder base area
Cube surface area = 6(182) = 1966
Cylinder lateral area = 8p (18) = 144p
Cylinder base area = 2p (16) = 32p
Surface area = 1966 + 144p - 32p =
1966 + 112p
3.3
1.Find the area of the 4 trapezoids that form the lateral surface of the frustum:
The surface area of one trapezoid is
1æ
3
bö
bö
æ
b + ÷ . The area of all four trapezoids is ç b + ÷ ! = b!.
ç
2è
2
2ø 2
2ø
è
Or find the lateral surface area of the whole pyramid and subtract the lateral surface area of
the small top pyramid.
1
3
æ1 öæ1 ö
2b! - 2 ç b ÷ ç
= 2b! - b! = b!
÷
2
2
è2 øè2 ø
2. Surface area of cone = p r! + p r 2
Surface area of truncated cone = surface area of cone - lateral surface area of little cone +
2
base of little cone = p r! + p r 2 - p
æ 3p r + 5p r 2 ö
r
ærö
´ + pç ÷ = ç
÷ø
è 2ø
2 2
4
è
3. Lateral surface area of big cone = p r!
Lateral surface area of little cone =
1
r
p ´ = p r!
4
2 2
The LSA of the big cone is 4 times greater than the LSA of the little cone. The LSA of the
3
3
truncated cone is the difference between the two: pr!; it is
the LSA of the big cone.
4
4
4. Diameter of cone: 6 2
Radius of cone: 3 2
Lateral area of cone: p (3 2 )(5.8) »
24.6p
Base of cone: 18p
Lateral area of pyramid: 2(6)(5) = 60
Base of pyramid: 62 = 36
Surface area = Lateral area of cone + base of cone + lateral area of pyramid - base of
pyramid
24.6p + 18p + 60 - 36 » 42.6p + 24 » 157.8
5.
Slantheightofleftcone:20
Slantheightofrightcone:13
Lateralsurfaceareaofleftcone:
p (12)20=240p
Lateralsurfaceareaofrightcone:
p (12)13=156
3.4
4 3 256p 3
pr =
; r = 64; r = 4;
3
3
SA = 4p r 2 = 4p (42) = 64p
1. V =
2. Surface area of Ganymede = 4p (1640)2 = 10,758,400p
Surface area of moon = 4p (1080)2 = 4,665,600p
Ratio: 10,758, 400p ÷ 4,665,600p » 2.3
Ratio of radii: 1640 ÷ 1080 » 1.52
1.522 » 2.3
The ratio of the surface areas is the square of the ratio of the radii.
3. A = 225p = p r 2; r2 = 225; r = 15;
SA = 4p r 2 = 4p (152) = 900
The surface area of a sphere is 4 times as great as the area of its great circle.
4. Surface area of can = 2p rh + 2p r 2 = 2p r(6r) + 2p r 2 = 12p r 2 + 2p r 2 = 14p r 2
Surface area of 3 balls = 3(4)p r 2 = 12p r 2
5. They are equal.
6. Lateral surface area of cylinder = 2p rh = 2p (6)(10) = 120p
Area of cylinder bases = 2p r 2 = 2p (62) = 72p
Surface area of hemisphere = 2p r 2 = 2p (32) = 18p
Base area of hemisphere = p r 2 = p (32) = 9p
Surface area of figure = 120p + 72p + 18p - 9p = 201p
7. Diameter of hemisphere = 4 2
Surface area of hemisphere = 3p r 2 = 3p (2 2 )2 = 24p
Lateral surface area of prism = 4(4)(1) = 16 (top and bottom bases of prism cancel each
other)
Surface area of figure = 24p + 16
4.1
1. Possible answer: The formula for the area of a triangle is A =
1
bh, where b is the length of
2
1
bh. Multiplying b by n and h by n gives
2
= n2 Ai .
the base and h is the height. The initial area is Ai =
Ac =
1
1
(bn)(hn), or Ac = n2bh. Therefore, Ac
2
2
2. Possible answer: The formula for the area of a circle is A = p r 2. The initial area is
Ai = p r 2. Multiplying r by n gives
Ac =p (nr)2 or Ac = p n2r 2. Therefore
Ac = n2 Ai .
3.
9 10
ft
2
4.
2 in.
5. 3 p cm
6. The volume is multiplied by 5.
7. The volume is multiplied by 8.
8. The height is divided by 7.
4.2
1. Possible answer: The formula for the area of a triangle is A =
1
bh, where b is the length of
2
1
bh. Multiplying b by n and h by n gives
2
= n2 Ai .
the base and h is the height. The initial area is Ai =
Ac =
1
1
(bn)(hn), or Ac = n2bh. Therefore, Ac
2
2
2. Possible answer: The formula for the area of a circle is A = p r 2. The initial area is
Ai = p r 2. Multiplying r by n gives
Ac =p (nr)2 or Ac = p n2r 2. Therefore
Ac = n2 Ai .
3.
9 10
ft
2
4.
2 in.
5. 3 p cm
6. The volume is multiplied by 5.
7. The volume is multiplied by 8.
8. The height is divided by 7.
4.3
1. 20 feet
2. depth 10 inches; width 30 inches; height 75 inches
3. 6 inches
4. 432 cubic feet
5. 3 feet
6. $58.88