Geometry
Semester 2
Unit 6 Lesson 3
10 May, 2017
Agenda 10 May, 2017
● Bulletin
● Homeworks
○ Return Homework worksheet - add notes (check at
home)
○ Volume of a Prism Homework page 19 - 20 check
○ Volume of Cylinders - pg 25 - 27 check
● Volume of Pyramid
○ Activity and video clip
○ Practice problems p. 38 - 40
● Volume of a Cone
● Homework
Volume and Surface area of rectangular prisms
At bottom of page - important to remember:
Volume = length x width x height
x height
Surface area = sum of all 6 face areas
=2(length*width)+ 2(l*h) + 2(w*h)
or
base
Volume and Surface area of rectangular prisms
1.V = 8.8yd x 2.5 yd x 7.8 yd = 171.6 yd3
A = 2(8.8ydx2.5yd) + 2(2.5yd x 7.8yd) + 2(8.8yd x 7.8yd)
= 44 yd2 + 39 yd2 + 137.28 yd2 = 220.3 yd2
2.V = 5.4 m x 2.4 m x 3.2m = 41.5 m3
A = 2[(5.4m x2.4m) + (2.4m x 3.2m) +(5.4m x 3.2m)]
= 75.8m2
Volume and Surface area of rectangular prisms
3.V = 4.9 in x 2.8 in x 3.1 in = 42.5 in3
A = 2[(4.9in x 2.8 in) + (2.8in x 3.1in) +(4.9in x 3.1in)]
= 75.2in2
4. V = 9.8km x 5.1 km x 5.8 km = 289.9 km3
A = 2[(9.8km x5.1km) + (5.1km x 5.8km) +(9.8km x 5.8km)]
= 272.8km2
Volume and Surface area of rectangular prisms
5. V = 5.5 mm x 3.4mm x 4.9mm = 91.6 mm3
A = 2[(5.5 mm x 3.4mm)+(3.4mmx 4.9mm) + (5.5mm x
4.9mm)] = 124.6 mm2
6. V = 3.7 cm x 1.1cm x 1.9cm = 7.7 cm3
A = 2[(3.7 cm x 1.1cm)+(1.1cm x 1.9cm)+ (3.7cm x 1.9cm)] =
26.4 cm2
Precise language: prisms - base supplies name
Prism - a polyhedron
with 2 congruent
parallel bases
Volumerectangular prism = Bh, where B is the area of the
base and h is the height of the prism.
Page 19
1) a) V= B h
V = (area of rectangle) (height)
V = (2cm x 10cm) (11cm)
V = 220cm3
Volumerectangular prism = Bh, where B is the area of the
base and h is the height of the prism.
1) a) V= B h
V = (area of rectangle) (height)
V = (10cm x 11cm) (2cm)
V = 220cm3
Volumerectangular prism = Bh, where B is the area of the
base and h is the height of the prism.
1) c) V= B h
V = (area of rectangle) (height)
V = (11cm x 2cm) (10cm)
V = 220cm3
What do you notice about the
volumes of these three examples?
Volumerectangular prism = Bh, where B is the area of the
base and h is the height of the prism.
In a rectangular prism, there are multiple pairs of
parallel congruent faces that could be considered
to be the bases of the prism.
Page 19
2) a) V= Bhrectangular prism + Bhtriangular prism
V = (7cm x 6cm)(8cm) + ½ (6cm x 7cm)(8cm)
V = 336cm3 + 168cm3
V = 504cm3
Page 19
2) b) V= Bhtriangular prism
V = ½ (6cm x 6cm)(8cm)
V = 144cm3
Page 19
2) c) V= Bhrectangular prism
V = (5cm x 6cm)(2cm)
V = 60cm3
Page 19
2) d) V= Bhtriangular prism
V = ½ (7cm x 3cm)(11cm)
V = 115.5cm3
Page 19
2) e) V= Bhrectangular prism
V = (5cm x 5cm)(11cm)
V = 275cm3
Page 19
2) f) V= Bhtriangular prism
V = ½ (12cm x 4cm)(14cm)
V = 336cm3
Page 19
2) g) V= Bhoblique rectangular prism
V = (6cm x 8cm)(7cm)
V = 336cm3
Page 19
2) i) V= Bhcube
V = (8cm x 8cm)(8cm)
V = 512cm3
Page 20 G.GMD.3
3) a) V= Bhtriangular prism
To find h (altitude)
of base
remember:
V = ½ (6cm x 3√3cm)(10cm)
3
V = 90√3cm3
Page 20 G.GMD.3
3) b) V= Bhtriangular prism
V = ½ (5cm x 12cm)(15cm)
V = 450cm3
Pythagorean
triple
132 = 52 + 122
Page 20 G.GMD.3
4
3) c) V= Bhhexagonal prism
2√3
V = ½ (perimeter*apothem)h
2
V = ½ (24cm)(2√3cm)(9cm)
V = 216√3cm3
Page 20 G.GMD.3
4
3) d) V= Bhsquare prism
3√2
V = (3cm)(3cm)(15cm)
V = 135cm3
3
Page 20 G.GMD.3
3) e) V= Bhoblique rectangular prism
6
V = (10cm)(8cm)(3√3cm)
3
V = 240√3cm3
Page 20 G.GMD.3
4
3) f) V= Bhhexagonal prism
2√3
V = ½ (perimeter*apothem)h
2
V = ½ (24cm)(2√3cm)(7cm)
V = 168√3cm3
Page 20 G.GMD.3
3) g) V= Bhtriangular prism
4
V =½ (4cm)(2√3cm) (8cm)
2
V = 32√3cm3
Page 20 G.GMD.3
3) h) V= Bhrectangular prism
70°
Tan 70° = 8/x
x
x = 8/tan 70° = 2.91176
V = (8cm)(2.91176cm)(10cm)
V = 232.94cm3
(says 2 dec.)
8cm
Page 20 G.GMD.3
3) i) V= Bhtriangular prism
V = ½ (6cm)(4cm)(7cm)
V = 84cm3
Lesson 5 Homework Volume of a Cylinder p. 25
1. Volume
2
= Area circle x height = ( r )h
V= (4cm)28cm
V = (16x8)cm3
3
V=128 cm
Lesson 5 Homework Volume of a Cylinder p. 25
2. Volume
2
= Area circle x height = ( r )h
r = 2.5cm h = 12cm
2
V= (2.5cm) 12cm
V = (6.25x12)cm3
V=75 cm3
Lesson 5 Homework Volume of a Cylinder p. 25
3. Volume
= ( r2)h
1/4 cylinder
= ¼ Area circle x height
V=¼ (8cm)210cm
3
V = (16x10)cm
3
V=160 cm
Then find Volumetriangular prism
Lesson 5 Homework Volume of a Cylinder p. 25
3. cont.
Volumetriangular prism= Base x height
17cm
8cm
x
√172 - 82= x
x = 15
Lesson 5 Homework Volume of a Cylinder p. 25
3. cont.
Volumetriangular prism= ½(15cmx8cm) x
10cm
8cm
10cm
15cm
Lesson 5 Homework Volume of a Cylinder p. 25
3. cont.
Volumetriangular prism= ½(15cmx8cm) x
10cm
= 600 cm3
Total Volume = (160 +600)cm3
Lesson 5 Homework Volume of a Cylinder p. 25
4. V
2
1
= ½ Area circle x height = ( r )h
r = 3cm h = 11cm
2
V1=½ (9cm) 11cm
V1 = (4.5x11)cm3
V1=49.5 cm3
Lesson 5 Homework Volume of a Cylinder p. 25
4. V
= Volume rectangular prism volume of half a cylinder
2
= (2cm x 6cm x 11cm) - ½ ( 22)11cm
V2=132cm3 - 22 cm3
3
Vtotal =(132 -22 +49.5 )cm
3
Vtotal= (132 + 27.5 )cm
Lesson 5 Homework Volume of a Cylinder p. 26
1a) Volume
2
= Area circle x height = ( r )h
V= (4cm)27cm
V = (16x7)cm3
3
V=112 cm
Lesson 5 Homework Volume of a Cylinder p. 26
1b) Volume
2
= Area circle x height = ( r )h
V= (3cm)28cm
V = (9x8)cm3
3
V=72 cm
Lesson 5 Homework Volume of a Cylinder p. 26
1c) Volume
= Area circle x height = ( r2)h
2
V= (5cm) 9cm
V = (25x9)cm3
V=225 cm3
Oblique cylinder - just use vertical height
from one base to the other
Lesson 5 Homework Volume of a Cylinder p. 26
1d) Volume
= Area circle x height = ( r2)h
d = 10cm, r = 5cm
V= (5cm)223cm
V = (25x23)cm3
3
V=575 cm
Lesson 5 Homework Volume of a Cylinder p. 26
1e) Volume
= Area circle x height = ( r2)h
h = 10cm = d = 10cm, r = 5cm
V= (5cm)210cm
V = (25x10)cm3
3
V=250 cm
Lesson 5 Homework Volume of a Cylinder p. 26
1f) Volume
= Area circle x height = ( r2)h
c = 6 cm = 2 r cm so r = 3cm
V= (3cm)23cm
V = (9x3)cm3
3
V=27 cm
Lesson 5 Homework Volume of a Cylinder p. 26
2a) V
2
Large
2
-Vsmall = ( r )h - ( r )h
V= (8cm)212cm - (4cm)212cm
V =768 cm3 - 192 cm3
3
V=576 cm
Radius split into 2 congruent parts, each
4cm
Lesson 5 Homework Volume of a Cylinder p. 26
2b) V
2
Large
2
-Vsmall = ( r )h - ( r )h
V= (2.5cm)28cm - (2cm)28cm
V =50 cm3 - 32 cm3
3
V=18 cm
Diameter of large cylinder given, radius
of inner one given.
Lesson 5 Homework Volume of a Cylinder p. 26
2c) V
2
Large
2
-Vsmall = ( r )h - ( r )h
V= (4cm)25cm - (1cm)25cm
V =80 cm3 - 5 cm3
3
V=75 cm
Diameters given
Lesson 5 Homework Volume of a Cylinder p. 27
3a) V
2
= ( r )h x 90/360 (fraction of whole)
V= (6cm)210cm x 1/4
V =¼ 360 cm
V=90 cm3
3
Lesson 5 Homework Volume of a Cylinder p. 27
3b) V
2
= ( r )h x 270/360 (fraction of
whole)
V= (2cm)25cm x ¾
V =¾ 20 cm3
3
V=15 cm
Half a cylinder on a rectangular prism
2cm
6cm
10cm
Lesson 5 Homework Volume of a Cylinder p. 27
3c) V
2
= ½ ( r )h + (l x w)h
V=½ (3cm)210cm + (6cm x 2cm x10cm)
3
V =45 cm +120cm
3
V = approx 261.3 cm3
Lesson 5 Homework Volume of a Cylinder p. 27
3d) V
2
= ½ ( r )h + (l x w)h + (½ ba)h
V=½ (3cm)212cm + (6cm x 4cm x12cm)
+ (½ 4cm x 6cm)12cm
= 54 cm3+ 288cm3 + 144cm3
V = (432 + 54 ) cm3
Lesson 5 Homework Volume of a Cylinder p. 27
3e) V
2
= ½ ( r )h
d = 8cm, r = 4cm
V=½ (4cm)212cm
V = 96 cm
3
Lesson 5 Homework Volume of a Cylinder p. 27
3f) Break it up:
2 rectangular prisms
2 (22cm)(6cm)(15cm) = 3960cm3
Lesson 5 Homework Volume of a Cylinder p. 27
3f) Break it up:
Half a cylinder with radius 9cm (6cm + ½ 6cm)
½ (9cm)2(15cm) = 607.5 cm3
Lesson 5 Homework Volume of a Cylinder p. 27
3f) Break it up:
Half a cylinder with radius 9cm (6cm + ½ 6cm)
½ (9cm)2(15cm) = 607.5 cm3
Half a cylinder cut out, radius 3cm ( ½ 6cm)
½ (3cm)2(15cm) = 67.5 cm3
Lesson 5 Homework Volume of a Cylinder p. 27
3f) so all together
V = 3960cm3 + 607.5 cm3 - 67.5 cm3
V = (3960 + 540 )cm3
Lesson 5 Homework Volume of a Cylinder p. 27
3g) V
= Volume cube - volume of hole
d = 2cm, r = 4cm
3
2
V = (4cm) - (1cm) 4cm
V=64cm3-4 cm3
V = (64- 4 ) cm
3
Lesson 5 Homework Volume of a Cylinder p. 27
3h) V
= Volume cylinder - volume of hole
d = 12cm, r = 6cm
2
2
V = (6cm) 9cm - (3cm) 9cm
V=324 cm3-81cm3
V = (324 - 81) cm
3
Lesson 5 Homework Volume of a Cylinder p. 27
3i) V
= Volume cylinder - volume of 2 holes
d = 16cm, r = 8cm
V = (8cm)213cm - 2 (3cm)213cm
V=832 cm3-234 cm3
V = 598 cm3
Volume of a Pyramid
Demonstration - live - follow oral
instructions
via video
Volume of a Pyramid - Lesson 7 p36 -
Vpyramid= ⅓ base x height
L- 7 HW Practice Finding Volume of a Pyramid p. 38
1. D 1. Slant height
Apex
F 2. Apex
E 3 Height
B 4 Lateral edge
Slant height
A 5 Face
C 6 Vertex
Vertex
Apex
Lateral edge
Slant height
Vertex
Face
P. 38
2. The slant height is the length we can measure
along a face of the pyramid from the apex to the
base, whereas the height of the pyramid is the
perpendicular distance from the base to the
apex. They will never be the same length.
P. 38
3.
3.
Square pyramid
3.
6 cm
8 cm
3.
rectangular pyramid
6 cm
8 cm
triangular pyramid
3.
rectangular pyramid
6 cm
8 cm
All 4 faces are
equilateral triangles
Regular
tetrahedron
4. Same base area and height, same volume.
Lesson 8 Volume of a cone p 45
Volume cone = ⅓ Base x height = ⅓ Bh =1/3 r2h
The Cone p 45
Volume cone = ⅓ Base x height = ⅓ Bh =⅓ r2h
3cm
8cm
Example #1
r= 3cm
h= 8cm
V = ⅓ Bh
V= ⅓ r2h
V= (3cm)28cm
V=24 cm3
G. GMD.3 Student Notes p. 45
Volume cone = ⅓ Base x height = ⅓ Bh =⅓ r2h
12cm
4cm
Example #2
Given diameter - find radius
before calculating volume.
G. GMD.3 Student Notes WS #5
Volume cone = ⅓ Base x height = ⅓ Bh =⅓ r2h
13cm
h
Example #3
Given slant height,
Use Pythagorean theorem to
find vertical height from base
first.
r = 5cm
Lesson 8 HW Volume of a cone p. 46
Volume cone = ⅓ Base x height = ⅓ Bh =⅓ r2h
Write equation along top of worksheet so you
can keep referring to it easily.
Circle 1a, 1b, 1c, 1d and 1e.
2a., 2c and 2e to do at home.
Homework: Volume of Pyramids and Cones
Volume pyramid = ⅓ Base x height = ⅓ Bh
Page 39-40 5 a, b, c, d, e. 6 a, c and e.
Volume cone = ⅓ Base x height = ⅓ Bh =⅓ r2h
Write equation along top of worksheet so you
can keep referring to it easily.
Page 46-47
1a, 1c, 1d and 1e. 2a., 2c, 2d and 2e
P. 39 Determine the volumes of the pyramids:
5. a) square pyramid
V = ⅓ (6cm x 6cm) x 14 cm
= 168cm3
b) rectangular pyramid height = √102 - 42 = √(100 - 16)
=
V = ⅓ (6cm x 8cm) x 9.165 cm
= 146.64cm3
9.165cm
P. 39 Determine the volumes of the pyramids:
5. c) regular hexagonal pyramid
V = ⅓ Bh = ⅓ (½ perimeter*apothem)h
V = ⅙ (48cm)(4√3cm)(20cm)
V= 640√3cm3
P. 39 Determine the volumes of the pyramids:
5. d) square pyramid
Right triangles to find h
V = ⅓ Bh
V = ⅓ (10cm)(10cm)(10.91cm)
V= 363.62cm3
Slant height2 + 52 = 132
Slant height = 12
Then
h2 + 52 = 122
h = √119 approx. 10.91
P. 40 Finding the volumes of the pyramids:
6. C. “composite figure”
Height of square pyramid =√102 - 62 =
= √(64) = 8 cm
10 cm
12 cm
12 cm
P. 40 Finding the volumes of the pyramids:
6. C. “composite figure”
Volume pyramid = ⅓ (12cmx12cm)x8cm
= 384 cm3
8 cm
12 cm
12 cm
P. 40 Finding the volumes of the pyramids:
6. C. “composite figure”
Volume pyramid = ⅓ (12cmx12cm)x8cm
= 384 cm3
8 cm
12 cm
12 cm
Volume of
rectangular prism =
12cm x 12cm x 5cm
= 720cm3
6. C. “composite figure”
Total volume = 384 cm3 + 720 cm3
= 1104 cm3
8 cm
12 cm
12 cm
6. e. 2 angles marked as right angles
Volume = ⅓ (½ x 4cm x 5cm) x 6 cm
= 20 cm3