Geometry
Semester 2
Unit 5 Lesson 3
20 May, 2016
Agenda 20 May, 2016
Review for final - start
Tues, including Stats test results 9:15 - 10:25
● Areas of triangles - quick correct pg 5-6
● Volume of Cylinders - pg16 -17
● Volume of Pyramid - Demonstration
○ Worksheet #4 p22 - 24
● The Cone
● Any Way You Spin It
● Homework
Calculate the area of each triangle p6
16. According to the values on the grid, one large square is 2
units.
Which makes each small square 0.4 units (⅖)
Area of triangle EFH = ½ base x height
= ½ x 5.2 units x 2.8 units
= 7.28 units2
Calculate the area of each triangle p6
16. According to the values on the grid, one large square is 2
units.
Which makes each small square 0.4 units (⅖)
Area of triangle GHF = ½ base x height
= ½ x 3.6 units x 2.8 units
= 5.04 units2
Calculate the area of each triangle p6
17. Any one crack it?
Calculate the area of each triangle p6
18. Area triangle FEG = ½ (5units)(10u)
= 25 units2
19. Area each triangle = ½ (5units)(15u)
= 37.5 units2
Worksheet #3
1a) Volume
2
= Area circle x height = ( r )h
V= (4cm)27cm
V = (16x7)cm3
3
V=112 cm
Worksheet #3
1b) Volume
2
= Area circle x height = ( r )h
V= (3cm)28cm
V = (9x8)cm3
3
V=72 cm
Worksheet #3
1c) Volume
2
= Area circle x height = ( r )h
V= (5cm)29cm
V = (25x9)cm3
3
V=225 cm
Oblique cylinder - just use vertical height
from one base to the other
Precise language: prisms - base supplies name
Prism - a polyhedron
with 2 congruent
parallel bases
Worksheet #3
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
Worksheet #3
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.
Worksheet #3
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
Worksheet #3 p 17
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
Worksheet #3 p 17
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
Worksheet #3 p 17
3c) V
2
= ½ ( r )h + (l x l x l)
V=½ (3cm)210cm + (10cm x 2cm x 6cm)
3
V =45 cm +120cm
3
V = approx 261.3 cm3
Volume of a Pyramid
Vpyramid = ⅓ base x height
Demonstration - live
And via video
Worksheet #4 pg 22
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
Worksheet #4 pg 22
Apex
Lateral edge
Slant height
Vertex
Face
Worksheet #4 pg 22
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.
Worksheet #4 pg 22
3.
Worksheet #4 pg 22
3.
Square pyramid
Worksheet #4 pg 22
3.
6 cm
8 cm
Worksheet #4 pg 22
3.
rectangular pyramid
6 cm
8 cm
triangular pyramid
Worksheet #4 pg 22
3.
rectangular pyramid
6 cm
8 cm
All 4 faces are
equilateral triangles
Regular
tetrahedron
Worksheet #4 pg 22
4. Same base area and height, same volume.
Worksheet #4 pg 24
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 9.2 cm
= 146.6cm3
9.2cm
Worksheet #4 pg 23
6. C. “composite figure”
Height of square pyramid =√102 - 62 =
= √(64) = 8 cm
10 cm
12 cm
12 cm
Worksheet #4 pg 23
6. C. “composite figure”
Volume pyramid = ⅓ (12cmx12cm)x8cm
= 384 cm3
8 cm
12 cm
12 cm
Worksheet #4 pg 23
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
Worksheet #4 pg 23
6. C. “composite figure”
Total volume = 384 cm3 + 720 cm3
= 1104 cm3
8 cm
12 cm
12 cm
Worksheet #4 pg 23
6. e. 2 angles marked as right angles
Volume = ⅓ (½ x 4cm x 5cm) x 6 cm
= 20 cm3
G. GMD.3 Student Notes WS #5
Volume cone = ⅓ Base x height = ⅓ Bh =1/3 r2h
G. GMD.3 Student Notes WS #5
Volume cone = ⅓ Base x height = ⅓ Bh =⅓ r2h
Example #1
r= 3cm
h= 8cm
V = ⅓ Bh
V= ⅓ r2h
V= (3cm)28cm
V=24 cm3
G. GMD.3 Student Notes WS #5
Volume cone = ⅓ Base x height = ⅓ Bh =⅓ r2h
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
Example #3
Given slant height,
Use Pythagorean theorem to
find vertical height from base
first.
G. GMD.3 Student Notes WS #5 page 26
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, 1c, and 1e.
2a., 2c and 2e to do at home.
5.2 Any Way you Spin It
“Solids of revolution”
G-GMD-4 Worksheet #1
1. Describe the solid that is formed by
rotating each of these figures about
the line m and sketch it.
https://www.youtube.com/watch?
v=3oAjcLD34kc
Homework
Worksheet #5 p. 26
1a, 1c, and 1e.
2a., 2c and 2e
Look at your videos again we will be watching them as
part of the review work next
week - and you will be able to
earn the final 10 points for the
project as we do.
Worksheet #1 p. 31- 32: 1, 2, 3
Page 33 1 & 2
Page 11 and 12: 5, 6, 7, 8, 9, &10.