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Course Title: EDD 5165A Information Technology in Education Instructor: Dr. LEE Fong Lok Group Members: Tiu Chui Yee (99039490) Li Chun Lan (99040120) Ching Yuk Shan (99043270 )

Course Title: EDD 5165A Information Technology in Education Instructor: Dr. LEE Fong Lok Group Members: Tiu Chui Yee (99039490) Li Chun Lan (99040120)

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Course Title: EDD 5165AInformation Technology in Education

Instructor: Dr. LEE Fong Lok

Group Members:

Tiu Chui Yee (99039490)

Li Chun Lan (99040120)

Ching Yuk Shan (99043270)

Form 3 and Band 2 students

Cone and Pyramid

Mathematics

Usage :

Title :

Subject :

Target Audience :

Lecturing

Prerequisite knowledge :

1. Pythagoras’ Theorem.

3. Ratio and proportion

2. Area of some plane figures

e.g. square, rectangle, triangle, circle, sector

Let the students know and apply the mensuration concepts

Objectives :

Distribution of workload

Tiu Chui Yee (99039490):

Content of Pyramid

Li Chun Lan (99040120):

Content of Cone

Ching Yuk Shan, Maggie (99043270):

Quiz

Egyptian Pyramid

These are pyramids

Vertex

Slant edges height

V

A

B

CD

xx

x

cube

Three congruent pyramids

31

x2 x =

= 31

base area height

Volume of the pyramid = 3

1x3

xx

x

For any pyramid,

Volume of pyramid =31 base area height

Example 1

The figure shows a pyramid with a rectangular base ABCD of area 192 cm2, VE = 15 cm and EF = 9 cm, find the volume of the pyramid.

V

A

CD

B

15 cm9 cmE

F

Solution : VF2 = (152 - 92 ) cm2

VF = 22 9 - 15 cm

= 12 cm

Volume of the pyramid

= 3

1 base area height

= ( 3

1 192 12) cm3

= 768 cm3

V

E F

15 cm

9 cm

Pyramid B

A frustum

Pyramid A

= Volume of Pyramid A -Volume of the frustum Volume of Pyramid B

B

= -

Example 2The base ABCD and upper face EFGH of the frustum are squares of side 16 cm and 8 cm respectively. Find the volume of the frustum ABCDEFGH.

V

A

E

DC

B

H G

F

6 cm

12 cm

Solution :

Volume of VEFGH = ( 31

( 8 8 ) 6) cm3

= 128 cm3

Volume of VABCD = ( 31

( 16 16 ) 12) cm3

= 1024 cm3

Volume of frustum ABCDEFGH

= (1024 - 128 ) cm3

= 896 cm3

V

D C

BA

V

BA

D C

VV

V

C

Base area The sum of of the area of all lateral faces+=

Total surface area of a pyramid

Total surface area of pyramid VABCD =

+ + + +

lateral faces

Base

V

D C

BA

Example 3

The figure shows a pyramid with a rectangular base ABCD of area 48 cm2. Given that area of VAB = 40 cm2 , area of VBC = 30 cm2, find the total surface area of the pyramid.

Solution :

Total surface area of pyramid VABCD

+ (Area VAB + Area VDC +

Area VBC + Area VAD )

+ (Area VAB 2) +

(Area VBC 2)

+ ( (40 2) + (30 2)) cm2

= 188 cm2

= Area of ABCD

= Area of ABCD

= 48 cm2

How to generate a cone?

…...

…...

How to calculate the curved surface area ?

Cut here

l

r

l

2πr

Curved surface area = πr lCurved surface area = πr l

Curved surface areaRemark : Area of sector = 1/2r2 (θ/2)= 1/2 r2 θ or 1/2 r l

Curved surface area = Area of the sectorCurved surface area = 1/2 ( l ) ( 2π r )

= π r l

After cutting the cone,

θ

r

l

Volume of a cone

r

h

r

h 31

Volume of a cone = πr2 h

Volume of a cone = πr2 h

13

How to calculate total surface area of a cone?

Total surface area =πr2 + πr l Total surface area =πr2 + πr l

+r

ll r

Examples1 a) If h = 12cm, r= 5 cm, what

is the volume?

Answer:

Volume = πr2h13

13

= π (52) ( 12)

= 314 cm3

b) what is the total surface area?Based Area = π52

= 25πcm2

Slant height

= 13 cm

Curved surface area = π(5) ( 13)= 65π cm2

Total surface area = based area + curved surface area

= 25π+65π= 90π

= 282.6cm2 (corr.to 1 dec.place)

= 122 + 5 2

Volume of Frustum

Volume of Frustum

= -

R r

= πR3 - π r3

31

31

31

π( R3 - r3 ) =

Volume of frustum = volume of big cone - volume of small cone

Start Now Exit

The volume of a pyramid of square base is 96 cm3. If its height is 8 cm, what is the length of a side of the base?

Q1

Answer is C

8cm

A. 2 cm

B. 2 3cm

C. 6cm

D. 12cm

E. 36cm

Help

Answer

To Q2

In the figure, the volumes of the cone AXY and ABC are 16 cm3 and 54 cm3 respectively, AX : XB =

Q2

Answer is A

A

X Y

B CA. 2 : 1

B. 2 : 3

C. 8 : 19

D. 8 :27

E. 3 16 : 3 38

Help

Answer

To Q3

V

D

C

A

B

M

Q3 In the figure, VABCD is a right pyramid with a rectangular base. If AB=18cm, BC=24cm and CV=25cm, find

a) the height (VM) of

the pyramid,

b) volume of the

pyramid. Help

Answer

To Q4

a) 20cm

b) 2880cm3

A

CB

50cm

48cm

Q4

The figures shows a right circular cone ABC. If AD= 48cm and AC= 50cm, find

(a) the base radius (r) of the cone,

(b) the volume of the cone.

(Take = )227

Help

Answera) 14cm

b) 704cm3

Let V is the volume of the pyramid and y be the length of a side of base

V = base area height13

96 = y2 813

288 = 8y2

36 = y2

y = 6

Therefore, the length of a side of base is 6 cm

Back to Q1

To Q28c

m

what is the length of a side of the base?

( )3 = ABAX 16

54

ABAX ( )3 = 8

27

AXAB = 2

3

AB = AX + XB and AX = 2, AB = 3

3 = 2 + XB

XB = 1

Therefore, AX : XB = 2 : 1

Hints: Using the concept of RATIOS

Back to Q2

To Q3

A

X Y

B C

AX : XB = ?

AC2 =182 + 242

AC2 = 900

AC = 30cm

252 = VM2 + MC2

625 = VM2 + 152

625 - 225 = VM2

VM2 = 400

VM = 20cm

MC = AC =15cm21

Therefore, the height (VM) of the pyramid is 20 cm

Volume of the pyramid is:

= ×18 ×24 ×2013

×base area ×height31

= 2880cm3

Therefore, the volume of the pyramid is 2880cm3

Back to Q3

To Q4

a) the height (VM) of the pyramid

b) volume of the pyramid.

The radius is r, therefore:

502 = 482 + r2

2500 = 2304 + r2

196 = r2

r = 14

The radius is 14cm.

The volume (V) of cone is:

V = r2 h31

= 142 4831 22

7

= 704 cm3

The volume is 704 cm3

Back to Q4

A

CB

50cm48

cm

(a) the base radius (r) (b) the volume of the cone

(Take = 22/7)