Egyptian Pyramidverulam.s3.amazonaws.com/resources/ks4/maths/Year 11 Maths Grade A... · 3 = 1 x2 x = base area height Volume of the pyramid = x3 x x x For any pyramid, Volume of

Egyptian Pyramid

These are pyramids

Vertex

Slant edges height

V

A

B

C

D

x

x

x

cube

Three congruent pyramids

3

1x2 x =

= 3

1 base area height

Volume of the pyramid =

3

1

x3 x

x

x

For any pyramid,

Volume of pyramid = 3

1 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

C D

B

15 cm 9 cm E

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 2 The 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

D C

B

H G

F

6 cm

12 cm

Solution :

Volume of VEFGH = ( 3

1 ( 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

B A

V

B A

D C

V V

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

B A

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 l

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

Curved surface area = Area of the sector

Curved 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 3

1

Volume of a cone = πr2 h

1 3

How to calculate total surface area of a cone?

Total surface area =πr2 + πr l

+

r

l l r

Examples

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

is the volume?

Answer:

Volume = πr2h 1 3

1 3

= 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

3

1

3

1

3

1

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

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 C A. 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

C B

50cm

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 = ) 22 7

Help

Answer

Let V is the volume of the pyramid and y be the length of

a side of base

V = base area height 1 3

96 = y2 8 1 3

288 = 8y2

36 = y2

y = 6

Therefore, the length of a

side of base is 6 cm

Back to Q1

To Q2

what is the length of a

side of the base?

( )3 = AB AX 16

54

AB AX

( )3 = 8 27

AX AB

= 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 =15cm 2 1

Therefore, the height

(VM) of the pyramid

is 20 cm

Volume of the pyramid is:

= ×18 ×24 ×20 1 3

×base area ×height 3 1

= 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 h 3 1

= 142 48 3 1 22

7

= 704 cm3

The volume is 704 cm3

Back to Q4

A

C B

50cm

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

(Take = 22/7)