VA VB − 9 15 −

20190521 F.5 MC Practice (Mensurations – Prisms, Pyramids, Cones and Spheres) Solution

1

Level 1: single concept, simple calculation, easier than HKDSE types

Level 2: one or two concept, some calculations, similar to HKDSE easy types

Level 3: involving high level, logical and abstract thinking skills, or with

complicated calculations, similar to HKDSE difficult types

** Please answer any 30 questions **

Level 1

1. Find the volume of the pyramid VABC in the figure.

A. 90 cm3 B. 112.5 cm3 C. 180 cm3 D. 270 cm3

Solution:

The answer is A.

In △VAB,

AB =22 VAVB − =

22 915 − cm = 12 cm

∴ Volume = 95122

1

3

1×

××× cm3 = 90 cm3

<end>

2. The figure shows a right pyramid VABCD, whose base is a rectangle ABCD. It is

given that the volume of the pyramid is 112 cm3. Find the length of AB.

A. 3

2cm B. 2 cm C. 4 cm D. 6 cm

Solution:

The answer is D.

Let x cm be the length of AB.

7)8(3

1××× x = 112

x = 6

∴ The length of AB is 6 cm.

15 cm

5 cm

8 cm


2

<end>

3. In the figure, ABCDEFGH is a cuboid and EACD is a pyramid. What is the ratio

of the volume of the pyramid to that of the cuboid?

A. 1 : 2 B. 1 : 3 C. 1 : 4 D. 1 : 6

Solution:

The answer is D.

cuboid theof Volume

pyramid theof Volume =

DEDCAD

DEDCAD

××

×

×××

2

1

3

1

=6

1

∴ The required ratio = 1 : 6

<end>

4. The total surface area of the regular pyramid in the figure is 184 cm2. The area of

each lateral face of the pyramid is 30 cm2. Find y.

A. 3.75 B. 7.5 C. 8 D. 8.5

Solution:

The answer is B.

Let x cm be the length of each side of the base.

4 × 30 + x2 = 184

x2 = 64

x = 8

∴ ×2

1 y × 8 = 30

y = 7.5

y cm


3

<end>

5. The figure shows a right pyramid ABCDE. Its base is a rectangle with dimensions

20 cm × 10 cm. The height AP of △ABC is 11 cm. Find the total surface area of

the pyramid.

A. 360 cm2 B. 380 cm2 C. 560 cm2 D. 625 cm2

Solution:

The answer is C.

Refer to the notations in the figure.

In △ACP,

PC = BC×2

1= cm 20

2

1× = 10 cm

AC =22 PCAP + =

22 1011 + cm = 221 cm

In △ACQ,

CQ = CD×2

1= cm 10

2

1× = 5 cm

AQ = 22CQAC − =

22 5)221( − cm = 14 cm

Total surface area of the pyramid

= 2 × area of △ABC + 2 × area of △ACD + area of rectangle BCDE

=

×+×××+××× 10201410

2

121120

2

12 cm2

= 560 cm2

<end>

11 cm

10 cm

A

B C

D E

P

Q

20 cm

11 cm

10 cm

A

B C

D E

P 20 cm


4

6. The figure shows a frustum of a regular pyramid. Its upper base and lower base

are squares of sides 7 cm and 14 cm respectively. VN = 25 cm. Find the volume

of the frustum.

A. 784 cm3 B. 817 cm3 C. 1 372 cm3 D. 1 429 cm3

Solution:

The answer is C.


XN = AB×2

1= cm 14

2

1× = 7 cm

In △VXN,

VX =22 XNVN − =

22 725 − cm = 24 cm

YM = FG×2

1= cm 7

2

1× = 3.5 cm

∵ △VYM ~ △VXN (AAA)

∴ VX

VY =

XN

YM

cm 24

VY =

cm 7

cm 5.3

VY = 12 cm

∴ Volume of the frustum = volume of pyramid VABCD – volume of pyramid VEFGH

=

××−×× 127

3

12414

3

1 22 cm3

= 1 372 cm3


5

<end>

7. Find the curved surface area of the right circular cone in the figure.

A. 270π cm2 B. 306π cm2 C. 450π cm2 D. 1 080π cm2

Solution:

The answer is B.

Slant height = cm 5.2212 22+ = 25.5 cm

Curved surface area = π × 12 × 25.5 cm2 = 306π cm2

<end>

8. Which of the following right circular cones has the greatest volume?

A. B.

C. D.

Solution:

The answer is B.

A: Volume = 105π3

1 2××× = π

3

183

B: Volume = 86π3

1 2××× = 96π

C: Volume = 6)69(π3

1 222×−×× = 90π

D: Volume = 222 6.9106.9π3

1−××× = 86.016π

∴ The circular cone in B has the greatest volume.

<end>

22.5 cm

12 cm

5

8


6

9. A right circular cone and a cylinder are of the same height and the same volume.

If the base radius of the cylinder is 6 cm, find the base radius of the cone.

A. 2 cm B. cm 12 C. cm 108 D. 18 cm

Solution:

The answer is C.

Let r cm be the base radius of the cone, and h cm be the heights of both solids.

hr 2π

3

1 = π × 62 × h

r2 = 108

r = 108

∴ .cm 108 is cone theof radius base The

<end>

10.

Fig. A shows a sector of area 6π cm2 with 135° as the angle of the sector. The

sector is folded into a right circular cone shown in Fig. B. Find the volume of the

cone obtained, correct to the nearest 0.1 cm3.

A. 2.8 cm3 B. 8.7 cm3 C. 9.4 cm3 D. 69.9 cm3

Solution:

The answer is B.

Let r cm and ℓ cm be the base radius and slant height of the cone respectively.

°

°××

360

135π

2ℓ = 6π

2ℓ = 16

ℓ = 4

∵ Circumference of the base of the cone = arc length of the sector

∴ 2πr =°

°××

360

13542π

r = 1.5

∴ Height of the cone =22 5.14 − cm

= 75.13 cm

Volume of the cone = 75.135.1π3

1 2××× cm3

Fig. A Fig. B

135°°°°


7

= 8.7 cm3, cor. to the nearest 0.1 cm3

<end>

11. The cone shown in the figure is cut along a slant height and then a sector is

obtained. Which of the following is/are true?

I. The radius of the sector is 24 cm.

II. The area of the sector is 175π cm2.

III. The angle of the sector is an acute angle.

A. I only B. II only C. I and III only D. II and III only

Solution:

The answer is B.

For I:

Slant height of the cone =

2

2

2

1424

+ cm = 25 cm

∴ Radius of the sector = 25 cm ≠ 24 cm

∴ I is not true.

For II:

Area of the sector = 252

14π ×× cm2

= 175π cm2

∴ II is true.

For III:

Let θ be angle of the sector.

°

××360

25π2 θ

= 175π

θ = 100.8°

> 90°

∴ The angle of the sector is not an acute angle.

∴ III is not true.

∴ Only II is true.

<end>


8

12. The figure shows a solid formed by a right circular cone and a cylinder with a

common base. It is given that the curved surface areas of the cone and the

cylinder are the same, and the height of the cone is 1.6 times that of the cylinder.

Find the total surface area of the solid.

A. 60π cm2 B. 120π cm2 C. 156π cm2 D. 276π cm2

Solution:

The answer is C.


∵ Curved surface area of the cone = curved surface area of the cylinder

∴ π × 6 × ℓ = 2π × 6 × h

ℓ = 2h

According to the question, we have H = 1.6h.

In △ABC,

H2 + 62 = 2ℓ

(1.6h)2 + 62 = (2h)2

2.56h2 + 36 = 4h2

1.44h2 = 36

h2 = 25

h = 5

Total surface area of the solid = [2 × (2π × 6 × 5) + π × 62] cm2

= 156π cm2

<end>

13. As shown in the figure, after putting a right circular cone into a cylindrical

container, the depth of water is one third of the height of the cone. Find the

volume of water inside the container.


9


Solution:

The answer is A.


cm 18cm 3

1127 =

−×=VA

∵ △VAB ~ △VPQ (AAA)

∴ PQ

AB =

VP

VA

cm 6

AB =

27

18

AB = 4 cm

Volume of the frustum under water

= volume of the cone VQR – volume of cone VBC

=

×××−××× 184π

3

1276π

3

1 22 cm3

= 228π cm3

Volume of water inside the container

= volume of the cylindrical part in contact with water – volume of the frustum under water

=

−

××× π22827

3

16π

2 cm3

= 96π cm3

<end>

6 cm

27 cm

V

A B

P Q

C

R

6 cm

27 cm


10

14. If the total surface areas of the hemisphere, the sphere and the cylinder as shown

in the figure are represented by p, q and r respectively, which of the following is

true?

A. p > q > r B. p > r > q C. q > p > r D. q > r > p

Solution:

The answer is B.

p = 22 4π44π2

1×+×× = 8π4

q = 23π4 ×× = π36

r = 24π2142π ××+×× = 0π4

∴ p > r > q

<end>

15. The figure shows a hemisphere of diameter d cm. Express its volume in terms of

d.

A. 3π

12

1d cm3 B. 3

π6

1d cm3 C. 3

π3

2d cm3 D. 3

π3

4d cm3

Solution:

The answer is A.

Volume =

3

2π

3

4

2

1

×××

dcm3 = 33 cm π

12

1d

<end>

16. A solid metal sphere of surface area 144π cm2 is melted and recast into 4 identical

small solid metal spheres. Find the volume of each small sphere.

A. 4.5π cm3 B. 36π cm3 C. 72π cm3 D. 288π cm3

Solution:

The answer is C.

Let r cm be the radius of the sphere before melting.

3

d cm


11

4πr2 = 144π

r2 = 36

r = 6

∴ The radius of the sphere before melting is 6 cm.

Volume of each small sphere = 36π3

4

4

1××× cm3

= 72π cm3

<end>

17. In the figure, a steel sphere is put into a cylindrical container of base radius

7.5 cm. The water surface just covers the sphere and the depth of water is 6 cm.

Find the original depth of water before putting the sphere into the container.

A. 0.64 cm B. 0.88 cm C. 4.8 cm D. 5.36 cm

Solution:

The answer is D.

Diameter of the sphere = depth of water

= 6 cm

Let h cm be the original depth of water.

∵ Volume of the sphere = volume of water that has risen

∴

3

2

6π

3

4

×× = π × 7.52 × (6 – h)

0.64 = 6 – h

h = 5.36

∴ The original depth of water is 5.36 cm.

<end>

18. In the figure, a test tube whose shape is a cylinder with a hemisphere at one end is

placed vertically. The height of the test tube is 7x, where x is the radius of the

hemisphere. If the volume of water in the test tube occupies half of the space in

the test tube, find the area of the surface of the test tube in contact with water.

7.5 cm


12

A. 2π

3

16x B. 7πx2 C. 2

π3

22x D. 2

π3

28x

Solution:

The answer is C.

Capacity of the test tube = volume of the hemisphere + volume of the cylinder

= )7(ππ3

4

2

1 23 xxxx −×+×

= 3π

3

20x

Volume of water = 3π

3

20

2

1x× = 3

π3

10x

Let h be the depth of water in the test tube.

)(ππ3

4

2

1 23 xhxx −×+× = 3π

3

10x

xhx −+3

2 = x

3

10

h = x3

11

∴ The required area = )(π2π42

1 2 xhxx −×+×

=

−×+ xxxx

3

11π2π2 2

= 2π

3

22x

<end>

x

7x


13

19. The height of a pyramid is 15 cm and its base area is 30 cm2. Find the volume of

the pyramid.

A. 150 cm3 B. 225 cm3 C. 325 cm3 D. 450 cm3

Solution:

The answer is A.

Volume of the pyramid 3cm 15303

1××=

3cm 150=

<end>

20. What is the volume of the pyramid in the figure?

A. 320 cm3 B. 230 cm3 C. 160 cm3 D. 80 cm3

Solution:

The answer is C.

Volume of the pyramid 3cm 161252

1

3

1×

×××=

3cm 160=

<end>

21. The figure shows a right pyramid VABCD, whose base ABCD is a square. Find

the total surface area of the pyramid.

A. 459 cm2 B. 864 cm2 C. 1 296 cm2 D. 1 404 cm2

5 cm

18 cm


14

Solution:

The answer is B.

OP = BC×2

1= cm 18

2

1× = 9 cm

In △VOP,

cm 15cm 912 2222=+=+= OPVOVP

Area of △VCD 2cm 15182

1××=

2cm 135=

Total surface area of the pyramid = (4 × 135 + 182) cm2 = 2cm 864

<end>

22. The height and the volume of a pyramid are 13 cm and 65 cm3 respectively. What

is the base area of the pyramid?

A. 5 cm2 B. 2

cm 65 C. 12 cm2 D. 15 cm2

Solution:

The answer is D.

Let A cm2 be the base area of the pyramid.

15

65133

1

=

=××

A

A

∴ The base area of the pyramid is 15 cm2.

<end>

23. The volume of a pyramid is 180 cm3. If the base of the pyramid is a square of side

6 cm, what is the height of the pyramid?

A. 90 cm B. 30 cm C. 15 cm D. cm 30

Solution:

The answer is C.

Let h cm be the height of the pyramid.

15

18063

1 2

=

=××

h

h

∴ The height of the pyramid is 15 cm.

<end>

24. In the figure, a solid right pyramid VABCD fits perfectly in a rectangular box

ABCDSPQR. What is the volume of the space left in the box?


15

A. 10 m3 B. 15 m3 C. 20 m3 D. 30 m3

Solution:

The answer is C.

Volume of the space left in the box = capacity of the box – volume of the pyramid

3m 5)23(3

1523

×××−××=

3

3

m 20

m ]1030[

=

−=

<end>

25. The figure shows the net of a right pyramid with a square base. What is the total

surface area of the pyramid?

A. 240 cm2 B. 340 cm2 C. 400 cm2 D. 2cm 3

300 1

Solution:

The answer is B.


13 cm

5 m

3 m

13 cm


16

NB = AB×2

1= cm 10

2

1× = 5 cm

In △VBN,

VN = cm 513 2222−=− NBVB = 12 cm

Total surface area of the pyramid 2cm 101012102

14

×+×××=

2

2

cm 340

cm )100240(

=

+=

<end>

26. The height of a right circular cone is 15 cm and its base diameter is 14 cm. Find

the volume of the circular cone.

A. 245π cm3 B. 630π cm3 C. 735 cm3 D. 980 cm3

Solution:

The answer is A.

Volume of the circular cone 3

2

cm 152

14π

3

1×

××=

3cm π245=

<end>

27. In the figure, the height of the right circular cone is 4 cm and its base radius is

3 cm. The base radius of the cylinder is 2 cm. If they have the same volume, find

the height of the cylinder.

A. 2 cm B. 3 cm C. 4 cm D. 6 cm

Solution:

The answer is B.

Let h cm be the height of the cylinder.

3

43π3

12π

22

=

×××=××

h

h

∴ The height of the cylinder is 3 cm.

<end>

3 cm


17

28. The figure shows a right circular cone. Its base radius and height are 9 cm and

12 cm respectively. Find the curved surface area of the circular cone.


Solution:

The answer is C.

Slant height of the circular cone cm 912 22+= = 15 cm

Curved surface area of the circular cone = π × 9 × 15 cm2

= 2cm π135

<end>

29. The base radius of a right circular cone is 1.5 cm and its height is 2 cm. Find the

total surface area of the circular cone.

A. 2.25π cm2 B. 3.75π cm2 C. 5π cm2 D. 6π cm2

Solution:

The answer is D.

Slant height of the circular cone cm 52cm 25.1 22 .=+=

Total surface area of the circular cone = (π × 1.5 × 2.5 + π × 1.52) cm2

= (3.75π + 2.25π) cm2

= 2cm 6π

<end>

30. A cylindrical container of base radius 6 cm and height 8 cm is filled up with

water. An empty cup in the shape of an inverted right circular cone of base radius

10 cm is placed vertically. Paul pours all the water from the container into the

cup. If the cup is just filled up, find the height of the cup.

(Give the answer correct to the nearest 0.1 cm.)

A 8.6 cm B. 10.1 cm C. 26.1 cm D. 86.4 cm

Solution:

The answer is A.

Let h cm be the height of the cup.

Volume of the conical cup = volume of the cylindrical container

h×××210π

3

1 = 86π

2××

12 cm

9 cm


18

h = 8.6, cor. to 1 d.p.

∴ The height of the cup is 8.6 cm.

<end>

31. In the figure, a rectangular tank of height 7 cm has a square base of side 10 cm

and is filled up with water. If the water in the tank can just fill up an empty

vertical right conical vessel of height 15 cm, what is the base radius of the

conical vessel?

(Give the answer correct to 3 significant figures.)

A. 6.68 cm B. 6.83 cm C. 11.8 cm D. 21.0 cm

Solution:

The answer is A.

Let r cm be the base radius of the conical vessel.

Volume of the conical vessel = volume of the rectangular tank

15π3

1 2××× r = 102 × 7

r2 =π

140

r =π

140

= 6.68, cor. to 3 sig. fig.

∴ The base radius of the conical vessel is 6.68 cm.

<end>

32. Find the volume of the right conical paper cup in the figure.

(Give the answer correct to 3 significant figures.)

A. 20.2 cm3 B. 63.4 cm3 C. 98.1 cm3 D. 190 cm3

15 cm

2.5 cm


19

Solution:

The answer is B.

Height of the conical paper cup cm 5210 22 .−=

cm 7593.=

Volume of the conical paper cup 32 cm 759352π3

1.. ×××=

3cm 463.= , cor. to 3 sig. fig.

<end>

33. The surface area of a sphere is 64π cm2. Find the volume of the sphere.

A. 3cm π3

64 B. 3cm π

3

128 C. 64π cm3 D. 3cm π

3

256

Solution:

The answer is D.

Let r cm be the radius of the sphere.

4πr2 = 64π

r2 = 16

r = 4

∴ The radius of the sphere is 4 cm.

Volume of the sphere 33 cm 4π3

4××=

3cm π3

256=

<end>

34. Find the volume of the hemisphere in the figure.


Solution:

The answer is C.

Volume of the hemisphere 33 cm 6π3

4

2

1×××=

3cm π144=

<end>

6 cm


20

35. The volume of a sphere is 36π cm3. Find its surface area.


Solution:

The answer is C.


3π

3

4r = 36π

r3 = 27

r = 3


Surface area of the sphere = 4 × π × 32 cm2

= 2cm π36

<end>

36. The surface area of a sphere is 576π cm2. Find its diameter.

A. 12 cm B. 24 cm C. 36 cm D. 48 cm

Solution:

The answer is B.


4πr2 = 576π

r2 = 144

r = 12


∴ Diameter = 2 × 12 cm

= 24 cm

<end>

37. In the figure, a solid sphere of radius 3 cm can just fit into a cylindrical container

of base radius 3 cm and height 6 cm. Find the volume of the remaining space in

the container.


3 cm


21

Solution:

The answer is D.

Volume of the remaining space = volume of the container – volume of the sphere

= 332 cm 3π3

463π

××−××

= (54π – 36π) cm3

= 18π cm3

<end>

38. The solid shown in the figure consists of a right circular cone and a hemisphere

with a common base. Find the volume of the solid.

A. 621π cm3 B. 891π cm3 C. 972π cm3 D. 1 377π cm3

Solution:

The answer is B.

Base radius of the cone = radius of the hemisphere

=2

18cm

= 9 cm

Height of the cone = (24 – 9) cm

= 15 cm

Volume of the solid = volume of the cone + volume of the hemisphere

= 332 cm 9π3

4

2

1159π

3

1

×××+×××

= (405π + 486π) cm3

= 891π cm3

<end>

18 cm


22

39. ABCDEFGH is a cuboid of length 15 cm, width 9 cm and height 10 cm.

DE is the height of pyramid ABCDE . Find the volume of the pyramid.

A. 375 cm3 B. 450 cm3 C. 675 cm3 D. 1 350 cm3

39. B

Volume of pyramid ABCDE

3

3

cm 450

cm ]10)915(3

1[

=

×××=

40. The figure shows a right pyramid with trapezium base ABCD . Find the

volume of the right pyramid.

A. 418 cm3 B. 832 cm3 C. 1 664 cm3 D. 2 496 cm3

10 cm

15 cm

9 cm

A B

CD

E

F G

H

12 cm

9 cm

10 cm

11 cm

A

CD

B

V


23

40. A

Base area

2

2

cm 5.104

cm ]11)109(2

1[

=

×+×=

Volume of the right pyramid

3

3

cm 418

cm )125.1043

1(

=

××=

41. In the figure, VABCD is a right pyramid of height 20 cm and the base

is a rectangle of dimensions 18 cm× 24 cm. Find the length of VE .

(Correct your answer to 1 decimal place.)

A. 21.9 cm B. 23.3 cm C. 25.0 cm D. 28.4 cm

41.A

cm 9

cm 18)2

1(

2

1

=

×=

= ADOE

24 cm

A

BC

D18 cm

20 cm

V

O E

20 cm

9 cmO

V

E


24

In ∆VOE ,

2VE 22

OEVO += (Pyth. theorem)

VE cm 920 22+=

cm 481=

cm 9.21= (corr. to 1 d.p.)

42. In the figure, the top and the base of the right frustum are rectangles.

Find the volume of the frustum.

A. abh3

127 B. abh

3

112

C. 150abh D. It cannot be found.

42.B

Volume of the frustum

abh

hbahhba

3

112

443

1)(88

3

1

=

×××−+×××=

43. In the figure, the top and the base of the right frustum are squares of

sides a and 6a respectively. If the height of the original right pyramid

is 6h and the height of the removed right pyramid is h, find

VPQRS dimaryp fo emulov

mutsurf eh tfo emulov

.

4a

4b

8a

8b

V

B

H

C

A

F

D

E

G

h

h

6h

6a

a

h

B

A

C

Q

P

R

S

V

D


25

A. 64

1 B.

64

63 C.

8

7 D.

216

215

43.D

Volume of pyramid VABCD

ha

ha

2

2

3

1

3

1

=

××=

Volume of pyramid VPQRS

ha

ha

2

2

72

6)6(3

1

=

××=

Volume of the frustum

ha

haha

2

22

3

215

3

172

=

−=

∴ VPQRS dimaryp fo emuloV

mutsurf eh tfo emuloV

216

215

72 2

2

3215

=

=ha

ha

44. The base of a right pyramid is a square. If the total surface area of the

pyramid is 420 cm2 and the area of one of its lateral faces is 56 cm2, find

the base area of the pyramid.

A. 84 cm2 B. 105 cm2 C. 196 cm2 D. 364 cm2

44. C

Area of all lateral faces = (4 × 56) cm2

= 224 cm2

∴ Base area of the pyramid = (420 − 224) cm2

= 196 cm2


26

45. The figure shows the right pyramid VABCD with a rectangular base. If the

total surface area of the right pyramid VABCD is 4 320 cm2, CD = 30 cm,

VP = 30 cm and VQ = 36 cm, find the length of BC.

A. 54 cm B. 63 cm C. 72 cm D. 108 cm

45. A

Let x cm be the length of BC.

54

3204080160

320430)302

1(2)3630

2

1(2

=

=+

=+×××+×××

x

x

xx

∴ The length of BC is 54 cm.

46. In the figure, the total surface area of the right pyramid VABCD is 408 cm2.

If the area of the square base is 144 cm2, find the height of the right

pyramid. (Correct your answer to 3 significant figures.)

A. 4.79 cm B. 9.22 cm C. 9.219 cm D. 12.5 cm

46. B

cm 12

cm 144

cm 144aera esaB

22

2

=

=

=

BC

BC

AD

CB

V

36 cm

30 cm

30 cm

P

Q

A

B

D

C

V


27

Let x cm be the height of ∆ABV.

Total surface area 2cm 408=

11

26424

408)122

1(4144

=

=

=×××+

x

x

x

∴ The height of ∆ABV is 11 cm.

Let h cm be the height of the pyramid.

22 )2

12(+h 2

11= (Pyth. theorem)

h 22.9= (corr. to 3 sig. fig.)

∴ The height of the pyramid is 9.22 cm.

47. The base radius of a right circular cone is 10 cm and its slant height is

11 cm. Find the volume of the right circular cone. (Correct your answer

to 1 decimal place.)

A. 230.4 cm3 B. 479.9 cm3 C. 691.2 cm3 D. 1 151.9 cm3

47. B

Let h cm be the height of the circular cone.

h2 = 112 − 102 (Pyth. theorem)

h = 21

∴ Volume of the circular cone 32 cm )21103

1( ××π×=

3cm 9.479= (corr. to 1 d.p.)

48. In the figure, the volumes of the cube and the right circular cone are

the same. Find the value of a . (Correct your answer to 1 decimal place.)

10 cm

11 cmh cm

a cm

a cm

9 cm


28

A. 26.7 B. 9.0 C. 8.9 D. 3.0

48.C

Volume of the cube = Volume of the right circular cone

93 = aa ××π×2

3

1

a = 8.9 (corr. to 1 d.p.)

49. In the figure, the top and the base of the right frustum are circles. Find

the volume of the frustum.

A. 2 500π cm3 B. 2 600π cm3 C. 2 700π cm3 D. π3

28162 cm3

49. B

Let MC = x cm, VN = y cm, ND = z cm.

13 cm

39 cm

V

A C

B N D

12 cm

M

12 cm

x cm

z cm

13 cm

39 cm

y cm

V

M C

DN


29

x2= 132 − 122 (Pyth. theorem)

x = 5

∵ ∆VMC ~ ∆VND

∴

36

39

1312

=

=

=

y

y

VD

VC

VN

VM

15

39

135

=

=

=

z

z

VD

VC

ND

MC

∴ Volume of the frustum

3

322

cm 600 2

cm )1253

13615

3

1(

π=

××π×−××π×=

50. In the figure, the inverted right circular cone is filled with water of

depth 12 cm. If 380 cm3 of water is added, find the rise in water level

in the cone. (Correct your answer to 1 decimal place.)

A. 0.6 cm B. 1.2 cm C. 2.7 cm D. 6.3 cm

50. C

Let MV = h cm and MD = r cm.

6 cm

12 cm

A B

C M D

E N F

V

6 cm

M D

N F

V

r cm

h cm

12 cm


30

∵ ∆VMD ~ ∆VNF

∴

2

126

hr

hr

NV

MV

NF

MD

=

=

=

3801263

1 2+××π× h

h××π×=

2)2

(3

1

380144 +π 3

12h

π=

h 70.14= (corr. to 2 d.p.)

∴ MV = 14.70 cm

∴ MN = (14.70 − 12) cm

= 2.7 cm (corr. to 1 d.p.)

VA VB − 9 15 −

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