SOLID MENSURATION REVIEWER (2 Semester 2011-2012)

PYRAMIDS

1. A vessel is in the form of an inverted regular square

pyramid of base edge 13 cm and the altitude is 25 cm,

how many liters are in it when the depth of the water is

15 cm?

A(small Square)

h(Water)

A(Big Square) = h(container)

2

A(small Square) 15

2

= 252

13

2

2

(15 )(13 )

2

A(small Square) =

25

2

A(small Square) = 60.84cm

1

V = 3 Bh

1

V= 3 (60.84)(15)

3

V=304.2 cm 0.304.2 L

2. The lateral faces of a regular square pyramid are

o

isosceles triangles whose base angles are 55 each. If the

lateral edge is 200 cm, find the lateral area and volume of

the pyramid.

2

2

2

l – slant height

h =(163.83) – (114.72)

l

o

sin55 =200

h=116.96 cm

o

l=200sin55

l=163.83 cm

e

o

cos55 =200

o

200cos55 =e

e=114.72

2e=229.4 cm

(163.83)(229.4)

2

2

S = 75,175.2 cm

S=

2

(229.4 )(116.96)

3

3

V= 2,051,181.2cm

V=

4. Find the volume and total surface area of a frustum of a

regular hexagonal pyramid if the base edges are 8 cm and

6 cm, respectively and the altitude is 20 cm.

hf

V= 3 [B+B1+sqrt.(BB1)]

hf 3 3

33

2

2

V= 3 2 (6 ) + 2 (8 ) +

()()

33

2

33

2

2

(6 ) +

V = 2563.47 cm

2

(8 )

3

3

r1 = 2 (6) = 5.2 cm

3

r2 = 2 (8) = 6.93 cm

r2-r1 = x

x=6.93-5.2

x=1.73

2

2

l =20 + 1.73

l= 20.07cm

*x- yung base ng right triangle

2

20.07[(6)(6)+(6)(8)]

2

2

S=842.94 cm

S=

CONES

2

1. The total area of right circular cone is 48π cm . If the altitude of the cone is equal to the diameter of the base,

find the slant height and R.

h=d

h=2R

2

2

l= h +R

2

2

l = 4R +R

l= 5 R

2

Atotal= πRl+πR

2

48π=πR( 5 R)+πR

2

2

48π= πR 5 +πR

2

48π=R (π 5 +π)

2

R =14.83 cm

R= 3.85 cm

2. A tool is made up of a cone on top of a cylinder. The

3. A pyramid having a altitude of 450 cm and a base with

2

an area of 930 cm is cut 120 cm from the vertex by a

plane that is parallel to the plane of the base. Find the

area of the section.

2

A(small)

h (small)

= h2

A(big)

(original height)

2

A(small) 120

=4502

930

2

(120 )(930)

2

A(small)= 4502

= 66.13 cm

cylinder has a height h of 15 cm and a radius of 5 cm. The

2

volume of the cone is 100 Π cm . O is the vertex of the

cone, AB is the diameter of the base of the cone and C its

Center. Points O, A, B and C are in the same plane. Find the lateral surface area.

12

V= 3 πr h

12

100= 3 (5 )πh

h= 12cm

2

2

S= πr h +r +2πrh

2

2

S= 5π 12 +5 +2π(5)(12)

2

S= 675.4 cm

4.

3.A right circular cone of slant height 5cm has a radius of 15 cm. Find the angleof sector of a circle of radius 10cm, if the area of the sector is equal to the lateral area of the cone. S=π(15)(5)

2

S=235.6 cm

S=Asector

x=

2

(235.6 cm )(360)=100πθ

o

θ=269.9

PRISMS

1. The lateral edge of a parallelepiped is 8cm and a

plane is passed cutting this edge at a right angle to

form a right section which is a square. If the lateral

2

area is 320cm , find the edge of the right section.

S = ePr

320 = (8)Pr

40 = Pr

40 = 4e

e = 10cm

2. A plane is passed through two opposite edges of a

cube forming a diagonal section bounded by the two

opposite edges and the diagonals of two opposite

CYLINDERS

1. When a body is immersed in water in a right circular

cylinder 40cm in diameter, the level of the water rises

20 cm. What is the volume of the body?

2

V1 = (πr ) x

2

V2 = (πr ) (x+20)

VBody = V2 – V1

2

2.

2

49 2 = e (e 2 )

2

3.

32

Abase = 4 e

3

2

Abase = 4 (12)

Abase = 36 3

h

Finding h: sin 30 = 12

h=6

2

VBody = ((πr )(x+20))- ((πr )x)

faces. If the area of the section is 49 2 cm . Find the

edge of the cube.

49 2 = e 2

2

e = 49

e=7

Find the volume of an oblique triangular prism

whose base in an equilateral triangle the side of

which is 12cm. The lateral edge of the prism is equal

to the side of the base and inclined to the base plane

o

at...