Solid Mensuration of Pyramids

nd

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 angle of 30 .
V = (Abase) (h)

d1  2+ d2  2
2  2 

x = 9.01
S = ePr
S = 13 (9.01) (4)
2
S = 468.52cm

θ
Asector=360 π(R)2 θ Asector=360 (100)

V = (36 3 ) (6)
3
V = 374.12cm
A right section of an oblique prism is a rhombus whose diagonals are 10cm and the lateral edge of the prism is 13cm. Find the lateral area.
S = ePr

3.

VBody = 400π (x+20) - 400πx
VBody = 400πx + 8000π - 400πx
3
VBody = 8000π cm
The diameter of the base of a circular cylinder is 8cm and the elements are inclined to the base at an angle o of 30 . If an element is 7cm long, find the volume and the lateral area of the cylinder.
V =? , S =? , h =? h = 7sin30
= 3.5
2
V = πr h
2
=π (4 ) (3.5)
= 56π
3
= 175.93cm
S = 2πrh
= 2π(4)(3.5)
= 28π
2
= 87.96cm
Find the lateral area and the total area of a cylindrical cell whose height equals its diameter if it has to hold
50kg of water. r =? , S =? , AT =?
*1kg = 1L

V = 50kg = 50L = 50 000cm h = d = 2r
2
V = πr h
2
V = π (r )2r
3
50 000 = 2πr

3

4.

50 000
2π
3 r = 7957.75 r = 19.96cm
S = 2πrh
3

r=

S = 2π (19.96) ((2)(19.96))
2

S = 5 006.46cm
AT = 2B + S
2
= 2 (πr ) + 5 006.96
2
= 2 (π19.96 ) + 5 006.96
2
= 7 510.19cm
SPHERES
1. A sphere of radius 8cm rests in a circular hole of radius 3cm. How far below the plane of the hole does the sphere extend?

2.

h = 8 - 82-32 h = 8 – 7.42 h = 0.58cm
A sphere of radius 6cm rests on 3 horizontal wires forming a plane triangle whose sides are 5cm, 12cm, and 13cm. Find the height of the top of the sphere above the plane of the wires.
2

2

h = 6 + 6 -r
1
2 rP = s(s-13)(s-5)(s-12)
P = 13+5+12 = 30
P
s = 2 = 15
1
2 rP = 15(15-13)(15-5)(15-12)
15r = 900
15r = 30 r=2 2

h = 6 + 6 -2

3.

2

h = 6 + 32 h = 11.66cm
A square is inscribed in a small circle of a sphere whose great circle has a radius of 9cm. If the plane of the square is 6cmfrom the center of the sphere, find the perimeter of the square.
Psquare = 4e
2
d
2
2
9 =6 + 2 

d = 13.42 d=e 2
13.42 = e 2 e = 9.49
Psquare = 4(9.49) = 37.96cm

An ornamental spherical ball made of bronze has an outer diameter of 1.2m and a uniform thickness of
3
13cm. If bronze weighs 8.5 mg/cm , find the weight of the ball.
V = V1 – V2
4 34 3
V = 3 π r1 - 3 π r2
4
34
3
V = 3 π60 - 3 π47
3

3

V = 469 885.91cm x 8.5 mg/cm
V = 3 994 030.235 mg
5.

6.

A hollow metal sphere 35cm in diameter is used as a float. If the metal sphere sinks to the depth of 12cm, what is the area of the wetted surface?
Z = 2πrh
Z = 2π (17.5) (12)
2
Z = 1 319.47cm
On a sphere of a diameter 125cm, two circles of the spheres whose planes are parallel have radii 20cm and 50cm, respectively. Find the area of the zone included between these cirlces.
Z = 2πrh h = h1 + h 2
2

2

2

2

h1 = 62.5 -20 h1 = 59.21

h2 = 62.5 -50 h2 = 37.5 h = 59.21+37.5 h = 96.71
Z = 2π (62.5) (96.71)
Z = 37 977.93