Solid Geometry

DAY 8 Submitted by:

April Grace L. Cabulong Submitted to:

Engr. Rex Jason H. Agustin



A circular piece of cardboard with a diameter of 1m will be made into conical hat 40 cm high by cutting a sector off and joining the edges to form a cone. Determine the angle subtended by the sector removed.

A.  B.  C.  D. 

144 148 152 154



= circumference of the circle



= circumference of the base of cone

 

C=



C = 2πr - 2πx



C = 2 π (50) - 2 π(30)



C = 40 π



C = rθ



40 π = (50)θ



θ=



θ = 144˚



What is the area in sq. m of the zone of a spherical segment having a volume of 1470.265 cu. m if the diameter of the sphere is 30 m?

A.  B.  C.  D. 

465.5 m 565.5 m 665.5 m 656.5 m

V=  1470.265 =  1404 = 45  h = 6m  A = 2πrh  A = 2π(15)(6)  A = 565.49 



A sphere having a diameter of 30 cm is cut into 2 segments. The altitude of the first segment is 6 cm. What is the ratio of the area of the second segment to that of the first?

A.  B.  C.  D. 

4:1 3:1 2:1 3:2



If the edge of a cube is increased by 30 %, by how much is the surface area increased?

A.  B.  C.  D. 

30 % 33 % 60 % 69 %



Each side of a cube is increased by 1 %. By what percent is the volume of the cube increased?

A.  B.  C.  D. 

1.21 % 2.8 % 3.03 % 3.5 %



Given a sphere of diameter, d. What is the percentage increase in its diameter when the surface area increases by 21 %?

A.  B.  C.  D. 

5% 10 % 21 % 33 %



Given a sphere of diameter, d. What is the percentage increase in its volume when the surface area increases by 21 %?

A.  B.  C.  D. 

5% 10 % 21 % 33 %



How many times do the volume of a sphere increases if the radius is doubled?

A.  B.  C.  D. 

4 times 2 times 6 times 8 times



A circular cone having an altitude of 9 m is divided into 2 segments having the same vertex. If the smaller altitude is 6 m, find the ratio of the volume of the small cone to the big cone.

A.  B.  C.  D. 

0.186 0.296 0.386 0.486



Find the volume of a cone to be constructed from a sector having a diameter of 72 cm and a central angle of 210°.

A.  B.  C.  D. 

12367.2 cm³ 13232.6 cm³ 13503.4 cm³ 14682.5 cm³



Let:



C=

 

 

C= 36 X = 21cm



h=



h=



h = 29.24 cm



v=



V = 13,503.44

= 2π

- 2π



Find the volume of a cone to be constructed from a sector having a diameter of 72 cm and a central angle of 150°.

A.  B.  C.  D. 

5533.32 cm³ 6622.44 cm³ 7710.82 cm³ 8866.44 cm³



Let:





C = length of arc of the sector



C=



rθ = 2πr - 2x

 

X = 15cm



h=



h=



h = 32.726 cm



v=



v = 7710.88



A conical vessel has a height of 24 cm and a base diameter of 12 cm. It holds water to a depth of 18 cm above its vertex. Find the volume (in cm³) of its content.

A.  B.  C.  D. 

188.40 298.40 381.70 412.60



By ratio and proportion:



r = 4.5



v=



v = 381.7

=



What is the height of a right circular cone having a slant height of sqrt.10x and a base diameter of 2x?

A.  B.  C.  D. 

2x 3x 3.317x 3.162x

=





10

= 9





h = 3x



The ratio of the volume to the lateral area of a right circular cone is 2:1. If the altitude is 15 cm, what is the ratio of the slant height to the radius?

A.  B.  C.  D. 

5:6 5:4 5:3 5:2

= πrL





V=

 

= 2 -> as given



V=2



rh = 6L



A regular triangular pyramid has an altitude of 9m and a volume of 187.06 cu.m. What is the base edge in meters?

A.  B.  C.  D. 

12 13 14 15

Note: θ = 60˚, since equilateral triangle. V= 



V=



187.06 =



X = 12m

h sin60˚



The volume of the frustum of a regular triangular pyramid is 135 m³. The lower base is an equilateral triangle with an edge of 9m. The upper base is 8m above the lower base. What is the upper base edge in meters?

A.  B.  C.  D. 

2 3 4 5



Note: θ = 60˚, since equilateral triangle.



Let: = area of the lower base



= area of the upper base



=



= 0.433

 

V=



Substitute:



135 =



50.625 = 35.074 + 0.433



0=



0=



X = 3 cm

+ 3.897x



What is the volume of a frustum of a cone whose upper base is 15 cm in diameter and lower base 10 cm in diameter with an altitude of 25 cm?

A.  B.  C.  D. 

3018.87 cm³ 3180.87 cm³ 3108.87 cm³ 3081.87 cm³

    



Substitute:



V = 3108.87



In a portion of an electrical railway cutting, the areas of cross section taken every 50 m are 2556, 2619, 2700, 2610 and 2484 m² Find its volume.

A.  B.  C.  D. 

522,600 m³ 520,500 m³ 540,600 m³ 534,200 m³





Note: Since the areas being cut is at the same distance, then the given solid is a prismatoid. And since there are five different areas being cut then, this solid is equivalent to two prismatoids.

Where:



= area of the first base (base 1)



= area of the second base (base 2)



= area of the middle section



h = distance between base 1 and base 2



Let:

 

= total volume of the two prismatoid



Determine the volume of a right truncated triangular prism with the following definitions: let the corners of the triangular base be defined by A, B and C. The length of AB = 10 ft, BC = 9 ft. And CA = 12 ft. The sides A, B and C are perpendicular to the triangular base and have the height of 8.6 ft., 7.1 ft. And 5.5 ft. Respectively.

A.  B.  C.  D. 

413 ft³ 311 ft³ 313 ft³ 391 ft³

 

S = 15.5

A = 44.039

V = 311



A circular cylinder with a volume of 6.54 m³ is circumscribed about a right prism whose base is an equilateral triangle of side 1.25 m. What is the altitude of the cylinder in meters?

A.  B.  C.  D. 

3.50 3.75 4.00 4.25



 



By cosine law:

    

1.5625 = 3 r = 0.72m v=π 6.54 = π h = 4m



A circular cylinder is circumscribed about a right prism having a square base one meter on an edge. The volume of the cylinder is 6.283 m³ Find its altitude in meters.

A.  B.  C.  D. 

4.00 3.75 3.50 3.25



d = 1.4142



h=4



The bases of a right prism are hexagons with one of each side equal to 6 cm. The base are 12 cm apart. What is the volume of the right prism?

A.  B.  C.  D. 

1211.6 cm³ 2211.7 cm³ 1212.5 cm³ 1122.4 cm³

Let:  A = area of one base  x = length of each side of the base 



V = 1122.4



Two vertical conical tanks are joined at the vertices by a pipe. Initially the bigger tank is full of water. The pipe valve is open to allow the water to flow to the smaller tank until it is full. At this moment, how deep is the water in the bigger tank? The bigger tank has a diameter of 6 ft and a height of 10 ft, the smaller tank has a diameter of 6 ft and a height of 8ft. Neglect the volume of water in the pipeline.

A.  B.  C.  D. 



Let:



= total volume of the bigger tank



= total volume of the smaller tank



V = volume left in the bigger tank V=

 

V = 18.849



By similar solids: = 94.247

h=



The central angle of a spherical wedge is 1 radian. Find its volume if its radius is 1 unit.

A.  B.  C.  D. 

2/3 1/2 3/4 2/5



A regular octahedron has an edge 2m. Find its volume (in m³).

A.  B.  C.  D. 

3.77 1.88 3.22 2.44



V = 3.77



A maximum compound of equal parts of two liquids, one white and the other black, was placed in a hemispherical bowl. The total depth of the two liquids is 6 inches. After standing for short time, the mixture separated, the white liquid settling below the black. If the thickness of the segment of the black liquid is 2 inches, find the radius of the bowl in inches.

A.  B.  C.  D. 

7.33 7.53 7.73 7.93



Let:



= volume of the black mixture



= volume of the white mixture



Substitute:



36(3r – 6) = 32(3r-4)



108r – 216 = 96r – 128



12r = 88



r = 7.33 in



The volume of water in a spherical tank having a diameter of 4m is 5.236 m³. Determine the depth of the water in the tank.

A.  B.  C.  D. 

1.0 1.2 1.4 1.8



An ice cream cone is filled with ice cream and a surmounted ice cream in the form of a hemisphere on top of the cone. If the hemispherical surface is equal to the lateral area of the cone, find the total volume (in cubic inches) of ice cream if the radius of the hemisphere is 1 inch and assuming the diameter of hemisphere is equal to the diameter of the cone.

A.  B.  C.  D. 

3.45 3.91 4.12 4.25



Let:



A cubical container that measure 2 inches on a side is tightly packed with 8 marbles and is filled with water. All 8 marbles are in contact with the walls of the container and the adjacent marbles. All of the marbles are of the same size. What is the volume of water in the container?

A.  B.  C.  D. 

0.38 in³ 2.5 in³ 3.8 in³ 4.2 in³



Let:



X= 4r



2 = 4r



The corners of a cubical block touched the closed spherical shell that encloses it. The volume of the box is 2744 cubic cm. What volume is cubic cm inside the shell is not occupied by the block?

A.  B.  C.  D. 

2714.56 3714.65 4713.56 4613.74



Let:



V = volume inside the sphere but outside the box



 



2744 =



X = 14 cm



d=



d=



d = 24.24 cm



r = 12.12 cm

V=

END.