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Volume of solids Questions
1. Metal cube of side 4.4cm was melted and the molten material used to make a sphere. Find to 3 significant figures the radius of the sphere (3 mks)
2. Two metal spheres of diameter 2.3cm and 3.86cm are melted. The molten material is used to cast equal cylindrical slabs of radius 8mm and length 70mm.
If 1/20 of the metal is lost during casting. Calculate the number of complete slabs casted. (4mks)
3. The volume of a rectangular tank is 256cm3. The dimensions are as in the figure.
¼ x
x-8
16cm
Find the value of x (3 marks)
4.
22.5cm
The diagram represent a solid frustum with base radius 21cm and top radius 14cm. The frustum is 22.5cm high and is made of a metal whose density is 3g/cm3 π = 22/7.
- Calculate
- the volume of the metal in the frustrum. (5 marks)
- the mass of the frustrum in kg. (2 marks)
- The frustrum is melted down and recast into a solid cube. In the process 20% of the metal is lost. Calculate to 2 decimal places the length of each side of the cube. (3 marks)
5. The figure below shows a frustrum
Find the volume of the frustrum (4 mks)
6. The formula for finding the volume of a sphere is given by. Given that V = 311 and
=3.142, find r. `(3 mks)
7. A right conical frustrum of base radius 7cm and top radius 3.5cm, and height of 6cm is stuck onto a cylinder of base radius 7cm and height 5cm which is further attached to a hemisphere to form a closed solid as shown below
Find:
(a) The volume of the solid (5mks)
(b) The surface area of the solid (5mks)
8. A lampshade is made by cutting off the top part of a square-based pyramid VABCD as shown in the figure below. The base and the top of the lampshade have sides of length 1.8m and 1.2m respectively. The height of the lampshade is 2m
Calculate
- The volume of the lampshade (4mks)
- The total surface area of the slant surfaces (4mks)
- The angle at which the face BCGF makes with the base ABCD. (2mks)
9. A solid right pyramid has a rectangular base 10cm by 8cm and slanting edge 16cm.
calculate:
(a) The vertical height
(b) The total surface area
(c) The volume of the pyramid
10. A solid cylinder of radius 6cm and height 12cm is melted and cast into spherical balls
of radius 3cm. Find the number of balls made
11. The sides of a rectangular water tank are in the ratio 1: 2:3. If the volume of the tank is
1024cm3. Find the dimensions of the tank. (4s.f)
12. The figure below represents sector OAC and OBD with radius OA and OB respectively.
Given that OB is twice OA and angle AOC = 60o. Calculate the area of the shaded region
in m2, given that OA = 12cm
13. The figure below shows a closed water tank comprising of a hemispherical part surmounted
on top of a cylindrical part. The two parts have the same diameter of 2.8cm and the cylindrical
part is 1.4m high as shown:-
- Taking = 22, calculate:
7
(i) The total surface area of the tank
(ii) the cost of painting the tank at shs.75 per square metre
(iii) The capacity of the tank in litres
(b) Starting with the full tank, a family uses water from this tank at the rate of 185litres/day
for the first 2days. After that the family uses water at the rate of 200 liters per day. Assuming
that no more water is added, determine how many days it takes the family to use all the water
from the tank since the first day
14. The figure below represents a frustrum of a right pyramid on a square base. The vertical height
of the frustrum is 3 cm. Given that EF = FG = 6 cm and that AB = BC = 9 cm
Calculate;
a) The vertical height of the pyramid.
b) The surface area of the frustrum.
c) Volume of the frustrum.
d) The angle which line AE makes with the base ABCD.
15. A metal hemisphere of radius 12cm is melted done and recast into the shape of a cone
of base radius 6cm. Find the perpendicular height of the cone
16. A solid consists of three discs each of 1½ cm thick with diameter of 4 cm, 6 cm and 8 cm
respectively. A central hole 2 cm in diameter is drilled out as shown below. If the density of
material used is 2.8 g/cm3, calculate its mass to 1 decimal place
17. A right conical frustrum of base radius 7 cm and top radius 3.5 cm and height 6 cm is stuck onto a
cylinder of base radius 7 cm and height 5 cm which is further attached to form a closed solid as
shown below.
Find;
a) The volume of the solid.
b) The surface area of the solid.
18. The figure below shows a frustrum
Find the volume of the frustrum
19. The diagram below shows a metal solid consisting of a cone mounted on hemisphere.
The height of the cone is 1½ times its radius;
Given that the volume of the solid is 31.5π cm3, find:
(a) The radius of the cone
(b) The surface area of the solid
(c) How much water will rise if the solid is immersed totally in a cylindrical container which
contains some water, given the radius of the cylinder is 4cm
(d) The density, in kg/m3 of the solid given that the mass of the solid is 144gm
20. A solid metal sphere of volume 1280 cm3 is melted down and recast into 20 equal solid cubes.
Find the length of the side of each cube.
21. The figure below shows a frustrum cut from a cone
Calculate the volume of the frustrum
Volume of solids Answers
1 | Volume of cube = 4.4×4.4×4.4 Volume of sphere | M1
M1
A1 | |
2. | Vol. of sphere = Remaining material No of slabs = |
M1
M1
M1
A1 |
Follow through of |
04 |
1. a) Length of diagonal = √ 102 + 82
= √164
Vertical height = √162 – (√164)2
2
= 14.66cm
b) Height of the slant surfaces
√162 – 42 = √240
√162 – 52 = √231
Area of slant surfaces
( ½ x 8 x √240 x 2) = 124.0 cm2
(½ x 10x √231 x 2) = 152.0cm2
Area of the rectangular base= 8 x 10 = 80cm2
Total surface area = 356cm2
c) Volume
= ( 1/3 x 80 x 14.66) = 391.0cm3
2. Volume of the cylinder
= (22/7 x 6 x 6 x 12)cm3 = 1357.71cm3
Volume of a sphere
= (4/3 x 22/7 x 3 x 3 x 3)cm3 = 113.14cm3
No. of spheres formed
= 1357.71
113.14cm3
= 12 spheres
3. Let the smaller length be x cm
Dimensions are x, 2x, 3x
x . 2x . 3x = 1024
6x3 = 1024
x3 = 1024
6
x= 3√1024
6
Dimensions are 5.547, 11.09, 16.64
4. (60/360 x 22/7 x 24 x 24) – (60/360 x 22/7 x12 x 12)
301.71 – 75.43 = 226.26
5. (a)(i) 2rh + 2r2 + r2
= 2 x 22/7 x 1.4 x 1.4) + 2 x 22/7 x 1.42) + ( 22/7 x 1.42)m2
= (12.32+ 12.32 + 6.16)m2= 30.8m2
OR r(2h + 2r + r)
= 22 x 1.4 (2x 1.4 + 3(1.4)= 30.8m2
(ii) shs. (75 x 30.8)= Shs.2,310
(iii) Total vol.
= 22/7 x 1.42 x 1.4) + ( ½ x 4/3 x 22/7 x 1.42)m3
= 8.624
4.106 = 12.7306m3
capacity = (12.7306 x 1000)liters= 12730.6litres
(b) First 2days = 185 x 2 = 370litres
Remaining amount = (12730.6 – 370)liters
= 12360.6litres
Days to use = 12,360.6
200
= 61.803days
In all it takes = (61.803 + 2)days = 63.803days
6. a) h + 3 = 9 √
h 6
6h + 18 = 9h
h = 6 cm√
height = 6 + 3 = 9 cm
- Base = 9 x 9 = 81 cm2
Top = 6 x 6 = 36 cm2
Sides = 3.67 x 15 x ½ x 4
= 110.15 cm2
Total = 227.15 cm2
- Vol. of bigger = 1/3 x 81 x 9
= 243
Vol of smaller = 1/3 x 36 x 6
= 72
Vol. of frustrum = 171 cm2
d) sin = 9
11.02
= 54.8o
7. Volume of a hemisphere
2πr3 = 2 x 22 x 12 x 12 x 12
3 3 7
= 176 x 144
7
= 3620.571429 = 3620.57
Volume of a cone
2/3πr2h
1 x 22 x 6 x 6 x h = 36.20.57
3 7
6 x 44h = 3620.57
7
264h = 3620.57 x 7
h =3620.57 x 7
264
= 95.9981 = 95.998
8. V = 22 x 2 x 2 1.5 + 22 x 3 x 3 x 1.5 + 22 x 4.4. x 1.5
7 7 7
= 132 + 297 + 528
7 7 7
V of hole = 22 x 1 x 1 4.5
7
= 99
7
V = 957 – 99 = 858
7 7 7 = 122.57 cm3
Mass = 2.8 x 122.57
= 343.196g
≃ 343.2g
9. Volume of hemisphere = ½ x 4 x 22 x 7 x 7 x 7
3 7
= 718.67 cm3
Vol. of cylinder = r2h = 22 x 7 x 7 x 5 = 770 cm3
7
Vol of frustrum = 1/3 x 22 x 7 x 7 x h1 –
7
1/3 x 22 x 3.5 x 3.5 x h2
7
Height of cone ⇒h1 = 7 but h1 = h2 + 6
h2 3.5
h2 + 6 = 7 ⇒ 7h2 = 3.5h2 + 21
h2 3.5
3.5 h2 = 21
h2 = 6 cm
h1 = 12 cm
∴ Vol. of frustrum = 1/3 x 22 x 7 x 7 x 12 –
7
1/3 x 22 x 3.5 x 3.5 x 6
7
= 616 – 77 = 539 cm3
Total volume = 718.67 cm3 + 770cm3 + 539 cm3
= 2027.67 cm3
- S.A of top = r2
22 x 3.5 x 3.5 = 38.5 cm2
7
S.A of curved part of frustrum = 22 x 7 x 13.89 –
7
22 x 3.5 x 6.945
7
305.580
– 76.395
229.185 cm2
S.A of curved part of cylinder = 2r x h = 2 x 22 x 7 x 5
7
= 2220 cm2
S.A of hemisphere = ½ x 4 r2 = 22 x 7 x 7 = 308 cm2
7
Total S.A = 795.685 cm2
10. L/S.F = 2.2/3.3 = 2/3
4.8/4.8 + h = 2/3
h= 24
volume of smaller cone
1/3 x 22/7 x 2.2 x 2.4
= 12.169
Volume of large cone
1/3 x 22/7 x 3.3 x 3.3 (4.8 + 2.2)
V of frustum
82.14 – 12.17 = 69.97 cm3
11. (a) Volume = 2
r3 + 1 r2 x 3 r = 31.5
3 3 2
4r3 + 3r3 = 31.5 x 6
r = 31.5 x 6
7
= 3cm
(b) slant height of con = 4.52 + 32
= 5.408cm
Surface are = 2 x 32 + x 3 x 5.408 = 107.5cm2
(c) Height = 31.5
42
= 1.969cm
(d) Density = 144
231.5
= 1.46g/cm3
12. Volume of cube side x cm = (xcm)3
∴ x3cm3 = 1280 cm3
20
x =3 1280
20
= 3 64
= 4 cm
13.
9/3 = 14 + h/h
9h = 42 + 3h
6h = 42
h = 7
volume of the frustrum = (1/3 x 22/7 x 9 x 9 x 21)cm3
= (1/3 x 22/7 x3 x 3x 7)cm3
= 1782 – 66 = 1716cm3