# Volume of Solids Questions and Answers - Form 2 Topical Mathematics

## Questions

1. A solid right pyramid has a rectangular base 10cm by 8cm and slanting edge 16cm.
calculate:
1. The vertical height
2. The total surface area
3. The volume of the pyramid
2. 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.
3. 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)
4. 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

5. 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:-

1. Taking π= 22/7, calculate:
1. The total surface area of the tank
2. the cost of painting the tank at shs.75 per square metre
3. The capacity of the tank in litres
2. 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 litres 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
6. 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:
1. The vertical height of the pyramid.
2. The surface area of the frustrum.
3. Volume of the frustrum.
4. The angle which line AE makes with the base ABCD.
7. 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
8. 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
9. 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;
1. The volume of the solid.
2. The surface area of the solid.
10. The figure below shows a frustrum

Find the volume of the frustrum
11. 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:
1. The radius of the cone
2. The surface area of the solid
3. 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
4. The density, in kg/m³ of the solid given that the mass of the solid is 144gm
12. 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.
13. The figure below shows a frustrum cut from a cone

Calculate the volume of the frustrum
14. 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)

1. Length of diagonal = √10² + 8²
= √164
Vertical height = √16²(√164)²/2
= 14.66cm
2. 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
3. Volume
= ( 1/3 x 80 x 14.66) = 391.0cm³
1. 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
2. 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
3. (60/360 x 22/7 x 24 x 24) – (60/360 x 22/7 x12 x 12)
301.71 – 75.43 = 226.26
1. 2πrh + 2rπ² +πr²
= 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
2. Shs. (75 x 30.8)= Shs.2,310
3. 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
1. 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
4.

1. (h + 3)/h = 9/6
6h + 18 = 9h
h = 6 cm√
height = 6 + 3 = 9 cm
2. 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
3. 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
4. sin θ = 9/11.02
θ = 54.8º
5. Volume of a hemisphere
2/3πr³ = 2/3 x 22/7 x 12 x 12 x 12
= 176/7 x 144
= 3620.571429 = 3620.57
Volume of a cone
2/3πr²h
1/3 x 22/7 x 6 x 6 x h = 36.20.57
(6 x 44h)/7 = 3620.57
264h = 3620.57 x 7
h =(3620.57 x 7)/264
= 95.9981 = 95.998
6. V = (22/7 x 2 x 2 1.5) + (22/7 x 3 x 3 x 1.5) + (22/7 x 4.4. x 1.5)
= 132/7 + 297/7 + 528/7
V of hole = 22/7 x 1 x 1 4.5
= 99/7
V = 957/799/7 = 858/7
= 122.57 cm³
Mass = 2.8 x 122.57
= 343.196g
≃ 343.2g
7.
1. Volume of hemisphere = ½ x 4/3 x 22/7 x 7 x 7 x 7
= 718.67 cm³
Vol. of cylinder = πr²h = 22/7 x 7 x 7 x 5 = 770 cm³
Vol of frustrum = 1/3 x 22/7 x 7 x 7 x h¹ − 1/3 x 22/7 x 3.5 x 3.5 x h²
Height of cone ⇒ h¹/h² = 7/3.5 but h¹ = h² + 6
h²/h² + 6 = 7/3.5 ⇒ 7h² = 3.5h² + 21
3.5h² = 21
h² = 6 cm
h¹ = 12 cm
∴ Vol. of frustrum = (1/3 x 22/7 x 7 x 7 x 12) − (1/3 x 22/7 x 3.5 x 3.5 x 6)
= 616 – 77 = 539 cm³
Total volume = 718.67 cm³ + 770cm³ + 539 cm³
= 2027.67 cm³
2. S.A of top = πr² 22/7 x 3.5 x 3.5 = 38.5 cm²
S.A of curved part of frustrum =(22/7 x 7 x 13.89) - (22/7 x 3.5 x 6.945)
305.580 - 76.395 = 229.185 cm²
S.A of curved part of cylinder = 2πr x h = 2 x 22/7 x 7 x 5
= 2220 cm²
S.A of hemisphere = ½ x 4 πr² = 22/7 x 7 x 7 = 308 cm²
Total S.A = 795.685 cm²
8. 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
1. Volume = 2/3 πr³ + 1/3πr² x 3/2 r = 31.5π
4r³ + 3r³ = 31.5 x 6
r = √(31.5 x 6/7 )
= 3cm
2. slant height of con = √4.5² + 3²
= 5.408cm
Surface area = 2π x 3² + π x 3 x 5.408 = 107.5cm²
3. Height = 31.5/4²π
= 1.969cm
4. Density = 144/231.5π
= 1.46g/cm³
9. Volume of cube side x cm = (xcm)³
∴ x³cm³ = 1280/20 cm³
x =3√(1280/20)
= 3√64
= 4 cm
10. 9/3 = 14 + h/

9h = 42 + 3h
6h = 42
h = 7
volume of the frustrum = (1/3 x 22/7 x 9 x 9 x 21)cm³

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Read 7566 times Last modified on Wednesday, 10 February 2021 06:55

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