Questions
 A solid right pyramid has a rectangular base 10cm by 8cm and slanting edge 16cm.
calculate: The vertical height
 The total surface area
 The volume of the pyramid
 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.
 The sides of a rectangular water tank are in the ratio 1: 2:3. If the volume of the tank is 1024cm^{³}. Find the dimensions of the tank. (4s.f)
 The figure below represents sector OAC and OBD with radius OA and OB respectively. Given that OB is twice OA and angle AOC = 60^{º}. Calculate the area of the shaded region in m^{²}, given that OA = 12cm
 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 π= ^{22}/_{7}, calculate:
 The total surface area of the tank
 the cost of painting the tank at shs.75 per square metre
 The capacity of the tank in litres
 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
 Taking π= ^{22}/_{7}, calculate:
 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: The vertical height of the pyramid.
 The surface area of the frustrum.
 Volume of the frustrum.
 The angle which line AE makes with the base ABCD.
 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
 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/cm^{³}, calculate its mass to 1 decimal place
 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; The volume of the solid.
 The surface area of the solid.
 The figure below shows a frustrum
Find the volume of the frustrum  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 cm^{³}, find: The radius of the cone
 The surface area of the solid
 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
 The density, in kg/m^{³} of the solid given that the mass of the solid is 144gm
 A solid metal sphere of volume 1280 cm^{³} is melted down and recast into 20 equal solid cubes. Find the length of the side of each cube.
 The figure below shows a frustrum cut from a cone
Calculate the volume of the frustrum  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)
Answers
 Length of diagonal = √10^{²} + 8^{²}
= √164
Vertical height = √16^{²} – ^{(√164)²}/_{2}
= 14.66cm  Height of the slant surfaces
√16^{²} – 4^{²} = √240
√16^{²} – 5^{²} = √231
Area of slant surfaces
( ½ x 8 x √240 x 2) = 124.0 cm^{²}
(½ x 10x √231 x 2) = 152.0cm^{²}
Area of the rectangular base= 8 x 10 = 80cm^{²}
Total surface area = 356cm^{²}  Volume
= ( ^{1}/_{3} x 80 x 14.66) = 391.0cm^{³}
 Length of diagonal = √10^{²} + 8^{²}
 Volume of the cylinder
= (^{22}/_{7} x 6 x 6 x 12)cm^{³} = 1357.71cm³
Volume of a sphere
= (^{4}/_{3} x ^{22}/_{7} x 3 x 3 x 3)cm^{³} = 113.14cm^{³}
∴ No. of spheres formed
= ^{1357.71}/_{113.14}cm^{³}
= 12 spheres  Let the smaller length be x cm
∴ Dimensions are x, 2x, 3x
x . 2x . 3x = 1024
6x^{³} = 1024
x^{³} = ^{1024}/_{6}
x= 3√(^{1024}/_{6})
Dimensions are 5.547, 11.09, 16.64  (^{60}/_{360} x ^{22}/_{7} x 24 x 24) – (^{60}/_{360} x ^{22}/_{7} x12 x 12)
301.71 – 75.43 = 226.26  2πrh + 2rπ^{²} +πr^{²}
= 2 x ^{22}/_{7} x 1.4 x 1.4) + 2 x ^{22}/_{7} x 1.42) + ( ^{22}/_{7} x 1.42)m^{²}
= (12.32+ 12.32 + 6.16)m^{²}= 30.8m^{²}
OR r(2h + 2r + r)
= 22 x 1.4 (2x 1.4 + 3(1.4)= 30.8m^{²}  Shs. (75 x 30.8)= Shs.2,310
 Total vol.
= ^{22}/_{7} x 1.42 x 1.4) + ( ½ x ^{4}/_{3} x ^{22}/_{7} x 1.42)m^{³}
= 8.624 4.106 = 12.7306m^{³}
capacity = (12.7306 x 1000)liters= 12730.6litres
 2πrh + 2rπ^{²} +πr^{²}
 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

 ^{(h + 3)}/_{h} = ^{9}/_{6}√
6h + 18 = 9h
h = 6 cm√
height = 6 + 3 = 9 cm  Base = 9 x 9 = 81 cm^{²}
Top = 6 x 6 = 36 cm^{²}Sides = 3.67 x 15 x ½ x 4
= 110.15 cm^{²}Total = 227.15 cm^{²}  Vol. of bigger = ^{1}/_{3} x 81 x 9
= 243
Vol of smaller = ^{1}/_{3} x 36 x 6
= 72
Vol. of frustrum = 171 cm^{²}  sin θ = ^{9}/_{11.02}θ = 54.8^{º}
 ^{(h + 3)}/_{h} = ^{9}/_{6}√
 Volume of a hemisphere
^{2}/_{3}πr^{³} = ^{2}/_{3} x ^{22}/_{7} x 12 x 12 x 12
= ^{176}/_{7} x 144
= 3620.571429 = 3620.57
Volume of a cone
^{2}/_{3}πr^{²}h
^{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  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}/_{7} – ^{99}/_{7} = ^{858}/_{7}= 122.57 cm^{³}Mass = 2.8 x 122.57
= 343.196g
≃ 343.2g 
 Volume of hemisphere = ½ x ^{4}/_{3} x ^{22}/_{7} x 7 x 7 x 7
= 718.67 cm^{³}Vol. of cylinder = πr^{²}h = ^{22}/_{7} x 7 x 7 x 5 = 770 cm^{³}Vol of frustrum = ^{1}/_{3} x ^{22}/_{7} x 7 x 7 x h_{¹} − ^{1}/_{3} x ^{22}/_{7} x 3.5 x 3.5 x h_{²}Height of cone ⇒ h_{¹/h²} = ^{7}/_{3.5} but h_{¹} = h_{²} + 6
h_{²/h²} + 6 = ^{7}/_{3.5} ⇒ 7h_{²} = 3.5h_{²} + 21
3.5h_{²} = 21
h_{²} = 6 cm
h_{¹} = 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 cm^{³}Total volume = 718.67 cm^{³} + 770cm^{³} + 539 cm^{³}= 2027.67 cm^{³}  S.A of top = πr^{²} ^{22}/_{7} x 3.5 x 3.5 = 38.5 cm^{²}
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 cm^{²}
S.A of curved part of cylinder = 2πr x h = 2 x ^{22}/_{7} x 7 x 5
= 2220 cm^{²}
S.A of hemisphere = ½ x 4 πr^{²} = ^{22}/_{7} x 7 x 7 = 308 cm^{²}
Total S.A = 795.685 cm^{²}
 Volume of hemisphere = ½ x ^{4}/_{3} x ^{22}/_{7} x 7 x 7 x 7
 ^{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 cm^{³}  Volume = ^{2}/_{3} πr^{³} + ^{1}/_{3πr²} x ^{3}/_{2} r = 31.5π
4r^{³} + 3r^{³ }= 31.5 x 6
r = √(^{31.5 x 6}/_{7} )
= 3cm  slant height of con = √4.5^{²} + 3^{²}= 5.408cm
Surface area = 2π x 3^{²} + π x 3 x 5.408 = 107.5cm^{²}  Height = ^{31.5}/_{4²π}= 1.969cm
 Density = ^{144}/_{231.5π}= 1.46g/cm^{³}
 Volume = ^{2}/_{3} πr^{³} + ^{1}/_{3πr²} x ^{3}/_{2} r = 31.5π
 Volume of cube side x cm = (xcm)^{³}
∴ x^{³}cm^{³} = ^{1280}/_{20} cm^{³}x =3√(^{1280}/_{20})
= 3√64
= 4 cm  ^{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)cm^{³}
Join our whatsapp group for latest updates
Tap Here to Download for 50/
Get on WhatsApp for 50/
Download Volume of Solids Questions and Answers  Form 2 Topical Mathematics.
Tap Here to Download for 50/
Get on WhatsApp for 50/
Why download?
 ✔ To read offline at any time.
 ✔ To Print at your convenience
 ✔ Share Easily with Friends / Students