1) What is the area of a concrete path 2m wide surrounding
a circular pond 12m in diameter.
Solution:
The structure of the question shows that every thing is all
circular.
Image
Representation:
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Area of pond = π r² = 6² × π = 36π
Area of pond + width(2cm) = 8² × π = 64π
Area of path = 64π - 36π = 28π (Approximately 88m²).
2) A circular table top of radius 1m has an inner section of
radius 75cm which is made of glass and an other section of
wood. If glass costs N2500 per m², and wood costs N1500
per m², what is the cost of the top of the table, correct to
the nearest naira
Exercise 4c New General Math SS3
Area of glass top = πr² = π × 0.75² = 1.7679m²
Area of wood section = Area of entire top - Area of glass top.
Area of entire top = π × 1² = 3.1429m²
Area of wood section = 3.1429 - 1.7679 = 1.375m²
1m² of glass = 2500
1.7679m² = 2500 × 1.7679 = N4419.75
1m² of glass = 1500
1.375m² = 1500 × 1.37 = N2062.5(Wood)
Total cost = N4419.75 + N2062.5 = N6,482.225
To the nearest naira N6482
3) A rectangular garden 30m by 20m consists of a lawn 28m
by 18m surrounded by a path. The path is to be made of
concrete of concrete to a depth of 8cm. Calculate the
volume of cement required in m³.
Solution.
Volume of cement required(PATH) =
Volume covered by garden - Volume covered by lawn
2
(30 × 20 × 0.08) - (28 × 18 × 0.08) = 7.68m³
b) The cement is delivered in bags each holding 1 cubic
metre. Calculate the cost if the cement costs N600 per bag
and the labor charges are N2000. The company states that
the cost will be N70. per square metre. Is this Correct?
[Exercise 4c. No. 5b].
Since each bag holds 1m³, About 8 cement bags will be
needed(7.68m³)
Cost = (8 × 600) + 2000 = N6,800.
Square metre = Area.
Area = (30 × 20) - (28 × 18) = 96m².
Testing: 70 × 96 = 6720 which is very close to 6,800 and
remember that 70 was an approximated value.
Yes they are correct.
4) A solid body consists of a cylinder surmounted by a
hemisphere of the same radius. The total length of the body
measured along the central axis of the cylinder is 10cm. If
the radius of the hemisphere is x cm, show that the volume
v, of the solid v = πx²/3 [30 – x]
Solution:
3
Height along central
axis = 10cm.
Th distance
between the
center of the
hemisphere and
the point where the
central axis touches its circumference = radius = x.
Refer to properties of hemispheres(Formed from a sphere)
Hence the height of the cylinder = 10 – x.
Volume of solid body = Volume of hemisphere + Volume of
cylinder.
Volume of cylinder = πr²h = πx²(10 – x)
Volume of hemisphere = ⅔πr³ = ⅔πx³.
Total volume = πx²(10 – x) + ⅔πx³
= πx²/3 [(30 - 3x) + 2x] = πx²/3[30 – x]
5) A lead mass consists of a hemisphere of diameter 2cm
surmounted by a right circular cone of the same diameter
and height 3cm. FInd:
a) The volume of the object in cm³. [Take π = 3.142]
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b) Its mass if the density of lead is 11.45g/cm³ [Ex. 4c No. 7].
Solution:
Volume of cone =
⅓πr²h = ⅓ × 3.142 × 1² × 2 = 2.095cm³
Height of cone = 3 - 1 = 2
Volume of hemisphere = ⅔ × 3.142 × 1³ = 2.095cm³
Total volume = 2.095 + 2.095 = 4.19cm³ respectively.
b) Mass = density × vol. = 4.19 × 11.45 = 48g
5
6) Blanks for steel bolts are of the form below:
Calculate the mass of 100 of these blanks, if the density of
steel is 7.83g/cm³
Solution:
Volume of blank = Volume of cylinder(A)+Volume of small
cylinder 2(B)
A = πr²h = π × 1² × 1 = 3.142cm³
B = πr²h = π × 0.5² × 4 = 3.142cm³
Volume of blank = 2(3.142) = 6.284cm³.
Unit mass = density × volume = 6.284 × 7.83 = 49.2g.
Mass of 100 blanks = 49.2 × 100 = 4920g or 4.92kg.
7) A hemispherical bowl has an external radius of 18cm and
is made of wood 3cm thick. Calculate:
a) The volume of wood in the bowl.
b) the mass of the bowl if the wood is of density 0.8g/cm³
6
Looking at the dimensions given the external radius = 18cm.
Since thickness = 3cm,
Internal radius = 18 - 3 = 15cm.
volume of wood = Ve – Vi
Ve = vol. at external radius, Vi = internal.
Ve = ⅔πr³ = ⅔ × π × 18³ = 12219.43cm³.
Vi = ⅔ × π × 15² = 7071.43cm³
Volume of wood = 12219.43cm³ – 7071.43cm³ = 5148cm³.
b) Mass of bowl = density × volume = 0.8 × 5148 = 4118.4g
Mass in kg is approximately 4.12kg.
8) The diagram below shows the vertical cross section of a
tent made of tarpaulin, in the shape of a right circular cone
placed above a right cylinder. The height of the tent is 6m.
Calculate:
a) Volume of the tent.
b) Area of the ground covered by the tent.
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c) Area of the tarpaulin used to make the tent. [Take π =
22/7]
a) Volume of the tent = Volume of cone + Volume of
cylinder.
Height of cone = 6 – 3½ = 2.5
Volume of cone = ⅓ × π × (12 ÷ 2)² × 2.5 = 94.29cm³
Volume of cylinder = π × 36 × 3.5 (πr²h) = 396cm³
Volume of tent = 396 + 94.29 = 490.26cm³
b) Area of ground = Area of circle = Base of cylinder
Radius = 6cn
Area of ground = π × 36 = 113.14m²
c) Area of tarpaulin =
how is a cylinder formed?.
By wrapping a rectangle.
So we need to get the distance round the base of the
cylinder.
= 2 × π × r = 2 × 6 × π = 37.71cm = Length.
Breadth = height of cylinder = 3.5m
Area of cylinder = L × B = 3.5 × 37.71 = 131.9m² = 132m²
Slant height of cone = √6² + 2.5² = 6.5
8
Area of cone = π × 6 ×6.5 = 122.57.
π × radius × length
Total Area = 122.57 + 132 = 254.57m²
9) A metal sphere 6cm in diameter is melted and cast into
balls of diameter ½ cm. How many smaller balls will there be
Solution:
Volume of larger sphere = 4/3 × π × 3³ = 36πcm³
Volume of smaller sphere = 4/3 × π × ¼³= 1/48π cm³
No. of bals = 36π ÷ 1/48 π = 1,728 balls.
10) A pendulum is 105cm long and swings about 1 end
through 5°on either side of the vertical. One complete swing
takes 2 seconds.
a) Through what angle does the pendulum swing in one
complete swing.
5°× 2(2) = 20° (Going and Coming 10° each).
b) How far does the tip travel in 1 hr.
1hr = 3,600 seconds.
With 105 as length and 20° as sector angle,
Length of arc = distance covered
= 20/360 × 2 × π × 105 = 36.66.
No. of swings = 3,600/2 = 1800 swings
Distance = No. of swings × 1 full distance
Distance = 1800 × 36.66 = 65999(appr. 66,000cm)
Distance = 6600m.
