SCHEME OF WORK
WEEK TOPIC
THEME: GEOMETRY (cont’d):
1. Revision of last term’s work.
2. Trigonometric Ratios: (a) Basic Trigonometric Ratios – (i) sine (ii) cosine (iii) tangent with respect to right-angled triangles.
3. Trigonometric Ratios: (b) Trigonometric ratio of; - (i) Angle 300 (ii) Angle 450 (iii) Angle 600 (c) Application of trigonometric ratios of special angles to simple problems.
4. Trigonometric Ratios: (d) Trigonometric ratios related to the unit circle. (e) Graphs of sines and cosines (f) Graphs of sines and cosines.
5. Mensuration: (a) Length of arcs of circle. (b) Perimeter of sectors and segments. (c) Areas of sectors of a circle. (d) Areas of segments of a circle.
6. Mensuration: (e) Relation between the sector of a circle and the surface area of a cone. (f) Surface area and volume of solids; (i) Cube, cuboids (ii) Cylinder (iii) Cone (iv) Prisms (v) Pyramids.
7. Mensuration:(g) Surface areas and volume of frustum of a cone and pyramid. (h) Surface area and volume of compound shapes.
THEME: STATISTICS:
8. Data Presentation: (a) Revision on collection, tabulation and presentation of data. (b) Frequency distribution.
9. Data Presentation: (c) Line graph, Bar graph and histograms.
10. Data Presentation: (d) Pie Chart. (e) frequency polygon.
11. Revision.
3RD TERM
WEEK 1
TOPIC: TRIGONOMETRIC RATIOS
CONTENT:
Basic trigonometric ratios (i) sine (ii) cosine (iii) tangent with respect to right-angled triangles.
These trigonometric ratios are applicable to right – angled triangle. A right – angle triangle is 900. Thus the remaining two angles add up to 900 since every triangle contains two right angles.
In ∆ABC, B + C = 900.
Such angles whose sum is 900 are said to be complementary angles. While capital letter are used for angles, small (lower case) letters are used for sides. Notice that the side opposite A is labelled a, the one opposite B is labelled b etc.
The side opposite the right angle is called the hypotenuse. Every right – angled triangle obeys the Pythagoras theorem. This theorem states that the square of the hypotenuse of any right angled triangle is equal to the sum of the square of the other two sides.
Thus in the above triangle, a2=b2+c2
Apply Pythagoras theorem to the right – angled triangles below to find the lettered sides.
There are six basic trigonometric ratios viz: sine, cosine, tangent, cosecant, secant and cotangent. The first three are commonly used.
They are applicable only to right – angled triangles. Their short forms are: Sin, Cos, tan, cosec, sec, and cot respectively.
https://youtu.be/bSM7RNSbWhM
https://youtu.be/r4YJuhS-1XE
SINE
The trigonometric ratio sine, is opposite divided by hypotenuse. Its reciprocal is cosecant.
Example
Example 1
(2) In right – angled triangle XYZ, with Z = 900. If l XY l = 5m and l YZ l = 3m. Find (i) Sin x (ii) y
(3) If a ladder of length 2m leans against a wall and makes 300 with the floor, how high above the floor does the ladder reach on the wall?
https://youtu.be/a5WQlcFTXyk
ASSIGNMENT
New General Maths SS1 Ex. 11a
No1,. 2a, b, d, 3a, b
COSINE
The trigonometric ratio cosine is adjacent divided by hypotenuse. Its reciprocal is secant.
In the triangle above, B and C are complementary Sin B = b/a = Cos C
Also Sin C = c/a = CosB
For complementary angles, the sine of one is the cosine of the other i.e
Sin θ = Cos (90 - θ)
Example: solve sin2x = cos3x
Solution
Since Sin θ = cos (90 - θ)
So, sin2x = cos (90 – 2x) (i)
But we are given that
Sin2x = cos3x (ii)
From the right hand sides of equation (i) and (ii)
We conclude that
900 – 2x = 3x
900 = 5x
So x = 90/5 = 180.
https://youtu.be/j3VLbjsWdHo
Example 1.
The angle of elevation of the top of a tree is 600. If the point of observation is 4m from the foot of the tree. How far is the point from the top of the tree?
Solution
Let the point of observation from the top of the tree be x metres.
Cos600 = 4/x
0.5 = 4/x
0.5x = 4
x = 4 ÷ 0.5
= 8m.
Example 2
Given that Cosx = 0.7431, 0 < x < 900, use tables to find the values of (i) 2sinx (ii) tan x/2
Solution
Cos x = 0.7431
X = cos-1 0.7431
= 420.
2Sinx = 2Sin420
= 2(0.6691)
= 1.3382
Tan x/2 = tan 42/2
= tan 210
= 0.3839
Evaluation
1. Find the unknown sides of the following triangles.
2. If sin θ = 0.3970 use the tables to find (i) cos θ (ii) tan θ
TANGENT
Tan = opposite/adjacent
https://youtu.be/57FTWh_E1ns
Example 1:
find y in the figure
Tan 32 = y/6
y = 6tan 32
= 6 x 0.6249
= 3.75
Example 2:
Example.
If sin x = 3/5 and x is an acute angle, find cosx + tanx
Solution
https://youtu.be/-XLueOZLJkA
Evaluation
1. Find the marked angles below:
2. If Cos P = 4/5 and P is an acute angle, what is the value of tanP?
3. Given that tanx = 8/15. What is the value of sinx + cosx?
4. A boy sets out to travel from A to C via B. From A he travels a distance of 4km on a bearing 030° to B. From B he travels a further 3km due east. Calculate how far is C
(i) North of A
(ii) East of A
Hence, or otherwise, calculate the distance AC correct to 1 decimal place.
ASSIGNMENT
New General Mathematics Ex. 11a. No 7, 14 and 15
CONTENT:
Basic trigonometric ratios (i) sine (ii) cosine (iii) tangent with respect to right-angled triangles.
These trigonometric ratios are applicable to right – angled triangle. A right – angle triangle is 900. Thus the remaining two angles add up to 900 since every triangle contains two right angles.
In ∆ABC, B + C = 900.
Such angles whose sum is 900 are said to be complementary angles. While capital letter are used for angles, small (lower case) letters are used for sides. Notice that the side opposite A is labelled a, the one opposite B is labelled b etc.
The side opposite the right angle is called the hypotenuse. Every right – angled triangle obeys the Pythagoras theorem. This theorem states that the square of the hypotenuse of any right angled triangle is equal to the sum of the square of the other two sides.
Thus in the above triangle, a2=b2+c2
Apply Pythagoras theorem to the right – angled triangles below to find the lettered sides.
There are six basic trigonometric ratios viz: sine, cosine, tangent, cosecant, secant and cotangent. The first three are commonly used.
They are applicable only to right – angled triangles. Their short forms are: Sin, Cos, tan, cosec, sec, and cot respectively.
https://youtu.be/bSM7RNSbWhM
https://youtu.be/r4YJuhS-1XE
SINE
The trigonometric ratio sine, is opposite divided by hypotenuse. Its reciprocal is cosecant.
Example
Example 1
(2) In right – angled triangle XYZ, with Z = 900. If l XY l = 5m and l YZ l = 3m. Find (i) Sin x (ii) y
(3) If a ladder of length 2m leans against a wall and makes 300 with the floor, how high above the floor does the ladder reach on the wall?
https://youtu.be/a5WQlcFTXyk
ASSIGNMENT
New General Maths SS1 Ex. 11a
No1,. 2a, b, d, 3a, b
COSINE
The trigonometric ratio cosine is adjacent divided by hypotenuse. Its reciprocal is secant.
In the triangle above, B and C are complementary Sin B = b/a = Cos C
Also Sin C = c/a = CosB
For complementary angles, the sine of one is the cosine of the other i.e
Sin θ = Cos (90 - θ)
Example: solve sin2x = cos3x
Solution
Since Sin θ = cos (90 - θ)
So, sin2x = cos (90 – 2x) (i)
But we are given that
Sin2x = cos3x (ii)
From the right hand sides of equation (i) and (ii)
We conclude that
900 – 2x = 3x
900 = 5x
So x = 90/5 = 180.
https://youtu.be/j3VLbjsWdHo
Example 1.
The angle of elevation of the top of a tree is 600. If the point of observation is 4m from the foot of the tree. How far is the point from the top of the tree?
Solution
Let the point of observation from the top of the tree be x metres.
Cos600 = 4/x
0.5 = 4/x
0.5x = 4
x = 4 ÷ 0.5
= 8m.
Example 2
Given that Cosx = 0.7431, 0 < x < 900, use tables to find the values of (i) 2sinx (ii) tan x/2
Solution
Cos x = 0.7431
X = cos-1 0.7431
= 420.
2Sinx = 2Sin420
= 2(0.6691)
= 1.3382
Tan x/2 = tan 42/2
= tan 210
= 0.3839
Evaluation
1. Find the unknown sides of the following triangles.
2. If sin θ = 0.3970 use the tables to find (i) cos θ (ii) tan θ
TANGENT
Tan = opposite/adjacent
https://youtu.be/57FTWh_E1ns
Example 1:
find y in the figure
Tan 32 = y/6
y = 6tan 32
= 6 x 0.6249
= 3.75
Example 2:
Example.
If sin x = 3/5 and x is an acute angle, find cosx + tanx
Solution
https://youtu.be/-XLueOZLJkA
Evaluation
1. Find the marked angles below:
2. If Cos P = 4/5 and P is an acute angle, what is the value of tanP?
3. Given that tanx = 8/15. What is the value of sinx + cosx?
4. A boy sets out to travel from A to C via B. From A he travels a distance of 4km on a bearing 030° to B. From B he travels a further 3km due east. Calculate how far is C
(i) North of A
(ii) East of A
Hence, or otherwise, calculate the distance AC correct to 1 decimal place.
ASSIGNMENT
New General Mathematics Ex. 11a. No 7, 14 and 15
WEEK 2
TOPIC: TRIGONOMETRIC RATIOS
Angles 300 and 600.
To consider the trigonometric ratios for the special angles 300 and 600, we shall consider an equilateral triangle of side 2units.
If the triangle is bisected we shall have 300
If the value of the altitude h, can be obtained using Pythagoras theorem.
h2 + 12=22
h2 = 4 – 1
h= √3 consequently, Sin300 = opp/hyp = ½
Cos300 = adj/hyp
= √3/2
Tan300 = opp/adj
= 1/√3
Sin 600 = √3/2
Cos 600 = ½
Tan 600 = √3/1
= √3
Example find x and y in the figure below
Look at the figure with respect to the given 600, the known side is adjacent. The unknown side x is the hypotenuse, the trigonometric ratio that connects the known (adjacent) and the unknown (hypotenuse) is cosine.
Thus we write
Cos 600 = 4/x
But cos 60 = ½
So ½ = 4/x
Hence x = 8.
To find y, y is the opposite side and 4 is the adjacent. Hence we use tangent
Tan 600 = y/4
But tan 600 = √3
√3=y/4,
hence y = 4√3
Do not use tables but leave surds in your answer.
https://youtu.be/73eNYjKlTd8
https://youtu.be/KYBhGQMKoAo
ASSIGNMENT
New General Mathematics SS 1, Ex 11d.Nos 3 – 8
ANGLE 450.
Consider a square one unit.
Suppose the length of its diagonal is x.
Using Pythagoras theorem,
x2 = 12 + 12
x2 = 2
x =√2
from this isosceles triangle obtained from above
To find x, use cosine
Example. Find x and y below
https://youtu.be/lBEXdR4WwEk
https://youtu.be/DCJZFXzvzA8
Evaluation
Find the marked sides in the triangles below. Do not use tables.
Angles 300 and 600.
To consider the trigonometric ratios for the special angles 300 and 600, we shall consider an equilateral triangle of side 2units.
If the triangle is bisected we shall have 300
If the value of the altitude h, can be obtained using Pythagoras theorem.
h2 + 12=22
h2 = 4 – 1
h= √3 consequently, Sin300 = opp/hyp = ½
Cos300 = adj/hyp
= √3/2
Tan300 = opp/adj
= 1/√3
Sin 600 = √3/2
Cos 600 = ½
Tan 600 = √3/1
= √3
Example find x and y in the figure below
Look at the figure with respect to the given 600, the known side is adjacent. The unknown side x is the hypotenuse, the trigonometric ratio that connects the known (adjacent) and the unknown (hypotenuse) is cosine.
Thus we write
Cos 600 = 4/x
But cos 60 = ½
So ½ = 4/x
Hence x = 8.
To find y, y is the opposite side and 4 is the adjacent. Hence we use tangent
Tan 600 = y/4
But tan 600 = √3
√3=y/4,
hence y = 4√3
Do not use tables but leave surds in your answer.
https://youtu.be/73eNYjKlTd8
https://youtu.be/KYBhGQMKoAo
ASSIGNMENT
New General Mathematics SS 1, Ex 11d.Nos 3 – 8
ANGLE 450.
Consider a square one unit.
Suppose the length of its diagonal is x.
Using Pythagoras theorem,
x2 = 12 + 12
x2 = 2
x =√2
from this isosceles triangle obtained from above
To find x, use cosine
Example. Find x and y below
https://youtu.be/lBEXdR4WwEk
https://youtu.be/DCJZFXzvzA8
Evaluation
Find the marked sides in the triangles below. Do not use tables.
WEEK 3
TOPIC: TRIGONOMETRIC RATIOS
CONTENTS
Trigonometric ratios related to the unit circle
Angles between 00 and 3600.
Consider a circle of radius one unit and centre at O, the origin, in the XY plane.
https://youtu.be/c819bGfH8FA
The circle is called a unit circle. Point P has co-ordinate (x,y) and lOPl = 1 unit.
Line Op makes an angle of θ with Ox.
In the right – angled triangles,
Sinθ = y/1 = y
Cosθ = x/1 = x
The values of x and y are a measure of Cosθ and Sinθ respectively.
Let the unit circle be graduated in degrees from 00 to 3600 as shown below. The graph of y = Sinθ can be drawn using the unit circle.
One can plot values of y against θ to get a wave-like curve.
For instance when θ = 3600, the corresponding y value is obtained by drawing dotted horizontal line from mark 300 on the unit circle to meet the y – axis.
When θ = 900, the dotted horizontal line for 900 on the circle will meet the y – axis where y = 1
https://youtu.be/1m9p9iubMLU
To draw the graph of Cosx, use the corresponding values of x and θ.
https://youtu.be/Vn4TnlrFpsw
For instance on the circle, when θ = 00, that corresponds to x = 1, when θ = 600, dotted line from point 600 on the unit circle vertically downwards will meet the x-axis at x = ½
(i.e when θ = 600, x = ½).
Plotting values of x against corresponding values of θ gives the graph of Cos θ . it is another wave-shaped curve.
As θ increases beyond 3600, the curves of Sin θ and Cosθ repeat themselves.
Notice that +700 = -2900 (same direction)
-300 = +3300
In what quadrant is (i) +1700 (ii) -1700 (iii) -450 (iv) -2600 (v)2350.?
What positive angle is equivalent to
(i) -1600 (ii) -1800 (iii) -2700?
What negative angle is equivalent to
(i) 1670 (ii) 2020 (iii) 2850
For angles in the first quadrant, 00<θ< 900.
In this quadrant, x is negative, y is positive.
https://youtu.be/wCOtV44IUw4
X is positive
Y is negative
Sin θ = -y/r = -Sin (3600 - θ). This is negative
Cos θ = x/r = Cos(3600 - θ). This is positive
Tan θ = -y/x = - tan(3600 - θ). This is negative
https://youtu.be/g0DXv_PXFng
https://youtu.be/5YFjBEFCUqo
Evaluation
Use tables to find the values of the following.
CONTENTS
Trigonometric ratios related to the unit circle
Angles between 00 and 3600.
Consider a circle of radius one unit and centre at O, the origin, in the XY plane.
https://youtu.be/c819bGfH8FA
The circle is called a unit circle. Point P has co-ordinate (x,y) and lOPl = 1 unit.
Line Op makes an angle of θ with Ox.
In the right – angled triangles,
Sinθ = y/1 = y
Cosθ = x/1 = x
The values of x and y are a measure of Cosθ and Sinθ respectively.
Let the unit circle be graduated in degrees from 00 to 3600 as shown below. The graph of y = Sinθ can be drawn using the unit circle.
One can plot values of y against θ to get a wave-like curve.
For instance when θ = 3600, the corresponding y value is obtained by drawing dotted horizontal line from mark 300 on the unit circle to meet the y – axis.
When θ = 900, the dotted horizontal line for 900 on the circle will meet the y – axis where y = 1
https://youtu.be/1m9p9iubMLU
To draw the graph of Cosx, use the corresponding values of x and θ.
https://youtu.be/Vn4TnlrFpsw
For instance on the circle, when θ = 00, that corresponds to x = 1, when θ = 600, dotted line from point 600 on the unit circle vertically downwards will meet the x-axis at x = ½
(i.e when θ = 600, x = ½).
Plotting values of x against corresponding values of θ gives the graph of Cos θ . it is another wave-shaped curve.
As θ increases beyond 3600, the curves of Sin θ and Cosθ repeat themselves.
Notice that +700 = -2900 (same direction)
-300 = +3300
In what quadrant is (i) +1700 (ii) -1700 (iii) -450 (iv) -2600 (v)2350.?
What positive angle is equivalent to
(i) -1600 (ii) -1800 (iii) -2700?
What negative angle is equivalent to
(i) 1670 (ii) 2020 (iii) 2850
For angles in the first quadrant, 00<θ< 900.
In this quadrant, x is negative, y is positive.
https://youtu.be/wCOtV44IUw4
X is positive
Y is negative
Sin θ = -y/r = -Sin (3600 - θ). This is negative
Cos θ = x/r = Cos(3600 - θ). This is positive
Tan θ = -y/x = - tan(3600 - θ). This is negative
https://youtu.be/g0DXv_PXFng
https://youtu.be/5YFjBEFCUqo
Evaluation
Use tables to find the values of the following.
WEEK 4
TOPIC: Mensuration
Content:
Definition and meaning of an arc sector and segment of a circle.
Length of arc of circle
Perimeter of sectors and segments
Areas of sectors and segments
Areas of sectors of a circle
Area of segments of a circle
DEFINITION AND MEANING, LENGTH OF ARCS OF CIRCLE
AN ARC: an arc of a circle is a part of the circumference of the circle.
Hence, an arc is a length or a distance along the circumference of a circle. It is never an area.
A SECTOR: a sector is a part or a fraction of a circle bounded by an arc and two radii.
Hence, an arc is a length whereas the sector covers an area of a circle.
A SEGMENT
The segment of a circle is the part cut off from the circle by a chord. A chord is the line segment AB.
https://youtu.be/viM7a4IUp9Y
LENGHT OF ARCS OF CIRCLES
If 5 sectors are cut off from 5 different circles and the lengths of the arcs l, radii r and angles θ measured and compared.
Then
i/2π = θ/360 or L = θ/360 x 2 πr = 2 πr θ/360
hence, in a circle of radius r, the length l of an arc that subtends angle θ at the centre is given by
L = θ/360 x 2 πr
Example 1:
Find the length of an arc of a circle of radius 5.6cm which subtends an angle of 600 at the centre of the circle (Take π = 22/7)
Solution
Length of arc = θ/360 x 2 πr
Given θ = 600, r = 5.6cm, π = 22/7
Substituting into the formula,
Length of arc AB = 60/360 x 2 x 22/7 x 5.6
= 1/6 x 2 x 22/7 x 5.6
= 1/6 x 2 x 22/7 x 0.8
= 17.6/3
= 5.8667cm
= 5.87cm (2 decimal places)
Example 2;
What angle does an arc 6.6cm in length subtends at the centre of a circle of radius 14cm. Use π = 22/7)
Solution
Length of arc xy = θ/360 x 2 πr
6.6 = θ/360 x 2 x 22/7 x 14
θx 2 x 22 x 14 = 6.6 x 360 x 7
θ = (6.6 x 360 x 7)/(2 x 22 x 14)
= (33 x 18)/11x2
= 3 x 9
= 270
θ = 270 , the angles subtend by the arc
Example 3:
An arc of length 12.57cm subtends an angle of 600 at the centre of a circle. Find
The radius of the circle
The diameter of the circle.
Solution
Arc= θ/360 x 2 πr
12.57 = 60/360 x 2 x 22/7 x r/1
12.57 x 360 x 7 = 60 x 2 x 22 x r
r = (12.57 x 360 x 7)/(60 x 2 x 22)
r = (12.57 x 6 x 7)/44
r = 527.94/44
r = 11.99
r = 12cm
Example 4:
An arc of a circle of diameter 28m subtends an angle of 1080 at the centre of the circle. Find the length of the major arc.
Solution
Minor arc angle = 1080
Major arc angle = 3600 – 1080
= 2520.
Arc = = θ/360 x 2 πr
= 252/360 x 2/1 x 22/7 x 14/1
= (252 x 44)/180
= 11088/180
= 61.6m
The length of major arc = 61.6m
https://youtu.be/C9z3FXS7nlo
Evaluation
Find the radius of a circle which subtends an angle of 1200 at the centre of the circle and is of length 2.8cm (π = 22/7)
In terms of π, what is the length of an arc of a circle of radius 31/2m?
https://youtu.be/SwrlKx52L5o
PERIMETER OF SECTORS AND SEGMENTS
The word perimeter simply means the distance round an object. So the perimeter of a sector of a circle is the distance round the circle.
hence the perimeter of a sector AOB is the sum of two radii (2r) and length of arc l, where r is radius and l = length of arc.
Perimeter of sector AOB = 2r + Lunits
Example
Find the perimeter of the sector of radius 3.5cm which subtends an angle of (i) 450 (ii)3150.
Solution
Length of arc = θ/360 x 2πr
Here θ = 450, r = 3.5cm, π = 22/7
Length of arc = 45/360 x 2 x 22/7 x 3.5
= 1/8 x 2/1x 22/7 x 7/2
= 22/8
= 2.75cm
The perimeter of the sector is 2r + l, here r = radius which is 3.5cm and l = 2.75cm.
Perimeter = 2r + length of arc
= 2r + 2.75cm
= (2 x 3.5) + 2.75cm
= 7.0 + 2.75cm
= 9.75cm
(ii) θ = 3150, π = 22/7 , r = 3.5cm
Length of arc = 315/360 x 2 x 22/7 x 3.5cm
= 63/72 x 2 x 22/7 x 7/2 cm
= 7/8 x 2 x 22/7 x 7/2 cm
= 77/4 = 19.25cm
Perimeter = 2r + 19.25cm
= (2 x 3.5) + 19.25
= 7 + 19.25
= 26.25cm
https://youtu.be/_y4E1cbga4Y
EVALUATION
Calculate the perimeter of a sector of a circle of radius 14cm, where the sector angle is 600. Take π = 22/7
Page 163, 15, 16, 23, 27
PERIMETER OF SEGMENT
The perimeter of a segment = length of arc + length of chord.
To find the length of chord AB, we bisect <AOB
Example 2:
The perimeter of a sector is 61.43cm. If the angle subtended by the sector at the centre is 1200. Find the radius of the sector.
Solution
Perimeter of a sector = Arc + 2r
61.43 = = θ/360 x 2πr + 2r
61.43 = = 120/360 x 2 x 22/7 x r + 2r
61.43 = = 1/3 x 2 x 22/7 x r + 2r
Multiply through by L.c.m = 21
21 x 61.43 = (211/3)x 2 x 22r)/ 3 x 7+ 2r)
1290.03 = 2 x 22r + 21 + 2r
1290.03 = 44r + 42r
1290.03 = 86r
86r = 1290.03
r = 1290.03/86
r = 15.00cm
https://youtu.be/3j8p1HW9GkI
Example 3
The perimeter of a sector is 61.43cm. If the radium of the sector is 15cm. Find the angle it subtends at the centre.
Solution
Perimeter of a sector = Arc + 2r
61.43 = θ/360 x 2 πr + 2r
61.43 = θ/360 x 2 x 22/7 x 15 + 2 x 15
61.43 = 11θ/(6 x 7) + 30
61.42 – 30 = 11θ/42
31.43/1 = 11θ/42
11θ = 42 x 31.43
θ = (42 x 31.43)/11
θ = 1200
https://youtu.be/vVAl1jyL8X0
https://youtu.be/umkOuw25nX0
EVALUATION
The angle of a sector of a circle radius 12.5cm is 680. Calculate the perimeter of the sector.
A rope of length 18m is used to form a sector of a circle of radius 2.5m on a school playing field. What is the size of the angle of the sector? Correct to the nearest degree?
The perimeter of a sector is 75.43cm. If the angle – subtended by the sector at the centre is 1350. Find the radius of the sector.
An arc of length 21.34cm subtends an angle 1010 at the centre of a circle. Find the diameter of the circle.
Using trigonometric ratios
AD/r = Sin θ/2
AD = r Sin θ/2
But, AD = DB
Hence
AB chord AB = AD + DB
= r Sin θ/2 + r Sin θ/2
= 2r Sin θ/2 units
Length of Chord = 2r Sin θ/2units
Thus,
Perimeter of segment = θ/360 x 2πr + 2r Sin θ/2
= θ/180 x πr + 2r Sin θ/2
Example 3:
AB is a chord of a circle with centre o and radius 4cm, AOB = 1200. Calculate the perimeter of the minor segment (π = 22/7).
Solution
Chord AB = rSinθ/2, r = 4cm, θ = 1200.
θ/2 = 120/2 = 600.
So chord AB = 2 x 4 Sin 600
= 8 x sin 600
= 8 x(√3)/2
= 4√3cm
= 6.92cm
Length of arc AB = 120/360 x 2 x 4 x 22/7
= 1/3 x 8 x 22/7
= 176/21
= 8.38cm (2 d.p)
Perimeter of minor segment = length of arc + chord
= 8.38 + 6.92cm
= 15.30cm
https://youtu.be/no64hhheEjs
EVALUATION
A sector subtends an angle of 840 at the centre of a circle of radius 5.6cm. calculate the perimeter of the sector to the nearest cm.
New General Mathematics for senior secondary school, Book 1 pages 154 exercise 12d, Nos 7 – 10
AREA OF SECTORS OF A CIRCLE
The area of a sector = θ/360 x area of circle
= θ/360 x πr2, where θ is the angle formed at the centre by the arc of the circle.
Example 1:
Find the area of the sector of a circle of radius 4.8cm which subtends an angle of 1350 at the centre.(Take π = 3.142)
Solution
The area of a sector = θ/360 x πr2
Here θ = 1350, r = 4.8cm, π = 3.142
The area of a sector = 135/360 x 3.142 x (4.8)2
= 135/360 x 3.142 x 4.8 x 4.8
= 27 x 3.142 x 0.4 x 0.8
= 27.14688cm2
Area of sector = 27.15cm2 (decimal places)
Example 2:
AB is an arc of a circle of length 9.2cm with centre 0 and the radius is 4.6cm. Find the area of the sector AOB.
Solution
Length of arc AB = θ/360 x 2πr
r = 4.6cm, length of arc = 9.2cm
Length of arc AB = θ/360 x 2 x π x 4.6
9.2 = θ/360 x 2 x π x 4.6
9.2 x 360 = θ x 2 x π x 4.6
θ = (9.2 x 360)/(2 π x 4.6)
θ = ( 360)/π
Area of sector AOB
= θ/360 x πr2
= θ/360 x πr2
= θ/360 x 1/360 x π x (4.6)2
= (4.6)2
= (4.6) x (4.6)
= 21.16cm2
https://youtu.be/7f9_U85_YX4
EVALUATION
Calculate the area of a sector of a circle which subtends an angle 450 at the centre of the circle, radius 14cm.
A sector of 800 is removed from a circle of radius 12cm. What area of the circle is left? Use 22/7.
New general mathematics for senior secondary school, Book 1 page 154 exercise 12d Nos 1 – 6.
SEGMENTS OF CIRCLES
Area of segments = area of sector – area of triangle
∴ Area of segment = θ/360 x πr2 - 1/2r2 Sin θ/2
= (r2/2) [πθ/180- Sin θ]
Example: The arc AB of a circle, radius 6.5cm, subtends an angle of 450 at the centre o. Find the area of the (i) Minor segment cut off by the chord AB (Take π = 22/7)
Solution
(i) Area of sector AOB = θ/360 x πr2
θ = 450, r = 6.5, π = 22/7
Area of sector AOB = 45/360 x 22/7 x (6.5)2cm2
= 1/8 x 22/7 x 6.5 x 6.5
= 1/8 x 22/7 x 42.25
= 16.598
= 16.60cm2 (2d.p)
(ii.) Area of segment = Area of sector – Area of ∆
Area of ∆ = 1/2 AO X OBSin450
= 1/2 X 0.5 x 6.5Sin450
= 42.25/2 x 1/√2
= (42.25√2)/4
= (42.25 x 1.414)/4
= 14.935375cm2
∴area of segment = 16.60 – 14.35375
= 1.66cm2
https://youtu.be/vVAl1jyL8X0
EVALUATION
An arc AB of a circle radius 4.8cm subtends an angle of 1580 at the centre O. Find
The area of the sector AOB
The area of the minor segment cut off by the chord AB (π = 3.142)
https://youtu.be/vYR9qicyWB4
GENERAL EVALUATION
What is the length of an arc which subtends an angle of 60° at the centre of a circle of radius 1/2m?
A chord AB of a circle radius 9.4cm is 12.8cm. find:
The angle subtended by the chord at the centre of the circle
The area of the minor segment cut off by this chord.
ASSIGNMENT
Solve questions from New General Maths for senior secondary schools 1 page 151. Ex 12c Qs 5 and 6.
Page 154, Ex 12d. Qs 5, 6, 7a, b and 11.
READING ASSIGNMENT
Read New General Maths for senior secondary schools 1 pages 150 – 155.
REFERENCE TEXTS
M. F. Macrae A. O e tal (2011). New General Mathematics for senior secondary school 1
Mrs Maria David – Osuagwuetal (2000). New General Mathematics for senior secondary schools.
Fundamental General Mathematics for senior secondary school by Idode G. O
Content:
Definition and meaning of an arc sector and segment of a circle.
Length of arc of circle
Perimeter of sectors and segments
Areas of sectors and segments
Areas of sectors of a circle
Area of segments of a circle
DEFINITION AND MEANING, LENGTH OF ARCS OF CIRCLE
AN ARC: an arc of a circle is a part of the circumference of the circle.
Hence, an arc is a length or a distance along the circumference of a circle. It is never an area.
A SECTOR: a sector is a part or a fraction of a circle bounded by an arc and two radii.
Hence, an arc is a length whereas the sector covers an area of a circle.
A SEGMENT
The segment of a circle is the part cut off from the circle by a chord. A chord is the line segment AB.
https://youtu.be/viM7a4IUp9Y
LENGHT OF ARCS OF CIRCLES
If 5 sectors are cut off from 5 different circles and the lengths of the arcs l, radii r and angles θ measured and compared.
Then
i/2π = θ/360 or L = θ/360 x 2 πr = 2 πr θ/360
hence, in a circle of radius r, the length l of an arc that subtends angle θ at the centre is given by
L = θ/360 x 2 πr
Example 1:
Find the length of an arc of a circle of radius 5.6cm which subtends an angle of 600 at the centre of the circle (Take π = 22/7)
Solution
Length of arc = θ/360 x 2 πr
Given θ = 600, r = 5.6cm, π = 22/7
Substituting into the formula,
Length of arc AB = 60/360 x 2 x 22/7 x 5.6
= 1/6 x 2 x 22/7 x 5.6
= 1/6 x 2 x 22/7 x 0.8
= 17.6/3
= 5.8667cm
= 5.87cm (2 decimal places)
Example 2;
What angle does an arc 6.6cm in length subtends at the centre of a circle of radius 14cm. Use π = 22/7)
Solution
Length of arc xy = θ/360 x 2 πr
6.6 = θ/360 x 2 x 22/7 x 14
θx 2 x 22 x 14 = 6.6 x 360 x 7
θ = (6.6 x 360 x 7)/(2 x 22 x 14)
= (33 x 18)/11x2
= 3 x 9
= 270
θ = 270 , the angles subtend by the arc
Example 3:
An arc of length 12.57cm subtends an angle of 600 at the centre of a circle. Find
The radius of the circle
The diameter of the circle.
Solution
Arc= θ/360 x 2 πr
12.57 = 60/360 x 2 x 22/7 x r/1
12.57 x 360 x 7 = 60 x 2 x 22 x r
r = (12.57 x 360 x 7)/(60 x 2 x 22)
r = (12.57 x 6 x 7)/44
r = 527.94/44
r = 11.99
r = 12cm
Example 4:
An arc of a circle of diameter 28m subtends an angle of 1080 at the centre of the circle. Find the length of the major arc.
Solution
Minor arc angle = 1080
Major arc angle = 3600 – 1080
= 2520.
Arc = = θ/360 x 2 πr
= 252/360 x 2/1 x 22/7 x 14/1
= (252 x 44)/180
= 11088/180
= 61.6m
The length of major arc = 61.6m
https://youtu.be/C9z3FXS7nlo
Evaluation
Find the radius of a circle which subtends an angle of 1200 at the centre of the circle and is of length 2.8cm (π = 22/7)
In terms of π, what is the length of an arc of a circle of radius 31/2m?
https://youtu.be/SwrlKx52L5o
PERIMETER OF SECTORS AND SEGMENTS
The word perimeter simply means the distance round an object. So the perimeter of a sector of a circle is the distance round the circle.
hence the perimeter of a sector AOB is the sum of two radii (2r) and length of arc l, where r is radius and l = length of arc.
Perimeter of sector AOB = 2r + Lunits
Example
Find the perimeter of the sector of radius 3.5cm which subtends an angle of (i) 450 (ii)3150.
Solution
Length of arc = θ/360 x 2πr
Here θ = 450, r = 3.5cm, π = 22/7
Length of arc = 45/360 x 2 x 22/7 x 3.5
= 1/8 x 2/1x 22/7 x 7/2
= 22/8
= 2.75cm
The perimeter of the sector is 2r + l, here r = radius which is 3.5cm and l = 2.75cm.
Perimeter = 2r + length of arc
= 2r + 2.75cm
= (2 x 3.5) + 2.75cm
= 7.0 + 2.75cm
= 9.75cm
(ii) θ = 3150, π = 22/7 , r = 3.5cm
Length of arc = 315/360 x 2 x 22/7 x 3.5cm
= 63/72 x 2 x 22/7 x 7/2 cm
= 7/8 x 2 x 22/7 x 7/2 cm
= 77/4 = 19.25cm
Perimeter = 2r + 19.25cm
= (2 x 3.5) + 19.25
= 7 + 19.25
= 26.25cm
https://youtu.be/_y4E1cbga4Y
EVALUATION
Calculate the perimeter of a sector of a circle of radius 14cm, where the sector angle is 600. Take π = 22/7
Page 163, 15, 16, 23, 27
PERIMETER OF SEGMENT
The perimeter of a segment = length of arc + length of chord.
To find the length of chord AB, we bisect <AOB
Example 2:
The perimeter of a sector is 61.43cm. If the angle subtended by the sector at the centre is 1200. Find the radius of the sector.
Solution
Perimeter of a sector = Arc + 2r
61.43 = = θ/360 x 2πr + 2r
61.43 = = 120/360 x 2 x 22/7 x r + 2r
61.43 = = 1/3 x 2 x 22/7 x r + 2r
Multiply through by L.c.m = 21
21 x 61.43 = (211/3)x 2 x 22r)/ 3 x 7+ 2r)
1290.03 = 2 x 22r + 21 + 2r
1290.03 = 44r + 42r
1290.03 = 86r
86r = 1290.03
r = 1290.03/86
r = 15.00cm
https://youtu.be/3j8p1HW9GkI
Example 3
The perimeter of a sector is 61.43cm. If the radium of the sector is 15cm. Find the angle it subtends at the centre.
Solution
Perimeter of a sector = Arc + 2r
61.43 = θ/360 x 2 πr + 2r
61.43 = θ/360 x 2 x 22/7 x 15 + 2 x 15
61.43 = 11θ/(6 x 7) + 30
61.42 – 30 = 11θ/42
31.43/1 = 11θ/42
11θ = 42 x 31.43
θ = (42 x 31.43)/11
θ = 1200
https://youtu.be/vVAl1jyL8X0
https://youtu.be/umkOuw25nX0
EVALUATION
The angle of a sector of a circle radius 12.5cm is 680. Calculate the perimeter of the sector.
A rope of length 18m is used to form a sector of a circle of radius 2.5m on a school playing field. What is the size of the angle of the sector? Correct to the nearest degree?
The perimeter of a sector is 75.43cm. If the angle – subtended by the sector at the centre is 1350. Find the radius of the sector.
An arc of length 21.34cm subtends an angle 1010 at the centre of a circle. Find the diameter of the circle.
Using trigonometric ratios
AD/r = Sin θ/2
AD = r Sin θ/2
But, AD = DB
Hence
AB chord AB = AD + DB
= r Sin θ/2 + r Sin θ/2
= 2r Sin θ/2 units
Length of Chord = 2r Sin θ/2units
Thus,
Perimeter of segment = θ/360 x 2πr + 2r Sin θ/2
= θ/180 x πr + 2r Sin θ/2
Example 3:
AB is a chord of a circle with centre o and radius 4cm, AOB = 1200. Calculate the perimeter of the minor segment (π = 22/7).
Solution
Chord AB = rSinθ/2, r = 4cm, θ = 1200.
θ/2 = 120/2 = 600.
So chord AB = 2 x 4 Sin 600
= 8 x sin 600
= 8 x(√3)/2
= 4√3cm
= 6.92cm
Length of arc AB = 120/360 x 2 x 4 x 22/7
= 1/3 x 8 x 22/7
= 176/21
= 8.38cm (2 d.p)
Perimeter of minor segment = length of arc + chord
= 8.38 + 6.92cm
= 15.30cm
https://youtu.be/no64hhheEjs
EVALUATION
A sector subtends an angle of 840 at the centre of a circle of radius 5.6cm. calculate the perimeter of the sector to the nearest cm.
New General Mathematics for senior secondary school, Book 1 pages 154 exercise 12d, Nos 7 – 10
AREA OF SECTORS OF A CIRCLE
The area of a sector = θ/360 x area of circle
= θ/360 x πr2, where θ is the angle formed at the centre by the arc of the circle.
Example 1:
Find the area of the sector of a circle of radius 4.8cm which subtends an angle of 1350 at the centre.(Take π = 3.142)
Solution
The area of a sector = θ/360 x πr2
Here θ = 1350, r = 4.8cm, π = 3.142
The area of a sector = 135/360 x 3.142 x (4.8)2
= 135/360 x 3.142 x 4.8 x 4.8
= 27 x 3.142 x 0.4 x 0.8
= 27.14688cm2
Area of sector = 27.15cm2 (decimal places)
Example 2:
AB is an arc of a circle of length 9.2cm with centre 0 and the radius is 4.6cm. Find the area of the sector AOB.
Solution
Length of arc AB = θ/360 x 2πr
r = 4.6cm, length of arc = 9.2cm
Length of arc AB = θ/360 x 2 x π x 4.6
9.2 = θ/360 x 2 x π x 4.6
9.2 x 360 = θ x 2 x π x 4.6
θ = (9.2 x 360)/(2 π x 4.6)
θ = ( 360)/π
Area of sector AOB
= θ/360 x πr2
= θ/360 x πr2
= θ/360 x 1/360 x π x (4.6)2
= (4.6)2
= (4.6) x (4.6)
= 21.16cm2
https://youtu.be/7f9_U85_YX4
EVALUATION
Calculate the area of a sector of a circle which subtends an angle 450 at the centre of the circle, radius 14cm.
A sector of 800 is removed from a circle of radius 12cm. What area of the circle is left? Use 22/7.
New general mathematics for senior secondary school, Book 1 page 154 exercise 12d Nos 1 – 6.
SEGMENTS OF CIRCLES
Area of segments = area of sector – area of triangle
∴ Area of segment = θ/360 x πr2 - 1/2r2 Sin θ/2
= (r2/2) [πθ/180- Sin θ]
Example: The arc AB of a circle, radius 6.5cm, subtends an angle of 450 at the centre o. Find the area of the (i) Minor segment cut off by the chord AB (Take π = 22/7)
Solution
(i) Area of sector AOB = θ/360 x πr2
θ = 450, r = 6.5, π = 22/7
Area of sector AOB = 45/360 x 22/7 x (6.5)2cm2
= 1/8 x 22/7 x 6.5 x 6.5
= 1/8 x 22/7 x 42.25
= 16.598
= 16.60cm2 (2d.p)
(ii.) Area of segment = Area of sector – Area of ∆
Area of ∆ = 1/2 AO X OBSin450
= 1/2 X 0.5 x 6.5Sin450
= 42.25/2 x 1/√2
= (42.25√2)/4
= (42.25 x 1.414)/4
= 14.935375cm2
∴area of segment = 16.60 – 14.35375
= 1.66cm2
https://youtu.be/vVAl1jyL8X0
EVALUATION
An arc AB of a circle radius 4.8cm subtends an angle of 1580 at the centre O. Find
The area of the sector AOB
The area of the minor segment cut off by the chord AB (π = 3.142)
https://youtu.be/vYR9qicyWB4
GENERAL EVALUATION
What is the length of an arc which subtends an angle of 60° at the centre of a circle of radius 1/2m?
A chord AB of a circle radius 9.4cm is 12.8cm. find:
The angle subtended by the chord at the centre of the circle
The area of the minor segment cut off by this chord.
ASSIGNMENT
Solve questions from New General Maths for senior secondary schools 1 page 151. Ex 12c Qs 5 and 6.
Page 154, Ex 12d. Qs 5, 6, 7a, b and 11.
READING ASSIGNMENT
Read New General Maths for senior secondary schools 1 pages 150 – 155.
REFERENCE TEXTS
M. F. Macrae A. O e tal (2011). New General Mathematics for senior secondary school 1
Mrs Maria David – Osuagwuetal (2000). New General Mathematics for senior secondary schools.
Fundamental General Mathematics for senior secondary school by Idode G. O
WEEK 5
TOPIC: Mensuration
Content:
Relation between the sector of a circle and the surface area of a cone.
Surface area of cubes, cuboids and cylinder
Surface area of prisms and pyramids
Volume of cubes, cuboids, cylinder, cone,
Prisms and pyramids
Relation between the sector of a circle and the surface area of a cone.
If a sector of a circle AOB is cut and folded into a come as shown in the diagram below.
The arc AB becomes the circumference of the base of the cone. The radius R becomes the slant edge l of the cone.
∴ Arc AB = 2πr
https://youtu.be/S8GB1RwMBps
Surface area of cone
From the above diagram.
The area of the sector = area of the curved surface of the cone
Length of arc AB = circumference of the circular base of the cone
Curved surface area of cone = θ/3600 x πl2
Also, θ/3600 x 2πl= 2πr
θ/3600 = r/l
Curved surface area of cone = r/l x πl2 =
Total surface area of cone = πrl + πr2
= πr(l + r).
Example 1: Calculate in terms of π, the total surface area of a cone of base diameter 12cm and height 10cm.
Solution
Using Pythagoras rule,
l2 = 102 + 62
l2 = 100 + 36
= 136
l= √136
Total surface area
= πrl + πr2
= πr(l + r).
= 6π(√136 + 6)cm2.
https://youtu.be/rd8tbD2eekM
2. A 2160 sector of a circle of radius 5cm is bent to form a cone. Find the radius of the base of the cone and its vertical angle.
Solution
Radius = r
Vertical angle = 2α
Circumference of base of cone = length of arc of sector
Substituting values
216/3600 x 2π x 5= 2πr
r = 216/3600 x 2π x 5 x 1/2π
r = 216/72 = 3
∴ r = 3
Sin α = 3/5 = 0.6000
α = 36.870
2α = 36.870 x 2
= 73.740
Radius of base = 3cm
Vertical angle = 73.70 (to 0.10)
https://youtu.be/gdg4czPlA84
EVALUATION
Find the curved surface and total surface areas of a closed cone of height 4cm and base radius 3cm (Take π= 22/7)
SURFACE AREA OF CUBES, CUBOIDS AND CYLINDER
CUBE: since a cube is solid with six faces
Area of one face is a x a = a2.
Total area = a x a x 6 = 6a2.
Surface Area = 6a2 units
Example 1:
What is the surface area of a cube of edges 12cm.
Solution
A = 6 x 12 x 12
= 6 x 144
= 864cm2
https://youtu.be/5xdKfAwDKWA
Example 2: calculate the surface area of a cube of edge 11cm.
Solution
A = 6a2
= 6 x 11 x 11
= 6 x 121
= 726cm2
CUBOIDS
Surface area of a cuboids = 2(lb + bh + lh)
Example: Given a cuboid of edges 3cm, 5cm, 8cm. calculate the surface area.
Solution
Surface area of a cuboids = 2(lb + bh + lh)
= 2(8 x 5 +8 x 3 + 5 x 3)
= 2(40 + 24 + 15)
= 2 x 79
= 158cm2
https://youtu.be/hi2QMbROemk
CYLINDER
Area of cylinder with one end closed = 2πrh + πr2
= πr(2h + r)
https://youtu.be/7mdSmvvOmF4
If two ends are closed.
Area = 2πrh + 2πr2
= 2πr(h + r), where r = radius of cylinder, h = height of cylinder
Example:
Find the surface area of a cylinder with height 14cm and base radius of 7cm consider all the three cases namely;
Open ended cylinder
One end open cylinder
Both ends closed cylinder
Solution
Open ended cylinder
Area = 2πrh
r = 7cm, h = 14cm
A = 2 x 22/2 x 7 x 14
= 44 x 14
= 616cm2
Open end open cylinder
Area = 2πrh + πr2
= (616 + 22/2 x 72 )cm2
= (616 + 154)cm2
= 770cm2.
https://youtu.be/PhA-7AE6MIU
Both ends closed cylinder
Area = 2πr(h + r)
= 2 x 22/2 x 7(14 + 7)cm2
= 44/2 x 7 x 21
= 924cm2.
https://youtu.be/gK9OgZ6eLx0
Evaluation
Calculate the surface area of a hollow cylinder which is closed at one end, if the base radius is 3.5cm and the height is 8cm. Take (Take π= 22/7)
Cylinder container, closed at both ends, has a radius of 7cm and height 5cm. Find the total surface area of the container (Take π= 22/7)
SURFACE AREA OF PRISMS AND PYRAMIDS
PRISMS: The total surface area is the sum of the surface of the five faces as shown below
https://youtu.be/xCdxURXMdFY
Solution
Area of ABCD
=20 x 10cm2
= 200cm2
Area of AFED = 20 x 6cm2
= 120cm2
Area of BCEF = 20 x 8cm2
= 160cm2
Area of ∆ABF = area of ∆DCE
= 1/2 X 8 X 6cm2
= 24cm2
Total surface area of prism = (200 + 120 + 160 + 24 + 24)cm2
= 528cm2
In the case of a rectangular prism or cuboid, the total surface area is determined by finding and summing up the areas of the four rectangular faces and the two end faces.
https://youtu.be/ZE41ljhfq2o
Examples 2: find the total surface area of the solid.
Area of A = 5 x 3cm2 = 15cm2
Area of B = 8 x 5cm2 = 40cm2
Area of C = 8 x 3cm2 = 24cm2
Area of D= 8 x 5cm2 = 40cm2
Area of E = 8 x 3cm2 = 24cm2
Area of F = 5 x 3cm2 = 15cm2
∴ Total surface area = (15 + 40 + 24 + 40 + 24 + 15)cm2
= 158cm2.
https://youtu.be/VwdMbDpMab4
PYRAMIDS: the total surface area of a pyramid is found by summing up areas of the common shapes that make up the pyramid.
Example 3:
Find the total surface area of a right pyramid with a rectangular base 6cm by 10cm, a height of 8.3cm and a slant edge of 10.14cm.
Solution
The total surface area is the sum of the surface areas of the five faces:
ABCD, VAB, VDC, and VAD
ABCD is a rectangle so AB = DC and AD = BC
If AB = 10cm, then DC = AB = 10cm
If AD = 6cm, then BC = AD = 6cm
Hence, area of ABCD = 10 x 6 = 60cm2
Area of ∆VAB = Area of ∆VDU and ∆VBC = Area of ∆VAD
Area of ∆VAB = 1/2 x AB x VP
By using Pythagoras rule
VP2 = VB2 – PB2
= (10.14)2 - 52
= 102.8 – 25
= 77.8
VP = √77.8 = 8.82cm
Area of ∆VAB = 1/2 x AB x VP
= 1/2 x 10 x 8.82
= 5 x 8.82
= 41.10cm2
Area of ∆VDU
Area of ∆VBC = 1/2 x AB x VF
But VF2 = VC2 – FC2
= (10.14)2 – 32
= 102.8 – 9
= 93.8
VF = √93.8 = 9.69cm
Area of ∆VBC = 1/2 x BC x VF
= 1/2 x 6 x 9.69
= 3 x 9.69
= 29.07cm2
Area of∆VAD
Area of rectangular base ABCD = 10 x 6 = 60cm2.
Total surface area of the pyramid = [(41.10 + 41.10 + 29.07 + 29.07 + 60)]
=(82.20 + 58.14 + 60)cm2
= 200.34cm2
https://youtu.be/vCf2yK4tzkk
Evaluation
Find the length of the slant edge of a right pyramid with
A rectangular base 3cm by 5cm and a height of 4.2cm
A rectangular base 6cm by 4.5cm and a height of 3.4cm
Find the surface area of a triangular prism 10.8cm long and having a triangular face of dimensions 8.8cm by 5.7cm by 6.8cm.
VOLUME, CUBES, CUBOIDS, CYLINDER, CONE, PRISMS AND PYRAMIDS
The Volume of cube
= s2 x s = s3
https://youtu.be/qJwecTgce6c
Example 1:
Find the volume of a cube whose edges are 7cm each.
Solution
Volume = s2 x s = s3
= 7 x 7 x 7cm3
= 49 x 7cm3
= 343cm3
https://youtu.be/a--mBK9XjLU
Volume of cuboids
The volume of a cuboid = Area x height
= lb x h
= lbhcm3
https://youtu.be/fw--FxHF6OA
Example 2:
What is the volume of a cuboids if
Height = 6cm, breadth = 10cm, length = 13cm?
Area of cross – section is 105m2 and height = 5cm?
Area of cross-section has the same numerical value with the height = 16cm?
Solution
Volume (V) = lbh = 13 x 6 x 10 = 780cm3
Volume = Ah = 10 x 5 = 525cm3
V = A x h
= 16 x 16
= 256cm3
https://youtu.be/PhLc6B5WXtc
VOLUME OF CYLINDER
With the cylinder the cross section is a circle.
Area = πr2, where is the radius of the circular base. Height = h
∴ Volume = Area x height
= πr2x h
= πr2hcm3
Example 4: what is the volume of a cylinder if the radius is 5cm and height is 20cm?
Solution
Volume = area of circular base x height
= πr2 x h
= 22/7 X 52 X 20cm2
= 22/7 x 5 x 5 x 20cm2
= 1571.43cm3
https://youtu.be/fxTsG4qkz1U
Volume of Cone
The volume of cone = 1/3 x volume of cylinder
= 1/3 πr2h where r = radius of the cone and h = height of the cone.
Example 5:
Find the volume of a cone if the perpendicular height is 9cm and radius 4cm.
Solution
Volume of cone = 1/3 πr2h
r = 4cm, h = 9cm
V = 1/3 x 22/7 x 4 x 4 x 9cm3
= 3168/21cm3
= 150.86cm3.
https://youtu.be/6ArZQFFKDHY
Volume of prisms
Volume of prisms = area of cross – section x distance between the end faces.
Example 6:
Find the volume of the triangular prism in the diagram below
Solution
Volume of prisms = area of cross – section x distance between the end faces.
Here, end face = ∆CDE
Area of ∆CDE = 1/2CE x DG
= 1/2 x 5 x 2.6cm2
= 13.5/2cm2
= 6.75cm2
Volume of prism = Area of ∆CDE x BC
= 6.75 x 4.5cm3
30.375cm3
Volume of triangular prism
= 30.4cm3
https://youtu.be/P79EC9qHXEg
Volume of pyramids
Volume of pyramid = 1/3 x base area x perpendicular height
Example 7:
Find the volume of right pyramid with vertex V and a rectangular base measuring 5.4cm by 4cm and a height of 8cm.
Solution
Volume of pyramid = 1/3 x base area x perpendicular height
= Volume of pyramid = 1/3 x 5.4 x 4 x 8cm3
Volume of pyramid = 1/3 x 5.4 x 32cm3
Volume of pyramid = 172.8/3cm3
= 57.6cm3
https://youtu.be/e7-am8JtREI
Evaluation
A right pyramid has a square base of side 8cm. The height of the pyramid is half the side of the square. Find the length of the sloping edge.
A rectangular tank is 76cm long, 50cm wide and 40cm high. How many litres of water can it hold? (WAEC)
Calculate in terms of π the total surface area of a cylinder of radius 3cm and height 4cm. (WAEC)
A conical container has radius 7cm and height 5cm. Calculate the volume of the container(WAEC)
The volume of a tank is 4.913cm3, what is the length of its edge.
What is the edge of a cube whose volume is 729cm3
A rectangular tank is 76cm long, 50cm wide and 40cm height. How many litres of water can it hold? (WAEC)
ASSIGNMENT
Solve the questions from New General Mathematics for SS1. Page 185.Ex 15a Qs 1a – e and 22.
REFERENCE TEXTS:
New School Mathematics for senior secondary schools by Mrs Maria – David (e tal)
Man Mathematics for senior secondary schools1
New General mathematics for senior secondary schools 1
Content:
Relation between the sector of a circle and the surface area of a cone.
Surface area of cubes, cuboids and cylinder
Surface area of prisms and pyramids
Volume of cubes, cuboids, cylinder, cone,
Prisms and pyramids
Relation between the sector of a circle and the surface area of a cone.
If a sector of a circle AOB is cut and folded into a come as shown in the diagram below.
The arc AB becomes the circumference of the base of the cone. The radius R becomes the slant edge l of the cone.
∴ Arc AB = 2πr
https://youtu.be/S8GB1RwMBps
Surface area of cone
From the above diagram.
The area of the sector = area of the curved surface of the cone
Length of arc AB = circumference of the circular base of the cone
Curved surface area of cone = θ/3600 x πl2
Also, θ/3600 x 2πl= 2πr
θ/3600 = r/l
Curved surface area of cone = r/l x πl2 =
Total surface area of cone = πrl + πr2
= πr(l + r).
Example 1: Calculate in terms of π, the total surface area of a cone of base diameter 12cm and height 10cm.
Solution
Using Pythagoras rule,
l2 = 102 + 62
l2 = 100 + 36
= 136
l= √136
Total surface area
= πrl + πr2
= πr(l + r).
= 6π(√136 + 6)cm2.
https://youtu.be/rd8tbD2eekM
2. A 2160 sector of a circle of radius 5cm is bent to form a cone. Find the radius of the base of the cone and its vertical angle.
Solution
Radius = r
Vertical angle = 2α
Circumference of base of cone = length of arc of sector
Substituting values
216/3600 x 2π x 5= 2πr
r = 216/3600 x 2π x 5 x 1/2π
r = 216/72 = 3
∴ r = 3
Sin α = 3/5 = 0.6000
α = 36.870
2α = 36.870 x 2
= 73.740
Radius of base = 3cm
Vertical angle = 73.70 (to 0.10)
https://youtu.be/gdg4czPlA84
EVALUATION
Find the curved surface and total surface areas of a closed cone of height 4cm and base radius 3cm (Take π= 22/7)
SURFACE AREA OF CUBES, CUBOIDS AND CYLINDER
CUBE: since a cube is solid with six faces
Area of one face is a x a = a2.
Total area = a x a x 6 = 6a2.
Surface Area = 6a2 units
Example 1:
What is the surface area of a cube of edges 12cm.
Solution
A = 6 x 12 x 12
= 6 x 144
= 864cm2
https://youtu.be/5xdKfAwDKWA
Example 2: calculate the surface area of a cube of edge 11cm.
Solution
A = 6a2
= 6 x 11 x 11
= 6 x 121
= 726cm2
CUBOIDS
Surface area of a cuboids = 2(lb + bh + lh)
Example: Given a cuboid of edges 3cm, 5cm, 8cm. calculate the surface area.
Solution
Surface area of a cuboids = 2(lb + bh + lh)
= 2(8 x 5 +8 x 3 + 5 x 3)
= 2(40 + 24 + 15)
= 2 x 79
= 158cm2
https://youtu.be/hi2QMbROemk
CYLINDER
Area of cylinder with one end closed = 2πrh + πr2
= πr(2h + r)
https://youtu.be/7mdSmvvOmF4
If two ends are closed.
Area = 2πrh + 2πr2
= 2πr(h + r), where r = radius of cylinder, h = height of cylinder
Example:
Find the surface area of a cylinder with height 14cm and base radius of 7cm consider all the three cases namely;
Open ended cylinder
One end open cylinder
Both ends closed cylinder
Solution
Open ended cylinder
Area = 2πrh
r = 7cm, h = 14cm
A = 2 x 22/2 x 7 x 14
= 44 x 14
= 616cm2
Open end open cylinder
Area = 2πrh + πr2
= (616 + 22/2 x 72 )cm2
= (616 + 154)cm2
= 770cm2.
https://youtu.be/PhA-7AE6MIU
Both ends closed cylinder
Area = 2πr(h + r)
= 2 x 22/2 x 7(14 + 7)cm2
= 44/2 x 7 x 21
= 924cm2.
https://youtu.be/gK9OgZ6eLx0
Evaluation
Calculate the surface area of a hollow cylinder which is closed at one end, if the base radius is 3.5cm and the height is 8cm. Take (Take π= 22/7)
Cylinder container, closed at both ends, has a radius of 7cm and height 5cm. Find the total surface area of the container (Take π= 22/7)
SURFACE AREA OF PRISMS AND PYRAMIDS
PRISMS: The total surface area is the sum of the surface of the five faces as shown below
https://youtu.be/xCdxURXMdFY
Solution
Area of ABCD
=20 x 10cm2
= 200cm2
Area of AFED = 20 x 6cm2
= 120cm2
Area of BCEF = 20 x 8cm2
= 160cm2
Area of ∆ABF = area of ∆DCE
= 1/2 X 8 X 6cm2
= 24cm2
Total surface area of prism = (200 + 120 + 160 + 24 + 24)cm2
= 528cm2
In the case of a rectangular prism or cuboid, the total surface area is determined by finding and summing up the areas of the four rectangular faces and the two end faces.
https://youtu.be/ZE41ljhfq2o
Examples 2: find the total surface area of the solid.
Area of A = 5 x 3cm2 = 15cm2
Area of B = 8 x 5cm2 = 40cm2
Area of C = 8 x 3cm2 = 24cm2
Area of D= 8 x 5cm2 = 40cm2
Area of E = 8 x 3cm2 = 24cm2
Area of F = 5 x 3cm2 = 15cm2
∴ Total surface area = (15 + 40 + 24 + 40 + 24 + 15)cm2
= 158cm2.
https://youtu.be/VwdMbDpMab4
PYRAMIDS: the total surface area of a pyramid is found by summing up areas of the common shapes that make up the pyramid.
Example 3:
Find the total surface area of a right pyramid with a rectangular base 6cm by 10cm, a height of 8.3cm and a slant edge of 10.14cm.
Solution
The total surface area is the sum of the surface areas of the five faces:
ABCD, VAB, VDC, and VAD
ABCD is a rectangle so AB = DC and AD = BC
If AB = 10cm, then DC = AB = 10cm
If AD = 6cm, then BC = AD = 6cm
Hence, area of ABCD = 10 x 6 = 60cm2
Area of ∆VAB = Area of ∆VDU and ∆VBC = Area of ∆VAD
Area of ∆VAB = 1/2 x AB x VP
By using Pythagoras rule
VP2 = VB2 – PB2
= (10.14)2 - 52
= 102.8 – 25
= 77.8
VP = √77.8 = 8.82cm
Area of ∆VAB = 1/2 x AB x VP
= 1/2 x 10 x 8.82
= 5 x 8.82
= 41.10cm2
Area of ∆VDU
Area of ∆VBC = 1/2 x AB x VF
But VF2 = VC2 – FC2
= (10.14)2 – 32
= 102.8 – 9
= 93.8
VF = √93.8 = 9.69cm
Area of ∆VBC = 1/2 x BC x VF
= 1/2 x 6 x 9.69
= 3 x 9.69
= 29.07cm2
Area of∆VAD
Area of rectangular base ABCD = 10 x 6 = 60cm2.
Total surface area of the pyramid = [(41.10 + 41.10 + 29.07 + 29.07 + 60)]
=(82.20 + 58.14 + 60)cm2
= 200.34cm2
https://youtu.be/vCf2yK4tzkk
Evaluation
Find the length of the slant edge of a right pyramid with
A rectangular base 3cm by 5cm and a height of 4.2cm
A rectangular base 6cm by 4.5cm and a height of 3.4cm
Find the surface area of a triangular prism 10.8cm long and having a triangular face of dimensions 8.8cm by 5.7cm by 6.8cm.
VOLUME, CUBES, CUBOIDS, CYLINDER, CONE, PRISMS AND PYRAMIDS
The Volume of cube
= s2 x s = s3
https://youtu.be/qJwecTgce6c
Example 1:
Find the volume of a cube whose edges are 7cm each.
Solution
Volume = s2 x s = s3
= 7 x 7 x 7cm3
= 49 x 7cm3
= 343cm3
https://youtu.be/a--mBK9XjLU
Volume of cuboids
The volume of a cuboid = Area x height
= lb x h
= lbhcm3
https://youtu.be/fw--FxHF6OA
Example 2:
What is the volume of a cuboids if
Height = 6cm, breadth = 10cm, length = 13cm?
Area of cross – section is 105m2 and height = 5cm?
Area of cross-section has the same numerical value with the height = 16cm?
Solution
Volume (V) = lbh = 13 x 6 x 10 = 780cm3
Volume = Ah = 10 x 5 = 525cm3
V = A x h
= 16 x 16
= 256cm3
https://youtu.be/PhLc6B5WXtc
VOLUME OF CYLINDER
With the cylinder the cross section is a circle.
Area = πr2, where is the radius of the circular base. Height = h
∴ Volume = Area x height
= πr2x h
= πr2hcm3
Example 4: what is the volume of a cylinder if the radius is 5cm and height is 20cm?
Solution
Volume = area of circular base x height
= πr2 x h
= 22/7 X 52 X 20cm2
= 22/7 x 5 x 5 x 20cm2
= 1571.43cm3
https://youtu.be/fxTsG4qkz1U
Volume of Cone
The volume of cone = 1/3 x volume of cylinder
= 1/3 πr2h where r = radius of the cone and h = height of the cone.
Example 5:
Find the volume of a cone if the perpendicular height is 9cm and radius 4cm.
Solution
Volume of cone = 1/3 πr2h
r = 4cm, h = 9cm
V = 1/3 x 22/7 x 4 x 4 x 9cm3
= 3168/21cm3
= 150.86cm3.
https://youtu.be/6ArZQFFKDHY
Volume of prisms
Volume of prisms = area of cross – section x distance between the end faces.
Example 6:
Find the volume of the triangular prism in the diagram below
Solution
Volume of prisms = area of cross – section x distance between the end faces.
Here, end face = ∆CDE
Area of ∆CDE = 1/2CE x DG
= 1/2 x 5 x 2.6cm2
= 13.5/2cm2
= 6.75cm2
Volume of prism = Area of ∆CDE x BC
= 6.75 x 4.5cm3
30.375cm3
Volume of triangular prism
= 30.4cm3
https://youtu.be/P79EC9qHXEg
Volume of pyramids
Volume of pyramid = 1/3 x base area x perpendicular height
Example 7:
Find the volume of right pyramid with vertex V and a rectangular base measuring 5.4cm by 4cm and a height of 8cm.
Solution
Volume of pyramid = 1/3 x base area x perpendicular height
= Volume of pyramid = 1/3 x 5.4 x 4 x 8cm3
Volume of pyramid = 1/3 x 5.4 x 32cm3
Volume of pyramid = 172.8/3cm3
= 57.6cm3
https://youtu.be/e7-am8JtREI
Evaluation
A right pyramid has a square base of side 8cm. The height of the pyramid is half the side of the square. Find the length of the sloping edge.
A rectangular tank is 76cm long, 50cm wide and 40cm high. How many litres of water can it hold? (WAEC)
Calculate in terms of π the total surface area of a cylinder of radius 3cm and height 4cm. (WAEC)
A conical container has radius 7cm and height 5cm. Calculate the volume of the container(WAEC)
The volume of a tank is 4.913cm3, what is the length of its edge.
What is the edge of a cube whose volume is 729cm3
A rectangular tank is 76cm long, 50cm wide and 40cm height. How many litres of water can it hold? (WAEC)
ASSIGNMENT
Solve the questions from New General Mathematics for SS1. Page 185.Ex 15a Qs 1a – e and 22.
REFERENCE TEXTS:
New School Mathematics for senior secondary schools by Mrs Maria – David (e tal)
Man Mathematics for senior secondary schools1
New General mathematics for senior secondary schools 1
WEEK 6
TOPIC: Mensuration
Content:
Surface area of frustum of cone and pyramids
Volume of frustum of a cone and pyramid
Surface area and volume of compound shapes.
TOTAL SURFACE AREA OF FRUSTUM OF CONE AND PYRAMIDS
A frustum is the remaining part of cone or pyramid when the top part is cut off as shown below. Daily examples of frustums are buckets lamps shades etc
Frustum of a cone
In surface area of the frustrum of a pyramid, we sum up all areas of the surfaces that make up the frustum.
For frustum of a cone.
Total surface area of a closed frustum = π(height x sum of radii) + area of top and base circles.
https://youtu.be/uhLrZvzKV9o
Example 1:
Find the total surface area of a bucket 36cm in diameter at the top and 24cm at the bottom.
The depth of the bucket is 30cm.
Solution
The total surface area of bucket = sum of curved part + area of bottom circle
= π x height x (sum of radii) + area of bottom
= π x 30 x (18 + 12) + π122
= π x 30 x 30 + π(144)
= 900π + 144π
= 1044π2
https://youtu.be/jHQhs9SfZM4
Example 2:
Find in cm2, the area of material required for a lamp shade in the form of a frustum of a cone of which the top and bottom diameters are 20cm and 30cm respectively and the vertical height is 12cm.
Solution
By Pythagoras’s theorem, y = 26 and z = 13
z2 = 122 + 52
= 144 + 25
= 169
Z = √169
= 13
Surface area of frustum
= π x 15 x 39 - π x 10 x 26cm2
= 13π(45 - 20)cm2
= 13π x 25cm2
= 1021cm2
Area of material required = 1021cm2. (3 s. f)
Evaluation
A bucket is 12cm in diameter at the top 8cm in diameter at the bottom and 4cm deep. Calculate its volume in cm3 in terms π. (JAMB)
SURFACE AREA OF FRUSTRUM OF PYRAMIDS
Example:
A frustum of a pyramid consists of a square base of length 10cm and a top square of length 7cm height of the frustum is 6cm. Calculate to the nearest whole number in the surface area.
Solution
Diagonals of the top and bottom
IPRI2 = IPQI2 + IQRI2
= 102 + 102
= 100 + 100
= 200
IPRI = √200
= 14.14
IYRI = 14.14/2
= 7.07cm
= 7.1cm
Diagonal AC
IACI2 = IABI2 + IBCI2
= 72 + 72
= 49 + 49
= 98
IACI2 = √98
= 9.89
BUT CX = 9.89/2
= 4.945
= 4.95cm
To calculate the height of the added pyramid.
Let the heights of the added pyramid be h by similar ∆
h/4.95 = (h+6)/7.07
7.07h = 4.95h + 29.7
7.07h – 4.95h = 29.7
2.12h = 29.7
= 29.7/2.12
= 14cm
∴height of bigger pyramid
= 13.8 + 6cm
= 19.8cm
To calculate the slant side of the smaller and bigger ∆s
Bigger ∆
IVRI2 = 7.12 + 19.82
= 50.41 + 392.04
= 442.45
∴IVRI2 = √442.45
= 21.04
= 21cm
https://youtu.be/jHQhs9SfZM4
Smaller ∆:
= 4.95 + 13.82
= 24.50 + 190.44
= 214.94
IVCI = √214.94
= 14.66cm
= 14.7cm
To find the surface area of frustum of pyramids using Hero’s formula for finding the area of the triangular face.
S = √((s(s-a)(s-b)(s-c))
Where s = 1/2(a + b + c) and a, b and c are sides of the triangle
s = 1/2(21+ 21 + 10)
= 1/2 x 52
= 26
Area of a ∆face of big pyramid = √((s(s-a)(s-b)(s-c))
S = √((26(26-21)(26-21)(26-10))
= √((26(5)(5)(16))
= √10400
= 101.98
= 102
Area of 4 ∆faces of the pyramid
= 4 x 101.98cm2
= 407.92cm2
= 408cm2
Area of 4 ∆faces of the small pyramid
s = 1/2(a + b + c)
s = 1/2(14.7 + 14.7 + 7)
s = 1/2 x 36.4
= 18.2
Area of a face = √((s(s-a)(s-b)(s-c))
= √((18.2(18.2-14.7)(18.2-14.7)(18.2-7))
= √((18.2(3.5)(3.5)(11.2))
= √2497.04
= 49.97cm2
Area of a 4face = 4 x 49.97
= 199.88
= 200cm2
Area of the base of the frustum = l x b
= 10 x 10
= 100cm2
Area of the top of the frustum = l x b
= 7 x 7
49cm2
Total surface area of the frustum = [(Area of the 4∆faces of the big pyramid – Area of the 4∆faces of the small pyramid )] + Bottom + Top area
= [(407.2 – 199.88) + 100 + 49]
= 208.04 + 149
= 357.04cm2
= 375cm2
https://youtu.be/bN9jHdWJzaI
Evaluation
A pyramid is on a square base of 25m side and 25m high. The top of the pyramids 10m high was cut off. Find the surface area of the four sides of the frustum formed.
VOLUME OF FRUSTUM OF CONE AND PYRAMIDS
Volume of frustum = volume of the full cone/pyramid – volume of teh part cut off.
Example:
A square cone 24cm high and of 18cm diameter at its base is cut off at half of its height. Calculate the volume of the remaining frustum. Leave π in your answer.
Solution
Volume of original cone = 1/3πr2h
= 1/3π x 92 x 24
= 1/3π x 81 x 24
= 27 π(24)
To find the radius r of the base of the cut off part, we use similarity of ∆s = 12/r = 24/9
r = (12 x 9)/24 = 9/2 = 41/2cm
Volume of small cone = 1/3π(9/2)2 x 12
= 81π
Hence volume of frustum (i) – (ii)
= 27 x 24π - 81π
= (81 x 8) π - 81π
= 81π(8 - 1)
= 567 πcm3
Example 2:
Find the volume of the frustum of a pyramid with 30cm square top and 50cm square base and height 20cm.Solution
If x is the height of the cut – off part and 15cm is half the side of its square base,
Then
x/(x+20) = 15/25
i.e 25x = 300 + 15x
10x = 300
x = 30.
Volume of entire pyramid
= 1/3(base x height)
= 1/3(base x height)
= 1/3(50 x 50) x (20 + 60)
= 1/3(2500 x 80)
= 1/3(200,000)cm3
But volume of top pyramid
= 1/3(base x height)
= 1/3(30 x 30 x 60)
= 1/3(900 x 60)
= 1/3(54,000)cm3
Volume of frustum = 1/3(200,000 – 54,000)
= 1/3(146,000)cm3
=48666 2/3cm3
https://youtu.be/Ig-Tv4NEqg4
Evaluation
What is the capacity of a bucket that is 42cm deep and inner radii of the base and topmost part of the bucket are 12cm by 20cm respectively?
Surface area of a Hut
A hut has:
The curved surface of cone
The curved surface of cylinder
The circular base of cylinder
Total surface area of a hut = 2πrh + πrl + π r2
= πr(2h + l + r)
Example 2:
A machine part is made up of a cylinder and a cone on each end, the dimension of which are shown below. Calculate the total surface area of the machine.
Solution
Surface area of 1st cone = πrl
By Pythagoras rule.
L2 = (62 + 42)cm2
= (36 + 16)cm2
= 52cm
l = √52
the surface area of the come
= π x 4cm x √52cm
= 4π√52cm2
Surface area of 2nd cone = πrl
l2 = 92 + 42
= 81 + 16
= 97cm
L = √97cm2
Area = 4π√97cm2
Therefore total surface area of the machine = surface area of cylinder + surface area of 1st cone + surface area of 2nd
= 64πcm2 + 4π√52cm2 + 4π√97cm2
= 4π(16+ √52+ √97)cm2
https://youtu.be/OteqamsO-Vc
Evaluation
Find the surface area of the bullet in the figure below, using those dimensions
https://youtu.be/ntoyqFJG52I
GENERAL EVALUATION
A cone of radius 7cm is 42cm deep. If the cone is 3/4 filled with water. How deep is the water in the cone (Take π = 22/7)
A square base of a pyramid of side 3cm has height 8cm. If the pyramid is cut into two parts by a plane parallel to the base midway between the base and the vertex, calculate the volumes of the two sections to the nearest centimetre.
ASSIGNMENT
New General Mathematics for senior secondary schools 1 page 191. Ex 15c Qs 1 – 3
REFERENCE TEXT:
New School Mathematics for senior secondary school 1.
Man Mathematics for Senior secondary schools
New General Maths for Senior secondary schools 1
Content:
Surface area of frustum of cone and pyramids
Volume of frustum of a cone and pyramid
Surface area and volume of compound shapes.
TOTAL SURFACE AREA OF FRUSTUM OF CONE AND PYRAMIDS
A frustum is the remaining part of cone or pyramid when the top part is cut off as shown below. Daily examples of frustums are buckets lamps shades etc
Frustum of a cone
In surface area of the frustrum of a pyramid, we sum up all areas of the surfaces that make up the frustum.
For frustum of a cone.
Total surface area of a closed frustum = π(height x sum of radii) + area of top and base circles.
https://youtu.be/uhLrZvzKV9o
Example 1:
Find the total surface area of a bucket 36cm in diameter at the top and 24cm at the bottom.
The depth of the bucket is 30cm.
Solution
The total surface area of bucket = sum of curved part + area of bottom circle
= π x height x (sum of radii) + area of bottom
= π x 30 x (18 + 12) + π122
= π x 30 x 30 + π(144)
= 900π + 144π
= 1044π2
https://youtu.be/jHQhs9SfZM4
Example 2:
Find in cm2, the area of material required for a lamp shade in the form of a frustum of a cone of which the top and bottom diameters are 20cm and 30cm respectively and the vertical height is 12cm.
Solution
By Pythagoras’s theorem, y = 26 and z = 13
z2 = 122 + 52
= 144 + 25
= 169
Z = √169
= 13
Surface area of frustum
= π x 15 x 39 - π x 10 x 26cm2
= 13π(45 - 20)cm2
= 13π x 25cm2
= 1021cm2
Area of material required = 1021cm2. (3 s. f)
Evaluation
A bucket is 12cm in diameter at the top 8cm in diameter at the bottom and 4cm deep. Calculate its volume in cm3 in terms π. (JAMB)
SURFACE AREA OF FRUSTRUM OF PYRAMIDS
Example:
A frustum of a pyramid consists of a square base of length 10cm and a top square of length 7cm height of the frustum is 6cm. Calculate to the nearest whole number in the surface area.
Solution
Diagonals of the top and bottom
IPRI2 = IPQI2 + IQRI2
= 102 + 102
= 100 + 100
= 200
IPRI = √200
= 14.14
IYRI = 14.14/2
= 7.07cm
= 7.1cm
Diagonal AC
IACI2 = IABI2 + IBCI2
= 72 + 72
= 49 + 49
= 98
IACI2 = √98
= 9.89
BUT CX = 9.89/2
= 4.945
= 4.95cm
To calculate the height of the added pyramid.
Let the heights of the added pyramid be h by similar ∆
h/4.95 = (h+6)/7.07
7.07h = 4.95h + 29.7
7.07h – 4.95h = 29.7
2.12h = 29.7
= 29.7/2.12
= 14cm
∴height of bigger pyramid
= 13.8 + 6cm
= 19.8cm
To calculate the slant side of the smaller and bigger ∆s
Bigger ∆
IVRI2 = 7.12 + 19.82
= 50.41 + 392.04
= 442.45
∴IVRI2 = √442.45
= 21.04
= 21cm
https://youtu.be/jHQhs9SfZM4
Smaller ∆:
= 4.95 + 13.82
= 24.50 + 190.44
= 214.94
IVCI = √214.94
= 14.66cm
= 14.7cm
To find the surface area of frustum of pyramids using Hero’s formula for finding the area of the triangular face.
S = √((s(s-a)(s-b)(s-c))
Where s = 1/2(a + b + c) and a, b and c are sides of the triangle
s = 1/2(21+ 21 + 10)
= 1/2 x 52
= 26
Area of a ∆face of big pyramid = √((s(s-a)(s-b)(s-c))
S = √((26(26-21)(26-21)(26-10))
= √((26(5)(5)(16))
= √10400
= 101.98
= 102
Area of 4 ∆faces of the pyramid
= 4 x 101.98cm2
= 407.92cm2
= 408cm2
Area of 4 ∆faces of the small pyramid
s = 1/2(a + b + c)
s = 1/2(14.7 + 14.7 + 7)
s = 1/2 x 36.4
= 18.2
Area of a face = √((s(s-a)(s-b)(s-c))
= √((18.2(18.2-14.7)(18.2-14.7)(18.2-7))
= √((18.2(3.5)(3.5)(11.2))
= √2497.04
= 49.97cm2
Area of a 4face = 4 x 49.97
= 199.88
= 200cm2
Area of the base of the frustum = l x b
= 10 x 10
= 100cm2
Area of the top of the frustum = l x b
= 7 x 7
49cm2
Total surface area of the frustum = [(Area of the 4∆faces of the big pyramid – Area of the 4∆faces of the small pyramid )] + Bottom + Top area
= [(407.2 – 199.88) + 100 + 49]
= 208.04 + 149
= 357.04cm2
= 375cm2
https://youtu.be/bN9jHdWJzaI
Evaluation
A pyramid is on a square base of 25m side and 25m high. The top of the pyramids 10m high was cut off. Find the surface area of the four sides of the frustum formed.
VOLUME OF FRUSTUM OF CONE AND PYRAMIDS
Volume of frustum = volume of the full cone/pyramid – volume of teh part cut off.
Example:
A square cone 24cm high and of 18cm diameter at its base is cut off at half of its height. Calculate the volume of the remaining frustum. Leave π in your answer.
Solution
Volume of original cone = 1/3πr2h
= 1/3π x 92 x 24
= 1/3π x 81 x 24
= 27 π(24)
To find the radius r of the base of the cut off part, we use similarity of ∆s = 12/r = 24/9
r = (12 x 9)/24 = 9/2 = 41/2cm
Volume of small cone = 1/3π(9/2)2 x 12
= 81π
Hence volume of frustum (i) – (ii)
= 27 x 24π - 81π
= (81 x 8) π - 81π
= 81π(8 - 1)
= 567 πcm3
Example 2:
Find the volume of the frustum of a pyramid with 30cm square top and 50cm square base and height 20cm.Solution
If x is the height of the cut – off part and 15cm is half the side of its square base,
Then
x/(x+20) = 15/25
i.e 25x = 300 + 15x
10x = 300
x = 30.
Volume of entire pyramid
= 1/3(base x height)
= 1/3(base x height)
= 1/3(50 x 50) x (20 + 60)
= 1/3(2500 x 80)
= 1/3(200,000)cm3
But volume of top pyramid
= 1/3(base x height)
= 1/3(30 x 30 x 60)
= 1/3(900 x 60)
= 1/3(54,000)cm3
Volume of frustum = 1/3(200,000 – 54,000)
= 1/3(146,000)cm3
=48666 2/3cm3
https://youtu.be/Ig-Tv4NEqg4
Evaluation
What is the capacity of a bucket that is 42cm deep and inner radii of the base and topmost part of the bucket are 12cm by 20cm respectively?
Surface area of a Hut
A hut has:
The curved surface of cone
The curved surface of cylinder
The circular base of cylinder
Total surface area of a hut = 2πrh + πrl + π r2
= πr(2h + l + r)
Example 2:
A machine part is made up of a cylinder and a cone on each end, the dimension of which are shown below. Calculate the total surface area of the machine.
Solution
Surface area of 1st cone = πrl
By Pythagoras rule.
L2 = (62 + 42)cm2
= (36 + 16)cm2
= 52cm
l = √52
the surface area of the come
= π x 4cm x √52cm
= 4π√52cm2
Surface area of 2nd cone = πrl
l2 = 92 + 42
= 81 + 16
= 97cm
L = √97cm2
Area = 4π√97cm2
Therefore total surface area of the machine = surface area of cylinder + surface area of 1st cone + surface area of 2nd
= 64πcm2 + 4π√52cm2 + 4π√97cm2
= 4π(16+ √52+ √97)cm2
https://youtu.be/OteqamsO-Vc
Evaluation
Find the surface area of the bullet in the figure below, using those dimensions
https://youtu.be/ntoyqFJG52I
GENERAL EVALUATION
A cone of radius 7cm is 42cm deep. If the cone is 3/4 filled with water. How deep is the water in the cone (Take π = 22/7)
A square base of a pyramid of side 3cm has height 8cm. If the pyramid is cut into two parts by a plane parallel to the base midway between the base and the vertex, calculate the volumes of the two sections to the nearest centimetre.
ASSIGNMENT
New General Mathematics for senior secondary schools 1 page 191. Ex 15c Qs 1 – 3
REFERENCE TEXT:
New School Mathematics for senior secondary school 1.
Man Mathematics for Senior secondary schools
New General Maths for Senior secondary schools 1
WEEK 7
TOPIC: Data Presentation
Sub-Topics: Revision on collection, tabulation and presentation of data.
Frequency distribution.
Statistics can be considered as a way or method used in the collection and organisation of data in order to interprets, predict and get other information required out of the data.
This data could be information about some people e.g..ages, weights, heights, examination scores, etc
https://youtu.be/T7KYO76DoOE
Collection of Data
Let us remind ourselves that there are two kinds of data namely:
(a) Discrete data (b) continuous data
Discrete data are those which are obtained by direct counting e.g numbers of persons born on a particular days of the week, month of the year, etc.
Continuous data are those which require measurement before counting e.g number of persons of some age, height, weight etc.
Activity 1
Collect and tabulate data on the days of the Week each member of the class was born.
Tabulation and Presentation of data.
For easy access to information, data are normally presented using frequency tables. This table marches each data with the number of times it appeared. The frequency table is prepared as follows:
1. Draw the columns
2. Represent the data given in the first column in ascending or descending order.
3. Represent each data with the use of a tally.
Raw data and frequency tables
When data is first collected and has not been organised in any way, it is called Raw Data. For example, in a game, a die was thrown several times. Below are the results of the scores.
2 3 4 4 2 1 3 2 6 5
3 2 1 1 2 5 2 1 4 4
6 5 6 1 6 5 4 5 4 3
6 5 5 3 5 2 1 4 5 2
4 5 4 6 3 1 5 6 6 5
The above is an example of raw data. One way we can organise the above data is to present it in a frequency distribution table (or frequency table for short) as shown below.
No of throws Tally Frequency
Note – Frequency table can be given with or without the tally column
https://youtu.be/3iWQQyTv77E
Example 2
In a test marked out of 10, a group of pupils obtained the following marks.
3 4 6 3 4 3 5 6 7 6
8 9 5 9 10 7 8 2 6 5
4 10 5 6 7 3 8 9 4 2
Prepare a frequency table for the distribution
https://youtu.be/L2krc0n8vOg
EVALUATION
New General Mathematics for senior secondary schools 1 page 170, questions 1 and 2
GENERAL EVALUATION
New General Mathematics for senior secondary schools 1 page 170, questions 3 -4
READING ASSIGNMENT
Study the presentation of data using the line graph, Bar graph and histogram
ASSIGNMENT
New General Mathematics for senior secondary schools 1 page 170, questions 5 - 6
Sub-Topics: Revision on collection, tabulation and presentation of data.
Frequency distribution.
Statistics can be considered as a way or method used in the collection and organisation of data in order to interprets, predict and get other information required out of the data.
This data could be information about some people e.g..ages, weights, heights, examination scores, etc
https://youtu.be/T7KYO76DoOE
Collection of Data
Let us remind ourselves that there are two kinds of data namely:
(a) Discrete data (b) continuous data
Discrete data are those which are obtained by direct counting e.g numbers of persons born on a particular days of the week, month of the year, etc.
Continuous data are those which require measurement before counting e.g number of persons of some age, height, weight etc.
Activity 1
Collect and tabulate data on the days of the Week each member of the class was born.
Tabulation and Presentation of data.
For easy access to information, data are normally presented using frequency tables. This table marches each data with the number of times it appeared. The frequency table is prepared as follows:
1. Draw the columns
2. Represent the data given in the first column in ascending or descending order.
3. Represent each data with the use of a tally.
Raw data and frequency tables
When data is first collected and has not been organised in any way, it is called Raw Data. For example, in a game, a die was thrown several times. Below are the results of the scores.
2 3 4 4 2 1 3 2 6 5
3 2 1 1 2 5 2 1 4 4
6 5 6 1 6 5 4 5 4 3
6 5 5 3 5 2 1 4 5 2
4 5 4 6 3 1 5 6 6 5
The above is an example of raw data. One way we can organise the above data is to present it in a frequency distribution table (or frequency table for short) as shown below.
No of throws Tally Frequency
Note – Frequency table can be given with or without the tally column
https://youtu.be/3iWQQyTv77E
Example 2
In a test marked out of 10, a group of pupils obtained the following marks.
3 4 6 3 4 3 5 6 7 6
8 9 5 9 10 7 8 2 6 5
4 10 5 6 7 3 8 9 4 2
Prepare a frequency table for the distribution
https://youtu.be/L2krc0n8vOg
EVALUATION
New General Mathematics for senior secondary schools 1 page 170, questions 1 and 2
GENERAL EVALUATION
New General Mathematics for senior secondary schools 1 page 170, questions 3 -4
READING ASSIGNMENT
Study the presentation of data using the line graph, Bar graph and histogram
ASSIGNMENT
New General Mathematics for senior secondary schools 1 page 170, questions 5 - 6
WEEK 8
TOPIC: DATA PRESENTATION (CONTINUOUS)
Contents:
LINE GRAPH
BAR GRAPH
HISTOGRAM
LINE GRAPH
Line graphs are used to show trends over a period of time and have the advantage that they can be extended. To draw a line graph, plot the given data as a series of points and then join the points together by straight lines. The lines can be drawn vertically or horizontally and have no thickness.
Example 1
A comparison of traffic densities in a number of countries produced the following results.
Present the data in a line graph
Example 2:
The table below shows the number of eggs laid in one month by 5 hens. Draw a horizontal line graph for the data.
Solution
The number of eggs laid in one month by 5 hens can be presented in the following horizontal line graph.
Evaluation:
The student population of the Federal Government college Kano was recorded as follows.
Represent the information on a line graph
BAR GRAPH/CHART
Bar charts are rectangular shapes of equal widths but different lengths drawn to represent the frequency. It is uniform in thickness. Bars can be drawn vertically or horizontally.
Example 1:
The table below shows the number of babies born to a number of women within a given age range to a number of women within a given age range.
Women ages - 24 25 26 27 28
No of babies - 1 4 2 3 2
Draw a bar chart to illustrate the above distribution.
Solution
Example 2:
The number of bottles of soft drinks sold in a restaurant one evening is given by the data in the table below.
Type of soft drink | No of bottles
Coke |12
Fanta | 10
Sprite | 6
Lemon | 4
Pepsi | 8
Draw a bar chart to display this information
Solution
Evaluation
New General Mathematics for Senior Secondary Schools 1, pages 173 – 171 14c question 1 – 2
HISTOGRAMS
Histogram is a rectangular block graph representing the frequency distribution of the analysed items. The blocks are joined to one another.
Example 1: the marks scored by 40 students in a particular subject are as follows:
38 74 28 32 10 31 49 34 50 19
30 92 50 42 38 64 24 65 91 77
18 35 12 87 41 27 8 90 22 21
42 43 52 59 72 70 90 91 29 28
Prepare a frequency table, using class interval: 1 – 20, 21 – 40, 41 – 60, .......
Use the table to draw a histogram.
Solution
(b)
Example 2:
Drawing histogram using class boundary, draw a histogram for the frequency distribution in the table below. (Use class boundaries to plot against the frequency).
NB: The use of class boundaries to plot against the frequency is important in using histogram to estimate the mode.
Solution
Evaluation 1
The table below shows the marks obtained by forty pupils in a Mathematics test.
Marks 0 - 9 10 - 19 20 - 29 30 - 39 40 - 49 50 - 59
No of pupils 4 5 6 12 8 5
Draw a histogram for the mark distribution
Evaluation 2
New General Mathematics for senior secondary schools 1, page 180 – 181, questions 1 and exercise 14e
General Evaluation.
Man Mathematics for senior secondary school 1 exercise B5, pages 246 – 247. Questions 2 and 8.
Reading Assignment
Study presentation of data using a pie chart and Frequency polygon.
Assignment
New General Mathematics for senior secondary schools 1 page 180 exercise 142e , questions
1 – 5. (Draw a histogram of the data only)
Man Mathematics for senior secondary schools 1, page 246, exercise B5, question 3 – 10.
Contents:
LINE GRAPH
BAR GRAPH
HISTOGRAM
LINE GRAPH
Line graphs are used to show trends over a period of time and have the advantage that they can be extended. To draw a line graph, plot the given data as a series of points and then join the points together by straight lines. The lines can be drawn vertically or horizontally and have no thickness.
Example 1
A comparison of traffic densities in a number of countries produced the following results.
Present the data in a line graph
Example 2:
The table below shows the number of eggs laid in one month by 5 hens. Draw a horizontal line graph for the data.
Solution
The number of eggs laid in one month by 5 hens can be presented in the following horizontal line graph.
Evaluation:
The student population of the Federal Government college Kano was recorded as follows.
Represent the information on a line graph
BAR GRAPH/CHART
Bar charts are rectangular shapes of equal widths but different lengths drawn to represent the frequency. It is uniform in thickness. Bars can be drawn vertically or horizontally.
Example 1:
The table below shows the number of babies born to a number of women within a given age range to a number of women within a given age range.
Women ages - 24 25 26 27 28
No of babies - 1 4 2 3 2
Draw a bar chart to illustrate the above distribution.
Solution
Example 2:
The number of bottles of soft drinks sold in a restaurant one evening is given by the data in the table below.
Type of soft drink | No of bottles
Coke |12
Fanta | 10
Sprite | 6
Lemon | 4
Pepsi | 8
Draw a bar chart to display this information
Solution
Evaluation
New General Mathematics for Senior Secondary Schools 1, pages 173 – 171 14c question 1 – 2
HISTOGRAMS
Histogram is a rectangular block graph representing the frequency distribution of the analysed items. The blocks are joined to one another.
Example 1: the marks scored by 40 students in a particular subject are as follows:
38 74 28 32 10 31 49 34 50 19
30 92 50 42 38 64 24 65 91 77
18 35 12 87 41 27 8 90 22 21
42 43 52 59 72 70 90 91 29 28
Prepare a frequency table, using class interval: 1 – 20, 21 – 40, 41 – 60, .......
Use the table to draw a histogram.
Solution
(b)
Example 2:
Drawing histogram using class boundary, draw a histogram for the frequency distribution in the table below. (Use class boundaries to plot against the frequency).
NB: The use of class boundaries to plot against the frequency is important in using histogram to estimate the mode.
Solution
Evaluation 1
The table below shows the marks obtained by forty pupils in a Mathematics test.
Marks 0 - 9 10 - 19 20 - 29 30 - 39 40 - 49 50 - 59
No of pupils 4 5 6 12 8 5
Draw a histogram for the mark distribution
Evaluation 2
New General Mathematics for senior secondary schools 1, page 180 – 181, questions 1 and exercise 14e
General Evaluation.
Man Mathematics for senior secondary school 1 exercise B5, pages 246 – 247. Questions 2 and 8.
Reading Assignment
Study presentation of data using a pie chart and Frequency polygon.
Assignment
New General Mathematics for senior secondary schools 1 page 180 exercise 142e , questions
1 – 5. (Draw a histogram of the data only)
Man Mathematics for senior secondary schools 1, page 246, exercise B5, question 3 – 10.
WEEK 9
Topic: DATA PRESENTATION
Content: Pie chart
Frequency
PIE CHART.
A pie chart is a circular shape which is divided into sections whose angles are proportional to the frequencies of the items.
Steps to Drawing a Pie Chart
Step 1.
Calculate the angles of sector of the pie chart using x/yx = 3600. Where x is the frequency of the total frequencies of all the items.
Step 2.
Using a pair of compasses, draw a circle with a suitable radius.
Step 3.
Partition it into the various angles of sectors obtained in step 1 using a protractor.
Step 4.
Label each sector showing the information which relates to the sector.
Step 5
Give a title to your pie chart, by stating the information the pie chart is representing.
https://youtu.be/foyPpC3XjhE
Example 1:
The table below shows the expenditure of a Bachelor in a given month
Items | Expenditure N1000
Food | 10
Rent | 5
Bills | 3
Clothing | 2
Savings | 6
Miscellaneous | 4
Draw a pie chart to illustrate the information
Solution
First, we shall calculate the angle of sector for each item as follows.
Title: Pie chart showing the expenditure of Bachelor in a certain month.
Example 2:
In a certain school, the lesson periods for each week are as itemized below.
English 10, Mathematics 7, Biology 3, Statistics 4, Ibo 3, others 9.
Draw a pie chart to illustrate this information
Title : pie Chart showing the lesson periods for each week in a certain school.
Example 3:
The pie chart illustrates the amount of private time a student spends in a week studying various subjects. Find the value of k.
Solution
Since the total angles of all sectors in a given circle is 3600.
1050 + 750 + (2k)0+ k0 + (3k)0 = 3600
1050 + 750 + (6k)0=3600
1800 + 6k =3600
6k = 3600 -1800
6k = 1800
Divide through by 6
6k/6 = 1800/6
K = 300.
Example 4:
The pie chart represents the fruits on display in grocery shop. If there are too oranges on display. How many apples are there?
Solution
The only sector that has useful information is that of oranges.
Number of oranges = 60
Sector angle of orange = 360 – (60 + 100 + 120)
= 3600 – 2800
= 800.
But 60/(total fruits) x 360/1 = 800.
60/(T.f)x 360/1 = 800
21600/(total fruits) = 80/1
Cross multiplying
21600 = 80(total fruit)
Dividing both sides by 80
21600/80 = (80(total fruit))/80
= 270
∴ Total number of fruit is 270.
Let the number of apples be x, then;
x/270 x 360/1 = 1200
x= (120 x 270)/3600
= 900
Example 5:
The Pie chart below shows the distribution of students in a certain school into some major ethnic groups in Nigeria.
In its simplest form, what fraction of the students are Igbos?
What percentage of the students are Hausas?
What is the ratio of the Binis to Hausas in the school in its simplest terms.
Solution
The angle of the sector representing Igbos is 700.
= 70/360 of the students are Igbos.
= 70/360 = 7/36
The angle of sector for Hausa is x0 and this can be obtained using
1100 + x0 + 700 + 600 + 900 = 3600
3300 + x0 = 3600
x = 300.
Taking the percentage of the fraction, the students who are Hausas use
= 30/ 300 x 100/1
= 1/12 x 100/1
= 100/12
8.3%
The angle of sector for binis = 900
The angle of sector for Hausas= 300
ratio of Binis to Hausas 900 : 300
= 3 : 1
Example 6:
The Pie chart below shows the weekly sales of a motor dealer in Lagos in 1999.
What fraction of the cars were Toyota?
What percentages of the Cars were Datsun?
If the dealer sold 16 Peugeots, How many BMW did he sell in a week?
Solution
The angle of the sector representing Toyota is 1200.
= 120/ 360
= 12/36
= 1/3
Taking the percentage of the fraction of the Datsun Cars
= 90/ 360 x 100/1
= 1/4 x 100/1
= 25%
To get the total number of Cars, we will first get the angle of sector for Peugeot.
70 + 120 + 90 + x + 360
280 + x = 360
X = 360 – 280
x = 800
Assuming the total number of Cars to be y, then
16/y x 360/1 = 800
5760/y = 80/1
Cross multiplying
5760 = 80y
Divide both sides by 80
5760/80 = 80y/80
y = 720.
To get the number of BMW sold in a week, let the number of the BMW sold be m
m/72 x 360/1 = 700
360m/72 = 70
Cross multiplying
360m = 5040
Dividing both sides by 360
360m/360 = 5040/360
m = 14
Evaluation
The pie chart below shows the allocation of money to the different departments in a secondary school.
If applied Science Department were allocated the sum of N120,000.00. What was the total allocation to Maths Department?
New General Mathematics for senior secondary schools 1. Page 174, exercise 14c, question 3.
MAN Mathematics for Senior secondary schools 1, page 248, exercise B6, question 1 - 3
Frequency polygon
The frequency polygon is obtained by joining the midpoints of the tops of the histogram rectangles with straight lines. Since the polygon is a closed figure, we take one interval below the lowest internal on the base axis. Then we join the ends of the polygon to the midpoints of the new intervals as shown below.
Example 1
The table shows the distribution of the masses of fifty logs exported in September 1981 by a timber and plywood company.
Mass (kg) No of Logs
150 -154 1
155 – 159 4
160 – 164 8
165 - 169 13
170 – 174 12
175 – 179 8
180 – 184 3
185 – 189 1
Construct the frequency polygon for the distribution
Solution
Mass (kg) Class Boundary Frequency
150 -154 149.5 – 154.5 1
155 – 159 154.5 – 159.5 4
160 – 164 159.5 – 164.5 8
165 - 169 164.5 – 169.5 13
170 – 174 169.5 – 174.5 12
175 – 179 174.5 -179.5 8
180 – 184 179.5 – 184.5 3
185 – 189 184.5 189.5 1
https://youtu.be/YOA344zHhIU
Evaluation
New General Mathematics for senior secondary schools 1, page 180, exercise 14e, questions 4 – 5
General evaluation
New General Mathematics for senior secondary schools 1, page 180, , questions 4 – 7 exercise 14e
Reading Assignment
Study Mean, Median and Mode.
Assignment
New General Mathematics for senior secondary schools 1, page 180, exercise 14e, questions 1 –8
MAN Mathematics for senior secondary schools 1, Miscellaneous exercises, pages 254 – 257 Questions 1 – 22.
Content: Pie chart
Frequency
PIE CHART.
A pie chart is a circular shape which is divided into sections whose angles are proportional to the frequencies of the items.
Steps to Drawing a Pie Chart
Step 1.
Calculate the angles of sector of the pie chart using x/yx = 3600. Where x is the frequency of the total frequencies of all the items.
Step 2.
Using a pair of compasses, draw a circle with a suitable radius.
Step 3.
Partition it into the various angles of sectors obtained in step 1 using a protractor.
Step 4.
Label each sector showing the information which relates to the sector.
Step 5
Give a title to your pie chart, by stating the information the pie chart is representing.
https://youtu.be/foyPpC3XjhE
Example 1:
The table below shows the expenditure of a Bachelor in a given month
Items | Expenditure N1000
Food | 10
Rent | 5
Bills | 3
Clothing | 2
Savings | 6
Miscellaneous | 4
Draw a pie chart to illustrate the information
Solution
First, we shall calculate the angle of sector for each item as follows.
Title: Pie chart showing the expenditure of Bachelor in a certain month.
Example 2:
In a certain school, the lesson periods for each week are as itemized below.
English 10, Mathematics 7, Biology 3, Statistics 4, Ibo 3, others 9.
Draw a pie chart to illustrate this information
Title : pie Chart showing the lesson periods for each week in a certain school.
Example 3:
The pie chart illustrates the amount of private time a student spends in a week studying various subjects. Find the value of k.
Solution
Since the total angles of all sectors in a given circle is 3600.
1050 + 750 + (2k)0+ k0 + (3k)0 = 3600
1050 + 750 + (6k)0=3600
1800 + 6k =3600
6k = 3600 -1800
6k = 1800
Divide through by 6
6k/6 = 1800/6
K = 300.
Example 4:
The pie chart represents the fruits on display in grocery shop. If there are too oranges on display. How many apples are there?
Solution
The only sector that has useful information is that of oranges.
Number of oranges = 60
Sector angle of orange = 360 – (60 + 100 + 120)
= 3600 – 2800
= 800.
But 60/(total fruits) x 360/1 = 800.
60/(T.f)x 360/1 = 800
21600/(total fruits) = 80/1
Cross multiplying
21600 = 80(total fruit)
Dividing both sides by 80
21600/80 = (80(total fruit))/80
= 270
∴ Total number of fruit is 270.
Let the number of apples be x, then;
x/270 x 360/1 = 1200
x= (120 x 270)/3600
= 900
Example 5:
The Pie chart below shows the distribution of students in a certain school into some major ethnic groups in Nigeria.
In its simplest form, what fraction of the students are Igbos?
What percentage of the students are Hausas?
What is the ratio of the Binis to Hausas in the school in its simplest terms.
Solution
The angle of the sector representing Igbos is 700.
= 70/360 of the students are Igbos.
= 70/360 = 7/36
The angle of sector for Hausa is x0 and this can be obtained using
1100 + x0 + 700 + 600 + 900 = 3600
3300 + x0 = 3600
x = 300.
Taking the percentage of the fraction, the students who are Hausas use
= 30/ 300 x 100/1
= 1/12 x 100/1
= 100/12
8.3%
The angle of sector for binis = 900
The angle of sector for Hausas= 300
ratio of Binis to Hausas 900 : 300
= 3 : 1
Example 6:
The Pie chart below shows the weekly sales of a motor dealer in Lagos in 1999.
What fraction of the cars were Toyota?
What percentages of the Cars were Datsun?
If the dealer sold 16 Peugeots, How many BMW did he sell in a week?
Solution
The angle of the sector representing Toyota is 1200.
= 120/ 360
= 12/36
= 1/3
Taking the percentage of the fraction of the Datsun Cars
= 90/ 360 x 100/1
= 1/4 x 100/1
= 25%
To get the total number of Cars, we will first get the angle of sector for Peugeot.
70 + 120 + 90 + x + 360
280 + x = 360
X = 360 – 280
x = 800
Assuming the total number of Cars to be y, then
16/y x 360/1 = 800
5760/y = 80/1
Cross multiplying
5760 = 80y
Divide both sides by 80
5760/80 = 80y/80
y = 720.
To get the number of BMW sold in a week, let the number of the BMW sold be m
m/72 x 360/1 = 700
360m/72 = 70
Cross multiplying
360m = 5040
Dividing both sides by 360
360m/360 = 5040/360
m = 14
Evaluation
The pie chart below shows the allocation of money to the different departments in a secondary school.
If applied Science Department were allocated the sum of N120,000.00. What was the total allocation to Maths Department?
New General Mathematics for senior secondary schools 1. Page 174, exercise 14c, question 3.
MAN Mathematics for Senior secondary schools 1, page 248, exercise B6, question 1 - 3
Frequency polygon
The frequency polygon is obtained by joining the midpoints of the tops of the histogram rectangles with straight lines. Since the polygon is a closed figure, we take one interval below the lowest internal on the base axis. Then we join the ends of the polygon to the midpoints of the new intervals as shown below.
Example 1
The table shows the distribution of the masses of fifty logs exported in September 1981 by a timber and plywood company.
Mass (kg) No of Logs
150 -154 1
155 – 159 4
160 – 164 8
165 - 169 13
170 – 174 12
175 – 179 8
180 – 184 3
185 – 189 1
Construct the frequency polygon for the distribution
Solution
Mass (kg) Class Boundary Frequency
150 -154 149.5 – 154.5 1
155 – 159 154.5 – 159.5 4
160 – 164 159.5 – 164.5 8
165 - 169 164.5 – 169.5 13
170 – 174 169.5 – 174.5 12
175 – 179 174.5 -179.5 8
180 – 184 179.5 – 184.5 3
185 – 189 184.5 189.5 1
https://youtu.be/YOA344zHhIU
Evaluation
New General Mathematics for senior secondary schools 1, page 180, exercise 14e, questions 4 – 5
General evaluation
New General Mathematics for senior secondary schools 1, page 180, , questions 4 – 7 exercise 14e
Reading Assignment
Study Mean, Median and Mode.
Assignment
New General Mathematics for senior secondary schools 1, page 180, exercise 14e, questions 1 –8
MAN Mathematics for senior secondary schools 1, Miscellaneous exercises, pages 254 – 257 Questions 1 – 22.