Pen-Write E-Library

SCHEME OF WORK
WEEK CONTENT
1. Percentiles and Quartiles from cumulative Frequency curve; Estimation of Quartiles, Percentiles and Range

2. Measures of Dispersion of ungrouped data

3. Bearings and Distances

4. Probability

5. The formula method for solving quadratic equation ; Solution of quadratic equation by graphical method

6. (a) Mensuration of plane shapes
i. Parts of a circle
ii. Calculation of length of an arc and perimeter of a sector
(b) Mensuration of solid shapes
i. Total surface area of solid shapes such as Cube, Cuboids, Cylinder, Triangle, prism, Cone and rectangular based pyramid.
ii. Volumes of solid shapes such as cube, cuboids, cylinder, triangular prism, cone and rectangular based pyramid.

7. Area of sectors, segments and triangles; Volumes of frustrums of cone, rectangular based pyramids and other pyramids

8. Trigonometric rations
i. Sine, Cosine, and tangent of acute angles
ii. Use of tables to find the values of given trigonometric rations
iii. Determination of length of chord using trigonometric rations

9. Geometrical construction:
i. Drawing and bisector of line segment
ii. Construction and bisection of angles 90o, 45o, 135o, 22 ½ o, 57 ½ o.
iii. Construction and bisection of angles 30o, 60o, 90o, 120o, 150o etc
iv. Revision of construction of triangles.

10. Construction:
i. Construction of quadrilaterals given certain conditions
ii. Construction of equilateral triangle
iii. Construction of Loci of points moving under given conditions.

11. TOPICAL REVISION FOR WAEC & NECO

12. MOCK PRACTICE TEST

Topic:
Percentile and Quartiles from cumulative frequency curve.
Percentiles: These are averages that divide a distribution into 100 equal parts. It can be estimated by interpolation method and by geometric method.

Quartiles: These are the values which divide the total frequency of a distribution into four equal parts namely:

First quartile, second quartile, third quartile and fourth quartile.
The first quartile is denoted by Q1 and obtained as ([sup]N[/sup]/[sub]4[/sub])[sup]th[/sup].

Second quartile = Q2 = ([sup]N[/sup]/[sub]2[/sub])[sup]th[/sup] ( this is the same as the median)

Third quartile = Q3 = ([sup]3N[/sup]/[sub]4[/sub])[sup]th[/sup]

Fourth quartile = Q4 = ([sup]4n[/sup]/[sub]4[/sub])[sup]th[/sup] =Nth ( This is the same as the total distribution ).

therefore, lower quartile = first quartile = Q1 = ([sup]N[/sup]/[sub]4[/sub])[sup]th[/sup]

Upper quartile = third quartile = Q3 = ([sup]3N[/sup]/[sub]4[/sub])[sup]th[/sup].

The Q1 and Q3 can be estimated by interpolation and geometrical method.

Interquartile range: This is the difference between the upper and lower quartiles

Interquartile range = Q3 – Q1

Semi interquartile range is the average of the interquartile range and obtained as

[sup]Q3 – Q1[/sup]/[sub]2[/sub]

Marks 0-9 10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99
Frequency 8 10 17 20 29 34 30 25 21 6


https://youtu.be/yhsobdD9W0Y

Draw a cumulative frequency curve and use your graph to determine the
I. lower and upper quartile
II. Interquartile range III. Semi-interquartile range IV 60th percentile.

Solution.

Mark F Cum. Freq Class boundaries
0-9 8 8 <9.5
10-19 10 18 <19.5
20-29 17 35 <29.5
30-39 20 55 <39.5
40-49 29 84 <49.5
50-59 34 118 <59.5
60-69 30 148 <69.5
70-79 25 173 <79.5
80-89 21 194 <89.5
90-99 6 200 <99.5

Masses(kg)
10-14
15-19
20-24
25-29
30-34
35-39
40-44
Frequency 3 7 9 5 11 6 9

a. Prepare a cumulative frequency table and use it to draw the ogive for the data.
b. Use your graph to estimate the I. Lower quartile and II. 70th percentile III. Median
c. Find the pass mark if 35 candidates passed.
https://youtu.be/t3SKZpTjbCo



Topic: ESTIMATION OF QUARTILES, PERCENTILES AND RANGE.
Quartiles: the lower and the upper quartiles can be estimated by interpolation.


L3 + Lower class boundary of the Q3 class.

Cfm3 = cumulative frequency of the class before the Q3 class.

Fm3 = frequency of the Q3 class.

C = Class width

N= Total frequency.

Percentiles: Percentiles can be calculated by the formula:
Pi = Percentile required

Li = Lower class boundary of the required percentile

Cfmi= cumulative frequency of the class before the percentile class

Fmi= Frequency of the percentile class

C = class width

Ni = Total frequency.

Example: The table below gives the distribution of ages of 120 students in an institution.

a. calculate the lower and upper quartiles of the distribution
b. calculate the percentiles
c. interquartile range



http://youtu.be/9QhU2grGU_E

EVALUATION
The table below shows the weeks profit in Naira from a mini market

Weekly Profit 1-10 11-20 21-30 31-40 41-50 51-60
Frequency 6 6 12 11 10 5

a. calculate for the distribution
I. the upper and lower quartile
II. semi interquartile range
III. 15th percentile.

Topic: MEASURES OF DISPERSION
-Range
- Mean Deviation
- Standard Deviation and Variance of Ungrouped data.32.
Measures of Dispersion is a statistical measure that shows how data having the same mean are dispersed or scattered around the mean. The major ones are the range, mean deviation, standard deviation etc.

Range: the range is the difference between the highest and lowers values of the variable. The range depends on the two extreme values and has a limited use.
https://youtu.be/1jqgwEvOzcg

Example: Find the range of the distribution 8.5,8.7, 9.2, 9.3, 9.5, 9.7.
Solution.
Highest value = 9.7, lowest value = 8.5.
:. Range = 9.7 - 8.5 = 1.2

Mean Deviation: This is the arithmetic mean of all absolute deviation from the mean. The mean deviation is defined by :

M.D = [sup]∑1x - x1[/sup]/[sub]N[/sub]

If the data involves frequency, the mean deviation is then defined by

M.D = [sup]∑f1x – x1[/sup]/ [sub]∑f[/sub].

Example: Find the mean deviation of the sets of data. 1,4,5,6,8,12.
Solution.
M.D = [sup]∑1x - x1[/sup]/[sub]n[/sub]

to find the mean i. e X


Standard Deviation: The most reliable and most widely used of the measures of dispersion is the standard deviation because it takes into account every value of the variable being discussed.
The standard deviation is defined by:


Example: Find the standard deviation of the set of numbers 12,15,19,22,25,27.
Solution.


VARIANCE
The square of the standard deviation is called variance and defined by:
Variance = ( s.d)2


https://youtu.be/SnBsw4Qgk_A

Example: The table below shows the distribution of masses of 35 students.

Masses(kg) 40 41 42 43 44 45
No of Student 5 10 6 4 7 3


Calculate correct to 3 significant figures
a. Range
b. The mean deviation
c. The standard deviation of the distribution.
Solution.


EVALUATION.
The table below shows the numbers of bottles of beer consumed by 30 people during a naming ceremony.

|No of bottles| 1| 2| 3| 4| 5|
-------------------------------------------------
|No of people | 6| 3| 7| 8| 6|



a. State the range of the distribution
b. Calculate the mean and standard deviation of the distribution correct to the nearest whole number

ASSIGNMENT
|No of Pages |15| 16| 17| 18| 19| 20|
|No of Magazines|3| 4| 7| 6| 5| 5|

Calculate correct to the nearest whole number, the mean deviation, standard deviation and variance of the distribution above.



RANGE, MEAN DEVIATION AND STANDARD DEVIATION OF GROUPED DATA
Mean Deviation of Grouped Data.

where X us the class mid-point of the class interval.
Standard Deviation of Grouped Data
When data are grouped into classes of equal interval, the standard deviation is defined by using the idea of the assumed mean.
Therefore,

Where d is the deviation obtained as
D = X – A,
A= assumed mean
X = class midpoint.
https://youtu.be/lp2nTFdYGec
Example: the table below shows the distribution of marks obtained by 100 students in a test

Using 54.5 as an assumed mean, calculate the I mean mark II standard deviation of the distribution.

Solution .

https://youtu.be/VyBfMDNRM9w

EVALUATION: The table below shows the salaries paid to 100 junior workers of a publishing company in Lagos.

Salaries 10-14 15-19 20-24 25-29 30-34 35-39
No of staff 22 15 18 30 9 6

Calculate for the data the
i. Mean ii. Standard Deviation if the assumed mean is 22.

ASSIGNMENT.
Find the mean grade and standard deviation of the distribution below if the working mean(assumed mean) is 40.5.


Grade
16-25 26-35 36-45 46-55 56-65 66-75
No of Competitors 4 11 19 14 0 2

Topic: BEARINGS AND DISTANCES
Bearings can be defined as the angular relationships between two or more places. Bearing are specified in two ways:
i. Cardinal points: It is specified in reference to the north and south.

Take O as the starting P
NOP = 60º is the bearing N60ºE
SOQ = 47º is S47ºW.

2. Three –digit notation : the bearing is specified in three-digit eg. 060º, 079º,180º etc


Example: Indicate the following bearing on the cardinal point at 075º (b) 190º
Solution.


Example 2: Write each of the following in three digit notation.


https://youtu.be/ACXWbkhDW44

EVALUATION: find the equivalent of the following bearing in three-digit notation.
1. S80oW 2. N 15oE.




BEARING OF ONE POINT FROM ANOTHER
Find the bearing of point A from B if B from A is 130º.
Solution .
A from B = ? B from A = 130º


2. if bearing of P from R is 065º, find bearing of R from P.
Solution.

https://youtu.be/S4Loo79JfH4

EVALUATION
Find the bearing of X from Y, if Y from X is 120º

The Sine rule and Cosine rule are the basic rule used to solve bearing problem.


Example.
A fly moves from a point U, on a bearing 060º to a point V, 20m away. It then moves from the point V, on a bearing of 130º, to a point W. If the point W is due east of U. Find the distance of the point V from W and U from W.


Distance of U from W = 29.3 (3.s.f).
https://youtu.be/oaq8ofHHrvg

ASSIGNMENT
Three towns A, B and C are such that the distance between A and B is 50km and the distance between A and C is 90km. If the bearing of B from A is 075º and the bearing of C from A is 310º, find the
a. distance between B and C
b. bearing of C from B.


Example: Town Q is on a bearing 210º from town P, town R is on a bearing 150º from town P and R is east of Q . The distance between R and P is 10km. Find the distance between R and Q.
Solution.
Since the angles are equal i.e 60º each, then the triangles is an equilateral triangle, hence, the sides must be equal.
:. Distance PQ = 10km

2. A village R is 10km from a point P on a bearing 025º from P. Another village A is 6km from P on a bearing 162o. Calculate
a. distance of R from A
b. the bearing of R from A.

Solution
A distance R from A using Cosine rule:
P2 = a2 + r2 – 2arCos P
P2 = 102 + 62 – 2 (10 x 6) Cos 137o
P2 =100 + 3 + (120 x 0.7314)
P2 = 136 + 87.768
P2 = 223.768
P = 223.768 = p = 14.96km
Distance R from A = 15km approximately.


b. Bearing of R from A
let the bearing of x
To find x, find A first

A = P
Sin A SinP
Sin A = 10 x sin 137o = 10 x `0.6820
14.96 14.96.
Sin A = 0.4559.
A= Sin-1 0.4559
A = 27.1o
But A = 180 + x
Xo = A -18
= 27 .1 – 18
x = 9.1º
The bearing of R from A is 009º.


https://youtu.be/0E1tpByA0F0

EVALUATION.
A boat sails 6km from a point x on a bearing of 065º and thereafter 13km on a bearing of 136º. What is the distance and bearing of the boat from X.

ASSIGNMENT
1. P, Q and R are points, in the same horizontal plane. The bearing of Q from P is 150º and the bearing of R from Q is 060º. If PQ = 5m and QR = 3m. Find the bearing of R from P correct to the nearest degree.


https://www.youtube.com/playlist?list=P ... NOncUdalJq

Topic: The formula method for solving quadratic equation

A. Derivative of the roots of the general form of quadratic equation.

The general formula of a quadratic equation is ax[sup]2[/sup] + bx + C = 0. The roots of the general equation are found by completing the square.

ax[sup]2[/sup] + bx + C = 0

Divide through by the coefficient of x[sup]2[/sup].



The square of half of the coefficient of x is


Take square roots of both sides of the equation :
X +
Take square roots of both sides of the equation :


https://youtu.be/IlNAJl36-10

Evaluation
Suppose the general quadratic equation is

Dy[sup]2[/sup] + Ey + F = 0

Using the method of completing the square, derive the roots of this equation

B. Using the formula methods to solve quadratic equations
Examples
Use the formula method to solve the following equations/ Give the roots correct to 2 decimal places:

i. 3x[sup]2[/sup] - 5x – 3 = 0

ii. 6x[sup]2[/sup] + 13x + 6 = 0

iii. 3x[sup]2[/sup] – 12x + 10 = 0

Solution

1. 3x[sup]2[/sup] – 5x – 3 = 0

Comparing 3x[sup]2[/sup] – 5x – 3 = 0

With ax[sup]2[/sup] + bx + C = 0

A = 3, b = -5, C = -3

Since


(2) 6x[sup]2[/sup] + 13x + 6=0

comparing 6x[sup]2[/sup] + 13x + 6=0

with ax[sup]2[/sup] +bx + C = 0

a= 6, b =13, C = 6

Since


(3) 3x[sup]2[/sup] – 12x + 10 = 0

comparing 3x[sup]2[/sup] – 12x + 10 = 0 with ax[sup]2[/sup] +bx + C = 0, then

a = 3, b= -12, c = 10.



https://youtu.be/393iXYiGv7Q

Evaluation
Use the formula method to solve the following quadratic equations .

1. t[sup]2[/sup] – 8t + 2 = 0

2. t[sup]2[/sup] + 3t + 1 = 0

Sum and Product of quadratic roots.
We can find the sum and product of the roots directly from the coefficient in the equation
It is usual to call the roots of the equation α and βIf the equation

ax[sup]2[/sup] +bx + C = 0 ……………. I
Has the roots α and β then it is equivalent to the equation
(x – α ) ( x – β ) = 0

x[sup]2[/sup] – βx – αβ = 0 ………… 2

Divide equation (i)by the coefficient of x[sup]2[/sup]


Examples

1. If the roots of 3x[sup]2[/sup] – 4x – 1 = 0 are αand β, find α + β and αβ

2. if α and βare the roots of the equation

3x[sup]2[/sup] – 4x – 1 = 0 , find the value of



Here, comparing the given equation, with the general equation,
a = 3, b = -4, C = - 1
from the solution of example 1 (since the given equation are the same ),



b. Since

( α-β) =α[sup]2[/sup] + β- 2 α β
but

α[sup]2[/sup] + β[sup]2[/sup] = ( α + β)[sup]2[/sup] -2 α β

:.(α- β)[sup]2[/sup] = ( α+ β )[sup]2[/sup] - 2αβ -2αβ

(α – β)[sup]2[/sup] = (α + β )[sup]2[/sup] - 4α β

:.( α – β) = √(α + β )[sup]2[/sup] - 4αβ

( α – β) =√([sup]4[/sup]/[sub]3[/sub] )[sup]2[/sup] – 4 (- [sup]1[/sup]/[sub]3[/sub] )

= [sup]√16[/sup]/[sub]9[/sub] + [sup]4[/sup]/[sub]3[/sub]

[sup]√16 + 12[/sup]/[sub]9[/sub]


[sup]√28[/sup]/[sub]9[/sub]

:. α - β = [sup]√28[/sup]/[sub]3[/sub]


Evaluation
If α and β are the roots of the equation

2x[sup]2[/sup] – 11x + 5 = 0, find the value of

a. α - β

b. [sup]1[/sup]/[sub]α + 1[/sub] + [sup]1[/sup]/[sub]B + 1[/sub]

Assignment
If α and β are the roots of the equation

2x[sup]2[/sup] – 7x – 3 = 0 find the value of :

1. α + β (a) [sup]2[/sup]/[sub]3[/sub] (b) [sup]7[/sup]/[sub]2[/sub] (c) [sup]2[/sup]/[sub]5[/sub] (d) [sup]5[/sup]/[sub]3[/sub]

2. α β (a) [sup]-3[/sup]/[sub]2[/sub] (b) [sup]2[/sup]/[sub]3[/sub] (c) [sup]3[/sup]/[sub]2[/sub] (d) [sup]– 2[/sup]/[sub]3[/sub]

3. α β[sup]2[/sup] + α[sup]2[/sup] β (a) [sup]21[/sup]/[sub]4[/sub] (b) [sup]4[/sup]/[sub]21[/sub] (c) [sup]– 4[/sup]/[sub]21[/sub] (d) [sub]-21[/sub]/[sub]4[/sub]

Solve the following equation using the formula method.

4. gp[sup]2[/sup] – 2p – 7 = 0

5. 3 = 8q – 2q[sup]2[/sup].

Theory
1. Solve the equation 2x[sup]2[/sup] + 6x + 1 = 0 using the formula method

2. If α and β are the roots of the equation 3x[sup]2[/sup] -9x + 2 = 0, find the values of

i. α β[sup]2[/sup] + α[sup]2[/sup]β

ii. α[sup]2[/sup] - αβ + β[sup]2[/sup]

Reading Assignment
New General Mathematics SS V|Bk2 pages 41-42 Ex 3e Nos 19 and 20 page 42.


Topic: Solution of quadratic Equation by graphical method.
The following steps should be taken when using graphical method to solve quadratic equation :
i. use the given range of values of the independent variable (usually x ) to determine the corresponding values of the dependent variable (usually y ) by the quadratic equation or relation given. If the range of values of the independent variable is not given, choose a suitable one

ii. From the results obtained in step (i), prepare a table of values for the given quadratic expression

iii. Choose a suitable scale to draw your graph

iv. Draw the axes and plot the points.

v. Use a broom or flexible curve to join the points to form a smooth curve.

Notes

1. The roots of the equation are the points where the curve cuts the x – axis because along he x- axis y = 0

2. The curve can be an inverted n – shaped parabola or it can be a v-shaped parabola.

It is n-shaped parabola when the coefficient of x[sup]2[/sup] is negative and it is V- shaped parabola when the coefficient of x[sup]2[/sup] is positive. Maximum value of y occurs at the peak or highest point of the n-shaped parabola while minimum value of y occurs a the lowest point of V-shaped parabola.

3. The curve of a quadratic equation is usually in one of three positions with respect to the x – axis.



In fig(3), the curve cosine of the x-axis at two clear points. These two points give the roots of the quadratic equation in fig (2), the two points are coincident, i.e their points are so close together that the curve touches the x axis at one point. This corresponds to an equation which has one repeated root
In fig(1), the curve does not cut the x axis . the roots of an equation which gives a curve in such a position are said to be imaginary
4. The line of symmetry is the line which divides the curve of the quadratic equation into two equal parts.

https://youtu.be/7C3f-sYMNCU

Examples

1(a) Draw the graph of y =11 = 8x – 2x[sup]2[/sup] from x = -2 to x = +6.

b. Hence find the approximate roots of the equation 2x[sup]2[/sup] – 8x – 11=0

c. From the graph, find the maximum value of y.

2a. Given that y = 4x[sup]2 [/sup]– 12x = 9 ,copy and complete the table below


b. Hence draw a graph and find the roots of the quaint 4x[sup]2[/sup] – 12x + 9 = 0

c. From the graph, what is the minimum value of y ?

d. From the graph, what is the line of symmetry of the curve?

Solutions

Y = 11 +8x -2x[sup]2[/sup] ,
from x =-2 to x = + 6

When x =-2
Y = 11 – 16 -2 ( +4)
Y =11 -16 – 8
Y = -5 – 9 = -13.

When x = -1

Y= 11 + 8 (-1) [sup]-2[/sup] (-1)[sup]2[/sup]

Y= 11 – 8 – 2 ( + 1)

Y = 11 – 8 -2

Y = 3 -2 = 1.

When x = 0

Y = 11 + 8 (0) – 2 (0)[sup] 2[/sup]

Y = 11 + 0 – 0 – 2 x 0

Y = 11 + 8 ( 1) [sup]-2[/sup] ( 1)[sup]2[/sup]

Y = 11 + 8 – 2 x 1

Y = 19 -2 = 17

When x 2

Y = 11 + 8 (2) -2 (2)[sup]2[/sup]

= 11 + 16 = 2 x 4

= 27 – 8 = 19

when x = 3

y = 11 + 8 ( 3) – 2 ( 3) [sup]2[/sup]

= 11 + 24 – 2 x 9

= 35 – 18 = 17

when x = 4

y = 11 + 8 (4) – 2 (4) [sup]2[/sup]

= 11 + 24 – 2 x 9

= 35 – 18 = 17

when x = 4

y = 11 + 8 (4) – 2 (4) [sup]2[/sup]

= 11 + 32 – 2 x 16

= 43 – 32 = 11

when x = 5

y = 11 + 8 (5) -2 ( 5)[sup]2[/sup]

= 11 + 40 -2 x 25

= 51 – 50 = 1

when x = 6

y = 11 + 8 ( 6) – 2 (6)

= 11 + 48 -2 x 36

= 59 – 72

-13

The table of values is given below
X -2 -1 0 1 2 3 4 5 6
Y -13 1 11 17 19 17 11 1 -13


Scale
On x axis, let 2cm = 1 unit on y axis, let 1cm = 2 units


b. From the graph, the approximate roots of the equation are the points where the curve cuts the x axis, this is so because

y = 11 + 8x – 2x[sup]2[/sup]

-1 x y = -1 x 11 + 8x – 1 -1 x – 2x[sup]2[/sup]

-y = -11 - 8x + 2x[sup]2[/sup]

-y = 2x[sup]2[/sup] – 8x – 11 = 0

-1 x – y = 0 x -1

I.e y = 0

Thus, from the graph, the roots of the equation 2x[sup]2[/sup] -8x – 11 = 0 are x = -1.1or x = 5.1

c. The maximum value of y = 19.

2 a. The completed table is given as follows

X -1 0 1 2 3 4
4x2 4 0 4 16 36 64
-12x 12 0 -12 -24 -36 -48
+9 9 9 9 9 9 9
Y 25 9 1 1 9 25


Scale
On x axis, let 2cm =1unit and y axis, let 1cm = 2units


From the graph, the roots of the equation is the points where the curve touches the x axis i.e x = 1.5 twice

c. from the graph, the minimum value of y = 0

d. from the graph, the line of symmetry of the curve is line x = 1.5

Evaluation
a. using a suitable scale, draw the graph of y = x[sup]2[/sup] – 2x from x = -2 to x = + 4

b. From the graph, find the approximate roots of the equation

x[sup]2[/sup] – 2x = 0
c. What is the minimum value of y ?

d. Find the values of x when y = 7.
https://youtu.be/7C3f-sYMNCU
https://youtu.be/Fap2g-lm8QE

Evaluation
a. prepare a table of values for the graph of

y = x[sup]2[/sup] + 3x – 4 for values of x from – 6 to + 3

b. Use a scale of 1cm to 1 unit on both axes and draw the graph.

c. Find the least value of y

d. what are the roots of the equation x[sup]2[/sup] + 3x – 4 = 0?

e. Find the values of x when y = 1

Reading Assignment

New General mathematics SS 1 pages 108- 113 by MF macrae et al ex 10g no 7 pg 111

CONTENT:

1. Part of a circle
2. Length of an arc and perimeter of a sector
3. Length of chord and perimeter of a segment

Parts of a circle.
The figure below shows circles and its parts.


The centre is the point at the middle of a circle.
The circumference is the curved outer boundary of the circle.
An arc is a curved part of the circumference.
A radius is any straight line joining the centre to the circumference. The plural of radius is radii.
A chord is any straight line joining two points on the circumference.
A diameter is a straight line which divides the circle into two equal part or a diameter is any chord which goes through the centre if the circle.





Region of a circle. The figure below shows a circle and its different regions.

A sector is the region between two radii and the circumference.
A semi-circle is a sector between a diameter and the Circumference i.e half of the circle. A segment of the region between a chord and the circumference.
https://youtu.be/cC0fZ_lkFpQ
https://youtu.be/O-cawByg2aA

EVALUATION:
Draw a circle and show the following parts on it. Two radii, a sector, a chord, a segment, a diameter, an arc; label each part and shade any regions.

Length of an arc and perimeter of a sector.
Given a circle centre O with radius r. The circumference of the circle is 2Пr. Therefore, the length, l, of arc xy is given as:

L = [sup]θ[/sup]/[sub]360[/sub] x 2 Пr

Where θ is the angle subtended at the centre by arc xy and r is the radius of the circle.

Also,
The perimeter of Sector XOYX = r + r + l
Where

L = length of arc xy
= [sup]θ[/sup]/[sub]360[/sub] x 2 Пr

Then
Perimeter of
Sector XOYX = r + r + l

= [sup]2r + θ[/sup]/[sub]360[/sub] x 2 Пr

https://youtu.be/Ff5qESGKJGo
https://youtu.be/He2-TZfS_rc

EXAMPLES
1. An arc of length 28cm subtends an angle of 240 at the centre of a circle. In the same circle, what angle does an arc of length 35cm subtend?

2. Calculate the perimeter of a sector of a circle of radius 7cm, the angle of the sector being 1080 if П is [sup]22[/sup]/[sub]7[/sub]

Solutions
1. L = length of arc xy
= [sup]θ[/sup]/[sub]360[/sub] x 2 Пr
When L = 28cm , θ = 240º, r = ?

Then

L = [sup]θ[/sup]/[sub]360[/sub] x 2 Пr


28 = [sup]24[/sup]/[sub]360º [/sub] x 2 x [sup]22[/sup]/[sub]7[/sub] x r

Cross-multiply:


Also


Thus, when the length of the arc is 35cm, the angle subtended at the centre is 300

2. Perimeter of a = 2r + [sup]θ[/sup]/[sub]360[/sub] x 2 Пr sector of a circle


EVALUATION
1. A piece o wire 22cm long is sent into an arc of a circle of radius 4 cm. What angle does the wire subtend at the centre of the circle of the?

2. Calculate the perimeter of a sector of a circle of radius 3.5cm, the angle of the sector being 1620 if it is [sup]22[/sup]/[sub]7[/sub]


Length of chord and perimeter of a segment.

Consider a circle centre O with radius

If OC is the perpendicular distance from O to chord AB and angle
AOB = 2 θ, then the length of chord AB can be found as follows:


AC = [sup]Sin θ[/sup]/[sub]r[/sub]

Cross multiply:
AC = r Sin θ
Since
AB = 2 x AC
AB = 2r Sin θ

Where
R = radius of the circle
Θ =Semi Vertical angle of the sector i.e half of the angle subtended at the centre by arc AB.
Also
The perimeter of segment ACBD
= Length of + length of chord AB arc ADB

= 2r Sin θ + [sup]2θ[/sup]/[sub]360[/sub] x 2 Пr
https://youtu.be/3j8p1HW9GkI

Examples
In a circle of radius 6 cm a chord is drawn 3cm from the centre.
(a) Calculate the angle subtended by the chord at the centre of the circle.

(b) Find the length of the minor arc cut off by the chord

(c) Hence find the perimeter of the minor segment formed by the chord and the minor arc.

Solution.




https://youtu.be/T-VqE63THL8

EVALUATION;
1a. A chord 4.8cm long is drawn in a circle of radius 2.6cm. Calculate the distance of the chord from the centre of the circle.

b. Calculate the angle subtended at the centre of the circle by the chord in Qu /Ca/ above

3. Hence find the perimeter of the minor segment of the minor segment formed by the chord and the minor arc of the circle.

ASSIGNMENT
1. A chord 4.2cm long is 2.8cm from the centre of a circle. Calculate the radius of the circle (A) 7cm (B) 3.5cm (C) 5.3cm (D) 10.6cm (E) 4cm

2. A chord 30cm long is 20cm from the centre of a circle. Calculate the length of a chord which is 24cm from the centre. (A) 35cm (B) 48cm (C) 25cm (D) 14cm (E) 28cm

3. What is the length of an arc which subtends an angle of 600 at the centre of a circle of radius ½ m? (A) 16/11m (B) 11/21 m (C) 2/3m (D) 1/3m (E) 3/4m.

4. Calculate the perimeter of a sector of radius 14cm, the angle of sector being 2160 if П is 22/7. (A) 48cm (B) 36cm (C) 24cm (D) 52cm (E) 54.4cm

5. The perimeter of a circle is 66cm. What is the diameter of the circle П = 22/7
(A) 12cm (B) 24cm (c) 25cm (D) 21cm (E) 31cm

THEORY
Complete the table below for arcs of circles. Make a rough sketch in each case.

Radius Angle at centre Length of arc

A 21cm ________ 22cm

B ____ 108o 132cm






Topic: Mensuration of plane shapes contd.

Area of sectors.
Area of a sector of a circle is given by the formula ;

Area of sector [sup]θ[/sup]/[sub]360[/sub] x πr[sup]2[/sup]

where r = radius of the circle θ = angle subtended at the centre bu XY or angle of the sector
https://youtu.be/7f9_U85_YX4

Examples
1. Calculate the area of sector of a circle which subtends an angle of 45º at the centre of the circle, diameter 28cm ( π= [sup]22[/sup]/[sub]7[/sub] )

2. The area of a circle PQR with centre O is 72cm[sup]2[/sup]. What is the area of sector POQ if POQ = 40º?

Solutions
1. Since the diameter of the circle = 28cm
d = 2r = 28
where d = diameter and r = radius
thus 2r = 28
2r = 28
r = 14cm

Area of sector = [sup]θ[/sup]/[sub]360[/sub] x πr[sup]2[/sup]

= [sup]45[/sup]/[sub]360[/sub] x 22 x ( 14 )[sup] 2[/sup]

[sup]1[/sup]/[sub]8[/sub] x [sup]27[/sup]/[sub]7[/sub] x 14 x 14 cm

= 77cm[sup]2[/sup]

Since the area of the whole circle PQR = 72cm[sup]2[/sup]

Then
Area of sector [sup]θ[/sup]/[sub]360[/sub] x πr[sup]2[/sup]

but πr[sup]2[/sup] = Area of the whole circle PQR = 72cm[sup]2[/sup]

:. Area of = [sup]40[/sup]/[sub]360[/sub] x 72cm[sup]2[/sup]

= 8cm[sup]2[/sup]
https://youtu.be/tD6wigYAYMk

Evaluation
complete the table below for areas of sectors of circles. make a rough sketch in each case.

Radius Angle of sector Area of sector

a.14cm - 462m[sup]2[/sup]

b. -- 140 99m[sup]2[/sup]


Area of segments
A segments of a circle is the area bounded by a chord and an arc of the circle, considering the figure below, we have a major segment and a minor segment .

= θ x πr[sup]2[/sup] - ½ r[sup]2[/sup] sine θ

Where
R = radius of the circle
Θ = angle subtended by the sector at the centre
Π= a constant = 22/7

Examples

1. calculate the area of the shaded segment of a circle :
https://youtu.be/cAOVS2DTU0U

3. Calculate the area of the shaded parts of a circle.
https://youtu.be/LmFXDBTSD4w

Solutions

1 Area of the given shaded =[sup] θ[/sup]/[sub]360[/sub] x πr[sup]2[/sup] - ½r[sup]2[/sup] Sin θ

= [sup]56[/sup]/[sub]360[/sub] x [sup]22[/sup]/[sub]7 [/sub] x ( 15 ) 2 - ½ x ( 15)[sup]2[/sup] sin 56º

= [sup]1[/sup]/[sub]45[/sub] x 22 x 15 x 15 - ½ x 15 x 15 sin 56

= 22 x 5 - ½ x 225 x sin 56º

= 110 - ½ x 225 x 0.4145

= 110 - 225 x 0. 4145

110 – 93.2625cm[sup]2[/sup]

110 – 93.2625cm[sup]2[/sup]

= 16.7375cm[sup]2[/sup]

16.7cm[sup]2[/sup] to 3 s. f



The arc in the given figure is part of a circle as shown in the figure above. Thus area of given shaded segment = Area of sector – area of

= [sup]θ[/sup]/[sub]360[/sub] x πr[sup]2[/sup] – ½ r[sup]2[/sup] sin θ

= [sup]90[/sup]/ [sub]360[/sub] x [sup]22[/sup]/[sub]7[/sub] x ( 14)[sup] 2[/sup] – ½ x (14) 2 sin 90º

¼ x [sup]22[/sup]/[sub]7[/sub] x 14 x 14 – ½ x 14 x 14 x 1

= 11 x 14 – 14 x 7 cm[sup]2[/sup]

= 154 - 98cm3[sup][/sup]

= 56cm[sup]2[/sup]


Evaluation
1. Calculate the area of a sector of a circle of radius 6cm which subtends an angle of 70º at the centre (π = 22/7)

A. 44cm[sup]2[/sup] B. 22cm[sup]2[/sup] C. 66cm[sup]3[/sup] D. 11cm[sup]2[/sup] E. 16.5cm[sup]2[/sup]

2. What is the angle subtended at the centre of a sector of a circle of radius 2cm if the area of the sector is 2.2cm2? (π = 22/7)

A. 1260 B. 31 ½ C. 43 D. 58 E. 63

3. What is the radius of a sector if a circle which subtended 140o at its centre and has an area of 99c\m2?
A. 18cm B. 27m C 9m D. 30m E. 24m

4. A sector of 80o is removed from a circle of radius 12cm What area of the circle is left?

A. 253cm[sup]2[/sup] B. 704cm[sup]2[/sup] C 176cm[sup]2[/sup] D. 125cm[sup]2[/sup] E. 352cm[sup]2[/sup]

5. Calculate the area of the shaded segment of the circle shown in the figure below:
( π = 22/7 )

A. 10.45cm[sup]2[/sup] B. 20.90cm[sub]2[/sub] C. 5.25cm[sup]2[/sup] D. 19.0cm[sup]2[/sup] E. 17.45cm[sup]2[/sup]

Theory
The figure below shows the cross section of a tunnel. It is in the shape of a major segment of a circle of radius 1m on a chord of length 1.6m. Calculate:

a. the angle subtended at the centre of the circle by the major arc correct to the nearest 0.10
b. the area of the cross section of the tunnel correct to 2d.p.

Topic: Properties of solid shapes
A Cube

A cube has the following properties.
1. It has 12 straight edges
2. It has 8 vertices
3. It also has 6 square faces
4. Its net consists of 6 square faces joined together

A Cuboid


A cuboid has the following properties.
1. It has 12 straight edges
2. It gas 8 vertices
3. It also has 6 rectangular faces
4. Its net consist of 6 rectangular faces


A Triangular Prism




A triangular prism has the following properties
1. It has 6 vertices
2. It has 9 straight edges
3. It also has 3 rectangular faces and two triangular faces which are the end faces
4. Its net consist of 3 rectangles and 2 triangles joined together

A Cylinder

1. A cylinder has 2 circular faces
2. It has 1 curved surface
3. It has 2 curved edges
4. Its net consist of two circular faces and 1 rectangular face i.e its net consist of 2 circles and 1 rectangle.



A Cone


A cone has the following properties
1. It has one vertex
2. It has 2 curved edges
3. It has 1 curved surface
4. It also has 1 circular face
5. Its net consist of a sector of a circle and a circle

Rectangular based pyramids


A rectangular based pyramid has the following properties
1. It has 8 straight edges
2. It has 5 vertices
3. It has 4 triangular faces
4. It has 1 rectangular face
5. Its net consists of 4 triangles and 1 rectangle
https://youtu.be/9yU_Q3qtGKo

Evaluation
1a. mention and draw 3 solid shapes that you know

b. Write down the properties of each of the solid shapes you mentioned in 1a above

c. List one real object for each of the solid shape mentioned in 1a above


Surface Area and Volume of Common Solid shapes
https://youtu.be/qJwecTgce6c
A prism is a solid which has uniform cross section. Cubes, cuboids, and cylinders are examples of prisms.
In general,
Volume of prism = area of uniform cross section X perpendicular height
=area of base x height

https://youtu.be/qJwecTgce6c


Triangular prism
Cube

Volume = l[sup]3[/sup]

Surface area = 6l[sup]2[/sup]



Cuboid
Volume = lbh
Surface area = 2 (lb + lb + bh)



Cylinder

Volume = πr[sup]2[/sup] h

Curved surface area = 2πrh

Total surface area = 2πrh + 2π r[sup]2[/sup]

= 2πr ( h + r)



Examples
1. Calculate the volumes of the following solids. All lengths are in cm.
https://youtu.be/w9O2kKfmgLo

In the figure above PQRS is a trapezium

2. Calculate the total surface area of the solid
Solutions
1a. Volume of prisms = area of uniform cross section X perpendicular height
= area of base X length of the prism

Area of PQRS = ½ ( 7 + 4) X QR cm2

Since |QR| = 1 X S1
Consider < P X S

|PX |[sup]2[/sup] + |XS|[sup]2[/sup] = 5[sup]2[/sup]

3[sup]2[/sup] + | XS|[sup]2[/sup] = 25

9 + |XS|[sup]2[/sup] = 25

|XS|[sup]2[/sup] = 25 – 9

|XS|[sup]2[/sup] = 16

|XS| = √4cm

Thus |XS| = |QR| = 4cm

Area of PQRS = ½ x ( 7 = 4) x |QR| cm[sup]2[/sup]

= ½ x 11 x 4 cm[sup]2[/sup]

= 22cm[sup]2[/sup]
Hence,
Volume of Prism = area of uniform cross section X length of prism

= 22cm[sup]2[/sup] x 12cm[sup]3[/sup]

= 264cm[sup]3[/sup]


(b) volume of given cylinder = πr[sup]2[/sup]h
from the given cylinder,

r = [sup]d[/sup]/[sub]2[/sub] = [sup]14[/sup]/[sub]2cm[/sub] = 7cm

h = 4cm
volume of given cylinder = π x (7) 2 x 4cm[sup]3[/sup]

[sup]22[/sup]/[sub]7[/sub] x 49 x 4cm

= 22 x 28cm[sup]3[/sup]

= 616cm[sup]3[/sup]

2. To calculate the total surface area of the solid shapes in 1a and b above.

2b. Total surface area of the given cylinder = 2πrh + 2πr[sup]2[/sup]

= 2πr ( h + r)

= 2 x [sup]22[/sup]/[sub]7[/sub] x 7 ( 4+ 7 ) cm[sup]2[/sup]

= 44 x 11cm[sup]2[/sup]

= 484 cm[sup]2[/sup]
https://youtu.be/P72Jfnr66Ac

Evaluation
1a A rectangular tank is 76cm long, 50cm wide and 40 cm high. How many litres of water can it hold?

b. Calculate the total surface area of the rectangular tank in question 1a above

Surface area of a Cone
A sector of a circle can be bent to form the curved surface of an open cone, In the figure below, the sector OA x B is of radius l and arc A x B subtends angle θ at O. This sector is bent to form a cone of base radius r and slant height


https://youtu.be/rd8tbD2eekM

The following points should be noted
1. The area of the sector is equal to the area of the curved surface of the cone .
2. The length of arc A x B in the 1st part of the figure above is the same as the circumference of the circular base of the cone in the 2nd part of the figure above

Curved surface area of cone = [sup]θ x πl[sup]2[/sup][/sup]/[sub]360[/sub] …………..0

Also,
[sup]θ x 2πl [/sup]/[sub]360[/sub] = 2 πr

Divide both sides by 2π



in equation (i) above

Curve surface area of cone = [sup]r[/sup]/[sub]L[/sub] x πl[sup]2[/sup]

Πrl
Hence,
Total surface area = curved surface area of a curve area of cone

= πrl +π r[sup]2[/sup]
= πr ( l + r)
https://youtu.be/K2ghejiUDXg

Examples
A paper cone has a diameter of 8cm and a height of 3cm

a. Make a sketch of the cone and hence use Pythagoras theorem to calculate its slant height.

b. Calculate the curved surface area of the cone in terms of π

c if the conies cut and opened out into the sector of a circle. What is the angle of
the sector?

d. assuming that the paper cone is closed at its base, what will be the total surface area of the closed paper cone?

Solutions.
From the given information about the paper cone,

Diameter = 8cm

:. Radius = [sup]diameter[/sup]/[sub]2[/sub]

= 8cm = [sup]4cm[/sup]/[sub]2[/sub]

using Pythagoras theorem in the right angled triangle OBC

l[sup]2[/sup] = |OB|[sup]2[/sup] + |BC|[sup]2[/sup]

l[sup]2[/sup] = 3[sup]2[/sup] + 4[sup]2[/sup]

l[sup]2[/sup] = 9 + 16

l[sup]2[/sup] = 25

Take square root of both sides

√ l[sup]2[/sup] =√ 25

l = 5cm
:.the slant height of the paper cone is 5cm

b. Curve surface area of the cone = πrl

= π x 4 x 5 cm[sup]2[/sup]

= 20 πcm[sup]2[/sup]


if the paper cone is cut and opened out into the sector of a circle as shown in the figure above, then
area of sector of circle = curved surface area of the cone

i.e [sup]θ[/sup]/[sub]360[/sub] x π x (5)[sup]2[/sup] = 20 x π

[sup]θ[/sup]/[sub]360[/sub] x π x 25 = 20 x π

72
5θ = 72 x 20
Divide both sides by 5

5θ = [sup]72 x 20[/sup]/[sub]5[/sub]

θ = 72 x 4

θ = 288
https://youtu.be/gdg4czPlA84

Evaluation
1. A 216 sector of a circle of radius 5cm is bent to form a cone. Find the radius of the base of the cone and tis vertical angle

2. Calculate (a) the curved surface area (b) the total surface area of the cone formed in question 1 above. Leave your answer in terms of y




Volume of pyramids and volume of cone
https://youtu.be/TDXSE4s9b84

In general,

Volume = [sup]1[/sup]/[sub]3[/sub] x base area x height



:. Volume of square based pyramid = [sup]1[/sup]/[sub]3[/sub] x b[sup]2[/sup] x h

volume of rectangular based pyramid = [sup]1[/sup]/[sub]3[/sub] x l x b x h

volume of cone = [sup]1[/sup]/[sub]3[/sub] x Πr[sup]2[/sup] x h

Examples
1.A pyramid 8cm high stands on a rectangular base 6cm by 4cm.Calculate the volume of the pyramid.
2. A right pyramid on a base 4cm square has a slant edge of 6cm.Calculate the volume of the pyramid.
3. Calculate the volume of a cone 14cm in base diameter and 24cm high.

Solutions

1. Volume of a rectangular based pyramid = 1/3 x l x b x h
= 1/3 x 6 x 4 x 8 cm3
= 8 x8 cm3
= 64cm3

Considering the square based ABCD

|DB| 2 = |DC| 2 + |CB|[sup]2[/sup]
Pythagoras rule

|DB|[sup]2[/sup] = 4[sup]2[/sup] + 4[sup]2[/sup]

|B|[sup]2[/sup] = 16 + 16.

:. √ |DB| = √ 32

|DB| = 4 √2 cm
but
|EB| = ½ |DB|
Since t is the midpoint of |DB|

Then |EB| = ½ X 4 X √ 2

= 2 √2 cm.

Now
Consider OEB

OE2 + EB 2 = ( OB)2
OE 2 + ( 2√2) 2 = ( 6) 2
OE 2 + 4 x 2 = 36
OE 2 + 8 = 36
OE 2 = 36 – 8
OE2 = 28

OE = √28

OE = √4 x 7
OE = √4 x 7
OE = 2 x √ 7 cm
OE = 2 √7cm
But OE =height of the pyramid = 2√7
:.volume of square of based pyramid = 1/3 x b2 x h

1/3 x 42 x 2 x √7cm3

1/3 x 16 x 2 x √7cm3

= 32 x √7 cm3
3
32 x 2.646cmm3
3
= 32 x.0.882cm3
= 28. 224cm3
= 28.2cm3 to 1 d.p.


Since
Diameter = 14cm

Radius = [sup]diameter[/sup]/[sub]2[/sub]

= [sup]14[/sup]/[sub]2[/sub] cm.

:. Volume of cone = [sup]1[/sup]/[sub]3[/sub] πr[sup]2[/sup] h

[sup]1[/sup]/[sub]3[/sub] x [sup]22[/sup]/[sub]7[/sub] x ( 7 )[sup]2[/sup] x 24

[sup]1[/sup]/[sub]3[/sub] x [sup]22[/sup]/[sub]7[/sub] x 49 x 24 cm[sup]3[/sup]

= 22 x 56cm[sup]3[/sup]

= 1232 cm[sup]3[/sup]
https://www.youtube.com/playlist?list=P ... tNQ93t01xf
https://youtu.be/fkdB0Zfme0w

Evaluation
1. A cone of height 9cm has a volume of n cm3 and a curved surface area of n cm3. Find the vertical angle of the cone

2. A right pyramid on a base 8cm square has a slant edge of 6cm. Calculate the volume of the pyramid

Assignment
1.Calculate the volume of a cylinder which has a radius of 21cm and height 6cm.

A. 8500cm3 B. 8316cm3 C. 7632cm3 D 7500cm3 E. 8000cm3
2. Calculate the total surface of the cylinder in question 1.
A, 5346cm2 B, 4653cm3 C. 3000cm2 D. 3564 cm2 E 3800cm2
3. Calculate the volume of a cone which has a base diameter of 7cm and a height of 6cm
A. 77cm3 B. 70cm3 C. 88cm3 D. 90cm3 E. 65cm3

4. Calculate the curved surface area of the cone in question 3 above.
A, 152cm2 B. 150cm2 C. 132cm2 D 142cm2 E. 160cm2

5. Calculate the total surface area of a cuboids which is 8cm by 5cm by 3cm.
A.198cm2 B. 178cm2 C 188cm2 D 168cm2 E. 158cm2.

Theory
1. A water tank is 1.2m square and 1.35m deep. It is half full of water . How many times can a 9 litre bucket be filled from the tank?

2. A measuring cylinder of radius 3cm contains water to a height of 49cm. If this water is poured into a similar cylinder of radius 7cm, what will be the height of the water column?.

Reading Assignment

NGM SS Bk 1 pg 166- 170 Ex 15a Nos 1 (d), 1(f), 2(b) and 29c) pages 168 -169.




TOPIC: ADDITION AND SUBTRACTION OF VOLUMES
Many composite solids can be made by joining basic solids together. In the figure below, the composite solids are made as follows:

(a) a cube and a square based pyramid.
(b) A cylinder and cone

Examples:
1. A figure shows a composite solid consisting of a cube of edge 28cm and a square-based pyramid of height 16cm. Calculate the volume of the solid
2. The outer radius of a cylindrical metal tube is R and t is the thickness of the metal.
(a) Show that the volume Y, of metal in a length, I units, of the tube is given by
V = II l t (2R – t)

(b) Hence calculate V when R = 7.5, t = 1 and 1 = 20

Solutions
1. From the diagram of the composite solid given
Volume of Composite solid = Vol. of square-based Pyramid + Vol. of cube

= [sup]1[/sup]/[sub]3[/sub] b[sup]2[/sup]h + l[sup]3[/sup]

= [sup]1[/sup]/[sub]3[/sub] x 28 x 28 x 16 + 283 cm[sup]3[/sup]

= 784 x 16 + 28 x 28 x 28 cm[sup]3[/sup]

= [sup]12544[/sup]/[sub]3[/sub] + 784 x 28 cm[sup]3[/sup]

= [sup]12544[/sup] /[sub]3[/sub] + 21952 cm[sup][sup]3[/sup][/sup]

= [sup]12544 + 65856/[sub]3[/sub] cm[sup]3[/sup]

= 26133 [sup]1[/sup]/[sub]3[/sub] cm[sup]3[/sup]

= 26133cm[sup]3[/sup]

Vol. of the Cylindrical metal Tube = Vol. of outside cylinder - Vol. of inside cylinder

= II R[sup]2[/sup] L - II r[sup]2 [/sup]L

But
R = t + r

Where
R = radius of outside cylinder
T = thickness of the cylindrical metal tube
R = radius of inside cylinder

From equation (2)
R = r - t
And substituting R – t for r in equ (1):

Vol of the cylindrical = ∏R[sup]2[/sup]l - ∏r[sup]2[/sup]l(metal tube)

= ∏R[sup]2[/sup]l – ∏(R – t)[sup]2[/sup]l

= ∏R[sup]2[/sup]l – ∏(R[sup]2[/sup] – 2Rt + t[sup]2[/sup]) l

= ∏R[sup]2[/sup]t – ∏R[sup]2[/sup]t + 2∏Rtl - ∏t[sup]2[/sup]l

= 2∏ Rtl – ∏t[sup]2[/sup]l

= ∏lt (2R – t)

(b) When R = 7.5, t = 1 and l = 20, then
Vol. of the cylindrical
Metal tube = ∏ l t (2R – t)
= 22/7 x 20 x 1 (2 x 7.5 - 1)
= 22/7 x 20 x (15 – 1)
= 22/7 x 20 x 14
= 44 x 20
= 880
https://youtu.be/IRQmu0aH-uI

EVALUATION
1. Calculate the approximate mass in kg of a 2-m length of cylindrical clay pipe of external and internal diameters 15cm and 12cm. The density of clay is 1.3g 1 cm3

2. A house is built in the shape of a cylindrical base with a conical roof. Calculate the volume of the house.




Volume of frustums of cone and pyramid
If a cone or pyramid standing on a horizontal table is cut through parallel to the table, the top part is smaller cone or pyramid. The other part is called a frustum.
To find the volume or surface area of a frustum, it is necessary to consider the frustum, as a complete cone (or pyramid) with the smaller cone (or pyramid) removed.
https://youtu.be/ja8yODSW5Ng

Examples:
1. Volume of a right circular cone is 5 litres. Calculate the volumes of the two parts into which the cone is divided by a plane parallel to the base, one-third of the way down from the vertex to the base. Give your answers to the nearest ml.

2. A right pyramid on a base 10m square is 15m high.
(a) Find the volume of the pyramid.
(b) If the top 6m of the pyramid are removed, what is the volume of the remaining frustum?

Solutions:
From the question,
h = 1
H 3


Also using similar triangles:
H = 3h

[sup]r[/sup]/[sub]R[/sub] = [sup]h[/sup]/[sub]H[/sub] = [sup]1[/sup]/[sub]3[/sub]

Thus:
R = 3r

Vol. of frustum Of cone = Vol. of big cone - Vol. of small cone

= [sup]1[/sup]/[sub]3[/sub] ∏ R[sup]2[/sup] l t - [sup]1[/sup]/[sub]3[/sub] ∏ r[sup]2[/sup] h

But vol. of big cone = 5 litres
= 5 x 1000 ml
Since 1 litre = 1 ml

i.e.
Volume of Big cone = 5000 ml

1 ∏R[sup]2[/sup] H = 500 ml

∏R[sup]2[/sup] H = 3 x 5000 ml

∏R[sup]2[/sup]H = 15000 ml - (i)
Also,
From 3h = H

h = [sup]H[/sup]/[sub]3[/sub]

and 3r = R

r = [sup]R[/sup]/[sub]3[/sub]

Thus, vol. of small Cone = [sup]1[/sup]/[sub]3[/sub] ∏r[sup]2[/sup]h



Vol. of the Small cone = Vol. of the big cone - Vol. of the frustum
= 5000ml - 4815 ml
= 185 ml

Hence:
Vol. of small cone = 185 ml
And vol. of frustum = 4815 ml

https://youtu.be/Ig-Tv4NEqg4

2.
(a)

And


:.Vol.of frustum of pyramid = [sup]1[/sup]/[sub]3[/sub]B[sup]2[/sup]H – [sup]1[/sup]/[sub]3[/sub]b[sup]2[/sup]H

= [sup]1[/sup]/[sub]3[/sub] x 10 x 15 – [sup]1[/sup]/[sub]3[/sub] x (4)[sup]2[/sup] x 6

= 100 x 5 – 16 x 2m[sup]3[/sup]

=500 – 32m[sup]3[/sup]

= 468m[sup]3[/sup]
https://youtu.be/jHQhs9SfZM4

Evaluation
1. A lampshade in the form of a frustrum of a cone has a height of 12cm and an upper and lower diameters of 10cm and 20cm.
a. what is the curved surface area of the frustum?
b. What is the volume of the frustum?
c. Give both answers in terms of II

3. A frustum of a pyramid is 16cm square at the bottom, 6cm square at the top and 12 cm high. Find the volume of the frustum.

Assignment
1. Calculate the volume in cm3 of the material in a cylindrical pipe 1.8m long, the internal and external diameters being 16cm respectively.

2. A composite solid consisting of a cone on top of a cylinder. The height of the cone is 25cm. the height and base diameter of the cylinder are 40cm and 30 respectively. Calculate to 3.s.f. the volume of the solid, taking II to be 3.14
3. A storage container is in the form of a frustum of a right pyramid 4m square at the top and 2.5m square at the bottom. If the container is 3m deep. What is its capacity in m3?

4. The cone in the figure below is exactly half full of water by volume. How deep is the water in the cone?

Reading Assignment

NGM SS Bk 1 pages 173-175 Ex 15c Nos 6 and 9 pg 175.

MAIN TOPIC: Mensuration
SPECIFIC TOPIC: The circumference of the circle
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 3
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
(1) Find the radius of the circle.
(2) Find the diameter of the circle.
(3) Calculate the circumference of the circle

CONTENT: RADIUS AND DIAMETER
A circle is a closed curved edge consisting of points which are at a constant distance from a fixed point.
The constant distance is called radius of the circle.
A diameter of a circle is a line segment joining any point of the circle to another, and passing through the centre. The length of a diameter is, therefore, twice the radius.

RADIUS;



The short line here is radius of the circle.
The line is diameter.
The circumference is 2∏r.
C = 2∏r
Example 1: Calculate the circumference of this circle, giving your answer to 2.d.p.
Solution;
C = 2∏r.
C =2∏ × 3.
C = 18.85.
The circumference is 18.85 cm.
Example 2; If the circumference of this circle is 12cm,calculate the radius, giving your answer to 2.d.p
Solution;
C = 2∏r.
r= C
2∏
r=12/2∏
r=1.91.
The radius is 1.91cm.
https://youtu.be/3af1pL-tUJk
https://youtu.be/5ewzElH9trk

EVALUATION: Calculate the circumference of each circle, giving your answer to 2.d.p.
(1)r=4cm
(2) r=3.5cm
(3) r=9.2cm
(2) Calculate the radius of a circle of the circle when the circumference is (a) 15cm (b) 4m (c) 8mm.

ASSIGNMENT
The wheel of a car has an outer radius of 25cm.Calculate; how far the car has travelled after one complete turn of the wheel.




SPECIFIC TOPIC: arc and chord
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on arc and chord of a circle.
CONTENT: ARC AND CHORD OF A CIRCLE.
An arc of a circle is a connected part of a circle. A semi-circle, which is half of a circle is an arc.
An arc smaller than a semi-circle is called a minor arc.
An arc bigger than a semi-circle is called a major arc.
https://youtu.be/rR9oCNnWlI0

Length of an arc of a circle of radius, r, and subtending, θ, at the centre is
2∏rθ or ( θ x 2∏r)
3600 360 1
Example 1: In a circle of radius 12cm,find the length of an arc which subtends angle 1050 at the centre. (take ∏ =22/7)
Solution
θ =1050
r=12cm
∏=22/7.

Length of arc = [sup] θ[/sup]/[sub]360º [/sub] x [sup] 2∏r[/sup]/[sub]1[/sub]

= [sup]105º[/sup]/[sub]360º[/sub] x 2x[sup]22[/sup]/[sub]7[/sub]x12

= 22cm.

CHORDS OF A CIRCLE
Length of chord = 2x 2rSin θ


Example 2; If the angle subtended at the centre by a chord of a circle of radius 6cm is 120º ,find the length of the chord.
Solution;
and θ=60º
sin 60º = x/6.
X = 6 sin 60º
X = 3 3
Length of chord =2x.
= 2x3 root3.
=6 root 3cm.
https://youtu.be/lz8uQYAQk5c

EVALUATION:
(1) If an arc of length 33cm subtends 70º at the centre of a circle, find the radius of the circle.
(2) A chord of a circle of diameter 42cm subtends an angle of 60º at the centre of the circle. Find the length of the minor arc.

ASSIGNMENT:
An arc of length 33cm subtends 70º at the centre of a circle, find the radius of the circle.

Topic
Trigonometric Ratios
Sine, Cosine and Tangent of acute angles.


Given a right angled triangle, the trigonometric ratio of acute angles can be found as show below


Examples
1. A triangle has sides 8cm and 5cm and an angle of 90º between them. Calculate the smallest angle of the triangle

2.A town Y is 200km from town X in a direction 040º. how far is Y east of X ?
3. In a given figure LK is 1 MN. Calculate MNL

Tan θ = [sup]8cm[/sup]/[sub]5cm[/sub] = 1.600

O = tan 1.6000
Θ = 58º.
The 3rd angle in the right angled triangle above = 90º – 58º = 32.
Hence, the smallest angle of the given triangle = 32º

From the diagram drawn, the distance = ZY
Y east of X
Using the right angled triangle XZY
Sin 40º = ZY
200km
ZY = 200km x sin 40º
= 200km x 0.6423
= 128.46km
= 128.5km to 1d.p

In the given diagram,

LK = [sup]Sin 70º[/sup]/[sub]7cm[/sub]

LK = 7cm x sin 70 ………… i
LK = 7cm x 0.9397
LK = 6.5779cm

In LKN

Lk = sin MNL
21

i.e Tan x 0.9397 = Sin MNL
21cm
0.9397 = Sin MNL
3
0.3132 = Sn MNL
0.3133 Sin 0.3132 = MNL
0.3134 18. 290 = MNL
0.3135 i.e MNL = 18.3

Evaluation
A ladder 20cm long rests against a vertical wall so that the foot of the ladder s 9m from the wall
(a) find, correct tot eh nearest degree, the angle that the ladder makes with the wall
(b) find correct to 1.dp the height above the ground at which the upper end of the ladder touches the wall..
use of tables of trigonometric ratios.

Determination of lengths of chords using trigonometric rations.
Trigonometric ratios can be used to find the length of chords of given circle. However, in some cases where angles are not given, Pythagoras theorem is used to find the lengths of chords in such cases Pythagoras theorem is stated as follows:


C[sup]2[/sup] = a[sup]2[/sup] + b[sup]2[/sup]
Pythagoras theorem states that in a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the square of the lengths of the other two sides.

Examples
1. A chord is drawn 3cm away from the centre of a circle of radius 5cm. Calculate the length of the chord.
2. In the figure below, O is the centre of circle, HKL. HK = 16cm, HL = 10cm and the perpendicular from O to the HK is 4cm. What is the length of the perpendicular from O to HL?
https://youtu.be/vLwJY4DHhXs

3.Given the figure below calculate the length of the chord AB.

Solutions
Area of the = Area of shaded segment sector -
From the diagram above in ABO:

AB[sup]2[/sup] + 3[sup]2[/sup] = 5[sup]2[/sup] ( Pythagoras theorem)

AB[sup]2[/sup] = 5[sup]2[/sup] - 3[sup]2[/sup]

AB[sup]2[/sup] = 25 – 9

AB[sup]2[/sup] = 16

AB = √16 = 4cm

Since B is the mid point of chord AC then

Length of chord AC = 2 x AB
= 2x 4cm =8cm

OH[sup]2[/sup] = HM[sup]2[/sup] + MN[sup]2[/sup]

OH[sup]2[/sup] = 8[sup]2[/sup] + 4[sup]2[/sup]

= 64 + 16
= 80
:. OH = √80

:. OH = √80cm

but OH = radius of the circle

I,e r = OH = OL = √80cm

In < ONL

OL[sup]2[/sup] = ON[sup]2[/sup] + NL[sup]2[/sup]

i.e ( √80)[sup]2[/sup] = x[sup]2[/sup] + 5[sup]2[/sup]

80- 24 = x[sup]2[/sup]

55 = x[sup]2[/sup]
Take square root of both sides

√55 = √x[sup]2[/sup]

√55 = x = 7. 416cm
:. The length of the perpendicular from O to HL is 7.416cm

3.The perpendicular from O to AB divides the vertical angle into 2 equal parts and also divides the length of chord AB into two equal parts.

In < ACO

[sup]AC[/sup]/[sub]OA[/sub] = [sup]Sin 29[/sup]/[sub]1[/sub]

Cross multiply

AC = OA x sin 29

AC = 14cm x Sin 29

AC = 14cm x 0. 4540

AC = 6.356cm

AB = 2 x 6.356 cm

AB = 12. 7cm
:. The length of the chord AB = 12.7cm to 1 d.p
https://youtu.be/u7-Ci45twC8

Evaluation
1. A chord 30cm long is 20cm from the centre of a circle . Calculate the length of the chard which is 24cm from the centre .

2. Q is 1.4km from P on a bearing 023o. R is 4.4 Km from P on a bearing 113o. Make a sketch of the positions of P, Q and R and hence calculate QR correct to 2 s.f.

Assignment
1. If Sin A = 4/5, what is tan A?
A 2/5 B. 3/5 C. ¾ D. 1 E. 4/3
2. Use tables to find the value of 8 Cos 77.
A. 5.44 B/ 6.48 C. 9.12 D 7,57 E. 1.80
3. If cos θ = Sin 33 o, find tan θ.
A. 1.540 B. 2.64 C.0.64 D. 1.16 E. 1.32
4. If the diagonal of a square is 8cm, what is the area of the square ?
A 16cm2 B. 2cm2 C. 4cm2 D. 20cm2 E. 10cm2
5. Calculate the angle which the diagonal in question 4 makes with any of the side of the square.
A. 65º B. 45º C. 35º D. 25º E. 75º

Theory
1. From a place 400m north of X, a student walks east wards to a place Y which is 800m from X. What is the bearing of X from Y?

3. In the figure below, the right angles and lengths of sides are as shown. Calculate the value of K .

Reading Assignment
NGM SS BK 1 pg 114- 123, Ex 11a .
NOS 10 and 25 pg 117 -118

Topic: Geometrical construction Part 1

Drawing and Bisection of line segments

To bisect a given line segment means to divide the given line segment into two parts of equal length. The steps to take to bisect a given line segment are as follows:

1. Draw the given line segment AB ( let AB = 8.6cm)

2. With centre A and radius of about ¾ of length of AB, draw an arc above and below the line AB

3. With centre B and the same radius used in step 2 above, draw arcs to cut the previous arcs in step 2.

4.Draw a line through the 2 points of inter section of the pair or arcs obtained from steps 2 and 3 . The line drawn in the perpendicular bisector of line AB.

Thus AE = EB = 4.3cm
https://youtu.be/2niufl8BvLk
https://youtu.be/LvKOtzWU52w

Evaluation
1. Draw a line CD = 11cm
Bisect the line CD
2. Construct the mid point M of the line drawn below
Where length CD = 10.6cm

https://youtu.be/2lP1NKYLKQw


Construction and Bisection of angles : 90º, 45º, 135º, 22½º , 67½º.
To construct angle 90º, take the following steps:
1. Draw a line BC and mark a point A at which the angle 90º is to be constructed .
2/ With centre B and any suitable radius draw an arc above line BC.
3. With centre C and the same radius used in step 2, draw an arc to cut the previous Arc at D.
4. Draw a line through points A and D. thus < DAB = <DAC = 90º
Since 45º = ½ of 90º, angle 90º is bisected to obtain angle 45º. This is shown in the figure below:



Thus < IGF = 45º
Also < HGI = 45º
https://youtu.be/pI1dztNXN_I
Similarly 22½º = ½ of 45º, By bisecting angle 45º, we can obtain angle 22½º as shown in the figure below:


Thus < EDB = 22½º. Also <EDB = 22½º

Also 135º = 90º + 45º. Thus by constructing angle 90º at a point on a line and bisecting the 90o on the other side, we can obtain angle 135º. This is shown in the figure below
https://youtu.be/O0zPb-5PemQ
Thus <FDE = 135º.
As explained above bisection of angle 135º will give angle 67½º


Bisection of a given Angle.
The step to take for bisecting a given angle are as follows.
1. Draw the given angle ABC i.e <ABC
2. With centre B and any suitable radius, draw an arc to cut AB at D and BC at E.
3. With centre D and any suitable radius, draw an arc
4. With centre E and the same radius as the one used in step (3) above, draw another arc to cut the previous arc at F.
5. Draw the line BF. Line BF is the bisector of ABC. This is shown in the figure below
https://youtu.be/U4q5rEw3ZpY
Thus < FBC = , ABF = ½ <ABC.

Evaluation
1. Construct angle 135º
2a. Construct angles 22½º
b. Construct angle 67½º

Construction and bisection of angle : 60º, 30º, 75º, 105º, 120º, 150º.

To construct angle 60º, the following steps must be taken:
1. Draw a line AB and mark the point A at which the angle 60o is to be constructed
2. With centre A and any convenient radius, draw an arc to cut line AB at C.
3. With centre C and the same radius used to draw the arc in step 2 above, draw another arc to cut the previous arc at D.
4. Draw line AD and extend it to E
5. Then ,EAB = 60o.



To construct angle 30º. Bisect angle 60º to give angle 30º, this is shown in the figure below:


Thus, <ABC = 30º

To construct angle 75º . Since 75º = 60 + ½ of 30º, then first construct angle 90º at a point on a straight line. Next construct angle 60º at the same point where angle 90º has been constructed. Then the angle 30º difference between the angle 90º and 60º is bisected to give 15º on either side . thus 60º + 15º = 75º. This is shown in the figure below:


Thus, <ABC = 75.
https://youtu.be/OVn0BI21-kA


Construction of angle 105º .
Angle 105º can be constructed by constructing 60º in the adjacent right angle at E and bisecting the remaining 30º. Thus 105 = 90 + ½ x 30. this is shown in the figure below:

Thus <FED = 105º.
https://youtu.be/so31TODhil0

To construct angle 120º. the following steps must betaken :

1. Draw a straight line AB and mark a point C on the line where the angle 120o is to be constructed.

2. With centre C and a suitable radius, draw a well extended arc to cut line CB at point D.

3. With centre D and the same radius used in step 2 above draw an arc to cut the extended arc in step 2 at point E.

4. With E as centre and the same radius, draw an arc to cut the extended arc at point F.

5. Draw line CF. Thus <FCB = 120. This is shown in the figure below:


https://youtu.be/x-U3ulR81-U


Construction of angle 150º.
Since 150 = 120º + ½ of 60º, first construct angle 120º on a straight line angle. Then bisect the adjacent 60º angle to get 30º. Thus 30º + 120º on the right hand side gives the required angle 150º. This is shown in the figure below:

<DCB = 150º
https://youtu.be/mswCUY71tQ4

Evaluation
1. Construct angle 70º
2a. Construct angle 105º
b. Construct angle 150º.

Construction of Triangle
Example
1. Construct ABC in which !AB ! = 7cm, !AC! = 9.5cm and ABC = 120o.Measure !BC!

2. Construct PQR in which !PQ != 5.5cm !QR! = 8..5cm and PQR = 75o . Construct M the midpoint of PR
Measure !QM


Solutions
1. First make a sketch of the triangle to be constructed .
Draw a line AB = 7cm
Then construct angle 120º at B with radius 9.5cm and centre A, draw an arc to cut the 120º at
C. Draw line AC
From the diagram !BC! = 4cm

2. First make a sketch of the triangle to be constructed.

a. Draw line PQ = 5.5cm
b. Construct angle 75o at Q
c. With centre Q and radius 8.5cm , draw an arc to cut the angle 75o at R.
d. Draw line QR.
From the diagram, !QM! = 5.5cm
https://youtu.be/NzcTKGxwCCE

Evaluation
a. Construct XYX in which !YZ! = 7.5cm XYZ = 60º and XZY = 45º
b measure !XY !and !XZ!

Assignment
1a. Use ruler and compasses to construct PQR in which Q = 90º, !QR! = 5cm and !PR! = 10C
c. Use Pythagoras theorem to check the result.
B. Measure !PQ!

2a. Construct ABC such that !AB! = 7cm, !BC! = 6cm and ABC = 60º
b. The bisector of C meets the perpendicular bisector of AC at X. Find the point X by construction
c. Measure !BX


Construction of Quadrilateral
Examples
1. Construct a quadrilateral ABCD in which AB is parallel to DC !AB 4cm, !BC != 5cm and !DC 7cm and !ADC = 105o..Measure the diagonal BD.

2. Use your ruler and compasses to construct the parallelogram PQRS in which !QR! = 5cm, !RS != 11cm and QRS = 134o.

b. Measure the length of the shorter diagram of PQRS

Solutions

First make a sketch of the quadrilateral to be constructed as shown in the figure below:


Steps of the required construction are stated as follows:

i. Draw DNC = 7cm with DN = 3cm and NC = 4cm

ii. Construct CDM = 105º

iii. With N as centre, radius 5cm draw an arc to cut Dm at A

iv. With A as centre and radius of 4cm draw an arc.

With C as centre and a radius of 5cm draw a second arc to cut the first arc at B

v. Join A to B and C to B to complete the quadrilateral ABCD.

By measurement , !BD!= 4.5cm
https://youtu.be/Pz64J1hJV8E

First make a sketch of the parallelogram PQRS

The step of the construction are stated as follows:

i. Draw line QR = 5cm

ii. Construct R = 135º

iii. With R as centre and radius 11 cm draw an arc to cut the angle 135o line at S.
iv. With S as centre and radius 5cm, draw an arc
v. With Q as centre and radius 11cm, draw a second arc to cut the arc of step iv/
This is point P
vi. Draw lines to join S to P and P to Q

vii. Draw dotted line through diagonal RP and measure it.

By measurement the length of the shorter diagonal PR is 8.7cm
Evaluation

a. Construct quadrilateral ABCD such that !AB! = 5cm !BD!= !DC! =8cm, ABD = 30o and BCD = 45o.
b. Measure the diagonal !AC!.



Construction of equilateral Triangle
An equilateral triangle is a triangle in which all the sides are of equal length and each of its angle is 60º.



Examples/
1. Construct an equilateral triangle XYZ such that !XY!= 5CM

2a. Construct an equilateral triangle ABC such that !AB!= 7cm
b. Construct the bisectors of A, B and C
c What do you observe?

Solutions
The required construction is

C. The bisectors of each angle meet each other at a point inside the equilateral triangle.
https://youtu.be/XBgwGROzzzk

Evaluation
a. Construct an equilateral triangle ABC such that !AB!= 8cm
b. Construct the midpoints of AB, BC, and CA
c. What do you observe?


Construction of Loci of moving points

1. Locus of points at a given distance from a fixed point.

In the figure below, O is a fixed point, Pi, P2 are at a constant distance x cm from O . The locus of the points is a circle of radius x cm.(see the figure below).

i. Locus of point at a given distance from a straight line

In the figure above AB is a straight line which continues indefinitely in both directions. Points Pi, P2, P3, P4 are each a distance x cm from AB. In two dimensions, the locus of the points consist of two straight lines parallel to AB, each at a distance x cm from AB.
Note that this locus consist of two separate lines.

iii. Locus of points equidistant from two given points.


In a given figure, x and y are two fixed points . Points Pi, P2, P3 are such that !P1X! = !P1Y!, !P2X!= !P2Y!and !P3Y. !. P1, P2, P3, lie on the perpendicular bisector of XY. The locus of the points is the perpendicular bisector of XY (shown in the figure above).

iv. Locus of Points Equidistant from two straight lines.
https://youtu.be/11jBBiF-INo

In a given figure, AB and CD are straight line which intersect at O. P1 is equidistant from AB and CD . Similarly, P2 is equidistant from the two lines. P1 and P2 lie on the bisector of the acute angle between the two lines .

In the figure above, P3 is equidistant from AB and CD.P3 lies on the bisector of the obtuse angle between the two lines.

Thus, the complete locus of points which are equidistant from two straight liens is the pair of bisectors of the angles between the lines.( see the figure below).
Note that the two parts of the locus intersect at right angles.

https://youtu.be/aCFJS6AlI5k

Example
Using ruler and compasses only
A, Construct ABC such that !AB! = 6cm, !AC! = 8.5cm and BAC = 120o

b. Construct the locus l1 of points equidistant from A and B,

c. Construct the locus l2 of points equidistant from AB and AC.

d. Find the points of intersection P1 and P2, of l1 and l2 and measure !P1P2!

Solution
a. Note the construction of BAC = 120.

b. l1 is the perpendicular bisector of AB

c. l2 is in two parts. AP1 is the bisector of BAC. Ap2 is perpendicular to Ap1, Note that points on AP2 are equidistant from AB and CA produced.
d. By measurement !P1P2! = 6.8cm

Evaluation
a. Construct a XYZ in which !YZ! = 8.2cm, XYZ = !45o and XZY = 75º.
measure !XY!.

b. Using ruler and compasses only, construct:

i. The locus of a point equidistant from Y and Z.
ii. A point Q on this locus, equidistant from YX and YZ.
Weekend Assignment

1. A circle centre O, radius 5cm is drawn on a sheet of paper. A point P moves on the paper so that it is always 2cm from the circle . The locus of O ___________

A. a circle, centre O, radius 3cm
B. two circles, centre O radii 3cm and 7cm
C. a circle, centre O, radius 6cm
D. two circles, centre O, radii 4cm and 6cm
E. a circle, centre O, radius 3.5cm.

2. XYZ is a straight line such that !XY! =!YZ!= 3cm .A point P moves in the plane of XYZ so that !PY! < !XY!, which of the following describes the locus of P?
A. line through X perpendicular to XZ
B. line through Y perpendicular to XZ
C. line through Z perpendicular to XZ
D. circular disc, centre X, radius 3cm
E. circular disc, centre 4, radius 3cm.

4. Describe the locus of a point which moves so that it is always 5cm from a fixed point O in a plane .
A. A rectangle which measures 10cm by 5cm
B. A square of side length 5cm
C. a parallelogram whose diagonals are 10cm and 5cm
D. a circle of radius 5cm, centre O
E. a circle of radius 10cm, centre O.

5. Describe the locus of a point which moves along a level floor so that it is 2m from a wall of a room.
A. One line, parallel to and 2m from the wall.
B. Two lines, one each side of, parallel to and 2m from the wall
C. A circle of radius 2m
D. A semi-circle of radius ½ m
E. Two perpendicular lines, each of length 2m

6. Describe the locus of a point which moves so that it is 3cm from a fixed line AB in a plane.
A. 2 lines parallel to AB and 6cm apart, joined by semi-circular ends.
B. 2 lines parallel to AB and 8cm apart; joined by semi-circular ends
C. 2 lines perpendicular to AB
D. A circle of radius 6cm
E. A circle of radius 3cm.

Theory
1. construct a trapezium ABCD in which AB is parallel to DC, AB =4cm BC = 8cm, CD = 11cm, DA = 6cm. (hint: in a rough figure, divide the trapezium into parallelogram AB X D and BCX. First construct BCX )

2. Using ruler and compasses only, construct ABC such that !AC! = 8.5cm and ACB = 135o.

b. Using any geometrical instruments, find a point p within ABC which is at a distance 2.8cm from AC and 6cm from B. Measure the length of AP.

Reading Assignment
NGM SS BK 1 pages 176-186 Ex 16e No.6 page 186.