Topic: LINEAR EQUATIONS Date: 12/01/2022 LEARNING OBJECTIVES: Understand the concept of ‘linear equation’ Evaluate linear equations without brackets and fractions When two expressions are equal, it can be mathematically stated through an equation. The two expressions which have formed the equation remain the same when; The same quantity is added to both sides of the ….  Read More

Topic: Linear Equations Date: 13/01/2022 Learning Objectives: Evaluate Linear equations without brackets and fractions Examples Solve the following equations 4x – 3 = 2x + 5 5x + 3 =3 13 + 4y = 6y + 18 Solutions: 4x -3 =2x + 5 Step 1: Collect like terms (4x and 2x are like terms, likewise ….  Read More

Topic: LINEAR EQUATIONS Date: 14/01/2022 LEARNING OBJECTIVE: • Evaluate linear equations with brackets If an equation involves brackets, remove the brackets first before collecting like terms in the process of removing the bracket, if any number is in front of a bracket, multiple each number in the bracket with the number outside the bracket to ….  Read More

Topic: LINEAR EQUATIONS WITH FRACTIONS Date: 17/01/2022 LEARNING OBJECTIVE: • Evaluate linear equations with fractions.   Always clear fractions before beginning to solve an equation. To clear fractions, multiply each term in the equation by the L.C.M. of the denominators of the fractions. Solve the equation ¾x – 1 2/3 = 2/3x Step 1: Express ….  Read More

Solution to Evaluation x/2 = x/3 + 1/2 Solution: Step 1: the L.C.M of 2 and 3 is 6. Multiply every term by 6 (6×x)/2 = (6×x)/3 + (6×1)/2 3x = 2x + 3 Step 2: collect like terms 3x – 2x = 3 x = 3 therefore, the solution to the equation is x ….  Read More

Topic: Word problems leading to linear equations Date: 19/01/2022 LEARNING OBJECTIVE: • Understand the concept of word problems as they lead to Linear equations Simple tips to use: (1 Understand the problem before you start solving (2 Find what the unknown is. (3 Choose a letter to represent the unknown (4 Translate each statement given ….  Read More

Solution: Let the numbers be x, (x+1) and (x+2) respectively. The sum of the  numbers are equal to 48. Therefore, x + (x+1) + (x+2) = 48 x + x +1 + x +2 = 48 Collect like terms x+x+x = 48 -1-2 3x = 45 Divide both sides by 3 to get x 3x/3 ….  Read More

Topic: Word problems leading to linear equations Date: 20//01/2022 LEARNING OBJECTIVE: • Understand better the concept of word problems leading to equations Further Examples (1 A mother is four times as old as her daughter. In 6 years time, she will be three times as old. How old is the mother now” Solution Let the ….  Read More

Topic: Linear equations: a review Date: 21/01/2022 .Solve the following equations: (1 3x + 2 = 8 (2 2y + 3(y-5) = 0 (3 x/2 + x/3 + x/4 = 2 (4 3(6+7t) +2(1-5t) = 42 Solutions: (1 3x +2 =8 Collect like terms 3x = 8-2 (‘8’ and ‘2’ are like terms) 3x = ….  Read More

Topic: Change of Subject of formula Date: 24//01/2022 Learning Objective: 1 . Understand the basics of change of subject of formula.   In an equation, change of subject of formula can only be done when the equation contains more than one variable. The essence of doing it is to make the quantity the subject to ….  Read More

Solution to Evaluation: (1.     x – a + c = b Since we are making ‘x’ the subject of the formula, lets carry all other terms except x over to the R.H.S x = b +a – c (Note: when ‘-a’ crosses over it becomes ‘+a’ when ‘+c” crosses over it becomes ‘-c’) Therefore, ….  Read More

Topic: Change of Subject of formula Date: 26/01/2022 Learning Objectives: Change of subject of formulae A formula is an equation which letters represent quantities. Example: Given an equation v= u + at. Make u, a and t the subject, respectively. Solution: Making ‘u’ the subject v = u +at (since we are making ‘u’ the ….  Read More

1. x = πab Divide both sides with coefficient of ‘b’ which is πa x/πa = πab/πa x/πa = b Therefore, b = x/πa   2. I = bE/(R+br) Cross-multiply to eliminate the fractions I × (R+br) = bE × 1 IR + Ibr = bE Since we are making ‘b’ subject of the formula ….  Read More

TOPIC: Change of subject of formula DATE: 27/01/2022 LEARNING OBJECTIVE: Apply change of subject of formula in solving mathematical problems   Example 1: v = 1/3πr²h is the volume of a circular cone of base radius r and height h. Make r the subject of the formula and find r when the volume is 898.3cm3 ….  Read More

Solutions: 1 a) x/a + y/b =1 Making y the subject of the formula The denominators are ‘a’ and ‘b’. The L.C.M of the denominators is ab Multiply each term in the equation with the áb’ to eliminate the fractions. x/a ×ab +y/b×ab = 1×ab xb + ya = ab carry xb over ya = ….  Read More

  TOPIC: Variation SUB-TOPIC: Direct Variation DATE: 28/01/2022 LEARNING OBJECTIVE: Solve problems involving direct variation A variation is a relationship between two or more quantities. If the cost of one school bag is ₦630, what would be the cost of 2, 3, 4,5 …. n bags? You will realize that as the number of bags ….  Read More

1) C ∝ d ⸫ C = kd Where C =14, d = 2, k =? 14 = k × 2 14 = 2k Divide both sides of the equation with 2 14/2 = 2k/2 7 = k ⸫ k = 7 Therefore, the relationship between C and d is: C = 7d (2) (a) ….  Read More

TOPIC: Variation SUB-TOPIC: Inverse Variation DATE: 31/01/2022 LEARNING OBJECTIVE: Solve problems involving inverse variation When the relationship between two variables is such that as one increases the other decreases and vice versa, they are said to be inversely related. If x varies inversely as y, it is written as x ∝ 1/y. ⇒ x = ….  Read More

Solutions: 1) y ∝ 1/x² y = k/x² y × x²= k Where, y =10, and x =2 10 × 22 = k 10 × 4 = k 40 = k y = 40/x² Finding y when x= 10 y = 40/10² (Note 10² = 100) y = 40/100 y = 2/5 Therefore, y = ….  Read More

TOPIC: Variation SUB-TOPIC: Joint Variation DATE: 02/02/2022 LEARNING OBJECTIVE: Solve problems involving joint variation   In joint variation, the relationship is between more than two quantities. Consider the volume of a cylinder, v=πr²h. The volume will be affected by the value of r and h. The volume, therefore, jointly varies as the square of the ….  Read More

Solution 1. x ∝ y/√z x = ky/√z Where; x = 300, y=75 and z= 25 300 = k×75/√25 (Note: √25 is equal to 5) 300 = 75k/5 300 = 75k/5 Cross-multiply 300×5 = 75k 1500 =75k Divide both sides of the equation with the coefficient of k 1500/75 = 75k/75 20 = k ⸫ ….  Read More

      TOPIC: Variation SUB-TOPIC: Partial Variation DATE: 03/02/2022 LEARNING OBJECTIVE: Solve problems involving Partial variation   In partial variation, the relationship is expressed as a sum of quantities. It is made up of two constants. Care must be taken to differentiate between joint and partial variation. In partial variation, assuming the constants are ….  Read More

    Solution 1. (a) The formula connecting C and u. Since C is partly constant and partly varies as u, C = a + ub Where C= 2500.00, u=50 ⸫ 2500 = a + 50b ……………………… (1) Where C= 3000, u=120 ⸫ 3000 = a + 120b ………………………..(2) Subtract equation (1) from equation (2) ….  Read More

      TOPIC: Quadratic Expressions DATE: 07/02/2022 LEARNING OBJECTIVE: (1 Identify quadratic expressions (2 Factorize quadratic expressions that have the coefficient of x2 as 1. Quadratic Expressions: A quadratic expression is of the form ax² +bx + c, where a, b amd c are constants and a ≠ 0, and the highest power of ….  Read More

      Solution to the evaluations: 1. The quadratic expressions are expressions: (a), (d), (e) and (f). they satisfy the conditions a≠0 and the highest power of the x² (or any other like designation) is 2. 2. Factorize (a) a² + 3a + 2 Step1: 1×2 = 2 Step2: two factors of 2, whose ….  Read More

  TOPIC: Quadratic Expressions DATE: 09/02/2022 LEARNING OBJECTIVE: Factorize quadratic expressions that have the coefficient of x² not equal to one (not unity). In this quadratic expression, the coefficient of x² is not 1. When factorizing, we follow the same steps as when the coefficient of x² is equal to one.   Example 1: Factorize ….  Read More

  Solution to Evaluation (a) 12d² –13d –14 Step1: 12×14 = 168 Step2: two factors of 168, whose sum is –13. The required factors are –21 and +8. Sum: –21 + 8 = –13. Step3: replace the middle term with the two factors. 12d² –21d +8d – 14 Step4: Factorizing the expression in step 3. ….  Read More

  TOPIC: Quadratic Expressions SUB-TOPIC: Difference of two squares. DATE: 10/02/2022 LEARNING OBJECTIVE: (a) Factorize quadratic expressions of the form x² – y² using the difference of two squares. (b) Introducing the solutions of quadratic equations. A quadratic expression of the form x² – y² is referred to as ‘difference of two squares’. This is ….  Read More

  Solution to Evaluation (a) 6t² – 54 = 6(t² – 9) = 6(t²–3²) = 6(t + 3) (t –3)   (b) 81d² – 64 =9²d² – 8² = (9d)²– 8² = (9d + 8) (9d –8) (c) 100 – (2a – 3b)² =10² – (2a – 3b)² = (10 + (2a – 3b)) (10 ….  Read More

  TOPIC: Quadratic Equations DATE: 11/02/2022 LEARNING OBJECTIVE: (a) Solve quadratic equations Example 1 Solve x² + 8x = 0 Solution: x² + 8x = 0 x(x + 8) = 0 Either x = 0 0r x+8 = 0 x = 0 – 8 x = –8 ⸫ x = 0, –8 Example 2: 169 ….  Read More

  Solution to Evaluation: 1 x² + 3x = 0 x² + 3x = 0 x(x + 3) = 0 Either x = 0 0r x+3 = 0 x = 0 – 3 x = –3 ⸫ x = 0, x = –3 2) 20p² – 320 = 0 20p² – 320 = 0 Divide ….  Read More

1st C.A Subject : Mathematics Instruction: Answer any four questions.   (1) Factorize the expressions: (a) t² – 12t + 36 (b) 4x² + 13x + 9   (2) Solve the equations: (a) 4 + 5x = 8 – 2x (b) 5(x+3) –2(3x –2) =2   (3) Given the equation 1/f = 1/u + 1/v, ….  Read More

TOPIC: Quadratic Equations DATE: 28/02/2022 LEARNING OBJECTIVE: (a) Solve quadratic equations Example 1: Solve x² + 11x +18 = 0 Solution: x² + 11x +18 = 0 Factorizing the equation 1× 18 = 18 +9 and + 2, are the factors of 18 whose sum is 11 x² + 9x +2x +18 = 0 (x² ….  Read More

Solution to Evaluation (1) p² + 5p +6 = 0 Solution: p² + 5p+6 = 0 Factorizing the equation 1× 6 = 6 +3 and + 2, are the factors of 6 whose sum is 5 p² + 3p +2p +6 = 0 (p² + 3p) +(2p + 6) = 0 Collect like terms in ….  Read More

  TOPIC: Quadratic Equations DATE: 02/03/2022 LEARNING OBJECTIVE: (a) Construct quadratic equations with given roots The roots of quadratic equation are solutions of that equation. Given that ‘u’and ‘v’ are the roots of a given quadratic equation where x is the variable or unknown variable, then: Either, x = u and x = v Then ….  Read More

  Solution: Note: the variable used does not invalidate the answer, any letter of the alphabet could be used as the variable. (a) Let the unknown variable be ‘y’ Either y = 3 or y = –4 Then, y – 3 = 0 or y + 4 =0 (y – 3) (y + 4) = ….  Read More

  TOPIC: Quadratic Equations DATE: 03/03/2022 LEARNING OBJECTIVE: Draw the graphs of quadratic equations The graphical method is another method of solving quadratic equations. In solving quadratic equations graphically, we have to consider the following: (1) Make a table of values for the function which contains the values of x and corresponding values of y ….  Read More

  TOPIC: Quadratic Equations DATE: 07/03/2022 LEARNING OBJECTIVE: Draw the graphs of quadratic equations Example 1: Solve graphically the equation y = x² + 2x – 5, for values of x from –4 to +3. Using any suitable scale of your choice. Solution: To plot a graph you need the x values and their corresponding ….  Read More

  TOPIC: Quadratic Equations DATE: 09/03/2022 LEARNING OBJECTIVE: Draw the graphs of quadratic equations   Example : (a)     Draw the graph of the function y = 2×2 – x – 10 for values of x from –3 to +3. (b)    Use the graph to find solutions to the questions below: (i)      What is the minimum ….  Read More

Solution:   Solution: When x =  –3 y = 3×2 –5x  – 7 y = 3(–3)2 – 5(–3) – 7 y = 3×9 + 15 – 7 y = 27 + 15 – 7 y =  35 Therefore, y is 35 when x = –3   When x =  –2 y = 3×2 – 5x  ….  Read More

TOPIC: Quadratic Equations DATE: 11/03/2022 LEARNING OBJECTIVE: (a) Solve word problems leading to quadratic equations.   Word problems leading to quadratic equations: In solving word problems that lead to quadratic equation, the first thing to do is to translate the word problem into a quadratic equation. Then use any of the methods of solving quadratic ….  Read More

  Solution: (1 Let the two consecutive positive numbers be ‘y’ and ‘y +2’ respectively. Their product y × (y +2) is 195 Therefore, y(y + 2) = 195 y² +2y = 195 y² + 2y – 195 = 0 Solving by factorization y² +2y – 195 = 0 y² + 15y – 13y – ….  Read More

TOPIC: Logical Reasoning DATE: 14/03/2022 LEARNING OBJECTIVE: (1 . Define Logic (2 . Identify Simple statements.   Logic is a branch of philosophy that is mainly concerned with the evaluation of arguments based on fundamental laws of scientific reasoning and thinking. The term ‘logic’ is derived from the Greek word logos, which means thought, reason ….  Read More

  1. Logic is a method or science of reasoning, ability to argue and convince.   2a. A statement is a sentence which is false or true but not both. 2b. All sentences are not statements because all sentences do not have a truth value. For a sentence to become a statement it must have ….  Read More

TOPIC: Logical Reasoning DATE: 16/03/2022 LEARNING OBJECTIVE: (1      Deduce the truth or falsity of a simple statement. (2      Form the negation of  simple statement.   True and false Statements The truth value of a statement is the truth or falsity of the statement. To determine the truth or falsity of a simple statement, one requires ….  Read More

Solution. (1 (a False F (b True T (c False F (d False F (e True T (2 The negations ( ∼) are as follows. (a David is not a sailor (b Abuja is not the capital of Nigeria (c 3 + 2 is not equal to 5 (d 8 is not greater than 15. ….  Read More

TOPIC: Logical Reasoning DATE: 17/03/2022 LEARNING OBJECTIVE: (1      Distinguish between simple statements and compound statements. (2     Understand Conjunctions and disjunctions     Compound Statements: We have considered simple statements which contain simple ideas or prepositions. However, two or more simple statements that contain more than one idea, and that can be separated into two ….  Read More

  Solution: (1 The truth table for P ˄ Q if P:     Ikeja is the capital of Lagos and Q: 2 + 5 = 7is shown below. P Q P ˄ Q T T T T F F F T F F F F   (2 Translating P and Q into P: He is Handsome ….  Read More

TOPIC: Logical Reasoning DATE: 18/03/2022 LEARNING OBJECTIVE: (1      Understand conditional statements   Implicative (or conditional) statements: In mathematics, many statements are of the form ‘if P, then Q’. Such statements are called conditional statements or implications. When two simple statements are combined by ‘If……………….. then’, such that the first statement implies the second. They are ….  Read More

Solution: (1 (a       Antecedent – If mathematics teachers work hard Consequent – Then they will be compensated (b      Antecedent – If Bayo is humble and prayerful Consequent – he will succeed (c       Antecedent – If you pass the matriculation examination Consequent – Then you will be admitted into the university.   2. P⇒Q : If ….  Read More

Assignment/ 2nd C.A Subject : Mathematics Instruction: Answer all  questions.   (1      (a)      Draw the graph of  the equation y = 3×2 –5x – 7 for values of x from –3 to 4. (b)    Using a scale of 2cm =1 unit along the x-axis and 2 cm= 5units along the y-axis, draw the ….  Read More

  Topic: Revision of Second term’s work Date: 09/05/2022 We are going to solve random mathematics problems that fall within our scheme of work for last time. (1) Solve the following equations (a) 7(5n –4) –10(3n –2) = 0 (b) 5(3m +4) = 3(4m + 7) Solution: (a) 7(5n – 4) – 10(3n –2) =0 ….  Read More

Topic: Basic Geometric Theorems Sub-Topic: Angles and Elementary theorems Date: 11/05/2022 A proof is a logical statement, using evidence to establish a fact, a hypothesis or an argument put forward. On the other hand, a theorem is a general conclusion in science or mathematics which makes assumption in order to explain some observations. Geometry is ….  Read More

Topic: Basic Geometric Theorems Sub Topic: Geometric Theorems of Triangles Date: 12/05/2021.   Theorem 1: The sum of angles of a triangle is 180º. Fig 1: The shape is attached to the lesson note.   Given:  triangle PQR To prove: angle P + angle Q + angle R = 180º Construction: Produce QR to point ….  Read More

1. Let the third angles be xº. (a) ∴ 69º + 46º + xº = 180º   (sum of angles in a triangle is equal to 180º 115º + xº = 180º Collect like terms xº = 180º – 115º xº = 65º Therefore, the third angle is 65º (b) 35º + 55º + xº  = ….  Read More

Topic: Basic Geometric Theorem Sub Topic: Theorems of Triangles Date: 16/05/2022 Note: For the purpose of this lecture, ˄P represents ‘angle P’, while P˄RS represents a triangle with the angle at P Theorem 2: The exterior angle of a triangle is equal to the sum of the opposite interior angles. Fig 1: Given: ∆PQR with ….  Read More

Solution: 1. 180º – 4x + x + 180º – 3x = 180º Collect like terms 180º + 180º – 4x + x – 3x = 180º 360º – 6x = 180º –6x = 180º – 360º –6x = 180º –6x/–6 = 180º/–6 x = 30º Therefore, x = 30º 4x = 4 × 30º ….  Read More

Topic: Basic Geometric Theorem Sub Topic: Theorems of Triangles Date: 18/05/2022   Theorem 3: Pythagoras Theorem In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Fig 1: Given: ∆XYZ with ˄X = 90º, lines YZ, XZ and XY are of length x,y ….  Read More

Topic: Basic Geometric Theorem Sub Topic: Polygons Date: 19/05/2022 A polygon is a plane figure with three (3) or more sides, none of which intersects eaxh other, but all form an enclosed geometric figure. A polygon is named according to the number of sides it has. The table attached to the leasol shows the names ….  Read More

Solution: (1). Sum of interior angles of a polygon = (2n – 4) right angles =(2n – 4)×90º = (2×20 –4)×90º          (Note: n is 20, because a 20 sided polygon has 20 sides =(40– 4)×90º =36× 90º = 3240º ∴ the sum of interior angles of a 20 sided polygon is 3240º ….  Read More

Topic: Basic Geometric Theorem Sub Topic: Isosceles and Equilateral Triangle Date: 20/05/2022   An isosceles triangle is one with two of its sides equal. The third side is called base. The two equal sides meet at a point called the vertex and the angle at the vertex is known as the vertical angle. An equilateral ….  Read More

  Solution (1 ) ^A= 75º A^BC =AB^C (base angles of isosceles triangle) ⸫ ^ABC + A^BC +AB^C = 180º (sum of angles in a triangle) 75º + A^BC + AB^C = 180º A^BC + AB^C = 180º – 75º A^BC + AB^C = 105º Then, A^BC = AB^C = 105º/2 = 52.5º (base angles ….  Read More

Topic: Trigonometry Sub Topic: Basic Trigonometric Ratios Date: 23/05/2022 Learning Objective: Define and derive the ratios of sine, cosine and tangent of angles   The basic trigonometric ratios are Sine, Cosine and Tangent abbreviated as ‘sin’, ‘cos’ and ‘tan’ respectively. A right angled triangle is a triangle that has one of its angles equals 90º. ….  Read More

Topic: Trigonometry Sub Topic: Basic Trigonometric Ratios Date: 25/05/2022 Learning Objective: Solve problems involving basic trigonometric ratios Example 1: Calculate (a) line PR i.e. z (b) line RS, i.e. y in fig 1 below. Give your answers correct to 3.S.f Solution: Line PR = x, Line PQ = z and line RS = y Therefore, ….  Read More

Solution: 1) From the triangle, Hypotenuse = x Adjacent = 15 Cos 60º = 15/x 0.5 = 15/x Cross-multiply 0.5x = 15 0.5x/0.5 = 15/0.5 x = 30 (2) Solving for x Hypotenuse = x Opposite = 15 Sin 30º = 15/x 0.5 = 15/x Cross-multiply 0.5x = 15 0.5x/0.5 = 15/0.5 x = 30

Topic: Trigonometry Sub Topic: Complementary Angles Date: 26/05/2022 Learning Objective: (1) Understanding Complementary Angles (2) Solve more problems on basic trigonometry ratios Complementary Angles Two angles are said to be complementary if their sum is 90º. In the triangle in fig 1, angle ^A and angle ^C are complementary angles. (i.e. ^A + ^C = ….  Read More

Topic: Trigonometry Sub Topic: Pythagoras Theorem Date: 30/05/2022 Learning Objective: (1) Introducing the Pythagoras Theorem Theorem: In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares of the other two sides. Fig 1 Given: Triangle ABC with right angle at A and sides line BC, line CA and ….  Read More

Solution Solving for x. Using the left side triangle Hypotenuse = 26. ‘10 and x’ are the other two sides. 26² = 10² + x² 676 = 100 + x² 676 – 100 = x² 576 = x² x² = 576 x = √576 x = 24 Solving for y. Using the right side triangle. ….  Read More

Lesson 12, 2021/2022. Topic: Trigonometry Sub Topic: Trigonometry ratios of special angles Date: 01/06/2022 Learning Objective: (1) Introducing the special angles. Deriving special angles is an approach which helps us to obtain trigonometric ratios of angles without using the trigonometric tables or calculators. Trigonometric ratios of 45º Consider an isosceles triangle with a unit length. ….  Read More

  Solution: 1. Solving for x. cos 60º = 6/x but cos 60º = 1/2 ∴ 1/2 = 6/x Cross-multiply 1 × x = 6×2 x = 12 Solving for y tan60º = y/6 But tan60º = √3 √3 = y/6 Cross-multiply √3 × 6 = y × 1 6√3= y y = 6√3 Therefore, x ….  Read More

Topic: Trigonometry Sub Topic: Trigonometry ratios of special angles. Date: 02/06/2022 Learning Objective: (1) Solving extensively more problems on trigonometric ratio. Example 1: A ladder of length 5m rests against a vertical wall and makes an angle of 60º with the ground. How far is the foot of the ladder from the wall? Solution Let ….  Read More

Topic: Mensuration of plane shapes Sub Topic: Introduction Date: 03/06/2022 Learning Objective: (1) Revision of the mensuration of plane shapes. Mensuration is the process of measuring and/or calculating lengths, areas and volumes of geometrical shapes. Areas of plane shapes; Circle: Fig 1: The formulae for calculating the area and circumference of a circle are: Area ….  Read More

  Solution 1. Area of a square = length × length Therefore, Area = 26.4 × 26.4 Area = 696.96 cm² Perimeter of a square = 4 × length Perimeter = 4 × 26.4 Perimeter = 105.6 Perimeter = 105.6 cm Therefore, area and perimeter are 696.96cm² and 105.6 cm respectively. 2. Area of a ….  Read More

Topic: Mensuration of plane shapes Sub Topic: Introduction Date: 06/06/2022 Learning Objective: (1) Solving more problems on mensuration of plane shapes Example 1: The length of a rectangular garden is twice its width. If the given perimeter is 72m, calculate the width and area of the rectangle. Solution: Length of the rectangular garden is twice ….  Read More

  Solution: 1. Area of square = area of rectangle Area of rectangle = L×W (i.e. length × width) Area of square = d×d = d² Therefore, d2 = L× W where. d = ?, L = 45cm, W= 5cm d²= 45 × 5 d² = 225 d = √225 d = 15 cm Therefore, ….  Read More

Topic: Mensuration of plane shapes Sub Topic: Length of an arc. Date: 08/06/2022 Learning Objective: (1) Evaluating problems involving the length of an arc. An arc of a circle is a fraction of the total distance around the circle (circumference). ­Example 1: Draw a circle with center 0 and any seven points on the circumference ….  Read More

Solution Length of arc, L = θ/360º × 2πr Where, θ = 105º , π = 3.142, r = 6cm Therefore, L = 105º/360º × 2 × 3.142 × 6 L = 0.2917 × 37.7 (note: 105º/360º equals 0.2917, while 2 × 3.142 × 6 equals 37.7 ) L = 10.9 cm L = 11cm ….  Read More

Topic: Mensuration of plane shapes Sub Topic: Perimeter of sectors and segments of a circle. Date: 09/06/2022 Learning Objective: Evaluating the perimeter of sectors. A sector  a part of a circle bounded by an arc and two radii. The shaded part of the circle in Fig 1 shows a sector, i.e. POQ. Perimeter of a ….  Read More

Solution Perimeter of a sector = 2r + θ/360o × 2πr Where, perimeter of a sector = ?  r = 35 cm              π= 3.142 θ = 150o ⸫   Perimeter of a sector = (2×35) + (150o/360 × 2×3.142× 35) = 70 + (0.4167 ×2 × 3.142 ×35) = 70 + (91.95) = 70 ….  Read More

  Topic: Mensuration of plane shapes Sub Topic: Area of sectors and segments Date: 10/06/2022 Learning Objective: Evaluating the area of sectors and segments of a circle.   Area of a sector: A sector of a circle is the area surrounded by two radii and an arc. The angle of the sector is θ. The ….  Read More

  Solution: (1) Area of a sector = θ/360o × πr2 Where, Area = ? r = 21 cm θ = 70o π = 3.142   Area of the sector = 70o/360o ×3.142×212 = 0.1944 ×3.142 ×441 = 269.34 = 269.5 cm2 Therefore, Area of the sector = 269.5cm2   2.   The area of ….  Read More

Topic: Surface area and volume of solid shapes Sub Topic: Surface area of solid shapes Date: 15/06/2022 Learning Objective: Find the surface area of solid shapes: cubes, cylinders, cuboids, pyramids, cones.   Solid shapes are three-dimensional objects. They have three dimensions – length, width, and height.   Surface area of solids Cube: This is a regular ….  Read More

Solution   The surface area of an open cuboid = The surface area of cuboid – Area of rectangle First find Total Surface Area (TSA) of whole cuboid which is , Given, Length (l) = 8cm Breadth/width (b) = 5cm Height (h) = 6cm   TSA of cuboid = 2(lb + bh + hl) = ….  Read More

Topic: Surface area and volume of solid shapes Sub Topic: Surface area of solid shapes Date: 16/06/2022 Learning Objective: Find the surface area of solid shapes: cubes, cylinders, cuboids, pyramids, cones.   Example 1: Calculate the surface area of a cylinder with height 35cm base radius 21 cm. use all the possible cases. Solution: There ….  Read More

Solution:   1. Curved surface area = πrl Where base radius of the cone, r = 3.5 cm Circle of radius/slant height of the cone, l = 8.20 cm π= 3.142 Curved surface area = πrl = 3.142 × 3.5 × 8.20 = 90.17 Curved surface area = 90.2 cm2.   Note: A sector of ….  Read More

Topic: Surface area and volume of solid shapes Sub Topic: Volume of solid shapes Date:22/06/2022 Learning Objective: Find the volume of solid shapes:prisms, pyramids, and spheres. Volume of prism The volume of any prism can be computed when the cross-sectional area is multiplied by the length or height. Volume of prism = area of cross-section ….  Read More

  Topic: Statistics Sub Topic: Date:23/06/2022 Learning Objective: 1. Collect data and present them in a tubular form. 2.  Construct frequency tables Introduction comes in different forms: words, pictures and numerically information that is presented in numerical form or as numbers called data. Statistics is the study and interpretation of information or data presented in ….  Read More

  Topic: Statistics Sub Topic: Bar Charts and Graphs Date:28/06/2022 Learning Objective: Draw line graphs, bar graphs, and bar charts.   Line Graphs Vertical or horizontal line graphs are one of the ways to represent data graphically. It is drawn after the construction of a frequency table. The lines are drawn on graph paper for ….  Read More

Topic: Statistics Sub Topic: Histogram Date:24/06/2022 Learning Objective: Draw histograms with equal and unequal sides.   A histogram is a series of rectangular blocks representing the frequency of a distribution. The blocks in a histogram are joined together and the height of each block is proportional to the frequency it represents. A histogram can be ….  Read More

Continuous Assessment Test Subject: Mathematics Date: 04/07/2022 Instruction: Answer any four questions.   (1a)   ABC is an isosceles triangle in which line AB = line AC = 5cm and line BC = 6cm. Calculate line AM, where M is the midpoint of line BC. (1b)   Solve the equation, cos Ɵ= sin(Ɵ + 40o)   (2a)   ….  Read More

YHWH YESHUA HAMASCHIAC School Summer/Spring term Examination Subject: Mathematics                                                      Session: 2021/2022 Class: SSS1                                                                    Time: 2hrs Date:     SECTION A Instruction: Answer all questions by selecting the correct option in each case   The edge of a cube whose volume is equal to the volume of a cuboid of dimensions 63 cm ×36 cm × ….  Read More