Excel in Mathematics - BOOK 8 Final (2080)_compressed

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 151 Vedanta Excel in Mathematics - Book 8 Creative Section - A 4. Solve: a) 5x – 2 = 23 b) 7y + 3 = 17 c) 5a – 1 = a + 11 d) 4x – 7 = 2x – 5 e) 3(x – 2) = 2 (x – 1) f) a + a 2 = 3 g) x 3 – x 4 = 1 h) p – 1 2 – p – 2 3 = 1 i) 2(x – 1) 3 – 3(x – 2) 4 = 1 j) y + 1 y – 1 = 3 4 k) x + 2 x + 3 = 1 2 l) 2 x = 3 x + 1 m) 1 x + 1 x + 1 = 2 x – 1 n) 2 y – 1 + 3 y + 1 = 5 y o) y – 1 y + 1 = 2y – 3 2y + 3 5. a) The sum of two numbers is 45. If one of them is 27, find the other number. b) The difference of two numbers is 6. If the smaller one is 15, find the greater number. c) The difference of two numbers is 10. If the greater one is 32, find the smaller number. 6. a) The sum of three consecutive odd numbers is 51. Find the numbers. b) The sum of three consecutive even numbers is 78. Find the numbers. 7. a) If the sum of two numbers is 37 and the greater number exceeds the smaller by 5, find the numbers. b) If the sum of two numbers is 56 and the smaller one is 8 less than the bigger one, find the numbers. c) If the sum of two numbers is 44 and their difference is 6, find the numbers. d) A sum of Rs 135 is divided into two parts. If the greater part exceeds the smaller by Rs 25, find the parts of the sum. e) A sum of Rs 190 is divided into two parts. If the smaller part is Rs 5 less than half of the greater part, find the parts of the sum. f) A sum of Rs 140 is divided into two parts. If the greater part is Rs 20 more than the double of smaller part, find the parts of the sum. g) A sum of Rs 105 is divided into two parts. If two times the greater part is 15 less than three times the smaller part, find the parts of the sum. It’s your time - Project Work and Activity Section 8. a)Let’s make linear equations in your own choice and solve each of them to get the following value of variables: (i) x = 2 (ii) x = 5 (iii) a = –1 (iv) y = –3 b)Let’s write the total number of students and the number of girls in your class. Let the number of boys be x. Then, make an equation and find x. c)Let’s write the total number of students and the number of boys in your class. Let the number of girls be y. Then make an equation to find y. 11.4 Linear equation with two variables Let’s consider an equation x + y = 7 This equation contains two variables x and y. Each variable has the power 1. Such equations are called the first degree equations and they are known as linear equations with two variables. Equation and Graph

Vedanta Excel in Mathematics - Book 8 152 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 11.5 Graph of linear equation Consider a linear equation x + y = 5. It can be written as y = 5 – x. It gives us the rule for taking values of x and using them to calculate values for y. For example, When x = 0, y = 5 – 0 = 5, When x = 1, y = 5 – 1 = 4 When x = 3, y = 5 – 3 = 2, When x = – 2, y = 5 – (– 2) = 7 Now, the results can be shown in a table given below. x 0 1 3 – 2 y = 5 – x 5 4 2 7 The values of x and y can be written in the form of coordinates, such as (x, y). Here, the values of (x, y) are (0, 5), (1, 4), (3, 2), and (– 2, 7). If these coordinates are plotted on graph and the points are joined, a straight line is obtained. Thus, a linear equation always gives a straight line. 11.6 Simultaneous equations Let’s consider a linear equation, y = x + 3 It is a linear equation on two variables x and y. The equation has as many pairs of solutions as we wish to find. The table given below shows a few pairs of solutions of the equation y = x + 3. x 0 1 3 6 – 1 – 3 – 5 y 3 4 6 9 2 0 –2 Thus, (0, 3), (1, 4), (3, 6), (6, 9), (– 1, 2), (– 3, 0), (– 5, – 2), ... are a few pairs of solutions that satisfy the equation y = x + 3. Again, let’s consider another linear equation, y = 4x – 6. A few pairs of solutions of this equation are shown in the table below. x 0 1 2 3 – 1 – 2 –3 y – 6 – 2 2 6 – 10 – 14 – 18 Thus, (0, – 6), (1, – 2), (2, 2), (3, 6), (– 1, – 10), (– 2, – 14), (– 3, – 18)... are a few pairs of solutions that satisfy the equation y = 4x – 6. If you observe the pairs of solutions of both equations, you will find that a pair (3, 6) is common to both equations. Here. (3, 6) satisfies both equations and it is the solution of both equations. Such pair of equations that have only one pair of solution are called simultaneous equations. (0,5) (1,4) (3,2) (–2,7) x + y = 5 X Y X' Y' O www.geogebra.org/classroom/sucjgfky Classroom code: SUCJ GFKY Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Equation and Graph

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 153 Vedanta Excel in Mathematics - Book 8 11.7 Methods of solving simultaneous equations There are various methods to solve simultaneous equations. Here, we discuss about three methods: (i) Graphical method In this method, we find a few pairs of solutions of each of two given equations in two separate tables. The pairs of solutions of each equation are plotted in a graph and by joining them two separate straight lines are obtained. The coordinates of the point of intersection of two straight lines are the solutions of the given equations. Worked-out Examples Example 1: From the given graph, find the solutions of the two equations represented by the straight lines. Solution: In the graph, the coordinates of the point of intersection of two straight lines l 1 , and l 2 are (2, 1). So, (2, 1) is the common solution of the equations represented by l 1 and l 2 . ∴ x = 2 and y = 1. Example 2: Solve graphically x + y = – 1 and 2x – y = – 11. Solution: Here, x + y = –1 or, y = –1 – x The points (0, – 1), (2, – 3) and (– 2, 1) are plotted on the graph and they are joined to get a straight line. Again, 2x – y = – 11 or, 2x + 11 = y or, y = 2x + 11 The points (– 3, 5), (– 4, 3) and (– 5, 1) are plotted on the graph and they are joined to get another straight line. In the graph, the coordinates of the point of intersection of two straight lines represented by the equations is (– 4, 3). ∴ (– 4, 3) is the common solution of the equations. So, x = – 4 and y = 3. Y' (2,1) Y X' X O l 2 l 2 Y' Y X' X (-4,3) O x + y = –1 2x – y = –11 x 0 2 – 2 y – 1 – 3 1 x – 3 – 4 – 5 y 5 3 1 Equation and Graph

Vedanta Excel in Mathematics - Book 8 154 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur EXERCISE 11.2 General Section - Classwork 1. Let’s say and write the solutions to each pair of simultaneous equations represented by straight line graphs x = ........, y = .......... x = ........, y = .......... x = ........, y = .......... Y' Y X' X O a) Y' Y X' X O b) Y' Y X' X O c) x 1 –1 y 4 0 x 0 5 y 0 4 x 1 5 y 0 2 x – 3 1 y 3 0 2. Let’s say and write the values of x and y, and complete the tables. a) y = x + 1 b) y = x – 3 c) y = 5 – x d) y = 4 + x Creative section 3. Copy and complete the table of values for x and y. Plot the coordinates separately from each table. Find the solutions of each pair of simultaneous equations. a) y = 7 – x y = x – 1 x 0 2 4 x 2 4 7 y y x = ........, y = .......... x = ........, y = .......... x = ........, y = .......... Y' Y X' X d) Y' Y X' O X e) Y' Y X' O X f) O Equation and Graph

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 155 Vedanta Excel in Mathematics - Book 8 b) y = 8 – x y = x + 2 x 0 x – 6 y 5 1 y 3 5 c) y = 2 – x y = x – 6 x 2 x 1 7 y – 1 3 y – 3 d) y = 10 – x y = 2x – 11 x x 1 2 6 y 9 6 2 y 4. Solve the following simultaneous equations graphically: a) y = 4 – x and y = x – 2 b) y = x + 3 and y = 7 – x c) y = 9 – x and y = x + 1 d) y = x – 4 and y = 10 – x e) y = 2x + 2 and y = x – 1 f) y = x + 3 and y = 3x – 1 g) y = x + 4 and y = 2x – 1 h) y = 2x – 4 and y = x – 3 i) x + y = 5 and x – y = 1 j) x + y = 8 and x – y = 2 k) 2x – y = 5 and x – y = 1 l) 3x + y = 7 and x = 2y m) 2x – y = 0 and 3x + 2y = – 14 n) 4x + 5y = 9 and 3x = 18 It’s your time - Project Work and Activity Section 5. Let’s make 5 pairs of simultaneous equations in your own choice and solve them graphically to get the following values of variables. (i) x = 2 y = 1 (ii) x = 3 y = 2 (iii) x = 1 y = 3 (iv) x = –1 y = –2 (v) x = 4 y = –3 (ii) Elimination method In elimination method, we add or subtract the given equations to eliminate one of the variables by making their coefficient equal. Then, we solve the equation with one variable and its solution is substitute in any simultaneous equation to get the value of another variable. Worked-out Examples Example 1: Solve x + y = 7 and x – y = 1. Solution: x + y = 7 .................... (i) x – y = 1 .................... (ii) Adding equations (i) and (ii) to eliminate y, x + y = 7 x – y = 1 2x = 8 or, x = 8 2 = 4 The coefficients of y are the same in both equations and they are of opposite signs. So, one equation is added to the other. Equation and Graph

Vedanta Excel in Mathematics - Book 8 156 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Substituting the value of x in equation (i), we get, 4 + y = 7 or, y = 7 – 4 = 3 ∴ x = 4 and y = 3 Example 2: Solve 2x + y = 5 and x – 2y = 5. Solution: 2x + y = 5 .................... (i) x – 2y = 5 ................... (ii) Multiplying equation (i) by 2 and adding equations (i) and (ii), we get, 4x + 2y = 10 x – 2y = 5 5x = 15 or, x = 15 5 = 3 Substituting the value of x in equation (i), we get, 2 × 3 + y = 5 or, 6 + y = 5 or, y = 5 – 6 = – 1 ∴ x = 3 and y = – 1. (iii) Substitution method In substitution method, a variable is expressed in terms of another variable from one equation and it is substituted in second equation. Then, the second equation containing one variable is solved and the value of the variable is substituted in any one equation to find the value of another variable. To eliminate y, the coefficients of y in both equations should be the same. So, equation (i) is multiplied by 2 to make the same coefficients of y. Example 3: Solve x + y = 4 and 2x + 3y = 6. Solution: x + y = 4 .................... (i) 2x + 3y= 6 .................... (ii) From equation (i), y = 4 – x .................... (iii) Substituting for y in equation (ii), we get, 2x + 3 (4 – x) = 6 or, 2x + 12 – 3x = 6 or, – x = 6 – 12 or, – x = – 6 or, x = 6 Now, substituting the value of x in equation (iii), we get, y = 4 – 6 = – 2 ∴ x = 6 and y = – 2. y is expressed in terms of x. Equation and Graph

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 157 Vedanta Excel in Mathematics - Book 8 11.8 Application of simultaneous equations We can use simultaneous equations to find unknown quantities asked in a verbal problem. From the two unknown quantities which are asked to find in a word problem, we make a pair of simultaneous equations under the two given conditions. By solving the equations, we find the unknown quantities. Worked-out Examples Example 1: The sum of two numbers is 27 and their difference is 3. Find the numbers. Solution: Let the greater number be x and the smaller number be y. From the first given condition, x + y = 27 or, y = 27 – x .................... (i) From the second given condition, x – y = 3 .................... (ii) Substituting the value of y from equation (i) in equation (ii), we get, x – (27 – x) = 3 or, x – 27 + x = 3 or, 2x = 3 + 27 or, 2x = 30 or, x = 30 2 =15 Substituting the value of x in equation (i), we get, y = 27 – 15 = 12 Hence, the required numbers are 15 and 12. Checking the answer: Here, 15 + 12 = 27 and 15 – 12 = 3, which are given in the question. Example 2: The difference of two numbers is 4. If six times the smaller number is equal to five times the bigger one, find the numbers. Solution: Let the smaller number be x and the bigger one be y. From the first condition, y – x = 4 or, y = x + 4 .................... (i) From the second condition, 6x = 5y .................... (ii) Substituting the value of y from equation (i) in equation (ii), we get, 6x = 5 (x + 4) or, 6x = 5x + 20 or, 6x – 5x = 20 or, x = 20 Substituting the value of x in equation (i), we get, y = 20 + 4 = 24 So, the required numbers are 20 and 24. When the bigger number is y, the smaller number is x and the difference is 4, then y – x = 4 6 times the smaller number is 6x 5 times the bigger number is 5y I can mentally check the answers ! 24 – 20 = 4 6 × 20 = 120 and 5 × 24 = 120 Equation and Graph

Vedanta Excel in Mathematics - Book 8 158 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 3: The total cost of 3 exercise books and 4 pens is Rs 230. If the cost of 2 exercise books is same as the cost of 5 pens, find the cost of 1 exercise book and 1 pen. Solution: Let the cost of 1 exercise book be Rs x and that of 1 pen is Rs y . From the first condition, 3x + 4y = 230 or, 4y = 230 – 3x or, y = 230 – 3x 4 .................... (i) From the second condition, 2x = 5y .................... (ii) Substituting the value of y from equation (i) in equation (ii), we get, 2x = 5(230 – 3x) 4 or, 8x = 1150 – 15x or, 8x + 15x = 1150 or, 23x = 1150 or, x = 1150 25 = 50 Substituting the value of x in equation (ii), we get, 2 × 50 = 5y or, 5y = 100 or, y = 100 5 = 20 Hence, the cost of 1 exercise book is Rs 50 and that of 1 pen is Rs 20. Checking the answer: Here, cost of 3 exercise book = 3 × Rs 50 = Rs 150 Cost of 4 pens = 4 × Rs 20 = Rs 80 Now, Rs 150 + Rs 80 = Rs 230 which is given in the question. Example 4: The sum of the age of a father and his son is 42 years. Six years ago, the father was 5 times as old as his son was. Find their present age. Solution: Let, the present age of the father be x years and that of the son is y years. From the first condition, x + y = 42 or, y = 42 – x .................... (i) From the second condition, x – 6 = 5 (y – 6) or, x – 6 = 5y – 30 or, x – 5y = – 30 + 6 or, x – 5y = – 24 .................... (ii) Substituting the value of y from equation (i) in equation (ii), we get, x – 5 (42 – x) = – 24 or, x – 210 + 5x = – 24 The present age of the father is x years. 6 years ago, his age was (x – 6) years. The present age of the son is y years. 6 years ago, age of the son was (y – 6) years. The cost of 1 exercise books = Rs x The cost of 3 exercise books = Rs 3x The cost of 1 pen = Rs y The cost of 4 pens = Rs 4y Equation and Graph

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 159 Vedanta Excel in Mathematics - Book 8 or, 6x = 210 – 24 or, 6x = 186 or, x = 186 6 =31 Substituting the value of x in equation (i), we get, y = 42 – 31 = 11 So, the present age of the father is 31 years and that of the son is 11 years. Example 5: The length of a rectangular ground is two times its breadth. If the perimeter of the ground is 96 m, find the length and breadth of the ground. Solution: Let, the length of the ground be x m and the breadth be y m. From the first condition, length = 2 × breadth x = 2y .................... (i) From the second condition, Perimeter = 96 m 2 (l + b) = 96 (x + y) = 96 2 x + y = 48 .................... (ii) Substituting the value of x from equation (i) in equation (ii), we get, 2y + y = 48 or, 3y = 48 or, y = 48 3 =16 Substituting the value of y in equation (i), we get, x = 2 × 16 = 32 Hence, the length of the ground is 32 m and the breadth is 16 m. EXERCISE 11.3 General Section - Classwork 1. Let’s say and write the values of the variables as quickly as possible. a) In x + y = 5, if x = 3, then y = .......... b) In x – y = 4, if x = 1, then y = ......... c) In 2x + y = 9, if y = 1, then x = ....... d) In 3x – 2y = 4, if y = 1, then x = ..... e) If (x1 , 2) satisfies the equation x + y = 6, then x1 = .............. f) If (3, y1 ) satisfies the equation x – y = –2, then y1 = .............. 2. Let’s say and tick the correct pair of solutions of the given pair of equations. a) x + y = 4 and x – y = 2 (i) (2, 2) (ii) (1, 3) (iii) (3, 1) b) x + 2y = 5 and 2x + y = 4 (i) (1, 2) (ii) (2, 1) (iii) (2, 2) c) 2x – y = 6, and x + y = 6 (i) (3, 0) (ii) (4, 2) (iii) (2, 4) Equation and Graph

Vedanta Excel in Mathematics - Book 8 160 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 3. Let’s say and write ‘True’ or ‘False’ whether the given ordered pair is a solution of the given pair of equations. a) x + y = 7, x – y = 3 → (5, 2) .......... b) x + y = 10, x – y = 4 → (6, 4) .......... c) y = x – 2, y = 2x + 2 → (–4, 6) .......... d) y = 2x + 1, y = x – 1 → (–2, –3) .......... Creative Section - A 4. Solve the following simultaneous equations by elimination method. a) x + y = 3 and x – y = 1 b) x + y = 5 and x – y = 1 c) x + y = 5 and x – y = 7 d) x + y = 6 and x – y = 4 e) x + y = 8 and x – y = 10 f) x + y = 9 and x – y = 3 g) x + y = 7 and 2x + y = 10 h) x – y = – 3 and y – 2x = 1 i) 2x – 3y = 1 and x + 2y = 18 j) 3x + 4y = 16 and x + 3y = 7 5. Solve the following simultaneous equations by substitution method. a) y = x – 2 and x + y = 4 b) y = x – 2 and x + y = 6 c) y = x + 2 and x + y = 8 d) y = x + 6 and x + y = 14 e) x – y = 5 and y = 11 – x f) 2x – y = 3 and y = 9 – x g) x + 2y = 19 and x + y = 13 h) 2x – y = 2 and x – y = – 3 i) 3x + 2y = 5 and x – y = – 10 j) 2x + 5y = 29 and x + y = 7 6. a) The sum of two numbers is 18 and their difference is 4. Find the numbers. b) The sum of two numbers is 39. If the greater number is 5 more than the smaller one, find the numbers. c) The sum of two numbers is 54. If the smaller number is 6 less than the greater one, find the numbers. d) The difference of two numbers is 6. If three times the smaller number is equal to two times the greater one, find the numbers. Creative Section - B 7. a) The total number of students in a class is 36. If the number of girls is 6 more than the number of boys, find the number of boys and girls. b) The total cost of a box and a pen is Rs 100. If the cost of a box is same as the cost of 3 pens, find the cost of a box and a pen. c) There are a few number of chairs and a few number of people sitting on the chairs in a room. There are 102 legs of chairs and people altogether. If there are 3 empty chairs, find the number of chairs and the number of people in the room. 8. a) The sum of the age of a father and his son is 40 years. If the father is 28 years older than his son, find their age. b) The sum of the age of two sisters is 20 years. If one of them is 4 years younger than other, find their age. c) Father is three times as old as his son. If the difference of their age is 24 years, find their age. d) The sum of the age of a father and his son is 48 years. Four years ago, the father was 9 times as old as his son was. Find their present age. Equation and Graph

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 161 Vedanta Excel in Mathematics - Book 8 e) A father is 30 years older than his son. After six years he will be 3 times as old as his son will be. Find their present age. f) If twice the son’s age in years is added to his father’s age, the sum is 70. But if twice the father’s age is added to the son’s age, the sum is 95. Find the present age of the father and the son. 9. a) The length of a rectangular garden is two times its breadth. If the perimeter of the garden is 72 m, find the length and breadth of the garden. b) The perimeter of a rectangular field is 120 m. If the field is 10 m longer than its breadth, find the length and breadth of the field. It’s your time - Project Work and Activity Section 10. a) Make the pair of simultaneous equations and solve them to get the following values of variables: (i) x = 2 y = 3 (ii) x = 5 y = –1 (iii) x = –3 y = 6 (iv) x = –1 y = –2 b) Let’s find the unit cost of each of the following pairs of items in your local market. Consider the unit cost of one item as Rs x and another item as Rs y. Then, make simultaneous equations with different number of items in each case. Solve the equations and find the values of x and y. (i) 1 pencil and 1 eraser (ii) 1 pen and 1 exercise book (iii) 1 kg of potatoes and 1 kg of tomatoes (iv) 1 kg of sugar and 1 kg of rice 11.9 Quadratic equation – Introduction Let’s consider an equation x + 5 = 0. Here, the highest power of the variable x is 1. So, it is called a first degree equation of one variable. A first degree equation is also called a linear equation. On the other hand, let’s consider another equation x2 + 7x +12 = 0. In this equation, the highest power of the variable is 2. So, it is called a second degree equation. A second degree equation of one variable is called a quadratic equation. The standard form of a quadratic equation is ax2 + bx + c = 0, where a, b, c ∈ R and a ≠ 0. 11.10 Solution of quadratic equations A quadratic equation is a second degree equation. Therefore, we obtain two solutions (or roots) of the variable from a given quadratic equation. There are many methods of solving quadratic equations. We discuss only about the factorisation method here. 11.11 Solving quadratic equations by factorisation method In this method the quadratic equation of the form ax2 + bx + c = 0 is factorised and expressed as the product of two linear factors. Then, each linear factor is separately solved to get the required solutions of the equation. Equation and Graph

Vedanta Excel in Mathematics - Book 8 162 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Worked-out Examples Example 1: Solve (x – 2) (x + 3) = 0. Solution: Here, (x – 2) (x + 3) = 0 Either x – 2 = 0 .................... (i) Or, x + 3 = 0 .................... (ii) Solving equation (i), x – 2 = 0 or, x = 2 Solving equation (ii), x + 3 = 0 or, x = – 3 Hence, x = 2 and – 3. Example 2: Solve x2 – 16 = 0 Solution: Here, x2 – 16 = 0 or, x2 – 42 = 0 or, (x + 4) (x – 4) = 0 Either, x + 4 = 0 .................... (i) or, x – 4 = 0 .................... (ii) Solving equation (i), x + 4 = 0 or, x = – 4 Solving equation (ii), x – 4 = 0 or, x = 4 Hence, x = – 4 and 3 or ± 4. Example 3: Solve x2 + x – 6 = 0 Solution: Here, x2 + x – 6 = 0 or, x2 + (3 – 2) x – 6 = 0 or, x2 + 3x – 2x – 6 = 0 or, x (x + 3) – 2 (x + 3) = 0 I’ve remembered ! a2 – b2 = (a + b) (a – b) So, x2 – 42 = (x + 4) (x – 4) I understood ! If a × b = 0, either a should be zero or b should be zero. Alternative process x2 – 16 = 0 or, x2 = 16 or, x2 = (± 4)2 removing square from both sides of the equation, we get or, x = ± 4 Equation and Graph

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 163 Vedanta Excel in Mathematics - Book 8 Checking the answer x2 + x – 6= 0, when x = –3, (–3)2 + (–3) – 6 = 0 9 – 3 – 6 = 0 0 = 0 When x = 2 22 + 2 – 6 = 0 4 + 2 – 6 = 0 0 = 0 Hence, –3 and 2 are correct solutions. Checking the answer When x = 6 2 × 62 – 9 × 6 – 18=0 or, 72 – 54 – 18 = 0 or, 18 – 18 = 0 or, 0 = 0 When x = – 3 2 2 × (– 3 2 ) 2 – 9 × (– 3 2 ) – 18 = 0 or, 9 2 + 27 2 – 18 = 0 or, 18 – 18 = 0 0 = 0 Hence, 6 and – 3 2 are correct solutions. Checking the answer Sum of 5 and 4 = 5 + 4 = 9 Product of 5 and 4= 5 × 4 = 20 or, (x + 3) (x – 2) = 0 Either, x + 3 = 0 .................... (i) or, x – 2 = 0 .................... (ii) Solving equation (i), x + 3 = 0 or, x = – 3 Solving equation (ii), x – 2 = 0 or, x = 2 Hence, x = – 3 or 2. Example 4: Solve 2x2 – 9x – 18 = 0 Solution: Here, 2x2 – 9x – 18 = 0 or, 2x2 – (12 – 3)x – 18 = 0 or, 2x2 – 12x + 3x – 18 = 0 or, 2x (x – 6) + 3 (x – 6) = 0 or, (x – 6) (2x + 3) = 0 Either, (x – 6) = 0 .................... (i) or, (2x + 3) = 0 .................... (ii) Solving equation (i), x – 6 = 0 or, x = 6 Hence, x = 6 and – 3 2 Solving equation (ii), 2x + 3 = 0 or, 2x = – 3 or, x = – 3 2 Example 5: If the sum of two numbers is 9 and the product is 20, find the numbers. Solution: Let, one of the numbers be x. Then, the other number is (9 – x) The product of these two numbers is 20. ∴ x (9 – x) = 20 or, 9x – x2 = 20 or, x2 – 9x + 20 = 0 or, x2 – (5 + 4)x + 20 = 0 or, x2 – 5x – 4x + 20 = 0 or, x (x – 5) – 4 (x – 5) = 0 or, (x – 5) (x – 4) = 0 Either, x – 5 = 0 .................... (i) or, x – 4 = 0 .................... (ii) Solving equation (i), x – 5 = 0 or, x = 5 Solving equation (ii), x – 4 = 0 or, x = 4 Hence, the required numbers are 5 and 4. The sum of two numbers is 9 and one of the numbers is x. So, other number will be 9 – x. Equation and Graph

Vedanta Excel in Mathematics - Book 8 164 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur EXERCISE 11.4 General Section - Classwork 1. Let’s say and write the values of variables as quickly as possible. a) x2 = 1, x = .............. b) y2 = 4, y = ................ c) p2 = 9, p = ............... d) a2 = 16, a = ........... e) x2 = 25, x = ............. f) y2 = 81, y = ............. g) x2 = 5, x = ........... h) m2 = 6, m = ............... i) x2 = 7, x = ............... 2. Let’s say and write the values of variables as quickly as possible. a) (x + 2) (x – 2) = 0, x = ................ b) (y – 1) (y – 3) = 0, y = ........., ......... c) (a + 2) (a – 3) = 0, a = .........., ......... d) (x + 4) (x + 7) = 0, x = ........., .......... Creative Section - A 3. Let’s solve the following equations: a) (x – 2) (x – 3) = 0 b) (x – 1) (x – 2) = 0 c) (x + 3) (x – 4) = 0 d) (x – 4) (x + 5) = 0 e) (x – 9) (x – 3) = 0 f) (x + 2) (x + 11) = 0 g) x (x + 3) = 0 h) x (x – 8) = 0 i) 2x (x – 7) = 0 4. Let’s solve the following equations: a) x2 – 1 = 0 b) x2 – 4= 0 c) x2 – 9 = 0 d) x2 – 16 = 0 e) x2 – 25 = 0 f) x2 – 36 = 0 g) x2 – 49 = 0 h) x2 – 64 = 0 i) x2 – 81 = 0 j) x2 – 100 = 0 k) 4x2 – 9 = 0 l) 16x2 – 25 = 0 m) 49 – 36x2 = 0 n) 100 – 81x2 = 0 o) 9x2 – 64 = 0 p) 64x2 – 121 = 0 5. Let’s solve the following equations: a) x2 – 3x + 2 = 0 b) x2 – 5x + 6 = 0 c) x2 – 5x + 4 = 0 d) x2 – 7x + 12 = 0 e) x2 – 12x + 36 = 0 f) x2 – 4x + 4 = 0 g) x2 – 9x + 18 = 0 h) x2 + 5x + 6 = 0 i) x2 + 7x + 12 = 0 j) x2 + 9x + 14 = 0 k) x2 + 13x + 42 = 0 l) x2 + 14x + 45 = 0 m) x2 – 9x – 70 = 0 n) x2 – 10x – 39 = 0 o) x2 – 3x = 10 p) x2 + 6x = 40 q) x2 + 10x = 75 r) x2 + 3x = 28 6. Let’s solve the following equations: a) 2x2 – x = 0 b) 3x2 – x = 0 c) 4x2 – x = 0 d) 2x2 – 3x = 0 e) 3x2 – 4x = 0 f) 4x2 – 5x = 0 g) 2x2 + 3x – 1 = 0 h) 2x2 – 3x + 1 = 0 i) 2x2 – x – 21 = 0 j) 2x2 – x – 6 = 0 k) 3x2 – 2x – 8 = 0 l) 3x2 – 10x + 8 = 0 m) 3x2 – 5x – 2= 0 n) 3x2 + 10x + 3 = 0 o) 4x2 + 12x + 9 = 0 p) 5x2 – 32x + 12 = 0 q) 6x2 + 17x + 12 = 0 r) 6x2 – x – 2 = 0 Equation and Graph

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 165 Vedanta Excel in Mathematics - Book 8 Creative Section - B 7. a) The sum of two numbers is 12 and their product is 32. Find the numbers. b) The sum of two numbers is 15 and their product is 54. Find the numbers. c) The difference of two numbers is 4 and their product is 60. Find the numbers. d) The difference of two numbers is 5 and their product is 84. Find the numbers. e) In a two digit number, the product of the digits is 18 and their sum is 9. Find the number. f) The sum of digits of two digit number is 7 and their product is 12. Find the number. g) A number exceeds than another number by 3 and their product is 40. Find the numbers. h) Divide 12 into two parts such that the product of the two parts is 35. i) If four times a number is added to the square of the number, the result is 21. Find the number j) The square of a number is four more than three times the number. Find the number. 8. a) The difference of the age of two sisters is 4 years and the product of their age is 45. Find the age of the two sisters. b) The sum of the age of two children is 13 years and the product of their age is 36. Find their age. 9. a) A room is 4 m longer than its breadth. If the area of the floor of the room is 96 sq. m, find the length and breadth of the room. b) The length of a room is 3 m longer than its breadth. If the area of its floor is 70 sq. m, find the length and breadth of the room. It’s your time - Project Work and Activity Section 10. a) Let’s write the quadratic equations in the form of x2 – a2 = 0, where a2 is a perfect square number. Then, solve them to get the following solutions: (i) x = ± 1 (ii) x = ± 2 (iii) x = ± 3 (iv) x = ± 4 (v) x = ± 5 Let’s check the equations replacing x by the given solutions. b) Let’s write any 3 quadratic equations of the for x2 – a2 = 0, where a2 is a square number. Solve each equation. c) Let’s write any 3 quadratic equations of the form ax2 – b = 0, where a and b are square numbers. Solve each equation. d) Let’s write any 4 quadratic equations of the following forms. (i) x2 + ax + b = 0 (ii) x2 + ax – b = 0 (iii) x2 – ax – b = 0 (iv) x2 – ax + b = 0 Solve each equations and find the solutions. Equation and Graph

Vedanta Excel in Mathematics - Book 8 166 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Assessment - III 1. Answer the following questions. (a) Which of the laws is called quotient law of indices? (i) am × an = am + n (ii) am ÷ an = am – n (iii) (am) n = amn (iv) ao = 1 (b) Simplify: (xa+ b) a – b × (x b + c) b – c ×(x c + a) c – a (c) If x = 10p + q, y = 10p – q, what is the value of xy? 2. (a) Write the expanded form of (a + b)2 . (b) For what value of K, the expression x2 + Kxy + 9y2 becomes a perfect square? 3. Given two algebraic expressions are x2 + 6x + 8 and x2 – 4. (a) Find the highest common factors of these expressions. (b) Find the lowest common multiples of these expressions. (c) For what values of x, the expression x2 – 4 equals to zero? (d) Prove that: 1 x2 + 6x + 8 + 2 x2 – 4 = 3 x2 + 2x – 8 4. Given rational expression is a2 + 3ab + 2b2 a2 – 4b2 – 4ab2 4a2 b + 8ab2 (a) Factorize the denominators. (b) Find the L.C.M. of the denominators. (c) Simplify to its lowest term. 5. The combined cost of umbrella and bag are shown in the given figure. Total cost = Rs 1,800 Total cost = Rs 2,100 Answer the following questions. (a) Assume the cost of an umbrella as x and the cost of bag as y then make the equations. (b) What is the rate of cost of umbrella and bag? (c) A shopkeeper sells a few number of umbrellas and 5 bags for Rs 7,500 in a day, how many umbrellas does he sells in the day? (d) What will be the total cost of 4 umbrellas and 3 bags when the rate of cost of the umbrella is increased by 10% and the rate of the cost of the bag is decreased by 15%? 6. (a) Define simultaneous linear equations. (b) Solve the equations x + y = 5 and 2x – y = 1 by graphical method.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 167 Vedanta Excel in Mathematics - Book 8 12.1 Transformation –Review The rule under which the position or size (or both) of an object (or a geometrical figure) may be changed is known as transformations: There are four fundamental transformation. (i) Reflection (ii) Rotation (iii) Translation (or Displacement) (iv) Enlargement Here, we shall discuss about reflection, rotation and displacement. Let's study the following illustrations and investigate the ideas about transformation. 12.2 Reflection The reflection of a geometrical figure means the formation of the image of the figure after reflecting about the line of reflection. The line of reflection is also called the axis of reflection. In the given figure, ∆ABC is reflected by the axis of reflection MM’ to form the image ∆A'B'C'. Properties of reflection (i) The geometrical figure and its image are at equal distance from the axis of reflection. (ii) The areas of the geometrical figure and its image are equal. (iii) The appearance of the image of a figure is opposite to the figure. ∆PQR is shifted from the position X to the position X' due to the rotation through 90° about O in anticlockwise direction. It is the transformation of ∆PQR due to rotation and ∆P'Q'R' is the image of ∆PQR. ∆ABC is shifted from the position P to the position P' reflecting by the axis of reflection MM'. It is the transformation of ∆ABC due to reflection and ∆A'B'C' is the image of ∆ABC. Reflection of ∆ABC Rotation of of ∆PQR A C B' C' A' P' M M’ P B M C B B' C' A' M' A Q' R Q P X O P' R' M’ X' Unit 12 Transformation

Vedanta Excel in Mathematics - Book 8 168 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Now, let’s learn to draw the image of a geometrical figure reflected about an axis of reflection. In the figure, ABC is a triangle. MM' is the axis of reflection. (i) Draw perpendiculars AP, BQ and CR from each vertex of ∆ABC on the axis of reflection. (ii) Produce AP, BQ, and CR. (iii) Measure the length of AP by using compasses and cut off PA' = PA. Similarly cut off QB' = BQ and RC' = CR. (iv) Join A', B' and C' by using a ruler. Thus, ∆A'B'C' is the image of ∆ABC. 12.3 Reflection of geometrical figure using coordinates Let’s learn to find the coordinates of the images of geometrical figures formed due to the reflection about x-axis and y-axis. Here, x-axis and y-axis are the axes of reflection. (i) x-axis as the axis of reflection Study the following illustrations and learn to find the coordinates of the image of a point in different quadrants when x-axis is the axis of reflection. From the above illustrations, it is clear that when a figure is reflected about x-axis, the coordinate of the image of vertex remains the same and the sign of y-coordinate of the image is changed. Therefore, P(x, y) → P'(x, –y) M' B' C' A' M P Q B R A C The point A is in the 1st quadrant. A (–4, 3)→ A' (–4, –3) ∴ P(–x, y)→ P' (–x, –y) Only the sign of y-coordinate is changed A (3, 5) → A' (3, –5) ∴ P (x, y) → P' (x, –y) Only the sign of y-coordinate is changed A (–2, –4)→ A' (–2, 4) ∴ P(–x, –y)→ P' (–x, y) Only the sign of y-coordinate is changed A (3, – 4)→ A' (3, 4) ∴P (x, – y)→ P' (x, y) Only the sign of y-coordinate is changed www.geogebra.org/classroom/ptecayee Classroom code: PTEC AYEE Vedanta ICT Corner Please! Scan this QR code or browse the link given below: The point A is in the 2nd quadrant. The point A is in the 3rd quadrant. The point A is in the 4th quadrant. Y' Y X' X O A'(3,-5) A(3,5) Y' X' Y X O A'(3,4) A(3,–4) Y' X' Y O X A'(-4,-3) A(-4,3) Y' X' Y X O A(-2,- 4) A'(-2,4) Transformation

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 169 Vedanta Excel in Mathematics - Book 8 Worked-out Examples Example 1: P(3, 6), Q (– 2, 4) andR(5, 1) are the vertices of∆PQR. Find the coordinates of its image when it is reflected about x-axis. Also, draw the graphs of this transformation. Solution: P (3, 6), Q (– 2, 4), and R (5, 1) are the vertices of ∆PQR. When ∆PQR is reflected about x-axis, P (3, 6) → P' (3, – 6) Q (– 2, 4) → Q' (– 2, – 4) R (5, 1) → R' (5, –1) ∴ P' (3, – 6), Q' (– 2, – 4) and R' (5, – 1) are the vertices of the image of ∆PQR. (ii) y-axis as the axis of reflection Study the following illustrations and learn to find the coordinates of the image of point in different quadrants when y-axis is the axis of reflection. From the above illustrations, it is clear that when a figure is reflected about y-axis, the coordinates of the image of vertex remains the same and the sign of x-coordinate of the image is changed. Therefore, P(x, y) → P'(–x, y) Example 2: D (0, – 2), E (6, – 4), and F (2, 5) are the vertices of ∆DEF. Find the coordinates of its image when it is reflected about y-axis. Also, draw the graphs of this transformation. Y Y' X' O Y P'(3,-6) Q'(-2,-4) R'(5,-1) R(5,1) Q(-2,4) P(3,6) The point A is in the 1st quadrant. A(– 3, 4) → A'(3, 4) ∴ P(– x, y) → P'(x, y) Only the sign of x-coordinate is changed A(–5, –4) → A'(5, –4) ∴ P(–x,–y)→ P'(x, –y) Only the sign of x-coordinate is changed A(4, 3) → A'(– 4, 3) ∴ P(x, y) → P'(– x, y) Only the sign of x-coordinate is changed A(5, –2)→ A'(–5, –2) ∴ P(x,–y)→ P'(–x, –y) Only the sign of x-coordinate is changed The point A is in the 2nd quadrant. The point A is in the 3rd quadrant. The point A is in the 4th quadrant. Y' Y' Y' Y' Y O X' X' X' X X' A'(-4,3) A(4,3) Y X A(-3,4) A'(3,4) O Y X A(-5,-4) A'(5,-4) O Y X A'(-5,-2) A(5,-2) O Transformation www.geogebra.org/classroom/cepkqf3n Classroom code: CEPK QF3N Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Vedanta Excel in Mathematics - Book 8 170 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Solution: D (0, – 2), E (6, – 4) and F (2, 5) are the vertices of ∆DEF. When ∆DEF is reflected about y-axis. D (0, – 2) → D' (0, – 2) E (6, – 4) → E' (– 6, – 4) F (2, 5) → F' (– 2, 5) ∴ D' (0, – 2), E' (– 6, – 4) and F' (– 2, 5) are the vertices of the image of ∆DEF. EXERCISE 12.1 General Section 1. The dotted lines are the axes of reflection. Let's draw the image of the following figures by using set-squares and ruler. 2. Let's say and write the coordinates of images as quickly as possible. a) X-axis is the axis of reflection. P (x, y) → .................... P (–x, y) → ......................... P (– x, – y) → ................... P (x, y) → .................... A (–3, 4) → ........................ B (3, – 4) → ..................... C (1, 5) → ................... D (– 1, – 5) → .................... E (0, – 2) → ..................... b) Y-axis is the axis of reflection. P (x, y) → ..................... P (– x, y) → ................... P (–x, – y) → ................... P (x, – y) → .................. A (2, – 7) → ................... B (8, 3) → ....................... C (– 4, – 5) → ............... D (– 1, 5) → .................... E (3, 0) → ....................... Creative Section - A 3. Copy the following figures in your own graph papers and draw their images under the reflection about x-axis. Also, write the coordinates of the vertices of images. Y' Y X' X F'(-2,5) E'(-6,-4) F(2,5) E(6,-4) D'(0,-2) O D(0,-2) B P C Q A R a) b) c) H E F G Transformation

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 171 Vedanta Excel in Mathematics - Book 8 4. Copy the following figures in your own graph papers and draw their images under the reflection about y-axis. Also, write the coordinates of the vertices of images. 5. Plot the following points in graph papers and draw triangles joining the points in order. Draw the image of each triangle under the reflection about x-axis. Write the coordinates of the vertices of images. a) P (3, 6), Q (5, 9), R (4, 2) b) A (– 2, 4), B (3, 3), C (– 5, 7) c) K (– 1, – 4), L (– 6, – 3), M (– 2, 4) d) D (3, – 5), E (2, 6), F (4, – 1) 6. Plot the following points in graph papers and draw triangles joining the points in order. Draw the image of each triangle under the reflection about y-axis. Write the coordinates of the vertices of images. a) A (1, 4), B (4, 7), C (6, 3) b) P (–2, 3), Q (–4, –6), R (–5, 5) c) E (3, -7), F (5, –2), G (0, –3) d) X (1, 4), Y (4, –5), Z (2, 6) 7. a) A (0, –4), B (– 3, 0) and C (2, 5) are the vertices of ∆ABC. Find the coordinates of its image under the reflection about x-axis. b) P (– 7, – 8), Q (4, 6) and R (– 6, 2) are the vertices of ∆PQR. Find the coordinates of its image under the reflection about y-axis. Creative Section- B 8. a) W (4, 2), X (– 3, 6) and Y (– 1, – 4) are the vertices of ∆WXY. Find the coordinates of the vertices of ∆W'X'Y' under the reflection about x-axis. Also, find the coordinates of ∆W''X''Y'' when ∆W'X'Y' is reflected about y-axis. b) R (7, – 2), S (– 4, – 5) and T (– 6, 3) are then vertices of ∆RST. Find the coordinates of the vertices of ∆R'S'T' under the reflection about y-axis. Also, find the coordinates of ∆R''S''T'' when ∆R'S'T' is reflected about x-axis. It's your time - Project Work and Activity Section 9. Let's draw a triangle in a squared graph paper. Write the coordinates of its vertices. Then reflect it about X-axis and Y-axis. Write the coordinates of its image. a) Y X’ X Y’ O A F E D C b) Y X’ X Y’ R O Q P L M c) Y X’ X X’ E O H G Q R S P F a) Y X’ X Y’ O Q R A C B b) Y X’ X Y’ O L F E G K M c) Y X’ X Y’ O A S R Q P D C B B K Transformation

Vedanta Excel in Mathematics - Book 8 172 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur To rotate a geometrical figure, following three conditions are required: (i) Centre of rotation (ii) Angle of rotation (iii) Direction of rotation We can rotate a figure in two directions. (i) Anti-clockwise direction (Positive direction) (ii) Clock-wise direction (Negative direction) In anti-clockwise direction, we rotate a figure in opposite direction of the rotation of second-hand of a clock. In clockwise direction, we rotate a figure in the same direction of the rotation of second-hand of a clock. Now, let’s learn to draw the image of a figure when it is rotated through the given angle in the given direction about the given centre of rotation. Rotation through 60° in anti-clockwise direction (i) Join each vertex of the figure to the centre of rotation with dotted lines. (ii) On each dotted line, draw 60° at O with the help of protractor in anti-clockwise direction. (iii) With the help of compasses, cut off OA' = OA, OB' = OB and OC' = OC. (iv) Join A', B', and C'. Thus, ∆A'B'C'is the image of ∆ABC formed due to the rotation through 60° in anti-clockwise direction about the centre O. +60° B' C' A' C A B O 12.4 Rotation Study the following illustrations and investigate the idea of rotation of a point. Rotation of a point P through 90° in anticlockwise direction about the centre of rotation O. Here, P' is the image of P. Rotation of a point P through 90° in clockwise direction about the centre of rotation O. Here, P' is the image of P. P' P 90° O P' P 90° O Rotation of a point P through 180° in anticlockwise direction about the centre of rotation O. Here, P' is the image of P. Rotation of a point P through 180° in clockwise direction about the centre of rotation O. Here, P' is the image of P. P' P 180° O P' P 180° O Transformation

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 173 Vedanta Excel in Mathematics - Book 8 Rotation through 60° in clockwise direction In this case, follow the steps as mentioned above. At O, draw 60° with the help of protractor in clockwise direction. In the adjoining diagram, ∆A'B'C' is the image of ∆ABC formed due to the rotation through 60° in clockwise direction about the centre O. Similarly, we can rotate the given geometrical figure through any given angle in anti-clockwise or clockwise direction about the given centre of rotation. 12.5 Rotation of geometrical figures using coordinates In this case, we shall discuss about the rotation of figures through some special angles such as 90° and 180° in anti-clockwise and clockwise directions about the centre at origin. The rotation through 90° is also called a quarter-turn and the rotation through 180° is called a half-turn. Rotation through 90° in anti-clockwise about the centre at origin www.geogebra.org/classroom/p75x3zwk Classroom code: P75X 3ZWK Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Thus, when a point is rotated through 90° in anti-clockwise direction (+ 90°) about the origin as the centre of rotation, the x and y-coordinates are exchanged by making the sign of y-coordinate just opposite. i.e., P (x, y) → P' (– y, x) The point A is in the 1st quadrant. A(2, 5) → A'(–5, 2) ∴ P(x, y) → P'(–y, x) The point A is in the 2nd quadrant. A(–3, 4) → A'(–4, –3) ∴ P(–x, y) → P'(–y, –x) The point A is in the 3rd quadrant. A(–4, –2) → A'(2, –4) ∴ P(–x, –y) → P'(y, – x) The point A is in the 4th quadrant. A(1, –5) → A'(5, 1) ∴ P(x, –y) → P'(y, x) O C C' A A' B' –60° B Y Y Y Y X' X' X' X' Y' Y' Y' Y' O O O O A'(-5,2) A'(-4,-3) A(-4,-2) A'(5,1) A(2,5) A(-3,4) A'(2,-4) A(1,-5) Transformation

Vedanta Excel in Mathematics - Book 8 174 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Thus, when a point is rotated through 90° in clockwise direction (– 90°) about the origin as the centre of rotation, the x and y-coordinates are exchanged by making the sign of x-coordinate just opposite. i.e. P (x, y) → P'(y, – x) Worked-out Examples Example 1: A(3, 1),B(4, – 2)and C (5, 3)are the vertices of∆ABC.Findthe coordinates of its image under the rotation through 90° in anti-clockwise direction about origin. Also, draw the graphs of this transformation. Rotation through 90° in clockwise about the centre at origin The point A is in the 1st quadrant. A(3, 2) → A'(2, –3) ∴ P(x, y) → P'(y, –x) The point A is in the 2nd quadrant. A(–4, 5) → A'(5, 4) ∴ P(–x, y) → P'(y, x) The point A is in the 3rd quadrant. A(–2, –4) → A'(–4, 2) ∴P(–x, –y) → P'(–y, x) The point A is in the 4th quadrant. A(1, –3) → A'(–3, –1) ∴ P(x, –y) → P'(–y, –x) Y Y Y Y X' X' X' X' Y' Y' Y' Y' O O O O A'(2,-3) A(3,2) A'(-4,2) A(-4,5) A'(5,4) A(-2,-4) A(1,-3) A'(-3,-1) Rotation through 180° in anti-clockwise and clockwise about origin When a point is rotated through 180° in anti-clockwise or in clockwise direction about origin as the centre of rotation, the coordinates of the image are same. Study the following illustration: The point A is in The point A is in The point A is in The point A is in the 1st quadrant. the 2nd quadrant. the 3rd quadrant. the 4th quadrant. A(5, 3) → A'(–5, –3) ∴ P(x, y) → P'(–x, –y) A(–2, 4) → A'(2, –4) ∴ P(–x, y) → P(x, –y) A(–3, –3) → A'(3, 3) ∴ P (–x, –y) → P(x, y) A(4, –3) → A'(–4, 3) ∴ P(x, –y) → P'(–x, y) Thus, when a point is rotated through 180° in anti-clockwise (– 180°) or in clockwise (+ 180°) direction, about the origin as the centre of rotation, the x and y-coordinates of the image remain the same just by changing their signs. i.e. P (x, y) → P' (– x, – y) Y X' Y' O A'(-5,-3) A(5,3) Y X' Y' O A'(-3,-3) A (3,3) Y X' Y' O A'(-4, 3) A (4, -3) Y X' Y' O A'(2,-4) A (-2, 4) Transformation

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 175 Vedanta Excel in Mathematics - Book 8 Solution: A (3, 1), B (4, – 2), and C (5, 3) are the vertices of ∆ABC. When it is rotated through + 90° about origin, A (3, 1) → A' (– 1, 3) B (4, – 2) → B' (2, 4) C (5, 3) → C' (– 3, 5) ∴ A'(– 1, 3), B'(2, 4), and C'(– 3, 5) are the vertices of the image of ∆ABC. Example 2: P (2, 3), Q (–2, 6), and R (– 4,2) are the vertices of ∆PQR. Find the coordinates of its image under the rotation through 180° in clockwise direction about origin. Also, draw the graphs of this transformation. Solution: P (2, 3), Q (– 2,6), and R (– 4, 2) are the vertices of ∆PQR. When ∆PQR is rotated – 180° about origin. P (2, 3) → P' (– 2, –3) Q (– 2, 6) → Q' (2, – 6) R (–4, 2) → R' (4 –2) ∴ P' (–2, – 3), Q' (2,– 6), and R' (4, – 2) are the vertices of the image of ∆PQR. Example 3: D(1, 4), E (–3, –5), and F (2, –3) are the vertices of ∆DEF. Find the coordinates of the vertices of ∆D'E'F' under the rotation through –90° about origin. Also, find the coordinates of ∆D''E''F'' when ∆D'E'F' is rotated through +180° about origin. Solution: D (1, 4), E (– 3, – 5), and F (2, –3) are the vertices of ∆DEF. When it is rotated through –90° about origin. D (1, 4) → D' (4, – 1) E (– 3, 5) → E' (5, 3) F (2, – 3) → F' (– 3,–2) ∴ D' (4, – 1), E' (5, 3), and F' (–3, – 2) are the vertices of ∆D'E'F' Again when DD'E'F' is rotated through +180° about origin. D' (4, –1) → D'' (– 4, 1) E' ( 5, 3) → E'' (– 5, – 3) F' (– 3, – 2) → F'' (3, 2) ∴ D'' (– 4, 1), E'' (–5, – 3) and F'' (3, 2) are the vertices of of ∆D''E''F'' Y X' X Y' A'(-1,3) C'(-3,5) C(5,3) A(3,1) B(4,-2) B'(2,4) Y X' X Y' Q(-2,6) P(2,3) R(-4,2) z Q'(2,-6) O R'(4,-2) P'(-2,-3) Transformation

Vedanta Excel in Mathematics - Book 8 176 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur EXERCISE 12.2 General Section - Classwork 1. Let's say and write the coordinates of images as quickly as possible. a) Rotation through 90° in anticlockwise direction P (x, y) → ................. P (– x, y) → ................. P (–x, –y) → ................ P (x, –y) → ................. A (–2, 3) → ................. B (2, –5) → ................. C (–4, –1) → ................. D (6, 7) → ................. E (0, –4) → ................. b) Rotation through 90° in clockwise direction. P (x, y) → ................. P (– x, y) → ................. P (–x, –y) → ................ P (x, –y) → ................. A (–2, –1)→ ................. B (3, –5) → ................. C (2, 3) → ................. D (–4, 5) → ................. E (–6, 0) → ................. c) Rotation through 180° in anticlockwise or clockwise direction P (x, y) → ................. P (– x, y) → ................. P (–x, –y) → ................ P (x, –y) → ................. A (–6, 5) → ................. B (3, –9) → ................. C (1, 3) → ................. D (–4, –2)→ ................. E (0, 4) → ................. Creative Section - A 2. Draw the images of the following figures rotating through the given angles in anti-clockwise and clockwise direction about the given centre of rotation. Rotation through 45° Rotation through 60° Rotation through 90° Rotation through 75° Rotation through 120° Rotation through 180° d) e) f) P R S Q M E P R O A C B D O a) b) c) A C B O E F G O P R Q O O Transformation

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 177 Vedanta Excel in Mathematics - Book 8 5. a) P (– 4, 2), Q (3, 7), and R (– 1, – 6) are the vertices of ∆PQR. Find the coordinates of its image under the rotation through 90° in (i) anti-clockwise (ii) clockwise direction about origin. b) A (– 5, – 8), B (2, – 3), and C (– 2, 9) are the vertices of ∆ABC. Find the coordinates of its image under the rotation through 180° in anti-clockwise direction about origin. Creative Section - B 6. a) K (1, 4), L (– 3, – 5), and M (5, – 2) are the vertices of ∆KLM. Find the coordinates of ∆K'L'M' under the rotation through –90° about origin. Also, find the coordinates of ∆K''L''M'' when ∆K'L'M' is rotated through +180° about origin. b) D (– 4, – 1), E (0, – 6), and F (5, 0) are the vertices of ∆DEF. Find the coordinates of ∆D'E'F' under the rotation through +90° about origin. Also, find the coordinates of ∆D''E''F'' under the reflection about x-axis. It's your time - Project Work and Activity Section 7. Draw a triangle in separate squared graph paper. Write the coordinates of its vertices. Rotate the triangle through (i) +90° (ii) –90° (iii) +180° (iv) –180°. Then write the coordinates of the vertices of image in each case. 4. Copy the following figures on your own graph papers. Draw their images rotating through 180° in anti-clockwise direction. Also, write the coordinates of the vertices of images. 3. Copy the following figures on your own graph papers. Draw their images rotating through 90° in (i) anti-clockwise (ii) clockwise directions about the centre at origin. Also, write the coordinates of the vertices of images. X’ X Y’ Y F A B C G E O c) X’ X Y Y’ K F E D L M O b) X’ X Y’ Y A B C P Q R O a) X’ X Y Y’ X’ X Y Y’ X’ X Y Y’ P B D F E G A C B E K F L M C A Q R O O O a) b) c) Transformation

Vedanta Excel in Mathematics - Book 8 178 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur ∆ABC is displaced to ∆A'B'C' Here, AA' = BB' = CC' Also, AA' // BB' // CC' Quadrilateral PQRS is displaced to P'Q'R'S 'Here, PP' = QQ' = RR' = SS' Also, PP' // QQ' // RR' // SS' Thus, the displacement of a geometrical figure has magnitude as well as direction. So, it is a vector quantity. Now, let’s learn to displace a given geometrical figure. 12.6 Displacement It is the transformation of geometrical figures in which each vertex of a figure is displaced by equal distance to the same direction. It is also called translation. Study the following illustrations. A A' B B' C C' P S R Q P’ S’ R’ Q’ 12.7 Displacement of geometrical figures using coordinates In the graph given below, A (2, 4), B (8, 2), and C (5, 9) are the vertices of DABC. When DABC is displaced by 7 units right and 3 units up, each of its vertices should be displaced by 7 units right and 3 units up. It means the x-coordinate of each vertex is added by 7 and y-coordinate is added by 3. If A', B', and C' are the images of the vertices A, B and C of DABC respectively, then A (2, 4) → A' (2 + 7, 4 + 3) → A' (9, 7) B (8, 2) → B' (8 + 7, 2 + 3) → B' (15, 5) C (5, 9) → C' (5 + 7, 9 + 3) → C' (12, 12) Thus, A' (9, 7), B' (15, 5), and C' (12, 12) are the vertices of the DA'B'C'. This is the case of right and up direction of displacement. Now, let's take the case of left and down direction of displacement. Y Y' X' X B' (15, 5) C' (12, 12) A' (9, 7) A (2, 4) C (5, 9) B (8, 2) O Transformation

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 179 Vedanta Excel in Mathematics - Book 8 In the given graph P (3, 7), Q (–4, 5), and R (1, 3) are the vertices of DPQR. Let's displace the triangle 6 units left and 5 units down. If P', Q', and R' be the images of the vertices P, Q and R of DPQR respectively, then, P (3, 7) → P' (3 – 6, 7 – 5) → P' (–3, 2) Q (–4, 5) → Q' (–4, – 6, 5 – 5) → Q' (–10, 0) R (1, 3) → R' (1 – 6, 3 – 5) → R' (–5, –2) Thus, in the case of left and down displacement, the number of units of left displacement is subtracted from x-coordinate of each vertex. Similarly, the number of units of down displacement is subtracted from y-coordinate of each vertex. In this way, if a be the number of units of right or left displacement and b be the number of units of up or down displacement, then, P (x, y) → P' (x + a, y + b) [Right and up displacement] P (x, y) → P' (x – a, y – b) [Left and down displacement] P (x, y) → P' (x + a, y – b) [Right and down displacement] P (x, y) → P' (x – a, y + b) [Left and up displacement] Worked-out Examples Example 1: Displace the adjoining ∆ABC in the magnitude and direction of vector p (p ). Solution: (i) From A, draw AA' // p and cut off AA' = p (ii) From B, draw BB' // p and the cut off BB' = p' (iii) From C, draw CC' // p and cut off CC' = p' . (iv) Join A', B' and C' (v) ∆A'B'C' is the required image of ∆ABC under the displacement of p . A A C B A’ B’ C’ B C p p Y Y' X' O X Q (–4, 5) P (3, 7) R (1, 3) P' (–3, 2) Q' (–10, 0) R'(–5, –2) Transformation

Vedanta Excel in Mathematics - Book 8 180 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 2: A (3, 6), B (–2, 5), and C (2, 8) are the vertices of DABC. Find the coordinates of the image of the triangle in the graph due to the displacement of 5 units to the right and 4 units down. Also, check your answer by using the right and down displacement rules. Solution: A (3, 6), B (–2, 5), and C (2, 8) are the vertices of DABC. When it is displaced by 5 units right and 4 units down, the coordinates of each vertices of its image are A' (8, 2), B' (3, 1) and C' (7, 4) Now, by applying the rules of right and down displacement, A (3, 6) → A' (3 + 5, 6 – 4) → A' (8, 2) B (–2, 5) → B' (–2 + 5, 5 – 4) → B' (3, 1) C (2, 8) → C' (2 + 5, 8 – 4) → C' (7, 4) Example 3: Find the coordinates of the image of the point P (–3, 5) when it is displaced by 4 units left and 6 units up. Also find the coordinates of the final image when it is again displaced by 6 units right and 9 units down. Solution: Here, P (–3, 5) is the given point When it is displaced by 4 units left and 6 units up, P (–3, 5) → P' (–3, – 4, 5 + 6) → P' (–7, 11) ∴ P' (–7, 11) is the coordinates of the image of the point P (–3, 5) due to the first displacement. Again, when P' (–7, 11) is displaced by 6 units right and 9 units down, P' (–7, 11) → P" (–7 + 6., 11 – 9) → P" (–1, 2) ∴ P" (–1, 2) is the coordinates of the final image due to the two successive displacements. Y Y' X' O X A (3, 6) C (2, 8) C' (7, 4) A' (8, 2) B' (3, 1) B (–2, 5) Transformation

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 181 Vedanta Excel in Mathematics - Book 8 EXERCISE 12.3 General Section Let's say and write the answers as quickly as possible. 1. a) When a point P (x, y) is displaced by a units right and b units up, coordinates of its image are ............................................................... b) When a point A (2, 4) is displaced by 1 unit right and 3 units up, coordinates of its image are ............................................................... c) When a point P (x, y) is displaced by a units left and b units down, coordinates of its image are ............................................................... d) When a point A (3, 1) is displaced by 3 units left and 2 units down, coordinates of its image are ............................................................... e) When a point P (x, y) is displaced by a units right and b units down, coordinates of its image are ............................................................... f) When a point A (–2, 5) is displaced by 3 units right and 2 units down, coordinates of its image are ............................................................... g) When a point P (x, y) is displaced by a units left and b units up, coordinates of its image are ............................................................... h) When a point A (3, –4) is displaced by 5 units left and 6 units up, coordinates of image are ............................................................... Creative Section 2. Displace the following geometrical figures on the magnitude and direction of p . A B F E G Q R C p p p a) b) c) P R A K E T I T I M R E p p p d) e) f) P Transformation

Vedanta Excel in Mathematics - Book 8 182 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 3. Plot the following points in the graph paper. Find the coordinates of the image of each point under the given displacements. a) A (1, 3) [2 units right and 4 units up] b) B (5, 2) [3 units left and 3 units down] c) C (–2, 4) [5 units right and 2 units down] d) D (–3, –5) [4 units left and 1 unit up] 4. a) P (3, 6), Q (6, 2) and R (5, 8) are the vertices of DPQR. Draw the triangle in the graph paper and find the coordinates of the image of its vertices when they are displaced by 7 units right and 5 units up. Also check your answer by using the displacement of right and up rules. b) Plot the points A (2, 5), B (–2, 0), C (–8, 4), and D (–5, 7) in the graph paper and join them. Displace the figure so formed by 4 units left and 6 units up. Find the coordinates of the vertices of the image so formed. Also, check your answer by using the displacement of left and up rules. 5. a) Find the units of displacement of the point (4, 2) to get the image of (7, 7). b) What must be the units of displacement so that the image of a point (–3, 5) can be (–8, 2)? 6. a) Find the coordinates of the image of the point M (–4, 7) when it is displaced by 3 units right and 5 units down. Also, find the coordinates of the final image when it is again displaced by 6 units left and 9 units up. b) A (1, 4), B (3, –2), and C (–4, –5) are the vertices of DABC. If it is displaced by 2 units right and 4 units down and then again 5 units left and 1 unit up, find the coordinates of the final image of DABC due to the two successive displacements. It's your time - Project Work and Activity Section 7. a) Let's draw a triangle in a squared graph paper. Write the coordinates of it's vertices. Displace the triangle under the different units of the following displacement of your own. Then, write the coordinates of the vertices of its image. (i) Right and up (ii) Right and down (iii) Left and up (iv) Left and down b) Let's draw a quadrilateral in a squared graph paper. Write the coordinates of its vertices. Displace the quadrilateral under the following displacements with your own units of displacements. Then, write the image of each vertex. (i) Right and up (ii) Right and down (iii) Left and up (iv) Left and down Transformation

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 183 Vedanta Excel in Mathematics - Book 8 13.1 Different pairs of angles – Looking back Classroom - Exercise 1. Let's say and write the pairs of adjacent angles, liner pairs, vertically opposite angles, complementary angles, and supplementary angles from the following figures. Adjacent angles Linear pairs Vertically opposite angle Complementary angles Supplementary angles ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ 2. Let's say and write the pairs of alternate angles, corresponding angles and co-interior angles. Alternate angles Corresponding angles Co-interior angles ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ ........and........ (i) Adjacent angles ∠AOB and ∠BOC are a pair of adjacent angles. They have a common vertex O and a common arm OB. www.geogebra.org/classroom/hpstegqm Classroom code: HPST EGQM Vedanta ICT Corner Please! Scan this QR code or browse the link given below: O A B C O A B C y x a c d b p q m n a b d e f h g c Unit 13 Geometry: Angle

Vedanta Excel in Mathematics - Book 8 184 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur (ii) Vertically opposite angles In the adjoining figure, ∠AOC and ∠BOD are vertically opposite angles formed by intersected line segments. They have a common vertex and they are lying to the opposite side of the common vertex. ∠AOD and ∠BOC are another pair of vertically opposite angles. Vertically opposite angles are always equal. ∠AOC = ∠BOD and ∠AOD = ∠BOC. (iii) Complementary angles A pair of angles are said to be complementary if their sum is a right angle (90°). In the figure, ∠AOB and ∠BOC are a pair of complementary angles. ∴ ∠AOB + ∠BOC = 90° Also, complement of ∠AOB = 90° – ∠BOC. Complement of ∠BOC = 90° – ∠AOB. (iv) Supplementary angles A pair of angles are said to be supplementary if their sum is two right angles (180°). In the figure, ∠AOB and ∠BOC are a pair of supplementary angles. ∴ ∠AOB + ∠BOC = 180°. Also, the supplement of ∠AOB = 180° – ∠BOC The supplement of ∠BOC = 180° – ∠AOB. Linear pair If the sum of a pair of adjacent angles is 180°, they are said to be linear pair. In the figure alongside, ∠AOB + ∠BOC = 180°. So, ∠AOB and∠BOC are the linear pair. B C A O D A O C B A B C O 90° C A B O 13.2 Experimental verifications of pair of angles formed by two intersecting lines Experiment 1: The sum of a pair of adjacent angles formed by two intersecting lines is 180°. Step 1: Draw three pairs of intersecting lines AB and CD intersecting at O. A B C D O A C B D O A B fig (i) fig (ii) fig (iii) C D O Geometry: Angle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 185 Vedanta Excel in Mathematics - Book 8 Step 2: Measure ∠AOC and ∠AOD (or ∠AOC and ∠BOC or ∠BOC and ∠BOD or ∠AOD and ∠BOD) with the help of protractor and write the measurements in the table. Fig. No. ∠AOC ∠AOD ∠AOC + ∠AOD Result (i) ∠AOC + ∠AOD = 180° (ii) (iii) Conclusion: The sum of pair of adjacent angles formed by two intersecting lines is 180°. [Note: The pair of adjacent angles whose sum is 180° are said to be the linear pair.] Experiment 2: Each pair of vertically opposite angles formed by two intersecting lines are equal. Step 1: Draw three pairs of intersecting lines AB and CD intersecting at O. Step 2: Measure ∠AOC and ∠BOD, ∠AOD and ∠BOC with the help of protractor. Write the measurements in the table. Fig. No. ∠AOC ∠BOD ∠AOD ∠BOC Result (i) ∠AOC = ∠BOD ∠AOD = ∠BOC (ii) (iii) Conclusion: Each pair of vertically opposite angles formed by two intersecting lines are equal. A B C D O A C B D O A B fig (i) fig (ii) fig (iii) C D O Worked-out examples Example 1: If x° and (x + 6)° are a pair of complementary angles, find them. Solution: Here, x° + (x + 6)° = 90° [The sum of a pair of complementary angles] or, 2x° = 90° – 6° or, x° = 84° 2 = 42° ∴ x° = 42° and (x + 6)° = 42° + 6° = 48°. Geometry: Angle

Vedanta Excel in Mathematics - Book 8 186 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 3 : Find the unknown sizes of angles. Solution: a) x + 2x + 3x = ∠AOB or, 6x = 90° or, x = 15° \ x = 15°, 2x = 2 × 15° = 30° 3x = 3 × 15° = 45° c) (x + 80°) + 5x° + (4x + 50°) = 360° or, 10x + 130° = 360° or, x = 23° \ x + 80° = 23° + 80° = 103° 5x° = 5 × 23 = 115° 4x + 50° = 4 × 23° + 50° = 142° O A a) B 3x 2x x A O B D C b) 3x x+30° x c) 5x° x+80° 4x+50° b) x + 3x + (x + 30°) = ∠BOA or, 5x + 30° = 180° or, x = 30° \ x = 30° 3x = 3 × 30° = 90° x + 30° = 30° + 30° = 60° EXERCISE 13.1 General Section- Classwork 1. Let's say and write the answers as quickly as possible. a) If x° and 100° are a linear pair, then x° = ..................... b) If y° and 50° are vertically opposite angles, then y° = ..................... c) If a° and 30° are a pair of complementary angles, then a° = ..................... d) If p° and 120° are a pair of supplementary angles, then p° = ..................... Creative Section - A 2. a) If x° and 50° form a linear pair, find x°. b) If 2x° and 3x° are adjacent angles in linear pair, find them. c) If y° and 48° are a pair of complementary angles, find y°. Example 2: If a pair of supplementary angles are in the ratio 4:5, find them. Solution: Let, the required supplementary angles be 4x° and 5x°. Now, 4x° + 5x° = 180° [The sum of a pair of supplementary angles] or, 9x° = 180° or, x° = 180° 9 = 20° ∴ 4x° = 4 × 20° = 80° and 5x° = 5 × 20° = 100°. Geometry: Angle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 187 Vedanta Excel in Mathematics - Book 8 d) If 2p° and (p + 15)° are a pair of complementary angles, find them. e) If x° and x° 4 are a pair of supplementary angles find them. f) If pair of complementary angles are in the ratio 4:11, find them. g) If a pair of supplementary angles are in the ratio 7:5, find them. 3. Find the value of x in the each of figures given below. a) b) c) d) x° 125° 3x° 2x° x° 58° 4x° 100° 3x°+50° 2x° e) f) g) h) 4x° 5x° x° x ° ) 2) x° x ° ) 4) 4. Find the unknown sizes of angles. a) b) c) d) x° 2x° (x+20)° x° 2x° 3x° e) f) g) h) z° x° y° 92° 3x° 2x° 4x° x° 2x° 2x+30° 2x–30° x+80° 4x+50° 5x° q° p° 3q° x° x+25° x+35° x° z° y° 3x° 5. a) In the given figure, calculate the values of angles p and q, where ∠p = 1 2∠q. b) In the figure, given alongside, if x = 3y, find the sizes of angles represented by a and b. Creative Section -B 6. a) In the adjoining figure, if ∠a + ∠b + ∠c = 180°, prove that ∠x = ∠b + ∠c. b) In the figure given alongside, ∠a = ∠x and ∠b = ∠y. Show that ∠x + ∠y + ∠z = 180°. a y x b q p x a b c a z x y b Geometry: Angle

Vedanta Excel in Mathematics - Book 8 188 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 13.3 Different pairs of angles made by a transversal with two straight lines In the adjoining figure, straight line segment EF intersects two line segments AB and CD at the point G and H respectively. Here, EF is called a transversal. ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8 are the angles made by the transversal with two line segments. Here, ∠1, ∠2, ∠7 and ∠8 are the exterior angles. ∠3, ∠4, ∠5 and ∠6 are the interior angles. Also, ∠4 and ∠6, ∠3 and ∠5 are two pairs of alternate angles. ∠1 and ∠5, ∠2 and ∠6, ∠4 and ∠8, ∠3 and ∠7 are four pairs of corresponding angles. ∠4 and ∠5, ∠3 and ∠6 are two pairs of co-interior angles. 13.4 Relation between pairs of angles made by a transversal with parallel lines Two line segments are said to be parallel if they do not intersect each other when they are extended to either directions. In the figure, AB and CD are two parallel lines. It is written as AB//CD. The perpendicular distance between two parallel lines is always equal. ∴ PQ = RS. Let's investigate the relation between the pair of alternate angles, pair of corresponding angles and the pair of co-interior angles formed due to the intersection of two parallel lines by a transversal. Experiment 3: Each pair of alternate angles formed by a transversal with two parallel lines are equal. Step 1: Draw three pairs of parallel lines AB and CD intersected by a transversal EF at G and H. A C B D P Q R S A C B 1 2 4 5 6 8 7 3 D E F H G 7. a) Draw a line segment AB. Mark a point O on AB and draw an angle BOC. Measure ∠BOC and ∠AOC. Verify that ∠BOC + ∠AOC = 180°. b) Draw two intersecting line segment AB and CD intersecting at O. Measure the size of each pair of vertically opposite angles. Verify that each pair of vertically opposite angles are equal. Geometry: Angle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 189 Vedanta Excel in Mathematics - Book 8 Step 2: Measure ∠AGH and ∠GHD, ∠BGH and ∠GHC with the help of protractor. Write the measurements in the table. Fig. No. ∠AGH ∠GHD ∠BGH ∠GHC Results (i) ∠AGH = ∠GHD ∠BGH = ∠GHC (ii) (iii) Conclusion: Each pair of alternate angles formed by a transversal with two parallel lines are equal. Furthermore, in the given figure, AB // CD and EF intersects AB and CD at G and H respectively. Here, ∠a and ∠d, ∠b and ∠c are two pairs of alternate exterior angles. They are the exterior angles lying towards the alternate sides of the transversal. The alternate exterior angles made by a transversal with parallel lines are always equal. ∴ ∠a = ∠d and ∠b = ∠c. Experiment 4: The sum of each pair of co-interior angles formed by a transversal with two parallel lines is 180°. Step 1: Draw three pairs of parallel lines AB and CD intersected by a transversal EF at G and H. fig (i) fig (ii) fig (iii) A B D C E G H F A C D B E F G H A C D B E F G H A C B a c d b G H F D A C B E G H F D A B D C E G H F A B D C E G H F fig (i) fig (ii) fig (iii) Geometry: Angle

Vedanta Excel in Mathematics - Book 8 190 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Step 2: Measure ∠AGH and ∠CHG, ∠BGH and ∠DHG with the help of protractor. Write the measurements in the table. Fig. No. ∠AGH ∠CHG ∠AGH + ∠CHG ∠BGH ∠DHG ∠BGH + ∠DHG Result (i) ∠AGH + ∠CHG = 180° ∠BGH + ∠DHG = 180° (ii) (iii) Conclusion: The sum of each pair of co-interior angles formed by a transversal with two parallel lines is 180°. Experiment 5: Each pair of corresponding angles formed by a transversal with two parallel lines are equal. Step 1: Draw three pairs of parallel lines AB and CD intersected by a transversal EF at G and H. fig (i) fig (ii) fig (iii) Step 2: Measure ∠AGE and ∠CHG, ∠BGE and ∠DHG, ∠AGH and ∠CHF, ∠BGH and ∠DHF with the help of protractor. Write the measurements in the table. Fig. No. ∠AGE ∠CHG ∠BGE ∠DHG ∠AGH ∠CHF ∠BGH ∠DHF Result (i) ∠AGE = ∠CHG ∠BGE = ∠DHG ∠AGH = ∠CHF ∠BGH = ∠DHF (ii) (iii) Conclusion: Each pair of corresponding angles formed by a transversal with two parallel lines are equal. The co-interior angles, alternate angles, and corresponding angles made by a transversal with two parallel lines appears in the following shapes. Co-interior angles a b a b a b a b a b a b ∠a and ∠b are co-interior angles. So, ∠a + ∠b = 180°. A B D C E G H F A C D B E F G H A B D C E G H F Geometry: Angle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 191 Vedanta Excel in Mathematics - Book 8 Alternate angles a a a a b b b b ∠a and ∠b are alternate angles. So, ∠a = ∠b. Corresponding angles a b b a a b b a a b b a ∠a and ∠b are corresponding angles. So, ∠a = ∠b. Worked-out Examples Example 1: Find the size of unknown angles in the following figures. Solution: i) x = 70° [Being alternate angles] y = 60° [Being alternate angles] 70° + z + 60° = 180° [Being the sum a straight angle] or, z = 180° – 130° = 50° a + 70° = 180° [Being the sum of co-interior angles] or, a = 180° – 70° = 110° b + 60° = 180° [Being the sum of co-interior anlges] or, b = 180° – 60° = 120° (ii) EF and GH parallel to AB and CD are drawn. ∠a = 40° [Being corresponding angles] ∠b = 160° [Being corresponding angles] ∠c = 80° – ∠a = 80° – 40° = 40° ∠c + ∠d = 180° [Being the sum of co-interior angles] or, 40° + ∠d = 180° or, ∠d = 180° – 40° = 140° ∠b + ∠d + ∠x = 360° [Being the sum a complete turn] or, 160° + 140° + ∠x = 360° or, ∠x = 360° – 300° = 60° a b x y (i) z 70° 60° (ii) 40° 80° 160° x E b d 160° a c x F A B G H C D Geometry: Angle

Vedanta Excel in Mathematics - Book 8 192 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur EXERCISE 13.2 General Section - Classwork 1. Let's say and write the answers as quickly as possible. a) Corresponding angle of ∠a is ................... b) Alternate angle of ∠x is ....................... c) ∠ a and ∠ d are a pair of ................................... angles. d) The sum of ∠ b and ∠ c is ................ . e) If ∠ b = 120°, then ∠ x = ........, ∠ c = .......... ∠ a = .......... ∠ d = ............ 2. If two non-paralled lines are intersected by a transversal, a) Are the alternate angles so formed equal ? .................... b) Are the corresponding angle so formed equal ? .................... c) Is the sum of a pair of co-interior angle 180° ? .................... Creative Section A 3. Let's find the unknown sizes of angles. b y z x a 100° d) y x 130° c) y x z 80° a) b) y z x 105° a b x d c b e) f) g) h) b a x y f e a d b c a x b z y a c 85° 115° 112° 78° 120° 105° a) (3x+10)° (4x–10)° b) (5x+12)° (6x–13)° d) (x–20)° (80–x)° c) 2x (x+45)° e) 3x° 2x° f) (2x–15)° (x+15)° h) (x+40)° 3x° g) x° (2x–30)° 4. From the figures given below, find the value of x. Geometry: Angle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 193 Vedanta Excel in Mathematics - Book 8 7. a) In the adjoining figure, PQ//MR ∠NMR= 150° and ∠QNM = 40°, calculate the value of x. P Q N x M R 150° 40° 5. Find the unknown sizes of angles. c) d) a) b) y x z a y z x y x z x 88° 130° 110° 96° y i) j) k) l) z y x x x a b 50° 58° 36° 45° 28° 35° 152° 60° x x x 40° 30° 135° 38° 72° 44° 88° 155° 162° 150° m) n) o) p) y x e) f) z y x y x x y a b y z x 120° 110° 60° 70° 60° 80° 20° g) h) y x d a a x b z w y x y c b z 110° 80° 30° 40° 55° 75° a) b) c) d) 6. Find the unknown sizes of angles in the following figures. Geometry: Angle

Vedanta Excel in Mathematics - Book 8 194 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur b) In the given figure AB//DC and BE//DF. Find the measures of x°, y° and z°. Creative Section -B 8. a) In the figure alongside, show that (i) ∠q = ∠d (ii) ∠f = ∠z (iii) ∠x = ∠h (iv) ∠a = ∠r (v) ∠p = ∠c (vi) ∠b = ∠s b) In the adjoining figure show that ∠x + ∠y + ∠z = 180°. c) In the figure alongside, show that ∠a + ∠b + ∠c + ∠d = 4 right angles. 9. a) In the adjoining figure, AB//CD. Show that, AC//BD. b) In the given figure, show that, PQ//TR. 10. Draw two parallel line segments. Intersect them with a transversal. a) Measure each pair of corresponding angles and verify that they are equal. b) Measure each pair of alternate angles and verify that they are equal c) Measure each pair of co-interior angles and verify that the sum of each pair is 180°. It's your time - Project Work and Activity Section 11. Let's put your ruler on a sheet of paper and draw two parallel lines through it's two opposite edges along length. Draw a transversal to intersect the parallel lines at two district points. Then measure eight angles so formed. Explore the relationships between corresponding angles, alternate angles and co-interior angles. A C B D E F 40° x° y° z° w z p s a d e h x y q r b c f g 100° 45° 35° P Q S R T 50° 70° 60° A C D B x y a b z y c b d a x Geometry: Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 195 Vedanta Excel in Mathematics - Book 8 14.1 Triangle - Looking back Classroom - Exercise 1. Let's select the appropriate types of triangles and write in the blank spaces. Scalene triangle, isosceles triangle, equilateral triangle, acute- angled triangle, obtuse-angled triangle, right-angled triangle a) d) b) e) c) f) 80° 52° 48° 90° 4 cm 5 cm 5 cm 4 cm 4 cm 5 cm 3 cm 3 cm 3 cm 140° 2. a) If x°, y°, and z° are the angles of a triangle, x° + y° + z° = ............... b) If a°, b°, and c° are the angles of a triangle and a° + b° = 100°, c° = ............... c) If p°, q°, and r° are the angles of an equilateral triangle, p° = q° = r°= ............... d) If m° and n° are the base angles of an isosceles triangle and m° = 50°, then n° = ................. Unit 14 Geometry: Triangle Geometry: Triangle

Vedanta Excel in Mathematics - Book 8 196 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 14.2 Experimental verification of properties of triangles Property 1: The sum of the angles of a triangle is always 180°. Activity Steps: (i) Draw a triangle on a chart paper and cut it out. (ii) Draw arcs from each vertex with same radius. (iii) Colour the angles (arcs) with different colours. (iv) Arrange the angles at a point as shown in the given diagram. In this activity, we observe that the angles of a triangle make a straight angle. Experimental verification Step 1: Draw three triangles ABC with different measurements. A A A B Fig (i) Fig (ii) Fig (iii) C B C B C Step 2: Measure the angles A, B and C with the help of protractor and write the measurements in the table. Fig. No. ∠BAC ∠ABC ∠ACB Result (i) ∠BAC + ∠ABC + ∠ACB = 180° (ii) ∠BAC + ∠ABC + ∠ACB = 180° (iii) ∠BAC + ∠ABC + ∠ACB = 180° Conclusion: The sum of the angles of a triangle is 180° (2 right angles). Theoretical proof Given: In ∆ABC, ∠ABC, ∠BCA, and ∠BAC are the angles of the triangle. To prove: ∠ABC + ∠BCA + ∠BAC = 180°. Construction: PQ//BC is drawn through A. www.geogebra.org/classroom/tfkspchg Classroom code: TFKS PCHG Vedanta ICT Corner Please! Scan this QR code or browse the link given below: P Q A B C Geometry: Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 197 Vedanta Excel in Mathematics - Book 8 Proof: (i) ∠PAB = ∠ABC [Being alternate angles] (ii) ∠QAC = ∠ACB [Being alternate angles] (iii) ∠PAB + ∠BAC + ∠QAC = 180° [The sum is a straight angle] (iv) ∠ABC + ∠BAC + ∠ACB = 180° [From (i), (ii) and (iii)] proved Property 2: The base angles of an isosceles triangle are equal. Activity Steps: (i) Draw an isosceles triangle ABC having AB = AC on a chart paper and cut it out. (ii) Draw the arcs at the vertices B and C with the same radius. (iii) Fold the paper and overlap angle B to angle C From this activity, we observed that the angles B and C exactly fit each other. Thus, the base angles of an isosceles triangle are equal. Experimental verification Step 1: Draw three straight line segments AB of different lengths. Step 2: From A and B, draw two arcs towards the same sides with the equal radius intersecting at C by using pencil compasses. Step 3: Join C, A and C, B. Now, ABC is an isosceles triangle in each figure. C A B Fig (i) A B C Fig (ii) B A C Fig (iii) Step 4: Measure the size of ∠A and ∠B in each triangle with the help of protractor. Write the measurement in the table. Fig. No. ∠BAC ∠ABC Result (i) ∠BAC = ∠ABC (ii) ∠BAC = ∠ABC (iii) ∠BAC = ∠ABC Conclusion: The base angles of an isosceles triangle are equal. Geometry: Triangle

Vedanta Excel in Mathematics - Book 8 198 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Property 3: Each base angle of an isosceles right angled triangle is 45°. Experimental verification Step 1: Draw three straight line segments AB of different lengths. Step 2: Draw ∠ABC = 90° at B with the help of protractor in each figure. Step 3: From B, cut off BC on BP such that BC = AB in each figure. Step 4: Join A, C. Now, ABC is an isosceles right angled triangle right angled at B in each figure. A B P C Fig (i) B A P C Fig (ii) B C P A Fig (iii) Step 5: Measure the sizes of ∠A and ∠C in each figure with the help of protractor. Write the measurement in the table. Fig. No. ∠BAC ∠ACB Result (i) ∠BAC = ∠ACB =45° (ii) ∠BAC = ∠ACB =45° (iii) ∠BAC = ∠ACB =45° Conclusion: Each base angle of an isosceles right angled triangle is 45°. Property 4: The line drawn from the vertex of an isosceles triangle to join the mid-point of the base is perpendicular to the base. Experimental verification Step 1: Draw three isosceles triangles ABC of different measurements. fig (i) fig (ii) fig (iii) Step 2: Find the mid-point P of the base BC of each triangle and join A, P. Step 3: Measure ∠APB and ∠APC with the help of a protractor and write the measurements in the table. A A A B B B C P P C P C Geometry: Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 199 Vedanta Excel in Mathematics - Book 8 Fig. No. ∠APB ∠APC Result (i) (ii) ∠APB = ∠APC = 90° (iii) Conclusion: The line drawn from the vertex of an isosceles triangle to join the mid-point of the base is perpendicular to the base. Property 5: The interior angles of an equilateral triangle are equal and each of 60°. Experimental verification Step 1: Draw three equilateral triangles ABC of different measurements. fig (i) fig (ii) fig (iii) Step 2: Measure ∠A, ∠B and ∠C with the help of a protractor and write the measurements in the table. Fig. No. ∠BAC ∠ABC ∠ACB Result (i) (ii) ∠BAC = ∠ABC = ∠ACB = 60° (iii) Conclusion: The interior angles of an equilateral triangle are equal and each of 60°. Property 6: The exterior angle of a triangle is equal to the sum of two opposite interior angles. Experimental verification Step 1: Draw three triangles ABC with different measurements. Step 2: Produce the side AB to D in each figure. C A B D Fig (i) A B C D Fig (ii) A B D C Fig (iii) A B C A B C A B C Geometry: Triangle

Vedanta Excel in Mathematics - Book 8 200 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Step 3: Measure the sizes of ∠CBD, ∠BAC and ∠ACB in each figure. Write the measurement in the table. Fig. No. ∠CBD ∠BAC ∠ACB Result (i) ∠CBD = ∠BAC + ∠ACB (ii) ∠CBD = ∠BAC + ∠ACB (iii) ∠CBD = ∠BAC + ∠ACB Conclusion: The exterior angle of a triangle is equal to the sum of two opposite interior angles. Worked-out Examples Example 1: Find the unknown sizes of angles in the following figures. Solution: (i) x + 95° + 35° = 180° [Being the sum of the angles of ∆ABC] or, x + 140° = 180° or, x = 180° – 140° = 40° (ii) ∠a = ∠b [Being the base angles of isosceles triangle PQR] ∠a + ∠b + 70° = 180° [Being the sum of the angles of ∆PQR] or, ∠a + ∠a = 180° – 70° or, 2∠a = 110° or, ∠a = 110 2 = 55° ∴ ∠a = ∠b = 55° (iii) ∠a + 100° + 30° = 180° [Being the sum of the angles of ∆ABC] or, ∠a + 130° = 180° or, ∠a = 180° – 130° = 50° Now, ∠b + 20° = ∠a [Being the exterior and opposite interior or, ∠b + 20° = 50° angles of ∆BDO] or, ∠b = 50° – 20° = 30° A B C P Q R D b a 35° b a 20° 30° 70° x 95° 100° B A E O C (ii) (iii) (i) Geometry: Triangle