Acme Mathematics 7 201B. Problem of Linear Equation on Two VariableIt is practical application of linear equations in two variable. In this case we will consider some practical problems involving unknown quantities. These problems are stated in words. They can be mathematically formulated as linear equations in two variable. The following steps can be followed to solve the word problems. (a) Read the problem carefully(b) Denote the unknown by some variable say x, y etc. (c) Translate the statement of the word problem step-by-step to a mathematical statement. (d) Make the equation. (e) Solve the equation for the unknown. After solving the equation, we get the solution to the given word problem.Solved ExamplesExample 1: The sum of two numbers is 21. If one of them is 9, find the numbers. Solution: Let, the first number be x, the second number = yNow, x + y = 21Let, y = 9or, x + 9 = 21or, x = 21 – 9or, x = 13Thus, the required numbers are : The first number be x = 13The second number = 9Hence, 13 and 9 are the numbers. Example 2: The length of a rectangle is 6 m more than its breadth. Find its length and breadth, if its perimeter is 28 m. Solution: Here, we need to find the length and breadth of the rectangle. Since length is given in terms of breadth, let the breadth of the rectangle be x meters and the length of rectangle = y m. According to the question,y = x + 6 x Perimeter28 mx + 6
202 Acme Mathematics 7Perimeter of rectangle = 28 m or, 2(length + breadth) = 28 or, 2(y + x) = 28or, 2(x + 6 + x) = 28 or, 2(2x + 6) = 28 or, 4x + 12 = 28 or, 4x = 28 – 12 or, 4x = 16 or, x = 164or, x = 4 Now, breadth of the rectangle = 4 m and the length of the rectangle = (4 + 6) m = 10 m.Classwork1. Make an equation for each of the following problems: (a) The sum of two number is 15. (b) The difference of two number is 10. (c) The product of two number is 27.(d) The quotient of a number when divided by 4 is 5. (e) Two times a number is 12. (f) Ten times of a number is 100. (g) Half times a number is 50. (h) A number divided by 11 gives 4. 2. Write in the sentence for the following mathematical equations. (a) x + 1 = 3 (b) x – 1 = 6 (c) 2x = 20 (d) 3x + 1 = 1 (e) 4x + 1 = 1 (f) 6x – 5 = 7 (g) 3x + x = 40 (h) 3x – x = 10 (i) x4 = 12
Acme Mathematics 7 203Exercise 4.11Solve the following word problems: 1. (a) The sum of two numbers is 20. If one of them is 15, find the other. (b) The difference of two numbers is 2. If one of them is 11, find the greater number. (c) The difference of two numbers is 5. If smaller number is 13, find the greater number.2. (a) If the sum of two consecutive numbers is 51, find them. (b) If the sum of two consecutive odd numbers is twenty four, find the number. (c) If the sum of two consecutive even numbers is 46, find the number. 3. (a) Divide Rs 2400 between A and B so that A gets double than B. (b) There are 1500 students in a school. If the number of girls is double the number of boys, find the number of boys and girls. (c) There are 44 students in a class. If the number of boys exceeds that of girls by 12, find the number of boys and girls. 4. (a) The breadth of a rectangular garden is half of its length. If the perimeter of the garden is 300 m, find its length and breadth.(b) The present age of a father is 4 times the age of his son. If the sum of their ages is 45 years, find the ages of son and father. (c) Mother is 30 years older than her daughter. Sum of their ages is 64 years. Find their ages.5. (a) \"When the sum of two number divided by 2, the quotient will be 12.\" (i) Write this in equation form. (ii) Solve it, when one number is double the next number.(b) The number of students in a class is 180 and the number of boys is double of the number of girls. (i) Express the fact in equation. (ii) Find the number of boys and girls.(c) There are 1500 students in a school. The number of girls is less than that of the boys by 90. (i) Convert the algebraic expression (ii) Find the number of boys and girls.(d) The sum of Bishal and two times of his son's age is 60 years. (i) Convert to the algebraic expression (ii) Find the age of son if Bishal is 42 years old.
204 Acme Mathematics 7B Inequality1. Trichotomy symbolsThese symbols; '=' (equal to), '<' (less than), and '>' (more than) are called trichotomy symbols. > is the symbol for 'is More than or equal to' < is the symbol for 'is Less than or equal to' We can compare the quantities using these symbols. For examples: 19 > 13 or, 13 < 19 44 = 44 or, x < 7 y > 7 If 'a' and 'b' are any two whole numbers then there are following possible relations: (i) a > b (ii) a < b (iii) a = b2. Negation of trichotomy symbols17 is greater than 15, Using the sign of trichotomy, we write it as 17 > 15. The negation of 17 > 15 is written as 17 >| 15. It is read as 17 is not greater than 15. The symbol >| is the negation of the symbol >. The symbol |> is the negation of the symbol <. The symbol ≠ is the negation of the symbol =.3. Trichotomy RulesStudy the following rules of trichotomy. Rule 1 Rule for additionLet 'a', 'b' and 'c' are any three numbers. If a > b and c is any integer, then a + c > b + c. If a < b and c is any integer, then a + c < b +c. Rule 2 Rule for subtraction If a > b and c is any integer, then a – c > b – c. If a < b and c is any integer, then a – c < b – c.
Acme Mathematics 7 205Rule 3 Rule for multiplication (i) When both sides of trichotomy symbols are multiplied by equal positive number. If a > b and c is positive number, then ac > bc. If a < b and c is positive number, then ac < bc.(ii) When both sides of trichotomy symbols are multiplied by equal negative number then symbol changes as follows: '<' change to '>' and '>' changes to '<'. If a > b and c is negative number, then ac < bc. If a < and c is negative number, then ac > bc.Rule 4 Rule for Division (i) When both sides of trichotomy symbols are divided by equal positive numberthen the symbol remains the same. If a > b and c is positive number, then ac > bc. If a < b and c is positive number, then ac < bc.(ii) When both sides of trichotomy symbols are divided by equal negative numbers then symbol changes as follows: '<' change to '>' and '>' changes to '<'. If a > b and c is negative number, then ac < bc. If a < and c is negative number, then ac > bc.The statement containing the symbols > or < is the inequality. x + 6 < 3 and 3x + 4 > 1 are the examples of an inequality.4. Solution of an inequalityConsider the inequality 2x – 5 > 9 and x ∈ W. It is true when we replace x by 8, 9, 10 or 11,… otherwise it is false. Hence x > 7 or, x = {8, 9, 10, 11,… } is called the solution set of the inequality. Inequality can be shown in the number line. 6 7 8 9x > 710 11 12 13 14 15The process of finding the solution of an inequality is same as solving the equation. is not coloured. So, 7 is not included in the solution.
206 Acme Mathematics 75. Graphical representation of an inequalityThe graphical representation of some inequalities are given below.(a) x > 2 –1 0 1 2 3 4 5 6 2 does not contain in x. So, x = {3, 4, 5, ...}.(b) x < 0 –2 –1 0 1 2 3 4 0 does not contain in x. So, x = {...., – 3, – 2, – 1}(c) x < 4 –1 0 1 2 3 4 5 64 contain in x. So, x = {...., –1, 0, 1, 2, 3, 4}(d) x > 4 –1 0 1 2 3 4 5 6 4 contain in x. So, x = {4, 5, 6, .....}(e) –2 < x < 2 –3 –2 –1 0 1 2 3– 2 and 2 does not contain in x.So, x = {–1, 0, 1}(f) 1 < x < 5 0 1 2 3 4 5 61 contain in x but 5 does not contain in x. So, x = {1, 2, 3, 4}(g) 0 < x < 4 –3 –2 –1 0 1 2 3 40 does not contain in x but 4 contain in x. So, x = {1, 2, 3, 4}(h) 2 < x < 8 0 1 2 3 4 5 6 7 82 and 8 both contain in x.So, x = {2, 3, 4, 5, 6, 7, 8}6. Solving an inequalityMethods of solving the inequality:(a) Method of Balancing (b) Method of transposition: The word transpose means 'changing the side'. The following rules are followed in this method: It is mostly common method. Rule - (i) + becomes – on changing the side Rule - (ii) – becomes + on changing the side Rule - (iii) × becomes ÷ on changing the side Rule - (iv) ÷ becomes × on changing the sideRight Hand Side (RHS)Left Hand Side (LHS)
Acme Mathematics 7 207Solved ExamplesExample 1: Solve the inequality 2x + 4 > 10 Solution: Here, 2x + 4 > 10 or, 2x + 4 – 4 > 10 – 4 [Subtracting 4 from both the sides]or, 2x > 6 or, 2x 2 > 62 [Dividing by 2 on both sides]or, x > 3 Hence, the value of x is greater than 3 or, x ={4,5,6,7,.......} Example 2: Solve: x3 < 2 Solution: Here, x3 < 2or, x3 × 3 < 2 × 3 [Multiplying both sides by 3]or, x < 6Hence, x = {6, 5, 4, 3, 2, 1, ………….} Example 3: Solve: –7x + 2 < 16 Solution: Here, – 7x + 2 < 16 or, – 7x + 2 – 2 < 16 – 2 [Subtracting 2 from both the sides]or, – 7x < 14 or, – 7x7 <147 [Dividing both sides by 7]or, – x < 2 or, x > – 2 [Multiplying both sides by –1]Hence, x = {– 2, – 1, 0, 1, ………….} Example 4: Solve and show the result in the number line 2(x + 1) < 14 Solution: Here, 2(x + 1) < 14 or, 2x + 2 < 14 or, 2x < 14 – 2 [Using rule (i)] or, 2x < 12 or, 2x2 <122 [Using rule (iii)] or, x < 6 Hence, x = {6, 5, 4, 3, 2, 1, ………….} The number line is given below: Remember!< changes to >0 1 2 4 5 6 7x < 63
208 Acme Mathematics 7Example 5: If three times of a number is added to 5, it is greater than or equal to 14. Find the numbers. Solution: Let the number be x, then; three times of x = 3x According to question, 3x + 5 ≥ 14or, 3x ≥ 14 – 5or, 3x ≥ 9or, 3x3 ≥ 93or, x ≥ 3Hence, x = {3, 4, 5, 6, 7, 8, .........}Therefore, the numbers are 3, 4, 5, 6, 7, ............ .Classwork1. Write True or False for the following statements.(a) 0 < – 3 (b) 3 – 4 = 4 – 3 (c) 5 < – 5(d) 6– 2 < – 41 (e) 2 + (–3) < 1 + 3 (f) 1 + 2 > 1 – 3(g) 1– 4 > 2– 5 (h) (–2) × 2 > 2 × (–3) (i) – 105 > 5– 102. Write the inequalities represented by the following number lines [take ‘x’ as variable].(a)–5 –4 –3 –2 –1 0 1 2 3 4 5(b)0 1 2 3 4 5 6 7 8 9 10(c)–5 –4 –3 –2 –1 0 1 2 3 4 5(d)0 1 2 3 4 5 6 7 8 9 10(d)–5 –4 –3 –2 –1 0 1 2 3 4 5(e)0 1 2 3 4 5 6 7 8 9 10Exercise 4.121. Find the solution set in the following inequalities where x ∈ W. (a) x + 2 > 8 (b) x – 1 > 5 (c) x + 6 < 10 (d) x – 8 < 0 (f) x + 2 ≥ 9 (h) x – 5 ≥ 7 (i) x + 4 ≥ 0 (j) x + 1 > 7
Acme Mathematics 7 2092. Draw the number lines for the following inequalities. (a) x > 7 (b) x < – 1 (c) x > 5 (d) x < – 2 (e) 3 < x < 8 (f) 5 < x < 10 (g) 4 > x > – 3 (h) 3 ≥ x > 1 (i) – 7 < x < – 2 3. Find the solution set in the given inequalities. (a) 6x + 1 > 2x + 9 (b) 5x – 10 > 4x + 18 (c) 4x – 9 < 3x + 2 (d) 2x – 9 < 4x +3 (e) x4 + 3 < 5 (f) x3 – 6 ≥ 4(g) 3(x – 2) < 0 (h) 2(x + 4) < 4(x – 1) (i) 4(2 – x) > 7 – 5x (j) 3 < x – 2 < 8 (k) 2 < x + 4 < 12 (l) 10 > 2x > 2 4. (a) Look at the adjoining figure and write the inequality in terms of x for the following conditions. (i) AB + BC > 3AD (ii) AD + DC < 2BC(iii) 2DC + AB > 4BC (b) (x + 2) cm, (x + 1) cm and (3 + x) cm are the measures of the three sides of the triangle ABC. Make three inequalities to represent the properties of the triangle according to these sides.5. Rewrite the following word problem in inequality and solve it.: The length of a rectangle is 6 cm longer than its breadth. Half of the perimeter of the rectangle is less than or equal to 50 cm. 6. Nine times a certain number is subtracted from 91. The result is greater than 10. (a) Make the inequality. (b) Find the solution set of the inequality (c) If possible, draw the graph of the inequality. 7. Four times the number added to 7 is less than 15. (a) Make the inequality (b) Find the solution set of the inequality (c) If possible, draw the graph of the inequalityDACB x + 7xx + 123x – 12
210 Acme Mathematics 7C Linear Equation and its graphThe rule discuss in the previous lesson is linear equation. y = x + 4, y = x – 3, y = 3x + 1 and y = 12x etc are linear equations. In the linear equations input gives the output. If we replaced the input by x and output by y as variables, then we get a relation in the algebraic form. Look at the following example:Input (x) 1 3 5 7 9Output (y) 5 7 9 11 13Here, 5 – 1 = 4, 7 – 3 = 4, 9 – 5 = 411 – 7 = 4, 13 – 9 = 4Now, we get a rule by which the given input and output are associated. The rule is y = x + 4 This y = x + 4 is a linear equation in two variables x and y. We can draw the graph of linear equation. To draw the graph, we plot the order pairs obtained from the table. In this case (1, 5), (3, 7), (5, 9), (7, 11) and (9, 13) are the ordered pairs. When we first plot the points on thegraph and joining the points in an order, we get the graph of the linear equation. The graph of the linear equation is always a straight line. The graph of y = x + 4 is given alongside.Classwork1. Complete the tables given below according to the given rule.(a) Rule: y = x + 4Input (x) 0 1 2 3 4 5 6Output (y) ........... ........... ........... ........... ........... ........... ...........(b) Rule: y = x – 3Input (x) – 3 – 2 – 1 0 1Output (y) ........... ........... ........... ........... ...........Y'YX' X(1,5)(3,7)(5,9)(7,11)(9,13)Input (x)Output(y)
Acme Mathematics 7 211(c) Rule: y = 3x + 1Input (x) 6 7 8 9 10 11Output (y) ........... ........... ........... ........... ........... ...........(d) Rule: y = 12xInput (x) 1 2 3 4 6Output (y) ........... ........... ........... ........... ...........2. Complete the following tables:(a) Rule: y = x + 3Input (x) 12 13 14 ........... 16 ........... ...........Output (y) ........... ........... ........... 18 ........... 20 213. Complete the following table and draw the graphs for each:(c) When, y = 2x x 0 1 2 3 4 5yExercise 4.131. Complete the following tables:(a) Rule: y = x – 10Input (x) 2 ........... ........... 8 10 12 ...........Output (y) ........... – 6 – 4 ........... ........... ........... 4(b) Rule: y = 6xInput (x) ........... ........... ........... 4 5Output (y) 6 12 18 ........... ...........(c) Rule: y = x5Input (x) – 20 ........... ........... – 5 5 ...........Output (y) ........... 2 3 ........... ........... 62. Complete the following table and draw the graphs for each:(a) When, y = x + 2 x 0 2 4 6 y(b) When, y = x – 5 xy
212 Acme Mathematics 7(c) When, y = x + 23x 1 4 7 10y3. Make the table and make 5 order pairs which satisfies the given linear equations and draw its graph. (a) y = 3x + 1 (b) y = 2x – 1 (c) y = x2 + 2 4. Draw the graph of the following linear equations: (a) y = – x (b) 4x + y = 7 (c) y – 3x = – 8 5. Study the graph and find the linear equations represented by the graph.(a)X'Y'YX O(1,2)(2,4)(3,6)(4,8)(5,10)Input (x)Output(y)(b)X'Y'YX O(–1,2)(0,4)(1,6)(2,8)(3,10)Input (x)Output(y)(c)X'Y'YX O(1,4)(0,3)(2,5)(3,6)Input (x)Output(y)6. Write the rule for each input and output. Express the rule as an equation.(a) Input (x) 0 3 6 9 12Output (y) 5 8 11 14 17(b) Input (x) – 3 1 5 9 13 17Output (y) – 6 – 2 2 6 10 14(c) Input (x) 4 5 7 10 14 20Output (y) 14 15 17 20 24 307. Study the graph given along side. AB represents the relation between the input (x) and output (y). (a) Express x and y as table. (b) Express the relation in mathematical form. 8. The relation between the input (x) and output (y) is given below:(a) Write its input sets. (b) Write its output sets. (c) Express it as table. (d) Draw its graph. X'Y'YX OBA01234567Input Output
Acme Mathematics 7 2131. Simplify:(a) (a + b)3 × (a + b)5(b) 10a4 × 15a575a8(c) If x + y + z = 0, then prove that: Px – y × Px + y × Py + z × Py – z× Pz + x × Pz – x = 12. (a) Find the value of 2x°.(b) Write the index form. (–2a) × (– 2a) × (–3b) × (– 3b) × (– 3b)3. Do the following questions.(a) Simplify: 27– 23(b) Prove that: xaxba + b . xbxcb + c . xcxac + a= 14. (a) Multiply: (5x – 2y) × (7x – 2y)(b) Divide: (x2 + 7x + 12) ÷ (x + 4)(c) The length (x + 2y)m and breadth (3x – y) m of a rectangular land is given, find the area of this land.5. (a) Expand the foumula of (a + b)2(b) By using formula, find the square of (z + 3).(c) If a + 1a = 10, then find the value of a2 + 1a2 .6. (a) Find the value of x from the given quadrilateral.(b) Find the quotient: (x3 + y3) ÷ (x + y)7. (a) Multiply: (5 m + 3 n) and (3 m – 5 n)(b) Calculate the area of given rectgular figure.8. (a) What is the expanded form of (a + b)2?(b) The length of a rectangular mat is 3x – y meter and its breadth is 3y – x meter. Find the area of the mat.3x 2yyxA BD C3x 4x2x xMixed Exercise
214 Acme Mathematics 79. On the occasion of birthday, Pemba distributed (p2 + 29p – 96) bottles of sanitizer equally among (p + 32) people of his village.(a) Find the number of bottles of sanitizer received by each people.(b) If p = 8, find the actual number of bottles of sanitizer, number of people, and share of each people.10. Two terms are given below.(2x + 3y)2m and (2x + 3y)2n(a) Multiply these two terms.(b) If m + n = 0, what will be the product?11. A polynomial in variable x and y is given x4 + 2x2y – 3xy2 + 7(a) Write degree of the polynomial.(b) If x = 2 and y = 3, find the value of the polynomial.12. Length and breadth of rectangle are (x – 3) and (x – 4) respectively.(a) Find its area.(b) If a + b = 3 and ab = 2, find the value of a2 + b2.13. (a) Define variable.(b) Write the type of polynomial: 3x + 4y – 6(c) Write the polynomial in standard form.5x + 3 + 7x2 + 3x – 6 + 3x2(d) Are the following polynomials equal?f(x) = 2x2 – 3x + 4 g(x) = – 62x + 82 + 2x214. The length of square is (3x + 4y) cm.(a) Find its area.(b) If x = 4 and y = – 1, find its actual area.(c) If M = (x2 + 2x + 4) and N = (x2 – 2x + 4), show that M.N = x4 + 4x2 + 16.15. Do the following:(a) Find the product of: (x – 2) and (x – 3).(b) If (a + b) = 5 and ab = 6, find value of a2 + b2.16. Study the given cube.(a) Calculate the total surface area of the cube.(b) If x = 4 and y = – 1, find its actual area.(c) If M = (x2 + 2x + 4) and N = (x2 – 4), show that M.N = x4 + 2x3 – 8x – 16.(x + 2y)(x + 2y)(x + 2y)
Acme Mathematics 7 21517. (a) Solve: 5x – 1 = 19(b) Solve the inequality: x3 ≤ 218. (a) Find the value of x from the given right angle triangle.(b) Complete the following table and draw the graph. When y = x + 2Input (x) 0 1 2 3Output (y)(c) Solve: x + 53 = x + 44 19. (a) Write down the inequality represented by the following graph.–10–9 –8 –7 –6 5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10(b) There are 40 students in a class. If the number of boys is 10 more than that of girls, find the number of boys and girls.20. (a) Solve: 3x + 13 = 9 + 5x(b) There are 410 students in a school. The number of boys exceeds that of the girls by 90. Find the number of boys and girls.21. The length and breadth of a playground are (2x + 3y + 5) m and (3x + 2) m respectively.(a) Find the area of that playground.(b) Find the actual area of playground if x = 2 m and y = 1 m.(c) Solve : 5 + 3x = 5x – 11.22. (a) Draw a graph for the following linear equation: 5x + y = 10(b) Divide : 3x2 – 5xy – 12y2 by x – 3y.23. Answer the following questions.(a) Write the degree of x2 + 5x + 6.(b) Solve and show the result in the number line.2(x + 1) < 14.(c) Find the product of : 3x3 – 4y3 and 2x2y(d) Divide : (4x2 – 6x – 6xy + 9y) by (2x – 3y)ABx6 cm C8 cm
216 Acme Mathematics 724. (a) Divide: (4a2 + 12a) ÷ (2a + 6)(b) Simplify: x2 + 7x + 12x + 425. The sum of length and breadth of a rectangle is 30 m.(a) Express the statements as equation.(b) If length of rectangle is 20 m, calculate its breadth.26. The age of father and twice the age of his son is 50 years.(a) Express the ages of father and his son as equation.(b) If the age of son is 5 years, find the age of father.27. In a class, there are 48 students.(a) Taking 'x' as number of boys and 'y' as number of girls, make a equation.(b) Complete the table.Possible no. of (x)Possible no. of (y)(c) Draw its graph.28. The sum of 3 times of number and seven is less than 10.(a) Express as inequality.(b) Solve the inequality.(c) Show on the number lines.29. Study the graph and find the linear equations represented by the graph.(a)X'Y'YX O(2,2)(3,4)(4,6)(5,8)(6,10)Input (x)Output(y)(b)X'Y'YX O(–1,4)(0,6)(1,8)(2,10)Input (x)Output(y) (–2,2)(c)X'Y'YX O(1,4) (3,4)(4,3) (5,2)(1,6)(0,3)(2,5)(3,6)Input (x)Output(y)
Acme Mathematics 7 217EvaluationTime: 43 minutes Full Marks: 181. An inequality is represented on a number line.(a) What does the black circle on the nuber line represents? Write it. [1](b) Table formed by the linear equation x + 2y = 8 with two variables is givenbelow. Draw the graph on the basis of the given table. [2]x 0 2 4y 4 3 22. (a) Show the inequality x > 2 on number line. [1](b) \"When the sum of 14 and double a number is divided by 3, the quotient will be 22.\" Write this in equation form. And solve it. [2]3. (a) An inequality is representedon a number line.What does the circle on the nuber line represents? Write it. [1](b) Table formed by the linear equation 2x - y = 8 with two variables is given below. Draw the graph on the basis of the given table. [2]x 0 4 6y -8 0 44. (a) An inequality is represented on a number line. Write the inequality represented by the number line. [1](b) The linear equation x + y = 11 is given. Complete the table from it. [2]xy5. The number of students in a class in 90 and the number of girls is half of the number of boys.(a) Express the fact in equation. [1] (b) Find the number of boys and girls. [2] 6. Sum of 4 and three times a number 'is equal to' or 'greater than' the difference of thatnumber and 2.(a) Express the given statement in inequality. [1](b) Solve the inequality and show on a number line. [2]-4 -3 -2 -1 0 1 2 3 40 1 2 3 4 5 _1 _2 _3 _4 _5x0 1 2 3 4 5 _1 _2 _3 _4 _5x
218 Acme Mathematics 75UNIT Geometry1. Name the lines:(a)ACBD(b)G FE H (c)A DCB2. Draw the following angles using protractor:(a) 30° (b) 40° (c) 100° (d) 120°3. Draw the line parallel to given lines.(a)A B(b)CD (c) PQ4. Measure the following angles.(a)OAB(b)OBC(c)OXY5. Calculate the value of 'x'.(a)50°x°(b)x°60°70°(c)125°x°6. Draw the line perpendicular to the given line through P.(a) A B P(b) C DP(c)X YP7. Measure the angles and write its types.(a)OAB(b)OBC(c) OXYWarm Up Test
Acme Mathematics 7 2195.1 Line and AngleA Construction of bisector(a) Construction of perpendicular bisector of a line segment using compass Draw a line segment of length 6 cm. AB is 6 cm. With A as foot of compass (center) draw an arc of radius more than half of length AB.Repeat the same precess from B also. The arcs cut each other at X and Y. Join X and Y. This line XY divides the line AB into two equal parts as AO = OB = 3 cm.XY is called the bisector of the line AB. ∠XOB is 90°.So, XY⊥AB.Hence, XY is called perpendicular bisector of the line AB.(b) Construction of bisector of angleBisector of the line is a line that divides a given line into two equal parts.Bisector of the angle is a line that divides a given angle into two equal parts.Let us bisect an angle of 60°. Draw an angle of 60° using set-square. ∠AOB = 60° Take O as foot of compass and take any arc cutting OA at P and OB at Q with compass.A 6 cm BA 6 cm BA 6 cm BA O BYX3 cm3 cmO B60°A
220 Acme Mathematics 7 Put foot of compass at P and draw an arc at D. Put foot of compass at Q and draw an arc at D. Join O and D.Now, line OD is the bisector of ∠AOB where ∠AOD = ∠DOB.B Construction of angles using Compass and Scale(a) Construction of 60° Draw a line AB = 5 cm. Put the foot of the compass at A and take an arc which cut AB at C. Taking the same arc, put the foot of compass at C and cut the first arc at D. Join A and D and extend AD to O using scale. Now, ∠OAB = 60°.(b) Construction of 120° Draw a line AB = 5 cm. Follow the steps of 60°, up to second and third steps. Put foot of compass at D and cut the arc CD at E. Such that arc CD = arc DE. Join A and E and extend it to F. Now, ∠FAB = 120°.O Q B60°PAO Q B60°PAA 5 cm B A C 5 cm BA C 5 cm B A 5 cm B60°CDOA CDA 5 cm B BA CD EBA CD EBF120° O Q BD D60°PAO Q B30°30°PA
Acme Mathematics 7 221(c) Construction of 90° Follow the steps of 120° up to second last step. Put foot of compass at E and draw an arc above the E. Put foot of compass at D and draw the arc above D which cut the first arc at F. Join A and F. Now, ∠FAB = 90°.(d) Construction of 45° Draw an angle of 90°. ∠FAB = 90° Bisect the angle FAB. Now, ∠GAB = 45°, half of 90°. Similarly, We can construct the angle 30°, 135° etc.We can bisect any angle following the above steps.(e) Construction of 30° Draw an angle of 60°, ∠CAB = 60° Bisect the angle CABNow, ∠DAB = 30°, half of 60°(f) Construction of 15° Draw an angle of 30°. Bisect the angle DAB.Now, ∠BAE = 15°, half of 30°A CE DBFA CE DBAFCE D90° BAFE DGC45° A BFCE D90° BCA B60°CA B30°DCA B30°DCA B15°DE
222 Acme Mathematics 7(g) Construction of 75° Draw an angle of 90° Draw an angle of 60°∠BOA = 90°, ∠COA = 60° Bisect the angle BOC. Now, ∠DOA = 75°(h) Construction of 105° Draw an angle of 90°∠BOA = 90° Draw an angle of 120°∠COA = 120° Bisect the angle BOC.Now, ∠DOA = 105°(i) Construction of 150° Draw an angle of 120°∠BOA = 120° Draw an angle of 180°∠COA = 180° Bisect the angle COB.Now, ∠DOA = 150°(j) Construction of 135° Draw an angle of 120°∠BOA = 120° Draw an angle of 150°∠COA = 150° Bisect the angle COB.Now, ∠DOA = 135°O AC B60°O AC B D75°O ACBO ACDB105°C O ABD150°C O AB120°OCAB120°OCABD135°
Acme Mathematics 7 223C Construction of angle equal to given angleLet ∠AOB be the given angle. With O as centre and a convenient radius, draw an arc XY, Cutting OA at X and OB at Y respectively. Now, take any line segment PQ. Taking P as centre and the same radius as before draw an arc CD cutting PQ at C. With C as centre and radius equal to XY draw an arc, cutting the arc CD at E. Join PE and extend it to R. ∠RPQ is equal to ∠AOB.D Pair of angles(a) Complementary angleIf the sum of two angle is 90°, then the angles are called complementary angle.In the figure, ∠AOB and ∠COB are complementary angles, as their sum is 90°.∠AOB and ∠PQR are complementary angles.AO70°BPQ20°R(b) Supplementary anglesIf the sum of two angles is 180°, then the angles are called supplementary angle.In the figure ∠AOC and ∠COB are supplementary angle as their sum is 180°.In the figure ∠AOB and ∠XYZ are supplementary angles.AO Y BXAO BRP C QEDACO30°60°BCA O110° 70°B
224 Acme Mathematics 7BA30°O YX150°Z(c) Adjacent anglesIn the figure OB is common arm of ∠AOB and ∠BOC. Such angles are called adjacent angles.(d) Vertically opposite angles (VOA)When two lines intersect, the angles formed opposite to each other as shown in figure alongside are called vertically opposite angles.∠AOD and ∠COB are VOA.∠AOC and ∠BOD are VOAE Experimental Verifications(a) The pair of vertically opposite angles are equal.Draw a pair of vertically opposite angles (V.O.A) in three different places as shown in the following figure, Measure each vertically opposite angle and tabulate the result in the table given below.(i)CAOBD (ii)C DBAO(iii)C DBAOFigure Pair of vertically opposite angles Result∠AOC ∠DOB ∠AOD ∠COB∠AOC = ∠DOB∠AOD =∠COB(i) 90° 90° 90° 90°(ii) 110° 110° 70° 70°(iii) 38° 38° 142° 142°Hence, vertically opposite angles are equal.(b) Sum of adjacent angle is 180° or sum of angles at a point and on the same side is 180°.Draw adjacent angles, in three different places as shown in the following figure. Measure each angles and tabulate the result in the table given below: ABO COCA DB
Acme Mathematics 7 225(i)A BCO(ii)A BCO(iii)A BCOFigure ∠AOC ∠COB ∠AOC + ∠COB Result(i) 112° 68° 180°∠AOC + ∠COB = 180° (ii) 65° 115° 180°(iii) 90° 90° 180°Hence, sum of adjacent angles is 180°.(c) Sum of angles at a point is 360°. Draw four angles at a point in three different places as shown in the following figure. Measure each angles and tabulate the result in the table given below :(i) (ii) (iii)Figure Angles at a points Sum of angles∠AOB ∠BOC ∠COD ∠AOD(i) 90° 50° 130° 90° 360°(ii) 70° 110° 70° 110° 360°(iii) 102° 78° 102° 78° 360°Hence, sum of angles at a point is 360°.Solved ExampleExample 1 Find the complementary angle of the following angles:(a) 74° (b) 2° (c) 90°Solution : (a) Complementary angle of 74° = 90° – 74° = 16°(b) Complementary angle of 2° = 90° – 2° = 88°(c) Complementary angle of 90° = 90° – 90° = 0°ADBOCBODACOB CD A
226 Acme Mathematics 7Example 2 Find the supplementary angle of the following angles:(a) 100° (b) 175° (c) 9°Solution : (a) supplementary angle of 100° = 180° – 100° = 80°(b) supplementary angle of 175° = 180° – 175° = 5°(c) supplementary angle of 9° = 180° – 9° = 171°Example 3 Look at the adjoining figure and find the value of x.Solution : Here, 2x + 3x + 30° = 180° (being straight angle)or, 5x + 30° = 180°or, 5x = 180° – 30°or, 5x = 150° or, x = 150°5 = 30°The value of x is 30° Example 4 In the adjoining figure, if ∠1 = 125°, find the measure of remaining angles.Solution : Here, ∠1 = 125°Now, (i) ∠1 + ∠2 = 180° [Straight angle]or, 125° + ∠2 = 180° or, ∠2 = 180° – 125° or, ∠2 = 55° (ii) ∠1 = ∠3 [VOA]or, 125° = ∠3 ∴ ∠3 = 125° (iii) ∠4 = ∠2 [VOA]or, ∠4 = 55° Hence, 55°, 125° and 55° are remaining angles.Example 5 Find the measure of angles 'a' and 'b' from the given figure. Solution : Here, a = 120° [VOA] a + b = 180° [Sum of adjacent angles]or, 120° + b = 180° or, b = 180° – 120° or, b = 60°30°3x2xA BFE1 23 4ba120°
Acme Mathematics 7 227Classwork1. Fill in the blanks. (Use figure given alongside) (a) ∠AOC and ∠DOB are ............ angles. (b) ∠AOC and ∠COB are ............. angles. (c) ∠1 and ∠3 are ................ angles. (d) ∠4 and ∠2 are ........... angles. (e) ∠8 and ∠7 are ............ angles. (f) The sum of ∠3 and ∠2 is ................ (g) The sum of ∠4 and ∠1 is ................ (h) The sum of ∠6 and ∠7 is ................2. Write true or false for the following statement.(a) Adjacent angles are equal.(b) V.O.A. are not equal in size.(c) Sum of supplementary angle is 180°(d) Sum of complementary angle is 90°(e) V.O.A. have same vertex and different arms.3. From the figure write the following angles. (a) Pair of V.O.A. angles. (b) Pair of adjacent angles. (c) Pair of supplementary angles. 4. Match the followings:(a) Angles 20° and 70° can be complementary(b) Two acute angles add up 360°(c) Two right angles are complementary(d) Two straight angles are supplementaryC DBOA1526374815483726
228 Acme Mathematics 7Exercise 5.11. Find the measure of unknown angles.(a)50° x(b)x 20°(c)y 20°(d)zx40°y(e)x 40°x(f)x 70° xa z y(g)a° a+50°(h)45°x+10° x–2°(i)(j)70° 2x°x–15°5x°(k)a3ax80° y(l)6x°131°x° x°2. Find the complementary angle of the following angles.(a) 80° (b) 53° (c) 7212° (d) 8314°3. Find the supplementary angle of the following angles.(a) 35° (b) 121° (c) 135.5° (d) 16614°4. Find the value of x in the following figures.(a)A105°B x(b)A B4x5x(c)A B80°2x–10°(d) (e) (f)60°x ab cy70°2x–50x+7CDAC B D2x° x+18°A B70°2x–10°
Acme Mathematics 7 2295. Construct the following angles using compass.(a) 15° (b) 75° (c) 105°(d) 135° (e) 150°6. Draw the angles equal to the size of given angle.(a) (b) (c)7. Measure the following angles and construct their supplementary angle.(a) (b) (c)(d) (e) (f)8. Measure the following angles and construct their complementary angle.(a) (b) (c)AO BVO N KO TMA TNO TPQ RMN OAB CRA TMA TNO TPQ R
230 Acme Mathematics 7(d) (e) (f)9. Draw given lines on your copy and construct given angle at given point.(a) At B, 63° (b) At X, 95° (c) At R, 105°10. Using protractor, construct an angle 160°. Name it. Bisect the angle and measure two parts of the angle. Write their measure.11. Using scale, construct an angle. Name it and find its measure also.12. Experimentally verify the following statements.(a) VOA are equal.(b) The sum of adjacent angle is 180°.(c) Angle of a point is 360°.13. Study the following figures.(i) (ii)(a) Measure the angles.(b) Write the equal angles.(c) Write the supplementary angles.(d) Write the angles, whose sum is 360°.MN ORA TAB CAB XYQRDCO A BSTP R Q
Acme Mathematics 7 2315.2 Triangle and QuadrilateralWarm Up Test1. Write the name of the triangle in 3 different ways.(a) (b) (c)2. Write the name of the quadrilateral in 2 different ways.(a) (b) (c)3. Classify the given triangles according to their angles.(a) (b)60°50°70°(c)110°40°30°4. Classify the given triangles according to their sides.(a)3 cm3 cm3 cm(b)3 cm5 cm5 cm(c)6 cm5 cm4 cmBCAPQRXYZBCDASPRQOPNM
232 Acme Mathematics 75. Measure the angles of the following triangles.(a)B CA (b)Q RP (c)YX Z6. Find the third angle of the following triangles.(a)60° 60°(b)40°(c) 40°20°7. Name the quadrilaterals.(a) (b) (c)(d) (e) (f)8. Measure the angles of the following quadrilaterals.(a)CB DA (b)QPRS (c) MNPO∠A = ...........∠B = ...........∠C = ...........∠D = ...........∠P = ...........∠Q = ...........∠R = ...........∠S = ...........∠M = ...........∠N = ...........∠O = ...........∠P = ...........
Acme Mathematics 7 233A. Construction of TrianglesWe know that a triangle has six parts Three sides and Three angles To construct a triangle we should have the measure of three parts of a triangle. Otherwise we can not construct the triangle. (a) Construction of triangle when three sides are given. To construct a triangle with a given measure, we should first draw a rough sketch to indicate the given measures. This rough sketch will help us to understand the steps to be followed in the construction.Solved ExampleExample 1 Construct a triangle, ABC, such that AB = 3 cm, BC = 4 cm and CA = 5 cm. Solution: The steps of constructions: (i) Draw a rough sketch of ΔABC and mention the given measure. (ii) Draw a line segment BC of length given.(iii) With C as centre and radius 5 cm, draw another arc intersecting the arcs at a point A(iv) With B as centre and radius 3 cm draw an arc above the point B.(v) Join A with B and C. ΔABC is the required triangle Remember the properties of triangle while constructing it.We must draw a rough sketch at first.B 4 cm C5 cmA3 cmRough SketchB 4 cm CBA4 cm C5 cm 3 cm
234 Acme Mathematics 7(b) Construction of triangle when two sides and included angle are given.Solved ExampleExample 1 Construct a triangle ABC, such that ∠B = 80°, BC = 6 cm and AB = 5 cm Solution: Steps of constructions: (i) Draw a rough sketch of ΔABC and mention the given measures. (ii) Draw a line segment BC of length 6 cm . (iii) Using protractor draw an angle of 80° at B. i.e. ∠XBC = 80° (iv) With B as centre and 5 cm as radius, cut the line BX at A. (v) Join A with C. ΔABC is the required triangle.(c) Construction of triangle when two angles and the included side are given.Solved ExampleExample 1 Construct a triangle ABC, such that BC = 7 cm, ∠B = 60° and ∠C = 70°.Solution: Steps of constructions: (i) Draw a rough sketch of ΔABC and mention given measure.Rough SketchB80°6 cm5 cmCAB 6 cm CB80°6 cm5 cmCAXRough Sketch60° 70°B 7 cm CAB80°6 cm CX
Acme Mathematics 7 235(ii) Draw a line segment BC of length 7 cm. (iii) Using protractor draw an angle of 60° at B and 70° at C.(iv) Extend the lines BP and CQ if needed, such that they meet at A.ΔABC is the required triangle. (d) Construction of right-angled triangle when length of hypotenuse and a side are given.Solved ExampleExample 1 Construct a right angled triangle ABC with ∠B = 90°, hypotenuse AC = 5 cm and BC = 4 cm. Solution: Steps of constructions: (i) Draw a rough sketch of ΔABC and mention the given measure. (ii) Draw a line segment of BC = 4 cm. (iii) Draw an angle 90° at B. i.e. ∠XBC = 90° (iv) With C as centre and CA = 5 cm as radius, draw an arc intersecting BX at A. (v) Join A and C. ΔABC is the required triangle.B 7 cm C60° 70°B 7 cm CAQ PRough SketchB90°4 cm5 cmACB 4 cm CB 4 cm C90°XBAX4 cm C90°5 cm
236 Acme Mathematics 7(e) Construction of a triangle when a side, angle on the side and angle opposite to given side are give.Solved ExampleExample 1 Construct a triangle ABC, such that BC = 7 cm, ∠B = 60° and ∠A = 50°.Solution: Steps of constructions: (i) Draw a rough sketch of ΔABC and mention the given measurement.(ii) Draw a line segment BC of length 7 cm. (iii) Using protractor draw an angle of 60° at B and 180° – (60 + 50) = 70° at C.i.e. ∠PBC = 60° and ∠QCB = 70° (iv) Extend the lines BP and CQ if needed, such that they meet at A where ∠A = 50°.ΔABC is the required triangle. Exercise 5.21. Construct ΔPQR with the following measurements. (a) PQ = 6 cm, QR = 5 cm and PR = 7 cm (b) PQ = 3.6 cm, QR = 4.5 cm and PR = 6 cm (c) PQ = 5 cm, QR = 7.5 cm and PR = 8 cm 2. Construct ΔABC with the following measurements. (a) AB = 6 cm, AC = 5 cm and ∠A = 62° (b) AB = 5.5 cm, AC = 6 cm and ∠A = 75° (c) BC = 7 cm, ∠B = 40° and BA = 5 cm Rough Sketch60°50°B 7 cm CAB 7 cm C60° 70°50°B 7 cm CAQ PAngle opposite to hypotenuse is 90°.
Acme Mathematics 7 2373. Construct ΔXYZ with the following measurements. (a) XY = 5 cm, ∠X = 50° and ∠Y = 60° (b) YZ = 6 cm, ∠Y = 40° and ∠Z = 55° (c) XZ = 5.5 cm, ∠X = 58° and ∠Z = 30° 4. Construct ΔMNO with the following measurements. (a) ∠N = 90 °, NO = 7 cm and ∠O = 45° (b) ∠M = 90°, MN = 6 cm and ∠N = 55° (c) ∠O = 90°, OM = 6.5 cm and ∠M = 30° 5. Construct an equilateral ΔABC with each side: (a) 4 cm (b) 5 cm (c) 6 cm (d) 5.5 cm 6. Construct an isosceles right angled triangle XYZ with ∠Z = 90° and XZ = 5.5 cm.7. Construct the right-angled triangle KPA in which: (a) PA = 4 cm and hypotenuse KA = 5.5 cm(b) KP = 5 cm and hypotenuse PA = 6.5 cm(c) AK = 3 cm and hypotenuse KP = 5 cm8. Construct ∆MKA with the following measurement.(a) AM = 6 cm, ∠A = 60° and ∠AKM = 90°(b) KM = 5 cm, ∠K = 70° and ∠KAM = 50°(c) KA = 6.5 cm, ∠K = 45° and ∠KMA = 80°9. Construct ∆ABC with the following measurement.(a) BC = 4 cm, BA = 4 cm and ∠B = 90°(b) BC = 5 cm, ∠B = 90° and ∠A = ∠C = 45°(c) BC = 6 cm, ∠B = 45° and ∠C = 45°
238 Acme Mathematics 7B. QuadrilateralsQuadrilateral is a closed figure enclosed by four sides. In the figure, ABCD is a quadrilateral. Quadrilateral is also called polygon with four sides. There are some special quadrilaterals. They are: 1. Parallelogram 2. Rectangle3. Square 4. Rhombus 5. Trapezium 6. Kite 1. Parallelogram: ABCD is a parallelogram. Its opposite sides are equal. i.e. AB = CD and BC = AD Its opposite sides are parallel. i.e. AB||DC and BC||AD Its opposite angles are equal. i.e. ∠A = ∠C and ∠B = ∠D. Its diagonal bisect to each other.2. Rectangle: ABCD is a rectangle. Its opposite sides are equal. i.e. AB = CD and BC = AD. Its opposite sides are parallel. i.e. AB||DC and BC||AD Its all angles are equal and 90°. i.e. ∠A = ∠B = ∠C = ∠D and all are 90°. Its diagonals are equal. i.e. Diagonal AC = diagonal BD. Its diagonal bisect to each other.i.e. AO = OC and BO = OD BBA CBACDBACDBACDBACOD
Acme Mathematics 7 2393. Square: ABCD is a square. Its all sides are equal. i.e. AB = BC = CD = DA Its opposite sides are parallel. i.e. AB||DC and BC||AD. It's all angles are 90°.i.e. ∠A = 90°, ∠B = 90°, ∠C = 90°, ∠D = 90° Its diagonals are equal. i.e. Diagonal AC = Diagonal BD. Its diagonals bisect each other at 90°. i.e. AO = OC and BO = OD. AC⊥BD. Diagonals of a square bisect the angle at the vertex. i.e. ∠BAC = ∠DAC4. Rhombus: ABCD is a rhombus. Its all sides are equal. i.e. AB = BC = CD = DA Its opposite sides are parallel. i.e. AB||DC and BC||AD. Its opposite angles are equal i.e. ∠A = ∠C and ∠B = ∠D Its diagonals are not equal. i.e. Diagonals AC ≠ Diagonal BD. Its diagonals bisect each other at 90°. i.e. AO = OC and BO = OD and AC⊥BD. Diagonals of a rhombus bisect the angle at the vertex.i.e. ∠BAC = ∠DAC5. Trapezium: If a quadrilateral has a pair of opposite sidesparallel then it is called trapezium. ABCD is a trapezium. Its two sides AD and BC are parallel. 6. Kite: If a quadrilateral has two distinct consecutive pairs of sides are equal then it is called kite. ABCD is a kite where AB = AD and BC = CD.BACDBAOCDBACDBOACDBACDCB DA
240 Acme Mathematics 7C. Properties of Parallelogram, Rectangle and Square(a) Properties of parallelogram(i) Opposite sides of a parallelogram are equal.Draw three parallelogram ABCD of different sizes by using set square.A(i)DBC(ii)DACB(iii)ADBCSides AB, DC and AD, BC are opposite sides.Now measure the sides of each parallelogram and complete the table.Figure AB DC AD BC Result(i) 2 cm 2 cm 1.6 cm 1.6 cm(ii) 1 cm 1 cm 1.6 cm 1.6 cm AB = DC and AD = BC(iii) 1.1 cm 1.1 cm 1.7 cm 1.7 cm(ii) Opposite angles of a parallelogram are equal.Draw three parallelograms ABCD of different sizes by using set square.A(i)DBC(ii)DACB(iii)ADBCOpposite angles are ∠A, ∠C and ∠B, ∠D.Now measure the angles of each parallelogram by using protractor and complete the table.Figure ∠A ∠C ∠B ∠D Result(i) 70° 70° 110° 110°(ii) 105° 105° 75° 75° ∠A = ∠C and ∠B = ∠D.(iii) 90° 90° 90° 90°Hence, opposite angles of a parallelogram are equal.Note : Sum of angles of any quadrilateral is 360°.
Acme Mathematics 7 241(iii) Diagonals of a parallelogram bisect to each other.Draw three parallelogram ABCD of different sizes by using set square.A(i)DBCO(ii)DACBO(iii)ADBCOIn the figures AC and BD are diagonals. O is the intersection of diagonals AC and BD. Now, measures the length of AO, OC, BO and OD in each parallelogram and complete the table.Figure AO OC BO OD Result(i) 1.5 cm 1.5 cm 1.3 cm 1.3 cm(ii) 1 cm 1 cm 1.2 cm 1.2 cm AO = OC and BO = OD(iii) 1.1 cm 1.1 cm 1.1 cm 1.1 cmHence, diagonals of a parallelogram bisect to each other.(b) Properties of Rectangle(i) Diagonals of a rectangle are equalDraw three rectangles ABCD of different size by using set squares.(i) ADBC(ii)DACB (iii)AD BCIn the figures AC and BD are diagonals.Now, measure the length of AC and BD in each rectangle and complete the table.Figure AC BD Result(i) 2.4 cm 2.4 cm(ii) 2.3 cm 2.3 cm diagonal AC = diagonal BD(iii) 2.4 cm 2.4 cmHence, the diagonals of a rectangle are equal.
242 Acme Mathematics 7(ii) All angles of rectangle are 90°.Draw three rectangle ABCD of different sizes by using set-square.(i)ADBC(ii)ADBC(iii)ADBCIn the figures ∠ABC, ∠BCD, ∠CDA and ∠DAB are angles of different rectangle.Now, measure all the angle by using protractor and complete the table.Fig. ∠ABC ∠BCD ∠CDA ∠DAB Result(i) 90° 90° 90° 90°(ii) 90° 90° 90° 90° ∠ABC = ∠BCD = ∠CDA = ∠DAB = 90°(iii) 90° 90° 90° 90°Hence, all angles of a rectangle are 90°.(iii) Opposite sides of rectangle are equal.Draw three parallelogram ABCD of different sizes by using set square.(i)ADBC(ii)ADBC(iii)ADBCSides AB, DC and AD, BC are opposite sides.Now measure the sides of each parallelogram and complete the table.Figure AB DC AD BC Result(i) 2.3 cm 2.3 cm 1.4 cm 1.4 cm(ii) 2.9 cm 2.9 cm 2 cm 2 cm AB = DC and AD = BC(iii) 2 cm 2 cm 2.5 cm 2.5 cm(c) Properties of Square(i) All angle of the square are equal.Draw three squares ABCD of different sizes by using set-square.
Acme Mathematics 7 243(i)ADBC(ii)ADBC(iii)ADBCIn the figures ∠ABC, ∠BCD, ∠CDA and ∠DAB are angles of different squares.Now, measure all the angle by using protractor and complete the table.Fig. ∠ABC ∠BCD ∠CDA ∠DAB Result(i) 90° 90° 90° 90°(ii) 90° 90° 90° 90° ∠ABC = ∠BCD = ∠CDA = ∠DAB = 90°(iii) 90° 90° 90° 90°Hence, all angles of a square are equal.(ii) All Diagonals of the square.Draw three squares ABCD of different sizes by using set-square.(i)ADBC(ii)ADBC(iii)ADBCIn the figures AB, BC, CD and DA are sides of squares ABCD.Now, measure the length of sides and complete the table. Fig. AB BC CD DA Result(i) 1.4 cm 1.4 cm 1.4 cm 1.4 cm(ii) 1.7 cm 1.7 cm 1.7 cm 1.7 cm AB = BC = CD = DA(iii) 2.1 cm 2.1 cm 2.1 cm 2.1 cmHence, all sides of the square are equal.
244 Acme Mathematics 7(iii) Diagonals of a square are equal.Draw three squares ABCD of different sizes by using set-square.(i)ADBC(ii)ADBC(iii)ADBCIn the figures AC and BD are diagonals.Now, measure the length of diagonals by using scale and complete the table.Fig. AC BD Result(i) 2 cm 2 cm(ii) 2.4 cm 2.4 cm Diagonals AC = Diagonals BD(iii) 3 cm 3 cmHence, diagonals of a square are equal.(iv) Diagonals of a square bisect each other at right angle.Draw three squares ABCD of different sizes by using set-square.(i)ADBCO O(ii)ADBCO(iii)ADBCIn the figures AC and BD are diagonals. O is the point of intersection of diagonals AC and BD.∠AOB, ∠BOC, ∠COD and ∠DOA are angles made by the diagonals.AO, OC, BO and OD are the parts of diagonals AC and BD.Now, measure the length by using scale and angles by protractor and complete the table.
Acme Mathematics 7 245Fig. AO OC BO OD ∠AOB ∠BOC ∠COD ∠DOA Result(i) 1 cm 1 cm 1 cm 1 cm 90° 90° 90° 90° AO = OC, BO = OD and ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°(ii) 1.2 cm 1.2 cm 1.2 cm 1.2 cm 90° 90° 90° 90°(iii) 1.5 cm 1.5 cm 1.5 cm 1.5 cm 90° 90° 90° 90°Hence, diagonals of a square bisect each other at right angle.(v) Diagonals of a square bisect the angle at the vertex.Draw the square of different size using set square.In the given squares AC and BD are diagonals. ∠ADB and ∠CDB are the parts of ∠ADC.(i)ADBCO O(ii)ADBCO(iii)ADBCSimilary, ∠DCA and ∠BCA are the parts of ∠DCB, ∠CBD and ∠ABD are the parts of ∠ABC. ∠DAC and ∠BAC are the parts of ∠BAD.Now, measure the angles by protractor and complete the table.Figure Vertex Angle Parts of the vertex angle Result(i)∠ADC = 90° ∠ADB = 45° ∠CDB = 45° ∠ADB = ∠CDB∠DCB = 90° ∠DCA = 45° ∠BCA = 45° ∠DCA = ∠BCA∠ABC = 90° ∠CBD = 45° ∠ABD = 45° ∠CBD = ∠ABD∠BAD = 90° ∠DAC = 45° ∠BAC = 45° ∠DAC = ∠BACHence, the diagonals of a square bisect the angle at the vertex.Note : Verify the result in the figures (ii) and (iii) yourself.D Exterior and Interior angles of triangle and quadrilateral(a) Interior angles: The angles x, y, z are the interior angles of the triangle. The angles 'a', 'b', 'c' and 'd' are the interior angles of the quadrilateral. x zydacb
246 Acme Mathematics 7(b) Exterior angles: The angles x, y and z are the exterior angles of the triangle. The angles 'a', 'b', 'c' and 'd' are the exterior angles of the quadrilateral.(c) Sum of interior and exterior angles of the triangle and quadrilateral We use (n – 2) × 180° to find the sum of interior angle of polygons, where ‘n’ is the number of sides and (n – 2) is the number of triangles. (i) Sum of interior angle of ‘n’ side polygon = (n – 2) × 180° (ii) One interior angle of ‘n’ side regular polygon = n – 2n ×180° (iii) Sum of Exterior angle of the polygon = 360° (iv) One exterior angle of the regular polygon = 360nFor example: zyxThe sum of angles x + y + z = 360° dcabThe sum of angles a + b + c + d = 360° Now, study the given table carefully. Polygons No. of Sides (n) No. of triangles Sum of all angles (n – 2) × 180° Triangle 3 1 180° Quadrilateral 4 2 360° zyxdcabAngles x, y and z are exterior angle
Acme Mathematics 7 247Solved ExampleExample 1 Find all the angles of the parallelogram ABCD. Solution: Here, ∠D = 6x, ∠A = 3x ABCD is a parallelogram. So, ∠B = ∠D = 6x [Opposite angles of the parallelogram.]∠C = ∠A = 3x [Opposite angles of the parallelogram.]Now, ∠A + ∠B + ∠C + ∠D = 360° [Angle sum of quadrilateral]or, 3x + 6x + 3x + 6x = 360° or, 18x = 360° or x = 20° ∠A = 3x = 3 × 20° = 60° = ∠C ∠D = 6x = 6 × 20° = 120° = ∠B Hence, angles are 60°, 120°, 60° and 120°. Example 2 If the angles of the quadrilateral are in the ratio 2: 3: 4: 6, find the measure of each angles. Solution: Let angles are 2x°, 3x°, 4x° and 6x° Now, 2x° + 3x° + 4x° + 6x° = 360° [interior angles of quadrilateral]or, 15x° = 360° or, x = 24° Hence, angles are: 2x = 2 × 24 = 48° 3x = 3 × 24 = 72° 4x = 4 × 24 = 96° 6x = 6 × 24 = 144° Example 3 Calculate the value of unknown angles, in the given figure. Solution: Here, 60° + z° + 40° = 180° [Being the sum of angles of the ΔABC]or, z° + 100° = 180° or, z = 80°or, z = a [Opposite angles of a parallelogram]or, 80° = a or, a = 80° ADBC3x6x4x°3x°6x°2x°BA60°x ay40° zCD
248 Acme Mathematics 7Again, ∠A = ∠Cor, 60° + x = 40° + yor, y – x = 20°or, y = x + 20° ............ (i)or, x + y + a = 180°or, x + y + 80° = 180° [ a = 80°]or, x + y = 100°or, x + x + 20° = 100° [ y = (x + 20°) from ...... (i)]or, 2x = 80°or, x = 40° and y = x + 20°or, y = 40° + 20° = 60°Hence, x = 40°, y = 60°, z = 80° and a = 80°.Classwork1. Fill in the blanks. (a) A quadrilateral has .............. diagonals. (b) In a trapezium, one pair of opposite sides are ............ .(c) In a rectangle, ........... sides are equal. (d) In a rhombus, ................ sides are equal. (e) If diagonals of a quadrilateral bisect each other at right angles then it is a ............ 2. State whether the following statements are true or false. (a) Opposite sides of a square are equal. (b) All the angles of a rhombus are right angle. (c) Opposite sides of a trapezium are equal. (d) Diagonals of a rectangles are equal. (e) If adjacent sides of a parallelogram are equal, then it is a rhombus. 3. Fill in the blanks: (a) A right angled-triangle has ………. corners. (b) Equilateral triangle has ………. equal. (c) The sum of interior angles of a quadrilateral is ………….. 4. Write any one properties of: (a) Square (b) Parallelogram.
Acme Mathematics 7 249Exercise 5.31. Draw the diagonals and name them in the following quadrilaterals.(a) (b) (c)(d) (e) (f)2. Measure the diagonals of given quadrilateral and write the result.(a) (b) (c)3. Name the opposite angles and measure them. Are they equals?(a) (b) (c)4. Name the opposite sides and measure them. Are they equals?(a) (b) (c)ADBCSPRQMNPOADBCPMONTQSRTQSRMNPOEFHGMPNOQRTSUVXWMNPOIJLKABDC
250 Acme Mathematics 75. Measure the marked angle. Draw a diagonal through marked angle. Measure the marked angles made by a diagonal. Write you conclusion.(a) (b) (c)6. Find the value of x and y in the following parallelograms.(a)yx15 cm10 cmA BD C(b) (c)y x + 22x – 1C 5 cm DF E(d)100°xy xEFDG(e)20°60°100°x2yE FH G(f)yO20°20°100° xF GI H(g)x+10°yy2x–10°G HJ I(h) (i)2x+20°80°3x–10°yTL KI J7. Solve the following problems: (a) If a°, 60°, 130° and 100° are the angles of the quadrilateral, find the value of 'a'. (b) If 4x, 3x, 2x and 50° are the angles of the quadrilateral, find the value of 'x'. 8. Calculate the size of each unknown angles.(a)3x°130°y°2x°50°(b) 2xa 4x4x80°(c)3x2x+12x x–5ADBCEFHGIJLK2y5x10 cm6 cmB CE Dy x89.5° x+2°H IK J