Algebra 201 Rule IV: Rules for dividing polynomials by polynomials Apply various steps as explained and illustrated in the following examples: A. Polynomial in one variable divided by a binomial 1. Divide the polynomial (4a2 – 4a + 1) by (2a − 1) Here, divisor, d = 2a − 1 and Dividend, D = 4a2 − 4a + 1 Step 1. Arrange the divisor and dividend in descending powers of a, leaving some space if there is some missing terms. Here, everything is in order. So, they are recopied as shown below: Divisor Dividend 2a – 1 4a2 − 4a + 1 Step 2. Divide the first term of the dividend 4a2 by the first term 2a of the divisor using Rule II, and write the quotient 2a above the line as shown below: 2a – 1 2a 4a2 − 4a + 1 Step 3. Multiply each term of the divisor (2a − 1) by the first term 2a of the quotient, and write the product below the dividend as shown below: 2a – 1 2a 4a2 − 4a + 1 4a2 – 2a Step 4. Subtract like terms and bring down one or more terms or all of the remaining terms. 2a 4a2 − 4a + 1 4a2 – 2a (–) (+) – 2a + 1 2a – 1 Step 5. Now, use the remainder − 2a + 1 as the new dividend and repeat step 2 to step 4. 2a 4a2 − 4a + 1 4a2 – 2a (–) (+) – 2a + 1 – 2a + 1 (+) (–) 0 (No Remainder R) 2a – 1 Quotient, Q Dividend, D
202 The Leading Mathematics - 7 Step 6. The remainder is zero, and we stop at this stage. Stop the process even if there is no term containing the variable or letter a or the exponent of a is less than that of the first term in the divisor. 2. Divide the polynomial (8a3 – 2) by (2a − 1) Here, divisor, d = 2a − 1, Dividend, D = 8a3 − 2 and two terms with the second and first powers of a are missing. So, we proceed as shown below: 4a2 + 2a + 1 8a3 + 0a2 + 0a – 2 4a3 – 4a2 (–) (+) + 4a2 + 0a – 2 + 4a2 – 2a (–) (+) + 2a – 2 + 2a – 1 (–) (+) –1 (Remainder, R) 2a – 1 8a3 – 2 = 8a3 + 0. a3 + 0.a – 2 We, thus, have (8a3 – 2) = (4a2 + 2a + 1) (2a − 1) + (− 1) Dividend = Quotient × Divisor + Remainder (D = Q × d + R) B. Polynomial in more than one variable divided by a binomial 1. Divide the polynomial (8a3 b − 8a2 b2 + 2ab3 ) by (2a − b) Here, divisor, d = 2a − b and Dividend, D = 8a3 b − 8a2 b2 + 2ab3 In such cases, we first have to arrange the divisor and dividend polynomials in descending order of the powers of a variable or literal factor and increasing powers of another variable or literal factor. In our case, it is already done. Thus, 4a2 b – 2ab2 8a3 b – 8a2 b2 + 2ab3 8a3 b – 4a2 b2 (–) (+) – 4a2 b2 + 2ab3 – 4a2 b2 + 2ab3 (+) (–) 0 (No Remainder) 2a – b Quotient Q Dividend D
Algebra 203 2. Divide the polynomial a3 + b3 + c3 − 3abc by a + b + c Arrange both the divisor and the dividend in descending powers of x accompanied by ascending powers of y and z. Then, a2 – ab – ac + b2 – bc + c2 a3 – 3abc2 + b3 + c3 a3 + a2 b + a2 c (–) (–) (–) – a2 b – a2 c – 3abc + b3 + c3 – a2 b – ab2 – abc (+) (+) (+) – a2 c + ab2 – 2abc + b3 + c3 – a2 c – abc – ac2 (+) (+) (+) + ab2 – abc + ac2 + b3 + c3 + ab2 + b3 + b2 c (–) (–) (–) – abc + ac2 – b2 c + c3 – abc – b2 c – bc2 (+) (+) (+) + ac2 + bc2 + c3 + ac2 + bc2 + c3 (–) (–) (–) 0 Divisor: a + b + c Hence, the quotient is a2 − ab − ac + b2 − bc + c2 = a2 + b2 + c2 − ab − bc − ca . Also, a3 − 3abc + b3 + c3 = (a2 − ab − ac + b2 − bc + c2 )(a + b + c) = (a + b + c)(a2 + b2 + c2 − ab − bc − ca) . EXERCISE 11.3 Your mastery depends on practice. Practice like you play. 1. Divide : (a) 6x2 y3 z7 ÷ 2xy2 z4 (b) 27x3 y–2z–1 ÷ 6x–2y4 z–2 (c) 36a4 b–2c–5 ÷ 12a3 b–1c–4 (d) 54p4 q–3r7 ÷ 12p2 q4 r–2
204 The Leading Mathematics - 7 2. Divide : (a) (a + b)3 by (a + b) (b) (2ax – by)5 by (2ax – by)2 (c) (3a – b + c)5 by (3a – b + c)4 (d) (4xy – 3z + 1)3 by (4xy – 3z + 1)3 3. Find the quotient of : (a) (3x2 y4 z5 + 6x3 y2 z3 ) ÷ 3x2 yz2 (b) (12a7 b–4c4 – 18a2 b2 c3 ) ÷ 6a2 bc–2 4. Divide : (a) (a3 – a2 b – ab + b2 ) by (a – b) (b) (x2 – xy – 2y2 ) by (x + y) (c) (6x2 – 7xy + 2y2 ) by (2x – y) (d) (4x2 – 6x – 6xy + 9y) by (2x – 3y) (e) (3x2 + x + 2xy + y – y2 ) by (x + y) (f) (2a2 – 6a + ab + 3b – b2 ) by (2a – b) (g) (2x2 – 3xy + 4x + 6y – 9y2 ) by (2x + 3y) (h) (12x2 + 12x – 11xy – 3y + 2y2 ) by (4x – y) 5. Divide : (a) (2x2 – x – 5xy + 4y – 12y2 ) by (2x + 3y – 1) (b) (x2 + 3x – 2xy + 6y – 8y2 ) by (x – 4y + 3) (c) (3x2 + 7x – 5xy + 4 – 6y + 2y2 ) by (3x – 2y + 4) (d) (2x2 + 2x + 5xy + 2zx + 3yz + 3y2 + y + 4z – 4) by (2x + 3y + 4) ANSWERS 1. (a) 3xyz3 (b) 9x5 z 2y6 (c) 3a bc (d) 9p2 r9 2q7 2. (a) (a + b)2 (b) (2ax – by)3 (c) (3a – b + c) (d) 1 3. (a) y3 z3 + 2xyz (b) 2a5 c6 b5 – 3bc5 4. (a) a2 – b (b) x – 2y (c) 3x – 2y (d) 2x – 3 (e) 3x – y + 1 (f) a + b – 3 (g) x – 3y + 2 (h) 3x – 2y + 3 5. (a) x – 4y (b) x + 2y (c) x – y + 1 (d) x + y + z – 1
Algebra 205 11.4 Establishment of (a ± b)2 At the end of this topic, the students will be able to: ¾ multiply the algebraic expressions. Learning Objectives What represents (a + b)2 ? Similarly, what represents (a − b)2 ? We know that a and b are both single units that means they represent the length of any things. Suppose a unit is the length of a line segment AB and b units, the length of the line segment CD. So, both a and b represent the lengths. A B a units C D b units When a and b are joined, what is represented by (a + b) units? (a + b) also represents the length of the line segment AD. Similarly, b is taken out from a, what is represented by (a − b) units? (a − b) also represents the length of the line segment AC. Now, do you know, what is represented by a2 and b2 ? a2 represents the area of the square with side a and b2 also represents the area of another square with side b. Similarly, (a + b) 2 and (a − b)2 represents the area of squares with sides (a + b) units and (a − b) units respectively. Now, we establish the formulae of (a ± b)2 below. Category A 1. (a + b)(a + b) = a (a + b) + b(a + b) = a2 + ab + ba + b2 . (ab = ba, why?) So, (a + b)2 = (a + b)(a + b) = a2 + 2ab + b2 . The geometrical significance of this formula is shown in the given diagram: ABCD = (a + b)(a + b) = (a + b)2 , AEFI = a × a = a2 , A (a + b) units D BC A(a − b) units C B D A B a units a2 sq. units b2 sq. units C D b units I J A a2 b2 ab ab a a a a b b b b F E D B G C a + b a + b
206 The Leading Mathematics - 7 EDJF = ab, IFGB = ab and FJCG = b × b = b2 . Also, Area ABCD = Area AEFI + Area EDJF + Area IFGB + Area FJCG So, (a + b)2 = a2 + ab + ab + b2 = a2 + 2ab + b2 , Which is expanded form of (a + b)2 . i.e., (a + b)2 = a2 + 2ab + b2 . Furthermore, a2 + b2 = (a + b)2 − 2ab 2. (a − b)(a − b) = a(a − b) − b(a − b) = a2 − ab − ba + b2 , a > b So, (a − b)2 = (a − b)(a − b) = a2 − 2ab + b2 . The geometrical significance of this formula is shown in the following diagram: ABCD = a × a = a2 , AEFI = (a −b) × (a − b) = (a − b)2 , EDJF = (a – b)b , IFGB = b(a – b) and FJCG = b × b = b2 . Also, AEFI = ABCD − EDJF − IFGB – FJCG (a − b)2 = a2 − (a − b)b − b(a − b) – b2 = a2 – ab + b2 – ab + b2 – b2 = a2 − 2ab + b2 Which is expanded form of (a − b)2 . i.e., (a − b)2 = a2 − 2ab + b2 . Furthermore, a2 + b2 = (a − b)2 + 2ab Also, (a + b)2 = a2 + 2ab + b2 − 2ab + 2ab = a2 − 2ab + b2 + 4ab = (a − b)2 + 4ab ∴ (a + b)2 = a2 + 2ab + b2 = (a − b)2 + 4ab. Again, (a − b)2 = a2 − 2ab + b2 − 2ab + 2ab = a2 + 2ab + b2 − 4ab = (a + b)2 − 4ab i.e., (a − b)2 = a2 − 2ab + b2 = (a + b)2 − 4ab. I J A b2b(a – b) a – b a – b a – b (a – b)2 a – b (a – b)b b b b b F E D B G C a a
Algebra 207 CLASSWORK EXAMPLES Example 1 Write the expanded form of the square of (x + 3) by the following ways: (a) multiplication (b) using formula (c) geometric figure Solution: (a) By multiplication; Square of (x + 3) = (x + 3)2 = (x + 3) (x + 3) = x (x + 3) + 3(x + 3) = x2 + 3x + 3x + 9 = x2 + 6x + 9. (b) By using formula; Square of (x + 3) = (x + 3)2 = x2 + 2 × x × 3 + 32 [ (a + b)2 = a2 + 2ab + b2 ] = x2 + 6x + 9. (c) By geometric figure; Square of (x + 3) = (x + 3)2 = x2 + x + x + x + x + x + x + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = x2 + 6x + 9. Example 2 Find the square of the following numbers by using the formulae of (a + b)2 and (a − b)2 : (a) 13 (b) 98 Solution: (a) Square of 13 = 132 = (10 + 3)2 = 102 + 2 × 10 × 3 + 32 [ (a + b)2 = a2 + 2ab + b2 ] = 100 + 60 + 9 = 169. x x 3 3 x + 3 x + 3 x x x x x2 x x x x 1 1 111 1 1 1 1 1 1 1 1 1 1
208 The Leading Mathematics - 7 (b) Square of 98 = 982 = (100 − 2)2 = (100)2 − 2 × 100 × 2 + 22 [ (a − b)2 = a2 − 2ab + b2 ] = 10000 − 400 + 4 = 9600 + 4 = 9604. Example 3 Find the square of the following algebraic expressions: (a) 2x + 3 2 (b) 3x − 5 2x (c) (x − y + z) Solution: (a) Square of 2x + 3 2 = 2x + 3 2 2 = (2x)2 + 2 × 2x × 3 2 + 3 2 2 = 4x2 + 6x + 9 4 . (b) Square of 3x − 5 2x = 3x − 5 2x 2 = (3x)2 − 2 × 3x × 5 2x + 5 2x 2 = 9x2 − 15 + 25 4x2 (c) Square of (x − y + z) = (x − y + z)2 = {(x − y) + z}2 = (x − y)2 + 2(x − y)z + z2 = x2 − 2xy + y2 + 2zx − 2yz + z2 = x2 + y2 + z2 − 2xy − 2yz + 2zx. Example 4 Simplify: (2x + y)2 – (2x – y)2 Solution: Here, (2x + y)2 – (2x – y)2 = (2x)2 + 2 × 2x × y + y2 – {(2x)2 – 2 × 2x × y + y2 } = 4x2 + 4xy + y2 – 4x2 + 4xy – y2 = 8xy. Example 5 If a + 1 a = 5, find the values of ; (a) a2 + 1 a2 (b) a − 1 a 2 Solution: Here, a + 1 a = 5
Algebra 209 (a) Now, a2 + 1 a2 = a + 1 a 2 – 2a . 1 a [ a2 + b2 = (a + b)2 – 2ab] = 52 – 2 = 25 – 2 = 23. "Alternatively " Here, a + 1 a = 5 or, a + 1 a 2 = 52 [ Squaring on both sides] or, a2 + 2a . 1 a + 1 a2 = 25 [ (a + b)2 = a2 + 2ab + b2 ] or, a2 + 2 + 1 a2 = 25 or, a2 + 1 a2 = 25 – 2 or, a2 + 1 a2 = 23. (b) a – 1 a 2 = a + 1 a 2 – 4a . 1 a [ (a – b)2 = (a + b)2 – 4ab ] = 52 – 4 = 25 – 4 = 21. "Alternatively " Here, a + 1 a = 5 or, a + 1 a 2 = 52 [ Squaring on both sides] or, a – 1 a 2 + 4a . 1 a = 25 [ (a + b)2 = a2 + 2ab + b2 ] or, a – 1 a 2 + 4 = 25 or, a – 1 a 2 = 25 – 4 or, a – 1 a 2 = 21.
210 The Leading Mathematics - 7 Example 6 If x + y = 7 and xy = 12, find the values of ; (a) x2 + y2 (b) x − y Solution: Here, x + y = 7 and xy = 12 (a) Now, x2 + y2 = (x + y)2 – 2xy [ a2 + b2 = (a + b)2 – 2ab ] = 72 – 2 × 12 = 49 – 24 = 25. (b) Now, (x – y)2 = (x + y)2 – 4xy [ (a – b)2 = (a + b)2 – 4ab ] = 72 – 4 × 12 = 49 – 48 = 12 . ∴ x – y = 1. EXERCISE 11.4 Your mastery depends on practice. Practice like you play. 1. Write the expanded form of the square of the following binomial by; (i) multiplication, (ii) using formulae and (iii) geometric figure. (a) x + 2 (b) a + 3 (c) x – 1 (d) 2a – 3 2. Find the square of the following numbers by using the formulae of (a + b)2 and (a − b)2 : (a) 23 (b) 47 (c) 62 (d) 99 3. Find the square of the following algebraic expressions: (a) 3x + 2 (b) 2x − 3 (c) 2a2 + 3 (d) 3p3 − 4 (e) 3x + 2 3 (f) 2x + 5 4 (g) x − 5 x (h) 6x − 2 3x (i) p + q + r (j) a − b + c (k) 2x + y - z (l) 3p − 2q − r
Algebra 211 4. Simplify: (a) (2a – 5b)2 – (5a – 2b) 2 (b) (3m – 2n)2 + (2m – 3n)2 (c) 2x + 3 2 2 − 3x − 4 3 2 (d) x – 1 2 2 − 2x − 3 2x 2 5. Express the following expressions into square: (a) x2 + 4xy + 4y2 (b) 9a2 b2 − 30abc + 25c2 (c) 9x2 − 15 + 25 4x2 (d) 4x2 + 6x + 9 4 (c) a2 + b2 + c2 − 2ab − 2bc + 2ca (d) 4x2 + y2 + z2 − 4xy − 2yz 4 4zx 6. (a) If a + 1 a = 7, find the values of; (i) a2 + 1 a2 (ii) (a − 1 a )2 (b) If x − 1 x = 9, find the values of; (i) x2 + 1 x2 (ii) (x + 1 x )2 7. (a) If x + y = 5 and xy = 6, find the values of; (i) x2 + y2 (ii) x − y (b) If a − b = 10 and ab = − 21, find the values of; (i) a2 + b2 (ii) a + b 8. If x − 1 x = 15, prove that; (a) x2 + 1 x2 = 227 (b) (x + 1 x )2 = 229 ANSWERS 1. (a) x2 + 2x + 4 (b) 9x5 z 2y6 (c) 3a bc (d) 9p2 r 9 2q7 2. (a) (a + b)2 (b) (2ax – by)3 (c) (3a – b + c) (d) 1 3. (a) y3 z3 + 2xyz (b) 2a5 c6 b5 – 3bc5 4. (a) a2 – b (b) x – 2y (c) 3x – 2y (d) 2x – 3 (e) 3x – y + 1 (f) a + b – 3 (g) x – 3y + 2 (h) 3x – 2y + 3 7. (a) x – 4y (b) x + 2y (c) x – y + 1 (d) x + y + z – 1
212 The Leading Mathematics - 7 CHAPTER 12 Equation, Inequality and Graph Lesson Topics Pages 12.1 Linear Equation of Two Variables 213 12.2 Inequality and Graph 218 What is the value of measuring cylinder in the beam balance? How much milk do you drink in a day? How much water do you drink in a day? How many members are in your family? What is the sign of equal? If x is equal to 3, write it in a mathematical notation. If the sum of two numbers is equal to 5, write it in a mathematical terms. What is the sign of unequal? x is never equal to 6. Write this statement in mathematical form. When two quantities are not equal, what possible cases can arise? a is greater than 3. Write this statement in symbolic form. p is less than 3. Write this statement in algebraic form. x is less than or equal to 10. Write this statement in algebraic form. x is greater than or equal to 10. Write the statement in algebraic form. WARM-UP x x 4 30
Algebra 213 12.1 Linear Equation of Two Variables At the end of this topic, the students will be able to: ¾ solve the problems on the linear function of two variables and draw their graphs. Learning Objectives Review on Linear Equation of One Variable Consider two algebraic expressions 2x – 1 ......................(i) x + 2 .......................(ii) For what value of x the expressions (i) and (ii) have the same value? If x = 3, the value of both expressions have the same value 5. Therefore, 2x – 1 = x + 2 for x = 3. This equality of two algebraic expressions is called equation. A mathematical statement of equality of two algebraic expressions for some values of the variables is called an equation. The value of the variable which satisfies the given equation, is called the solution or root of the algebraic equation. Linear Equation of Two Variables The alongside figure is the grinder. We put wheat as input (x) in the grinder, we get flour as output (y) by grinding. Let's consider, an algebraic machine in which every number becomes adding by 2. This adding process is the rule of machine. If input = 1, output = 1 + 2 = 3 If input = 3, output = 3 + 2 = 5 If input = 5, output = 5 + 2 = 7 If input = x, output = x + 2 Suppose the output is y then we get a relation y = x + 2, whose x and y are both variable. But the value of y Wheat Flour Grinder +2 1 x 3 x + 2 7 5 3 5
214 The Leading Mathematics - 7 depends on the value of x. This algebraic relation or equation y = x + 2 is called a function. This function y = x + 2 can be expressed by the following methods : 1 3 5 x 3 5 7 x+2 Mapping or Arrow diagram. x 1 3 5 x y 3 5 7 x + 2 Tabulation Method Graphical Method O X Y (1, 3) (3, 5) (5, 7) CLASSWORK EXAMPLES Example 1 Compute the values of the given function machine. Solution: In the given function machine, Input = {2, – 3, 4} Function = Three times ∴ Output = Three times of input = {2 × 3, – 3 × 3, 4 × 3} = {6, – 9, 12} Hence, the required values of output in the given function are 6, – 9 and 12 for the values of input as 2, – 3 and 4 respectively. Example 2 Complete the given table for the function y = 2x – 3. x 2 4 7 – 2 0 y 5 – 1 – 7 Solution: Here, the given function is, y = 2x – 3 or, y + 3 = 2x or, x = y + 3 2 If x = 2, then y = 2 × 2 – 3 = 4 – 3 = 1 Three times 4 –3 2
Algebra 215 If x = 4, then y = 2 × 4 – 3 = 8 – 3 = 5 If x = 7, then y = 2 × 7 – 3 = 14 – 3 = 11 If x = –2, then y = 2 × (– 2) – 3 = – 4 – 3 = – 7 If x = 0, then y = 2 × 0 – 3 = 0 – 3 = – 3 If y = 5, then x = 5 + 3 2 = 8 2 = 4 If y = – 1, then x = – 1 + 3 2 = 2 2 = 1 If y = – 7, then x = – 7 + 3 2 = – 4 2 = – 2 Now, filling the missing values in the given table, x 2 4 7 – 2 0 4 1 – 2 y 1 5 11 – 7 – 3 5 – 1 – 7 Example 3 Draw the graph from the function 2x + y = 3 by drawing table. Solution: Here, the given function is, 2x + y = 3 or, y = 3 – 2x If x = 1, y = 3 – 2 × 1 = 3 – 2 = 1 If x = 2, y = 3 – 2 × 2 = 3 – 4 = – 1 If x = 3, y = 3 – 2 × 3 = 3 – 6 = – 3 If x = 4, y = 3 – 2 × 4 = 3 – 8 = – 5 x 1 2 3 4 y 1 – 1 – 3 – 5 Making Table Drawing Graph O X Y (4, – 5) (3, – 3) (2, – 1) (1, 1)
216 The Leading Mathematics - 7 EXERCISE 12.1 Your mastery depends on practice. Practice like you play. 1. If the inputs for the following function machines are – 2, 0, 3, 7 and 9, find the values of output. (a) (b) (c) (d) (e) (f) (g) (h) (i) 2. (a) If input (x) = {3, 5, –2, – 4}, find the values of output (y) for the adjoining function machine. (b) If input (x) = {– 2, – 1, 0, 3, 5, 6}, find the value of output (y) from the given function machine. 3. Complete the following table for the given functions: (a) y = 2x – 1 – 1 0 2 3 5 7 9 + 2 + 5 – 2 × 4 × 3 + 1 × 2 – 3 ÷ 3 + 2 ÷ 4 – 1 ÷ 3 + 3 3x – 1 5x – 4
Algebra 217 (b) y = 3x + 2 – 3 – 1 1 2 3 4 (c) y = 2x – 3 3 2 0 1 5 7 (d) 2x + 3y = 8 4 1 – 2 0 – 2 – 4 4. Draw the graph for the following functions: (a) y = x (b) y = 2x (c) y = x + 2 (d) y = x – 3 (e) y = 2x – 1 (f) y = 3x + 4 (g) y = x 2 + 1 (h) y = 4x – 3 2 (i) x + y = 4 (j) 2x + y = 5 (k) 2x – y = 3 (l) x + 2y = 7 (m) 2x + 3y = 3 (n) 3x – 4y = 1 (o) 3x + 5y = 8 (p) 3x – 2y = 4 (q) 3x + 4y = 12 (r) 5x – 2y = 10 ANSWERS 2. (a) {8, 14, – 7, – 14} (b) {– 14, – 9, – 4, 11, 21, 26} 3. (a) – 3, – 1, 3, 5, 9, 13, 17 (b) – 7, – 1, 5, 8, 11, 14 (c) 3, 1, – 3, 2, 4, 5 (d) 0, 2, 4, 4, 7, 10
218 The Leading Mathematics - 7 12.2 Inequality and Graph At the end of this topic, the students will be able to: ¾ draw the linear inequality in the number line. Learning Objectives Consider an equation of one variable 2x + 3 = 7 .............(i) For what value of x the left part of the equation (i) will not equal to 7? If x = 2, the value of 2x + 3 is 7. So, the left part of the equation (i) will not equal to 7 for x ≠ 2. Therefore, the value of x will be either greater than 2 or less than 2 i.e. x > 2 or x < 2. The negation of x > 2 and x < 2 are x 2 (x is not greater than 2) and x 2 (x is not less than 2) respectively. x 2 means either x = 2 or x < 2 i.e. x ≤ 2 (x is equal or less than 2) and x 2 means either x = 2 or x > 2 i.e. x ≥ 2 (x is equal or greater than 2). If x < 2, x ∈ Z, what are the values of x ? ∴ x = 1, 0, – 1 , – 2, .......... If x > 2, x ∈ Z, what are the values of x ? ∴ x = 3, 4, 5, 6, 7, .......... If x ≤ 2, x ∈ Z, x = 2, 1, 0, – 1, .......... If x ≥ 2, x ∈ Z, x = 2, 3, 4, 5, 6, .......... Number Line A straight line in which the numbers specially integers are shown in equal interval on right and left from the zero point is called a number line. In the number line, the right part from the zero indicates positive numbers and that of left indicates negative numbers. Graphical Representation of Inequalities (a) Inequality excluding equality sign : (i) x > 2 means x represents the numbers that are just greater than 2, but not equal to 2. So, we represent x > 2 in the number line as follows : – 4 – 3 – 2 – 1 0 1 2 3 4 – 4 – 3 – 2 – 1 0 1 2 3 4 OR – 3 – 2 – 1 0 1 2 3
Algebra 219 (ii) x < 2 means x represents the numbers that are just less than 2, but not equal to 2. So, we represent x < 2 in the number line as follows : – 4 – 3 – 2 – 1 0 1 2 3 4 – 4 – 3 – 2 – 1 0 1 2 3 4 OR (b) Inequality including equality sign : (i) x ≥ 2 means x represents the numbers that are equal or greater than 2, but not equal to 2. So, we represent x ≥ 2 in the number line as follows : – 4 – 3 – 2 – 1 0 1 2 3 4 – 4 – 3 – 2 – 1 0 1 2 3 4 OR (ii) x ≤ 2 means x represents the numbers that are equal or less than 2, but not equal to 2. So, we represent x ≤ 2 in the number line as follows : – 4 – 3 – 2 – 1 0 1 2 3 4 – 4 – 3 – 2 – 1 0 1 2 3 4 OR (c) Inequality including two inequality sign : The inequalities –2 < x < 3, 2 > x > –1, 0 ≤ x < 3, – 1 < x ≤ 2, – 1 ≤ x ≤ 5 and 3 ≥ x ≥ –1 are presented in the number lines as follows : – 4 – 3 – 2 – 1 0 1 2 3 4 – 4 – 3 – 2 – 1 0 1 2 3 4 – 2 < x < 3 2 > x > – 1 or – 1 < x < 2 – 4 – 3 – 2 – 1 0 1 2 3 4 – 4 – 3 – 2 – 1 0 1 2 3 4 0 ≤ x < 3 – 1 < x ≤ 2 – 4 – 3 – 2 – 1 0 1 2 3 4 – 4 – 3 – 2 – 1 0 1 2 3 4 1≤ x ≤ 4 3 ≥ x ≥ – 1 CLASSWORK EXAMPLES Example 1 Find the inequalities for the following number lines : (a) (b) Solution: (a) From the given number line, the required inequality is x > –2. (b) From the given number line, the required inequality is –2 ≤ x < 2. – 3 – 2 – 1 0 1 2 3 – 3 – 2 – 1 0 1 2 3
220 The Leading Mathematics - 7 EXERCISE 12.2 Your mastery depends on practice. Practice like you play. 1. Represent the following inequalities in the number lines : (a) x < 4 (b) x < – 2 (c) x > 0 (d) x ≤ 3 (e) x ≤ – 2 (f) x ≥ 1 (g) x ≥ – 3 (h) x > – 2 (i) x ≠ 2 (j) x ≠ – 1 (k) x 2 (l) x – 2 (m) 1 < x < 5 (n) – 2 < x ≤ 4 (o) – 1 ≤ x < 4 (p) – 2 ≤ x ≤ 2 (q) 2 > x > – 3 (r) 4 > x ≥ – 1 (s) 5 ≥ x > 0 (t) 6 ≥ x ≥ – 1 (u) 5x < 3x – 1 ≤ 8 2. Find the inequalities from the following number lines : (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) ANSWERS 2. (a) x < 1 (b) x < – 1 (c) x > – 1 (d) x > 1 (e) x < 0 (f) x < – 1 (g) x < 0 (h) x > – 2 (i) 3 > x > – 2 (j) – 2x < x < 3 (k) 1 > x > – 3 (l) – 2 < x < 3 (m) – 3 < x < 2 (n) – 3 < x < 0 (o) x ≠ 0 (p) x ≠ – 1 – 3 – 2 – 1 0 1 2 3 – 3 – 2 – 1 0 1 2 3 – 3 – 2 – 1 0 1 2 3 – 3 – 2 – 1 0 1 2 3 – 3 – 2 – 1 0 1 2 3 – 3 – 2 – 1 0 1 2 3 – 3 – 2 – 1 0 1 2 3 – 3 – 2 – 1 0 1 2 3 – 3 – 2 – 1 0 1 2 3 – 3 – 2 – 1 0 1 2 3 – 3 – 2 – 1 0 1 2 3 – 3 – 2 – 1 0 1 2 3 – 3 – 2 – 1 0 1 2 3 – 3 – 2 – 1 0 1 2 3 – 3 – 2 – 1 0 1 2 3 – 3 – 2 – 1 0 1 2 3
Algebra 221 Read, Understand, Think and Do 1. Two numbers are given below 64 and 128 (a) Factorize 64 and 128. (b) Can we write the factors of 64 as the form of indices in terms of 2? Write it if yes. (c) Justify that: 128 ÷ 64 = 27 ÷ 26 = 27 – 6 = 2 2. Two roots are given below. 216 3 and 27 3 (a) Multiply: 216 3 × 27 3 (b) Divide: 216 3and 27 3 (c) Can we add 216 3and 27 3 ? Give reason. 3. Two terms are given below. (2x + 3y)2m and (2x + 3y)2n (a) Write the value of x°. (b) If m + n = 0, prove that: (2x + 3y)2m × (2x + 3y)2n = 1 4. If x + y + z = 0, prove that: (a) a (x + y + z) = 1 (b) ax – y × ax + y × ay – z × ay + z × az – x × ax + z = 1 5. In a square, the length of a side is (3x + 2) cm. [Area of square = l × l and perimeter = 4l] (a) Find the area of square in terms of x. (b) If the value of x = 2, prove that its perimeter is 32 cm. MIXED PRACTICE–IV
222 The Leading Mathematics - 7 6. The area of rectangular surface is (9x2 – 4y2 )cm2 and its length is (3x + 2y) cm with x > y. [Area of rectangle = l × b] (a) Find the breadth of rectangle in terms of x. (b) If the value of x = 2, how much length is greater than its breadth? 7. In a rectangle, the length of a side is (3x + 2) cm and its area is (6x2 + 13x + 6)cm2 . [Where, x > 0; x ∈ N and Area of rectangle = l × b] (a) Find the breadth of the rectangle in terms of x. (b) Expand: (3x + 2y)2 8. A number x and its reciprocal is 1 x . (a) If sum of x and 1 x is 3, prove that: x2 + 1 x2 = 7 (b) If x – 1 x = 3, prove that: x2 + 1 x2 = 11 9. The given algebraic tiles geometrically represent the statement (x + 3)2 . (a) Find the length and breadth of square box in terms of x. (b) Copy this square box and fill by x and 1 in the box. (c) Prove that: (x + 3)2 = x2 + 6x + 9 10. Two algebraic expressions are given below. (2x2 + 7x – 4) and (x – 3) (a) Divide (2x2 + 7x – 4) by (x – 3) (b) For what value replaced in 4 of the above expression, the remainder is equal to 0? 11. The sum of two natural numbers is 7 and their product is 12. (a) Make two equations supposing two numbers x and y. (b) Find the sum of square of the two numbers. (c) Prove that difference of two number is 1. 1 4
Algebra 223 ANSWERS 1. (b) Yes, 26 2. (a) 18 (b) 2 (c) Yes 3. (a) 1 5. (a) 9x2 + 12x + 4 (b) 32 cm 6. (a) 3x – 2y (b) 4 times of y 7. (a) (3x + 2) cm (b) 9x2 + 12x + 4y2 9. (a) x + 3 10. (a) quotient 2x + 13, reminder 35 (b) 39 11. (a) 3, 4 (b) 25 12. (a) a2 + 2ab + b2 13. (a) 2x + 4 < 10 (c) – 1, 0,1, 2 14. (a) x, x + 2 (b) 13 15. (a) 3x, x (b) 45 years (c) 30 years 16. (a) divided by 3 12. (a) Expand (a + (b)2 . (b) Show the geometrical representation of (a + (b)2 . 13. A number 2x + 5 is less than 10. (a) Write it in the form of inequality. (b) Show this inequality in number line. (c) Write any 4 possible values of x. 14. A sum of two consecutive odd number is 24. (a) Write it in mathematical form supposing a number x. (b) Find the greatest number. 15. The current age of father is three times the age of his son and sum of their age is 60. (a) Write it in mathematical form by supposing the age of son x. (b) Find the age of father. (c) After how many years will the age of the son be the same as the present age of the father? 16. A function is given below. 6x + 3y = 12 (a) What should you do in the given equation to make y as single or with numerical coefficient 1? (b) Draw the graph.
224 The Leading Mathematics - 7 1. Two cubic roots are given below. 32 5 and 243 5 (a) Write a × a × a × a × a × a in the form of indices. [1] (b) Multiply the given two cubic roots. [1] (c) Prove that: xa xb a + b xb xc b + c xc xa c + a = 1 [2] 2. Observe the given rectangle. (a) Find the perimeter of the given rectangle in terms of x. [2] (b) Find the area of the given rectangle in terms of x. [2] (c) Write the degree of the perimeter and area of the rectangle. [1] 3. (a) Find the square of (2x – 3y). [2] (b) If x − 1 x = 9, find the values of x2 + 1 x2 . [2] (c) If x + y = 5 and xy = 6, find the values of x2 + y2 . [2] 4. (a) Find the values of the function y = 3x + 1 for the input x = {2, 4, – 2, 0}. [2] (b) Write the inequality for the given number line. [1] (c) Represent the inequality 5 > x > – 2 in the number line. [2] 2x – 4 x2 – 3 – 2 – 1 0 1 2 3 Attempt all the questions. FM : 20 Time : 40 Min. CONFIDENCE LEVEL TEST IV ALGEBRA
Geometry GEOMETRY 225 COMPETENCY Construction the geometric plane and solid shapes and test their properties and facts. Solution of the problems with use of the concept of coordinates, Pythagoras theorem and transformation. CHAPTERS 13. Lines and Angles 14. Plane Shapes 15. Congruency 16. Solid Objects 17. Coordinates 18. Symmetry and Tessellation 19. Transformation 20. Bearing and Scale Drawing LEARNING OUTCOMES construct the angles 15°, 75°, 105°, 135°, 150° by using a compass. construct the angle equal to the given angle. introduce the pair of angles. test geometric facts on pair of angles. construct triangle under the given conditions. search and test the properties of the given quadrilateral (parallelogram, rectangle and square). establish the relation among the base, perpendicular and hypotenuse of a right-angled triangle according to Pythagoras theorem. identify the congruent shapes. prepare the hollow models of the tetrahedron, octahedron, cone and cylinder. establish the relation among edge, surface and corner of the tetrahedron, cube, octahedron and dodecahedron. find the coordinates of the point in the given graph and plot the points on the graph. find the distance between two points on the lines parallel to the axes. separate point and line symmetry of the plane figures. search the tessellations made by triangles. prepare the tessellation from the triangles. introduce transformation. reflect the point and line segment on x-axis and y-axis on the graph. translate the point and line segment on the vertical and slanting line. introduce bearing and scale drawing. read the direction in the map. GEOMETRY UNIT V
226 The Leading Mathematics - 7 CHAPTER 13 Lines and Angles Lesson Topics Pages 13.1 Construction of Standard Angles 227 13.2 Construction of Equal Angles 233 13.3 Introduction to Pair of Angles 235 What are line and line segment ? Can you measure the line and line segment ? How are the edges of the room, board, book and ruler ? How are the corners of the board, book and classroom ? Can you find the measure of the angles ? Can you draw the given angles by using protractor ? How can you identify the greater and smaller angles ? Why do we use set-squares ? Why do we use compass ? WARM-UP
Geometry GEOMETRY 227 13.1 Construction of Standard Angles At the end of this topic, the student will be able to: ¾ construct the standard angles 15°, 75°, 105°, 135°, 150°, etc. Learning Objectives Angle An angle is said to be formed when two rays have a common endpoint. The common endpoint is called the vertex of the angle and the two rays its arms. The symbol '∠' is used to represent an angle. In the adjoining figure, OA and OB represent the arms and O the vertex. The angle itself is denoted by ∠AOB or ∠BOA. For simplicity, it is also denoted by ∠O or ∠x. The angle ∠x can be constructed as the turning of line from the position OA to the position OB or vice versa. In this case, we say that an angle is nothing other than an amount of turn. An angle is measured in terms of some standard of measurement. There are different standards. We follow one particular standard based on the unit called degree. Bisection of Angle by Using a Compass Draw an angle AOB of 60° in your copy by using a protractor. With the centre at O, draw an arc of a suitable radius cutting the arms OA and OB at C and D respectively as shown in the adjoining figure. With the centre at C, draw an arc of the same radius in interior region of ∠AOB and with the centre at D, draw another arc which intersects the previous arc at a point, say P. Join OP and produce, if necessary. Now, measure the angles AOP and BOP. Are they both of 30°? Do the same process to bisect angle of any size. Review on Construction of Angles Construction of Straight Angle 180o Let's recall how an angle of 180° is drawn at a given point on a line. Vertex Arm Arm O B A x O C P A B D 30° 30° 60°
228 The Leading Mathematics - 7 Steps 1) Draw a line. 2) Take a point O on the line. 3) Fix the needle of a compass at O. Draw an arc of any size cutting the given line on both sides of O at A and B respectively. Hence, we obtain the straight angle ∠AOB = 180°. Without changing the opening size of compass, draw arcs from A to B and say the intersecting points as C and D as shown in the adjoining figure. What are the measures of ∠AOC, ∠COD and ∠DOB? Construction of angle 60o and 120o Steps (i) Draw a ray OX. (ii) With the centre at O, draw an arc with suitable radius cutting OX at P. (iii) With the centre at P and radius the same as in the step (ii) draw an arc cutting the arc in step (ii) at Q. (iv) Join OQ and produce it to Y. Then ∠XOY is the required 60° angle. (v) Again, draw an arc with the centre Q cutting the arc in the step (ii) at R in the same direction. (vi) Join OR and produce it to Z. Then ∠XOZ is the required 120°. Construction of Angles 30° and 150° Steps (i) Draw a ray OX. (ii) With the centre at O and suitable radius draw an arc cutting OX at S (or at S' cutting OX' drawn opposite to OX). O 180° B O A D C O P X R Q 120° Z O P X Y Q 60° Check you construction using a protractor.
Geometry GEOMETRY 229 (iii) With the centre at S (or S') and radius the same as in step (ii) draw an extended arc cutting the arc in step (ii) at T (or T'). (iv) Bisect the angle T or T' of 60° (v) Join OP and produce it to Y (or Y'). Then XOY and XOY' are the required 30° and 150° angles. X' S' Y' T' P' O 30° 150° T P S X Y Check your construction using a protractor. Oh! 30° is the half of 60°. i.e., the average of 0° and 60° and 150° is the average of 120° and 180°(or 180° – 30°). Construction of Angle 90° and 15° Steps (i) Draw a ray OX. (ii) With the centre at O, draw an arc cutting OX at P. (iii) With the centre at P and radius the same as in the step (ii) draw an arc cutting the arc in the step (ii) at Q and with the centre at Q draw an arc cutting the same arc at R. (iv) Again, with the centers Q and R draw the arcs intersected at S between Q and R up or down the arc in step (ii) (v) Join OS and produce it to Y. Then ∠XOY is the required 90° angle. 90° Y S R Q O XP Oh! 90° is the average of 60° and 120°.
230 The Leading Mathematics - 7 Construction of Standard Angles 45° and 135° Steps (i) Draw a ray OX. (ii) With the centre at O and suitable radius, draw a semi-circular arc cutting OX at S for 45° (or OX' drawn opposite to OX at S' for 135°). (iii) With the centre at S (or S' for 135°) and radius the same as in the step (ii) draw an arc cutting the arc in step (ii) at T. (iv) With the centre at T and radius the same as in the step (iii) draw an arc cutting the arc in step (iii) at Q. (v) Join OQ and produce it to Y. Then angle XOY is 90°. (vi) With the centre at S (or S' ) draw an arc on the side of OX for 45° (on the side of OX' for 135°). (vii) With the centre at Q and radius the same as in Step (vi), draw an arc on cutting the arc in the step (vi) at P on the side of OX for 45° (on the side of OX' for 135°). (viii) Join OP and produce it to Z (or OP' and produce it to Z'). Then, XOZ (or X'OZ') is the required angle 45° (or 135°). 45° 135° S' O S X X' P' P T' T Y Q Z' Z If ∠SOZ = 45°, then ∠X'OZ = 180° – 45°. ∴ ∠X'OZ = 135°. Alternative Method 1. Draw an angle of 90°. 2. Bisect it. You get an angle of 45°. 3. Draw 180° and bisect it. You get two anglse of 90°. 4. Bisect these two right angles, you get an angle of (90° + 45°) = 135°.
Geometry GEOMETRY 231 Construction of Standard Angles 75° and 105° Steps (i) Construct 60°, 120° and 90° as the process mentioned above on the same figure. i.e., ∠AOC = 60°, ∠AOD = 120° and ∠AOE = 90° as shown in below: (ii) Bisect ∠COE and obtain an angle 75°. i.e., ∠AOF = 75°. (iii) Similarly, bisect ∠DOE and obtain an angle 105°. i.e., ∠AOG = 105°. Oh! 75° is the average of 60° and 90°. Similarly, 105° is the average of 90° and 120°. B O A D G E F C 75° 105° Note: (i) Can we construct 15° and 165° ? Discuss. (ii) Can we construct an angle of 7 1 2 °? Discuss. EXERCISE 13.1 Your mastery depends on practice. Practice like you play. 1. Construct the following angles using the compass. (a) 180° (b) 60° (c) 120° (d) 90° (e) 30° (f) 150° (g) 45° (h) 135° (i) 75° (j) 105° (k) 15° (l) 165° 2. Draw the following angles using compass and bisect them. Also, measure the bisecting angles: (a) 45° (b) 15° (c) 60° (d) 75° (e) 120° (f) 150° (g) 135° (h) 105°
232 The Leading Mathematics - 7 3. Draw the following angles at the given point of the given line: (a) 45° at A (b) 75° at P (c) 135° at M (d) 15° at R (e) 105° at C (f) 150° at H (g) 90° at A (h) 60° at B (i) 22 1 2 ° at P ANSWERS Consult with your teacher. Project Work 13.1 Observe the following instruction and write the answer of the given questions. Draw a line segment AB = 5 cm. Construct the angles 60° and 90° at the points A and B respectively. (a) Are the arms that make the angles 60° and 90° with AB intersected at any point? (b) Show the diagram obtained by the above instruction in your classroom. A P M R C H A B P
Geometry GEOMETRY 233 13.2 Construction of Equal Angles At the end of this topic, the student will be able to: ¾ construct angle equal to the given angle. Learning Objectives Construction of An angle equal to a given angle by using compass Suppose ∠BAC is the given angle. (i) With the centre at A, draw an arc of a suitable radius cutting the arms AB and AC at D and E respectively. (ii) Draw a ray OX. (iii) With the centre at O and the same radius as in the step (i), draw an arc cutting OX at S. S T Y O ∴∠ X O X O X XOY = ∠BAC (iv) With the centre at S and radius equal to DE of the given ∠A, draw an arc cutting the arc in the step (iii) at T. Join OT and produce, if necessary, to Y. Then XOY is the required angle, which is of equal measure to the given angle BAC. i.e., ∠BAC = ∠XOY. EXERCISE 13.2 Your mastery depends on practice. Practice like you play. 1. Copy the following angles on your notebook by using compass and measure both angles. Are they equal ? (a) (b) P O Q (c) D E C A B B O A P O Q
234 The Leading Mathematics - 7 (d) (e) (f) 2. Trace the given angles in your notebook and construct the same angle at the given point. (a) (b) (c) 3. Study the given instructions and answer the given questions. (a) Construct an angle of 75° and name it. (b) Copy the constructed angle in the number (a) by using compass, but do not use protractor. (c) Now, measure the both angles. Write our conclusion. ANSWERS Consult with your teacher. Project Work 13.2 Take an A4 paper and draw a line segment of any length. Construct an angle at one end of them and then copy it by using compass on the same paper. Now, measure the both angles and show their results in the table. Hence, write your conclusion. Prepare a report about it and present it in your classroom. L M O X O Y O Y X P B O A M B O A N X Y O
Geometry GEOMETRY 235 13.3 Introduction to Pair of Angles At the end of this topic, the student will be able to: ¾ categorise the pair of angles at a point. Learning Objectives Pair of Angles at a Point Two lines or rays in a plane may (i) cut or intersect each other at a point, (ii) be parallel (or do not cut or intersect each other) or, (iii) overlap or coincide. Two intersecting lines make two special pairs of angles. They have special names. They are: (a) Adjacent angles, (b) Vertically opposite angles These angles have a common point as the vertex of the angle. (a) Adjacent Angles Two angles are said to be adjacent angles if they have; (a) the same vertex, (b) a common arm and (c) arms on either side of the common arm. In the following figure, the line segment OC is the common arm for the angles ∠AOC and ∠BOC with the same vertex O. Here the angles AOC and BOC form a pair of adjacent angles. In this case, these different lines are intersected at the point O. In case, two straight lines intersect at a point the pair of angles on the same side of one of the straight lines are also adjacent angles. In this case, the adjacent angles are supplementary to each other. A B D C A C D B A C DB B O C A Common arm Adjacent angles Common vertex B D O C A Common arm Adjacent angles Vertex
236 The Leading Mathematics - 7 (b) Complementary Angles Two adjacent angles whose sum is one right angle (90°) are said to be complementary to each other. Thus, 60° and 30° are complementary to each other. In the figure, ∠AOC and ∠COB are complementary angles. Obviously, ∠AOC = 90° − ∠COB and ∠COB = 90° − ∠AOC. (c) Supplementary Angles Two adjacent angles whose sum is one hundred and eighty degrees (180°) are said to be supplementary to each other. Thus, 60° and 120° are supplementary to each other. Here, the angles AOC and COB form a pair of adjacent angles. Also, ∠ AOC = 180° − ∠COB and ∠ COB = 180° − ∠ AOC. Pairs of angles defined above have many interesting properties. They form the building blocks for further study in geometry. Now, we discuss some of them theoretically and wherever possible, experimentally. (d) Vertically Opposite Angles The angles on either side of any angle of two intersecting straight lines are called vertically opposite angles. In the adjoining figure, the straight lines AB and CD intersect at the point O. Here, the angles BOC and AOD are on either side of the same angle AOC (or ∠BOD) So, they form a pair of vertically opposite angles (VOA). In the same way, the angles AOC and BOD form another pair of vertically opposite angles. These angles are on either side of the common angle BOC or AOD. B O A C Complementary angles A O B C Supplementary angles B D O C A Angle of reference Vertically opposite angles
Geometry GEOMETRY 237 Experimental Verification Statement: The measures of vertically opposite angles are equal. Draw three sets of the straight lines AB and CD intersecting at O in the figure below. C D A B O Fig. (i) Fig. (ii) C D A B O Fig. (iii) C A D B O In each figure, (i) ∠AOC and ∠BOD (ii) ∠AOD and ∠BOC are two pairs of vertically opposite angles. Now, measure ∠AOC, ∠BOD, ∠AOD and ∠BOC. Fill the results in the table below. Fig. ∠AOC ∠BOD ∠AOD ∠BOC Remarks (i) (ii) (iii) Conclusion: From the above table, it is clear that the measures of vertically opposite angles are equal. (e) Angles in a Straight Line Look at the semi-circular protractor. The angle of semicircle is 180°. We use the semi-circular protractor to measure the angles from 0° to 180°. The angles x, y and z are added up to a straight line. So, the sum of angles formed at a point of the straight line is 180°. i.e., x + y + z = 180°. Thus, a straight line has 180° at any point. 0 180 170 10 160 20 30 150 140 40 130 50 110 100 80 100 110 70 60 50 40 30 20 10 0 180 170 160 150 140 130 120 90 80 70 120 60 90° 90° z y x
238 The Leading Mathematics - 7 Experimental Verification Statement: The sum of the measures of adjacent angles at a point of a straight line is 180°. Draw three sets of the straight line AB. Take a point O in AB and draw OC as shown in the figures below. A B C O Fig. (iii) A B C O Fig. (ii) A B C O Fig. (i) In each figure, ∠BOC and ∠AOC are adjacent angles on the straight line AOB. Now, measure ∠BOC and ∠AOC. Fill the results in the table below. Fig. ∠BOC ∠AOC ∠AOC + ∠BOC Remarks (i) (ii) (iii) Conclusion: From the above table, it is clear that the sum of the measures of adjacent angles at a point of a straight line is 180°. (f) Angles in a Circle Look at the figure of two semi-circular protractors. These two semi-circular protractors form a circular protractor as a circle. Since we already know that the measure of a semi-circular is 180°, so the measure of circular protractor is 180° + 180° = 360°. We know from the circular protractor that there are 360° in a circle. In the figure, the angles a, b and c are added up to a complete one full circle. Then the angles add up to 360°. So, the sum of angles formed at a point is 360°. i.e., a + b + c = 360°. 0 180 170 10 160 20 30 150 140 40 130 50 110 100 80 100 110 70 60 50 40 30 20 10 0 180 170 160 150 140 130 120 90 80 70 120 60 180 0 170 10 160 20 30 150 140 130 40 110 50 100 80 100 110 70 60 50 40 30 20 10 180 0 170 160 150 140 130 120 90 80 70 120 60 180° 180° a b c
Geometry GEOMETRY 239 Experimental Verification Statement: The sum of angles around a point is 360°.Draw three sets of figures in which the angles are formed at the same point O. A O B C Fig. (i) Fig. (ii) A O D B C Fig. (iii) A D E O B C Now, measure all the angles in each figure. Fill the results in the table below. Fig. ∠AOB ∠BOC ∠AOC ∠COD ∠AOD ∠DOE ∠AOE (i) (ii) (iii) Remarks: In fig. (i), ∠AOB + ∠BOC + ∠AOC = ............................................................ = 360° In fig. (ii), ∠AOB + ∠BOC + ∠COD + ∠AOD = ................................................. = 360° In fig. (iii), ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠AOE = .................................. = 360° Conclusion: From the above table, it is clear that the sum of angles around a point is 360°. CLASSWORK EXAMPLES Example 1 Observe the given figure. (a) Find the value of x in the given figure. (b) Find the value of ∠AOB by using the value of x. Solution: (a) From the figure, ∠AOB = ∠COD [ Vertically opposite angles] or, 2x + 32° = 62° or, 2x = 62° – 32° or, x = 30 2 ° or, x = 15° B D C O A 2x + 32° 62°
240 The Leading Mathematics - 7 (b) Here, ∠AOB = 2x + 32° or, ∠AOB = 2 × 15° + 32° or, ∠AOB = 30 + 32° or, ∠AOB = 62° Example 2 Find the value of x in the adjoining figure. Solution: Since ∠AOB and ∠BOC are complementary angles so, ∠AOB + ∠BOC = 90° or, x + 52° = 90° or, x = 90° – 52° or, x = 38°. Example 3 What is the supplementary angle of 98°? Solution: Since the sum of supplementary angles is equal to 180°, so x + 98° = 180° or, x = 180° – 98° or, x = 82° Therefore, the supplementary angle of 98° is 82°. Example 4 Find the size of the angle x in the given figure. Solution: From the given figure, x + 200o + 81o = 360o or, x + 281o = 360o or, x = 360o – 281o or, x = 79o . EXERCISE 13.3 Your mastery depends on practice. Practice like you play. 1. From the given diagram, name the following angles: (a) Right angle (b) Acute angle (c) Obtuse angle (d) Straight angle (e) Complementary angle of ∠BOC (f) Supplementary angle of ∠AOB C O B A x 52° 81° 200° X C B D O A
Geometry GEOMETRY 241 2. (a) In the given figure, ∠x and ∠y are not vertically opposite angles. Why? (b) The angles AOB and OBC are not adjacent angles. Why? (c) Why ∠AOB and ∠AOC are not adjacent angles? (d) The angles 91° and 90° are not supplementary angles. Why? (e) The angles 41° and 48° are not complementary angles. Why? 3. Find the complementary angles of the following angles: (a) 40° (b) 30° (c) 53° (d) 67° (e) 75° 4. Find the supplementary angles of the following angles: (a) 140° (b) 125 (c) 83° (d) 67° (e) 175° 5. Calculate the size of the angle x in each of the following figures: (a) (b) (c) (d) 6. Calculate the size of the angle 'a' in each of the following figures: (a) (b) (c) (d) A O B C O A B C x 42° 2x 232° x + 12° 3x – 34° 5x – 12° x + 108° a 27° 67° a 34° a 20° 3a 2a a Y x y X O Z W
242 The Leading Mathematics - 7 7. Observe the following figures and answer the given questions. (a) (b) (c) (d) (i) What ar complementary angles? Define it. (ii) Calculate the size of the angle x in each of the figures. (iii) Find the measures of the given unknown angles. 8. Observe the following figures and answer the following questions. (a) 135° x (b) (c) 2x – 9 95° 30° 4x x (d) 3x – 7 92° 80° x 35° x (i) What are the supplementary angles? Define it. (ii) Calculate the size of the angle x in each of the given figures. (iii) Find the measures of the given unknown angles. 9. Observe the following figures. (a) (b) (c) 45° a c 2b – 9° (d) a + 15° 2b – 17° 3c + 14 113° (i) What are the relation between the vertically opposite angles. (ii) Calculate the size of the angles a, b and c in each of the given figures. (iii) Find the measures of the given unknown angles. 10. (a) If two complementary angles are in the ratio 4 : 5, find the angles. (b) If two complementary angles are in the ratio 3 : 2, find the angles. (c) If two supplementary angles are in the ratio 5: 1, find the angles. (d) If two supplementary angles are in the ratio 4 : 5, find the angles. 135° x 55° x 60° x 3x + 10° 30° – x x – 4° 2x 85° 57° a 150° b c 2a b 2c – 15° a + 20°
Geometry GEOMETRY 243 11. (a) If one angle is triple of its supplementary angle, find the angles. (b) If one angle is double of its complementary angle, find the angles. (c) If an angle is 35° less than its complementary angle, find the angle. (d) If an angle is 111° greater than its supplementary angle, find the angle. 12. (a) If the difference of the measure of complementary angles is 15°, find the angles. (b) If the difference of the measure of complementary angles is 40°, find the angles. (c) If the difference of the measure of supplementary angles is 40°, find the angles. (d) If the difference of the measure of supplementary angles is 37°, find the angles. ANSWERS 3. (a) 50° (b) 60° (c) 37° (d) 23° (e) 15° 4. (a) 40° (b) 55° (c) 97° (d) 113° (e) 5° 5. (a) 42° (b) 116° (c) 23° (d) 30° 6. (a) 63° (b) 23° (c) 36° (d) 15° 7. (a) 45° (b) 125° (c) 30° (d) 48° 8. (a) 135° (b) 64° (c) 22° (d) 32° 9. (a) 30°, 150°, 30° (b) 20°, 140°, 77.5° (c) 135°, 72°, 135° (d) 52°, 65°, 33° 10. (a) 40°, 50° (b) 54°, 36° (c) 150°, 30° (d) 80°, 100° 11. (a) 135°, 45° (b) 60°, 30° (c) 55° (d) 69° 12. (a) 52.5°, 37.5° (b) 65°, 25° (c) 110°, 70° (d) 108.5°, 71.5°
244 The Leading Mathematics - 7 CHAPTER 14 Plane Shapes Lesson Topics Pages 14.1 Construction of Triangles 245 14.2 Quadrilateral and Its Types 249 What types of closed figure do you form from three line segments or sticks? Name the objects in the form of triangular shapes. When the ends of the arms of two angles are joined in the opposite, what type of figure is formed ? What types of closed figure do you form from four line segments or sticks? Name the objects in the form of quadrilateral shapes. Tell the name of the triangular and quadrilateral objects in your bag. Discuss the surface of the above objects. WARM-UP
Geometry GEOMETRY 245 14.1 Construction of Triangles At the end of this topic, the student will be able to: ¾ construct the triangles from the given measures. Learning Objectives Take three points A, B and C on the same plane surface. Join these points by the straight lines. How many straight lines are formed by these three points? These three lines AB, BC and AC form a closed plane figure, called triangle ABC. It is denoted by ∆ABC. The line segments that formed triangles are called the sides of triangle and these three points are called the vertices of triangle. In ∆ABC, each pair of sides forms an angle. There are three angles ∠BAC(∠A), ∠ABC (∠B) and ∠ACB (∠C) in ∆ABC. Now, what is the meaning of triangle? Triangle = Tri + Angle = Three angles. Types of Triangles In general, there are six types of triangles based on sides and angles. Based on Sides Scalene triangle: All sides are not equal in a triangle. A B C Isosceles triangle: Any two sides are equal in a triangle. A B C Equilateral triangle: All sides are equal in a triangle. A B C Based on Angles Acute-angled triangle: All angles are acute in a triangle. Q R P Obtuse-angled triangle: Any one angle is obtuse in a triangle. Q R P Right-angled triangle: Any one angle is right in a triangle. Q R P B C A
246 The Leading Mathematics - 7 Construction of Triangles To construct a triangle we should first draw a rough sketch of the triangle with the given measures. This rough sketch helps us to understand the steps of construction. i. When two sides and the angle between them are given (SAS) Construct a triangle PQR in which PQ = 4.5 cm, QR = 3.2 cm and ∠PQR = 30°. Steps i. Make the rough triangle PQR with the given measures. ii. Draw a line segment PQ of the length 4.5 cm. iii. Draw an angle of 30° at Q by using compass. iv. Cut the arm that makes 30° by an arc of 3.2 cm from Q to R. v. Join PR. P Q R 4.5 cm 30° 3.2 cm Thus, the required triangle PQR is constructed. ii. When two angles and the side between them are given (ASA) Construct a triangle ABC in which AB = 4.5 cm, ∠BAC = 60° and ∠ABC = 45°. Steps i. Make a rough triangle ABC from the given instruction. ii. Draw a line segment AB having the length 4.5 cm. iii. Construct an angle 60° at A by using compass. P Q R 4.5 cm 30° 3.2 cm Rough Sketch A B C 4.5 cm 60° 45° Rough Sketch
Geometry GEOMETRY 247 iv. Construct an angle 45° at B by using compass. v. Name the point of intersection of the arms by C. A B C 4.5 cm 60° 45° Thus, the required triangle ABC is formed. iii. When three sides are given (SSS) Construct a triangle ABC in which AB = 3.5 cm, AC = 3.2 cm and BC = 4.2 cm. Steps i. Make a rough sketch from the given measures. ii. Draw a line segment AB of the length 3.5 cm. iii. With the centre A and radius 3.2 cm in the compass, draw an arc. iv. With the centre B and radius 4.2 cm in the compass, draw a second arc that cuts its first arc at C. v. Join AC and BC. Thus, the required triangle ABC is formed. B C A 3.5 cm 3.2 cm 4.2 cm Rough Sketch A B C 3.5 cm 4.2 cm 3.2 cm
248 The Leading Mathematics - 7 EXERCISE 14.1 Your mastery depends on practice. Practice like you play. 1. Construct ∆ABC in each of the following measures: (a) AB = 4.3 cm, BC = 3.5 cm and CA = 5.1 cm (b) AB = 4.5 cm, BC = 4.8 cm and CA = 5.3 cm. (c) AB = 5.2 cm, BC = 6 cm and CA = 5.5 cm. (d) AB = 4.8 cm, BC = 4 cm and CA = 5.2 cm. 2. Construct ∆PQR in each of the following measures: (a) PQ = 4.5 cm, QR = 5.5 cm and ∠Q = 60°. (b) PQ = 4.5 cm, QR = 5 cm and ∠Q = 45°. (c) QR = 5.2 cm, PR = 4.5 cm and ∠R = 90°. (d) QR = 4.8 cm, PR = 5.2 cm and ∠R = 30°. 3. Construct ∆LMN in each of the following measures: (a) MN = 5 cm, ∠M = 60° and ∠N = 30°. (b) LM = 5.2 cm, ∠L = 60° and ∠M = 45°. (c) LN = 4.5 cm, ∠N = 75° and ∠L = 60°. (a) MN = 5.2 cm, ∠M = 90° and ∠N = 30°. 4. (a) Draw a line segment CD = 5.2 cm. (b) Construct the angles 60° and 30° at C and D respectively. (c) Produce the arms that make these angles with CD. (d) What type of triangle do you find ? Give reason. 5. (a) Draw a line segment PR = 6.5 cm. (b) Construct the angles 75° and 30° at P and R respectively. (c) Produce the arms that make these angles with PR. (d) What type of triangle do you find ? Give reason. ANSWERS Consult with your teacher.
Geometry GEOMETRY 249 14.2 Quadrilateral and Its Types At the end of this topic, the student will be able to: ¾ verify the properties of quadrilaterals experimentally. Learning Objectives Draw two line segments (not in the same straight line) AB and BD at one endpoint. From the free endpoint D of BD, we draw another line segment DC. Finally, we join the free-end C of this line segment with the free-end A of the line segment AB. Then this line CA may or may not intersect BD as shown in the figures (i) and (ii). We note that in the Fig (i) BD and AC intersect at a point other than the endpoints. In the Fig (ii), no two line segments intersect at points other than endpoints. We also note that no two line segments are collinear. The second type of figure is known as a quadrilateral. A precise definition of a quadrilateral is as follows: A quadrilateral is a plane figure bounded by four line segments such that i. no two line segments intersect at points other than the endpoints, and ii. no two line segments having a common endpoint is collinear. The line segments forming a quadrilateral are called the sides and the endpoints its vertices. The angle determined by any two sides is called an angle of the quadrilateral. A line segment joining one vertex to another vertex is called its diagonal, if it is distinct from its sides. In the Fig (iii), AB, BC, CD and AD are the sides of quadrilateral ABCD. A, B, C and D are its vertices. ∠A, ∠B, ∠C and ∠D are its angles. AC and BD are its diagonals. Quadrilaterals may be of various types. Some of them are (a) Trapezium (b) Parallelogram (c) Rectangle (d) Rhombus (e) Square C D Fig (ii) A B Fig (i) C D A B Fig (iii) C D A B
250 The Leading Mathematics - 7 Note: In each of the following figures below, four line segments enclose a certain region of the plane. Even then, none of them represents a quadrilateral. A quadrilateral may or may have its opposite sides parallel. Accordingly, we have different types of quadrilaterals. Trapezium A quadrilateral having only one pair of parallel sides is called a trapezium. The sides other than parallel sides of the trapezium are called legs. In the adjoining figure, ABCD is a trapezium in which AB//DC. AD and BC are its legs. i. Parallelogram A quadrilateral having both pairs of opposite sides parallel is called a parallelogram. In the adjoining figure, PQRS is a parallelogram. Some of the properties of a parallelogram are: i. The opposite sides of a parallelogram are equal. ii. The opposite angles of a parallelogram are equal. iii. The diagonals of a parallelogram bisect each other. Parallelograms are again classified based on sides and angles. ii. Rectangle A rectangle is an equiangular parallelogram. To be more specific, a rectangle may be defined as a parallelogram having each of its angle equal to a right angle. In the adjoining figure, ABCD is a parallelogram with ∠A = 90°. Hence, ABCD is a rectangle. In addition to the properties of a parallelogram, a rectangle has equal diagonals. D A C B P R Q S D C A B