Acme Mathematics 8 151Example 3: Simplify : a – ba + b – 2abb2 – a2Solution : Here, a – ba + b – 2abb2 – a2= a – ba + b – 2ab– (a2 – b2)= a – ba + b + 2aba2 – b2= a – ba + b + 2ab(a – b) (a + b)= (a – b) (a – b) + 2ab(a + b) (a – b)= a2 – ab – ab + b2 + 2ab(a + b) (a – b)= a2 + b2(a + b) (a – b)Example 4: Simplify: 3x2 + 3x + 2 – 5x2 + 4x + 3 Solution : Here, 3x2 + 3x + 2 – 5x2 + 4x + 3 = 3(x + 1) (x + 2) – 5(x + 3) (x + 1) = 3(x + 3) – 5(x + 2)(x + 2) (x + 1) (x + 3) = 3x + 9 – 5x – 10(x + 1) (x + 2) (x + 3) = –2x – 1(x + 1) (x + 2) (x +3)Classwork1. Add the following rational expressions: (a) 3x + 4x (b) xx + 1 + 2xx + 1(c) 7 – xx – 4 + x + 7x – 4 (d) x – 1x – y + 3 y – x It is LCM of denominatorsWe have: x2 + 3x + 2 = (x+1) (x+2) x2 + 4x + 3 = (x+1) (x+3)
152 Acme Mathematics 82. Subtract the following rational expressions: (a) 9x – 7x (b) 3 x x – y – 3yx – y3. Simplify: (a) 45x2 + 3x2 (b) 3 + 2x 4 + 4 – 3x5x (c) px + qy (d) 3 + 2x 4x + 4 – 3y5yExercise 4.81. Add the following rational expressions: (a) 2x + 3y 4x – 3y + x + y 3y – 4x (c) 4x – 1 + x 1 – x (c) x x2 – y2 + y x2 – y2 (d) (x – y)2(x + y)2 + 2xy (x + y)2 2. Subtract the following rational expressions: (a) x2x + y – y2x + y (b) 7 x 2x + 3y – y2x + 3y(c) 4x x – y – 4yy – x (d) x x2 – 4 – 24 – x2 3. Simplify: (a) 1x + 1 + 1 x – 1 (a) x – yx + y + 2xy x2 – y2(c) 10x + 2 – 1 x2 – 4 (d) x(x + y) – xy (x + y)2(e) 8x – 2 – 8 x(x – 2) (f) 43(x – 2) + 5 4(x – 2) 4. Simplify:(a) x + 1x + 4 + x – 2 x2 + 2x – 8 (b) x + 3x + 1 – 4 – x x2 – x – 2(c) x x – 1 + 2x + 3 x2 + 2x – 3 (d) x + 1x(x + 1) – 2 x2 – x – 2(e) x – 2x2 – 25 – x + 5 x2 – 5x (f) x + 4 2x – 2 + x2 – 4 2x2 – 3x + 1(g) 3x + 43x2 – 2x – 8 + x–1 3x2 + x – 4 (h) 3x x2 – 7x + 10 – 2x x2 – 8x + 15(i) 2x x – y – xy x2 – y2 (j) 5xy x2 – y2 – 3y y – x(k) 6a (2 – x)2 – a2 – x (l) 1 x – 1 + 2x x2 + 1
Acme Mathematics 8 153(d) Multiplication of rational expression Rational algebraic expressions can be multiplied in the same way as rational numbers.For example: The product of 23 and 45 = 23 × 45= 2 × 43 × 5= 8 15Similarly, x4 × 5y = x × 54 × y = 5x 4ySolved ExamplesExample 1: Multiply: bxay and 3x4ySolution : Here, bxay × 3x4y= bx × 3xay × 4y= 3bx24ay2Example 2: Multiply: xy × x3ySolution : Here, xy × x3y= x × x y × 3y = x23y2Example 3: Multiply: x – 2 2x + 1 × 4x2 – 1 3x2 – 6xSolution : Here, x – 2 2x + 1 × 4x2 – 1 3x2 – 6x= x – 22x + 1 × (2x)2 – 123x(x – 2)= x – 22x + 1 × (2x – 1) (2x + 1)3x(x – 2)= (x – 2) (2x – 1) (2x + 1)(2x + 1) × 3x(x – 2) = 2x – 1 3x
154 Acme Mathematics 8(e) Division of rational expressions:Division is the inverse operation of multiplication. In case of division, we multiply the dividend by the reciprocal of the divisor and then simplify. Study the given examples carefully.Solved ExamplesExample 1: Divide: xy by 4ySolution : Here, xy ÷ 4y= xy × y4= x × yy × 4 = x4 Hence, xy ÷ 4y = x4 Example 2: Simplify: x – 22x + 1 ÷ 3x2 – 6x(2x + 1)Solution: Here, x – 22x + 1 ÷ 3x2 – 6x(2x + 1)= x – 22x + 1 × 2x + 13x2 – 6x= (x – 2) × (2x + 1)(2x + 1) × 3x(x – 2)= 13xExample 3: Simplify: x2 – x – 6 x2 – 7x + 12 ÷ x2 – 3x – 10 x2 – 2x – 8Solution: Here, x2 – x – 6 x2 – 7x + 12 ÷ x2 – 3x – 10 x2 – 2x – 8= (x – 3) (x +2) (x – 4) (x – 3) ÷ (x – 5) (x + 2) (x –4) (x + 2) [by factoring]= (x – 3) (x +2) (x – 4) (x – 3) × (x – 4) (x + 2) (x – 5) (x + 2)= x +2x – 5It is reciprocal of 4y .4y is divisor.
Acme Mathematics 8 155Classwork1. Find the reciprocal fraction of the following.(a) 4y3x (b) xyab (c) x – yx + y2. Multiply: (a) 75x and 5x y (b) 4 3x + 4 and 3x + 4 y3. Simplify: (a) 3x2 x2 – 1 × (x – 1)6x2 (b) x2 – 4 x – 2 × x2 + 3 x2 + 2xExercise 4.91. Find the reciprocal fraction of the following.(a) x2 + 2x2 –3 (b) a2 – ab + b2a – b (c) 7x + y(d) 5x + 73x – 7 (e) a + b + ca2 + bx (f) 1x2 + 52. Multiply: (a) 3 x + 2 and x2 – 4 2 (b) 3 x + y and x2 – y2 3x – 3y(c) 7x28z , 16z2 14x and 2y26y (d) x2 a2z2 , ay3 bx and b3z3 x2y23. Simplify: (a) x – y x + y × x2 – y2 (x – y)2 (b) x2 – 2x + 1 3(x– 1) × x– 2 x2 + 2 (c) x2 + 2x – 35 3x2 – 2x × 9x2 – 4 7x + 49 (d) x2 + 4x + 4 6(x – 2) × x2 – 2x x2 – x – 64. Divide:(a) x2 y by x2 y2 (b) x – y y + x by y – x x + y(c) xy210a by x2y2z2 20a2 (d) 16x2y2z3 by 2y3xyz(e) 3x2y4z416a5 by 9x5y5z5 8abc3 (f) x3 – x2 (x + 2)x by (x – 1) (x – 2) x2 –2x
156 Acme Mathematics 85. Simplify:(a) x + 2 x + 3 ÷ (x + 2)2 x2 – 9 (b) 2x – 6 x + 2 ÷ x2 – 9 4x(c) x2 – 16 x + 5 ÷ x – 4 x2 – 25 (d) x2 – xy x2 + xy ÷ x – y x + y(e) x2 – xy xy + y2 ÷ xy – y2 x2 + xy (f) x2 + xy x2y – xy2 ÷ x2 + 2xy + y2 x2y – y36. Simplify: (a) (2x – 1) (x + 2) (x – 4) (x – 3) ÷ (2x – 1) (x – 1) (x – 3) (x – 4) (b) x2 + 4x – 12 x ÷ x2 – 5x + 6 x (c) x 2 – 1 x + 4 ÷ x 2 + 3x –4 2x + 8 (d) 3x2 + 14x – 5 x2 – 3x + 2 ÷ (2x + 1) (3x – 1) 2x2 – 3x – 2 (e) x 2 – 8x + 15 x2 + 4x – 45 ÷ x 2 + 2x – 15 x2 + 8x – 9 (f) x 2 – 2x – 3 x2 – 6x + 9 ÷ x 2 – 3x + 2 x2 – 5x + 6 (g) x 2 – 11x + 30 x2 – 3x – 10 ÷ x 2 – 8x + 12 x2 + 3x + 2 7. Simplify:(a) 3x 5y × 3y5 + y2(b) xx – 1 – 1x + 1 ÷ x + 2 x – 1 (c) 1x – 2 + 1x + 2 ÷ x – 1 x + 2 (d) x + yx – y – x – yx + y × x2 – y22xy8. If p = x2 + 2x – 3 and q = x – 1x , express each of the following as a rational expression.(a) p + q (b) p – q(c) p × q (d) p ÷ q
Acme Mathematics 8 1571. Factorize:(a) x4 – 4(b) 2x2 – 3x – 20(c) 81 – x4(d) x (a – b) + y(b – a) (e) 4a2 + 5ab – 6b22. PQRS is a square field of length 50 m in which a square pond ABCD of length 20 m is to be made.(a) Write the formula of a2 – b2. (b) Calculate the area of the remaining land after the pond is duged? 3. (a) Write the expanded form of (a + 3)2. (b) The area of a land is (x2 – 2x – 15) m2. Find the length and breadth. 4. (a) If x = 1, y = 2 and z = – 2, then find the value of x2y2(xz)2.(b) Factorize: x2 – y + x – xy. 5. Answer the following questions:(a) If (x + y) = 5 and xy = 3 find the value of x2 + y2 and (x + y)2? (b) Simplify: 2 – 1a2 – 2 + 1a2 6. The two algebraic expressions are x2 – 6x + 5 and x3 – 25 x(a) Find the factors of given expressions.(b) Find the LCM of the given expressions. 7. (a) Define an algebraic term with example.(b) Simplify: 3x x+1 +3 x+1(c) Factorise : x5 + x4 + x3 + 2x2 + 2x + 2AS RD CA BD50 m20 mMixed Exercise
158 Acme Mathematics 88. (a) Find the area of shaded region.(b) Represent geometrically a2 – b2. 9. (a) Simplify: 5x2y 10 xy2(b) Factorize: 9x2 – 16y2 (c) Fill in the blank to make it perfect square. x2 + ...... + 16. 10. An expression is given x2 – 5x + 6.(a) If the product of two number is 6 and its sum is – 5, what are these two numbers.(b) Factorizes given expression. 11. The process of spliting an algebraic expression into its factors is called factorization.(a) Find the factors of the following expressions.(i) 3a2 + 6ab (ii) x2 + 2x – 8 (b) Find the expression whose factors are (x + 2) and (x – 2). 12. (a) The highest common factors of the given algebraic expression is called HCF in shortcut. Find HCF of following expressions.x2 – 2x, x2 – 4, x2 – 6x + 8(b) Simplify: x2x –3 + 6x + 9x –313. x3 – xx2 + x is a algebraic fraction. Answer the following questions.(a) What are numerator and denominator of this fraction.(b) Simplify it.14. (a) Write the formula of a2 – b2 and multiply: (x + y)(x – y)(b) Simplify : x2 – y2y2 ÷ x2 + xyxy 15. (a) Write the expanded form of (a + b)2.(b) Factorize: x2 + 8x + 1616. If a + 1a = 3, answer the following questions.(a) Is a + 1a a polynomial? Give reason.(b) Find the value of: (i) a2 + 1a2 (ii) a + 1a2 AP QS RDa cmc cmd cm b cmBB
Acme Mathematics 8 15917. Two expressions are: x2 – 3x + 2 and x2 – 4x + 3(a) Find LCM of above expressions.(b) Show it in the venn diagram. 18. (a) Simplify: 2xyx+y – xyx+y (b) Prove that: a – 2a2 – 5a + 6 + a – 5a2 – 8a + 15 = 2a–3 19. If two algebraic expressions are x2 – 4x + 3 and x2 – 3x + 2(a) What is the full-form of HCF?(b) Find the HCF of the given algebraic expressions.(c) At what value of x, the value x2 – 4x + 3 is zero?20. Following are the given algebraic expressions.(x2 + 3x – 18) and (x2 – 9)(a) Divide first expression by second expression.(b) Find the LCM of given expression.21. (a) Factorize: (x – y)2 – 9(x – y) – 22(b) Simplify: a2 + 5a + 6a2 – 1 ÷ a2 – 9a2 – 2a – 322. Answer the following question on the basis of given picture.(a) Write the length and breadth of given rectangle.(b) Draw the picture on your copy and write the area of each rectangles inside it. (c) Add the each area and find the total area of the big rectangle. 23. In the figure square 'A' is inside the square 'B'.(a) Find the area of B excluding A by using a2 – b2.(b) Express (a2 – b2) geometrically. 1x 1x1 116 cmAB 6 cm
160 Acme Mathematics 8EvaluationTime: 50 minutes Full Marks: 211. (a) Select the correct value of (?)0,? ≠ 0 from the given alternatives. [1](i) 0 (ii) 1 (iii) 2 (iv) 3 (b) Simlify: 3x – a + 4x + a [2]2. (a) If ax and ay are any terms with the same base 'a', what is the value of ax × ay ? [1](b) Simplify: x + 3x – 5 – x + 5x – 3 [2]3. (a) Write the expanded from of (a+2)2. [1](b) Simplify: (xa–b)2. (xb–a)2 [2]4. (a) If x = 3o + 2, find the value of x. [1](b) Simplify: k + 3k2+6k+9 + k – 3k2–6k+9 [2]5. (a) Factorise: z2 – b2. [1](b) Simplify: xa–b × xb–c × xc–a [2]6. (a) Simplify: 3x – 2x2+5x+6 – 2x + 5x2+5x+6 [1](b) Simplify: xaxb a+bxbxc b+c xaxb c+a [2]7. (a) What is the value of: (x+1)o [1](b) Simplify: 2x – y+ 3x + y [2]
Acme Mathematics 8 161Evaluation EvaluationTime: 58 minutes Full Marks: 241. (a) Find the highest common factors (HCF) of the given expressions.a2 – 25 and a2 + 2a – 15 [2](b) Solve: 4x2 – 36 = 0 [2] 2. (a) Simplify: aa + b + aa – b [2](b) Find the L.C.M. of the given algebraic expression:a2 + 5a + 6 and a2 – 4 [2]3. (a) Find the H.C.F.: x2 + 6x + 8, x + 4 [2](b) At what value of x, the value of (x2 – 16) becomes zero ? [2]4. Two algebraic expressions x2 + 6x + 8 and x2 – 4 are given. (a) Find the LCM of given expressions. [2](b) For what values of x, the value of x2 + 6x + 8 becomes zero ? [2]5. (a) Find the H.C.F. of: x2 + 7x + 12 and x2 – 16 [2](b) What is the value of p, the value of the equation p2 – 5p + 6 is zero. [2]6. (a) Simplify: 1a + b + 1a – b [2](b) Find the L.C.M. of: (a + b) and a2 + 2ab + b2 [2]
162 Acme Mathematics 84.3 Equation and GraphREVISIONA. Linear Equation in one variable Let us consider the following statements. x + 6 = 14 ....................(i) 3x – 7 = 20 ..................(ii) 4x = 100 ......................(iii) x 7 = 1 ..........................(iv) We observe that the symbol '=' appears in each of the statements. These are equations. A statement which contain variable and equality ('=' ) sign is called an equation. All equations are linear equations. If the degree of the variable in the equation is 1, the equation is called a linear equation. In the above examples there is only one variable x so these are linear equation in one variable.Equation has two sides: Left hand side (LHS) and Right hand side (RHS). In equation (i), x + 6 is LHS and 14 is RHS.B. Solution of equation Consider the equation x – 15 = 9. It is true when we replace x by 24 otherwise it is false. Here 24 is called the solution of the equation. The process of finding the solution of an equation is called solving the equation.C. Solving the equation Methods of solving the equations (a) Method of Balancing An equation may be compared with a balance used for weighing. Its sides are two pans and equality (=) tells us that the two pans are in equilibrium.
Acme Mathematics 8 163The following rules help us while solving the equation. Rule 1 We can add the same number or quantity to both sides. Rule 2 We can subtract the same number or quantity from the both sides. Rule 3 We can multiply both sides of the equation by the same non-zero number or quantity. Rule 4 We can divide both sides of the equation by the same non-zero number or quantity Now we can solve the linear equations using the above four rules 1, 2, 3 and 4.Solved ExamplesExample 1: Solve: x – 19 = 1Solution : Here, x – 19 = 1 or, x – 19 + 19 = 1 + 19 → using rule 1 or, x = 20 Hence, the value of x is 20.Example 2: Solve: x + 15 = 6Solution : Here, x + 15 = 6 or, x + 15 – 15 = 6 – 15 → using rule 2or, x = – 9Hence, the value of x is – 9.Example 3: Solve: y2 = 3 Solution : Here, y2 = 3 or, y2 × 2 = 3 × 2 → using rule 3or, y = 6Hence, the value of y is 6.Example 4: Solve: 14x = 70Solution: Here, 14x = 70or, 14x14 = 7014 → using rule 4or, x = 5 Hence, the value of x is 5.
164 Acme Mathematics 8Example 5: Solve: 3(x + 1) = 2x + 14Solution : Here, 3(x + 1) = 2x + 14or, 3x + 3 = 2x + 14or, 3x + 3 – 3 = 2x + 14 – 3 → subtracting 3 from both sidesor, 3x = 2x + 11or, 3x – 2x = 2x – 2x + 11 → subtracting 2x from both sides or, x = 11Hence, the value of x is 11.Example 6: Solve: 5x – 76 = 3x – 54Solution : Here, 5x – 76 = 3x – 54or, 4(5x – 7) = 6(3x – 5) → by cross multiplicationor, 20x – 28 = 18x – 30or. 20x – 28 + 28 = 18x – 30 + 28 → adding 28 on both sidesor, 20x = 18x – 2or, 20x – 18x = 18x – 18x – 2 → subtracting 18x from both sidesor, 2x = – 2or,2x 2 = – 22 → dividing both sides by 2or, x = – 1Hence, the value of x is –1(b) Method of transposition: It is mostly common method used for solving equation. The word transpose means 'changing the side'.The following rules are followed in this method: (i) + becomes – on changing the side (ii) – becomes + on changing the side(iii) × becomes ÷ on changing the side (iv) ÷ becomes × on changing the side
Acme Mathematics 8 165Example 7: Solve 5(x + 1) + 3 = 2x + 14Solution : Here, 5(x + 1) + 3 = 2x + 14 or, 5x + 5 + 3 = 2x + 14 or, 5x – 2x + 8 = 14 → transposing 2x from RHS to LHSor, 3x + 8 = 14 or, 3x = 14 – 8 → transposing 8 from LHS to RHS or, 3x = 6 or, x =63Hence the value of x is 2.Classwork1. Solve for x. (a) 6(x – 1) = 2x + 2 (b) 5x – 2 = 2(x + 20)(c) 9(x – 1) = 3(x + 3) (d) 20(x – 1) + 5 = 14x + 32. Solve for x.(a) 7x – 14 = 15 (b) x2 + 4x2 = 16Exercise 4.101. Solve for x. (a) 1 – 3(x + 2) = 9x + 4 (b) 8 – 6(x + 3) = 6x + 13 (c) (2 – x) + 12 = 2(x + 7) (d) 14x + 10 = 10x + 30(e) 4x – 1 = 10x +10 (x + 2) (f) 3( x + 6) – 8 = 2(x + 2)(g) 2(x + 5) + 8 = 7x + 8 (h) 4( x + 3) – 8 = 2(x + 2) (i) 4(x + 3) = 52 (x – 2) (j) 4(x –3) – 3 = 1 2. Solve for x.(a) x – 2 2 = x – 13 (b) x + 2 4 = x + 35 (c) x – 3 2 = 3x – 14 (d) x + 4 3 = x + 84 (e) x + 4 3 – x 4 = x – 6 (f) 4 5 = 2x 3 – 1 5
166 Acme Mathematics 8Project WorkObjective : To solve the equation in one variableMaterials required :(i) linear equation and any one geometrical shapeA pair of green and blue strips or singles represent zero. Green strips / singles represent negative. Blue strips / singles represent positive. The strips represent the unknown x and the singles represent the constraints.Activity : Solve the equation and fill the results in the given circles. Add the numbers along each line. The numbers written along each line of a star have the same sum (magic sum).(a) 4x – 10 – x = 8(b) x2 = 2 12(c) 4(2x – 3) = 12(d) x5 – 2 = 0(e) 2(5 – x) = 8(f) 16 = – x + 20(g) 2x3 – 3 = 3(h) 12x – (– 6) – 4x = 22(i) 3(x – 2) – 2(x + 1) = 0(j) x3 + x2 = 10abij dec fhg
Acme Mathematics 8 167Project WorkObjective : Solving the one variable equation by using strips and unitsMaterials required :(i) Strips - in two colours - black and white (ii) Singles - in two colours - black and whiteStrips unitsA pair of black and white strips or units represent zero. Black strips / units represent negative. White strips / units represent positive. The strips represent the unknown 'x' and the units represent the constraints.Activity : a. Consider the equation 4x + 1 = 2x + 3Using the number strips and units, we build the equation as shown. =Removing one unit and two strips (common) from both the side, we get,Therefore, the equation becomes 2 = 2x.=By comparing this, we get x = 1. VerificationLHS = 4 × 1 + 1 = 4 + 1 = 5RHS = 2 × 1 + 3 = 4 + 1 = 5∴ LHS = RHS = 5
168 Acme Mathematics 8b. Consider the equation 3x – 5 = x – 1.Using the number strips and units we build the equation.=Removing one single and one strip (common) from both sides, we get, 2x – 4 = 0.=Now, adding 4 white units to each side , we get,=Since a pair of black and white units represents zero.=∴ Now we have =By comparing this, we get x = 2 VerificationLHS = 3 × 2 – 5 = 6 – 5 = 1RHS = 2 – 1 = 1∴ LHS = RHS = 1Now, try to solve some more linear equations in one variable by using the method of strips and units.
Acme Mathematics 8 169D. Simultaneous Equation in two variables Here some examples of simultaneous equations in x and y are given x + y = 5 ………………. (i) and x – y = 2 ………………………….. (ii)3x + 5y = 20 and 5x – 2y = 7 etc.Here, x and y are variables. The height power of x is 1, and y is also 1. These equations are linear equations. A system of equation is composed of two equations considered simultaneously. Equations (i) and (ii) is a system of two linear equations in two variables x and y. The solution set of this system consists of all order pairs that make both equations true. Solving system of Equation Graphically Graph of linear equations x + y = 4 and x – y = 2 are the two straight lines. The intersection point is a solution of both equations.Solved ExamplesExample 1: Solve the equation graphically. x + y = 4, x – y = 2 Solution : Here, x + y = 4or, y = 4 – x ...............(i) x – y = 2or, – y = – x + 2 or, y = x – 2 ...........(ii) Table of (i), y = 4 – xx 0 3 4y 4 1 0Pair of points are (0, 4), (3, 1) and (4, 0)Table of (ii), y = x – 2 x 0 2 4y – 2 0 2Pair of points are (0, – 2) (2, 0) and (4, 2).Take at least 3 pair of values
170 Acme Mathematics 8Now, plotting the points on the graph we get the following graph.Y'YX' XA(3,1)Ox + y = 4x – y = 2From the graph point A is the intersecting point.Coordinates of A are (3, 1) Hence, the value of x = 3 and y = 1Example 2: Solve the problem graphically. The sum of the numbers x and y is 5. If the difference of six times 'x' and five times 'y' is 8, find the numbers.Solution : Here, the sum of the numbers x and y is 5.Now, x + y = 5or, y = 5 – x...............(i)The difference of six times 'x' and five times 'y' is 8.So, 6x – 5y = 8or, y = 6x – 85 ................(ii)Table of (i), y = 5 – xx 0 1 5y 5 4 0Pair of points are (0, 5), (1, 4) and (5, 0)Table of (ii), y = 6x – 85x – 2 – 7 3y – 4 – 10 2Pair of points are ( – 2, – 4), (– 7, – 10) and (3, 2).(2,0) (4,0)(0,–2)
Acme Mathematics 8 171Now, plotting the points on the graph we get the following graph.Y'YX' XA(3,2)Ox + y = 5 6x – 5y = 8From the graph point A is the interesting point Coordinates of A are (3, 2)Hence, the numbers are 3 and 2. Classwork1. Complete the table and graph the line for each equation.(a)x – 2 – 1 0 1yy = 4x (b)x – 5 – 3 0 5yy = x + 1Y'YX' X OY'YX' X O
172 Acme Mathematics 8(c) (d)x – 4 – 3 – 2 1yy = x + 6x 1 2 3 4yy = x – 1Y'YX' X OY'YX' X O2. Complete the table and graph the line. Write the value of x and y at point of intersection of lines.O X' XY'(a) Yx y7510y = x – 4 (0, 6)(5, 1)Y'YX' X O(b)x y3127y = x – 6(–2, 0)(2, –4)Exercise 4.111. Draw the graph of the following linear equation:(a) x + y = 7 (b) x – y = 5 (c) 2x + y = 1 (d) y = 2x 2. Solve each of these pairs of equations graphically.(a) y = 2x + 1 (b) y = x – 2y = x + 2 y = 10 – 3x
Acme Mathematics 8 173(c) y = 2x – 3 (d) y = 7 – x x + y = 0 2x + 3y = 15(e) y = 4 – x (f) 2y + 3x = 162y + 3x = 7 y = 4x – 3(g) 2x + y = 10 (h) x + y = 16 4x + 3y = 24 5x + 3y = 603. Solve the following word problems. (a) Divide 8 into two parts such that the greater part is three times the smaller part.(b) The length of a rectangle is three times its breadth. Find the length and breadth of the rectangle if its perimeter is 16 m. (c) Students of a class - 8 have a total of 22 pens. Some of the pens are gel pens andothers have fountain pens. The number of fountain pens is 2 more than three times the number of gel pens. What is the number of gel pens and non-gel in theclass. (d) One number is 4 less than three times another number. If their sum is increased by 5, the result is 25. Find both the numbers. (e) In a class of 35 students, the number of girls is two-fifth of the number of boys.Find the number of boys and girls in the class. (f) The difference between two angles is 5° and there sum is 9°. Find the angles. (g) One angle is twice the other angles. If difference between these angles is 3°, find the two angles(h) The length of a rectangle is 8 m more than its breadth. If its perimeter is 24 m, find its length and breadth.(i) Find two consecutive odd numbers whose sum is 20.
174 Acme Mathematics 8F. Quadratic EquationAn equation like 3x2 + 2x + 2 = 0 is called quadratic equation. It is second degree equation. x2 – 4 = 0 or x2 = 36 etc are called pure quadratic equations. A quadratic equation is an equation of the form ax2 + bx + c = 0, where a, b and c are real numbers and a ≠ 0.While solving the equation, remember the following points. If a = b is true then a + c = b + c is also true. If a = b is true then ac = bc is also true. If ab = 0 is true then a = 0 or b = 0 is also true. (It is zero property) If x2 = r2 is true then x = r or x = – r is also true.Solved ExamplesExample 1: Solve: x2 = 4 Solution : Here, x2 = 4or, x2 = (± 2)2or, x = ± 2 or, x = + 2 or – 2 Alternative method : x2 = 4or, x2 – 4 = 0 or, x2 – 22 = 0or, (x + 2) (x – 2) = 0 Either, x + 2 = 0 .............(i) x – 2 = 0 .............(ii) from (i), x + 2 = 0 or x = – 2 from (ii), x – 2 = 0 or x = + 2 Hence, x = + 2 or – 2 The value of x are + 2 or – 2. + 2 or – 2 are called the roots of x. Don't forget to write 'or' in LHS part while solvingIt is like ab = 0.a and b are two factors.
Acme Mathematics 8 175Example 2: Solve: x2 – 4x = 0 Solution : Here, x2 – 4x = 0or, x(x – 4) = 0 Either, x = 0 ...................(i)or, x – 4 = 0 .............(ii) from (ii), x – 4 = 0 or x = 4 Hence, x = 0 or 4 Example 3: Solve: x2 – 3x = 4 Solution : Here, x2 – 3x = 4 or, x2 – 3x – 4 = 0 or, x2 – (4 – 1)x – 4 = 0 or, x2 – 4x + x – 4 = 0 or, x(x – 4) + 1(x – 4) = 0 or, (x – 4) (x + 1) = 0 Either, x – 4 = 0 .............(i) OR, x + 1 = 0 .................(ii) from (i), x – 4 = 0 or, x = 4 from (ii), x + 1 = 0 or, x = – 1 Hence, x = – 1 or 4Example 4: Solve: 16x2 – 24x + 9 = 0Solution : Here, 16x2 – 24x + 9 = 0or, 16x2 – (12 + 12)x + 9 = 0 or, 4x(4x – 3) – 3(4x – 3) = 0or, (4x – 3) (4x – 3) = 0 or, (4x – 3)2 = 0
176 Acme Mathematics 8or, 4x – 3 = 0 → by removing square both sides. or, 4x = 3or, x = 34Hence, x = 34[Note: In this case roots are repeated]Example 5: Solve: x2 – 5 = 0 Solution : Here, x2 – 5 = 0 or, x2 – ( 5 )2= 0or, (x + 5) (x – 5) = 0 Either, x + 5 = 0 ................(i)or, x – 5 = 0 .................(ii)from (i), x = – 5from (ii), x = + 5Hence, x = ± 5Remember ! Every quadratic equation has two roots. Roots may be equal or unequal.Example 6: The length and breadth of rectangular land is (2x + 3) m and (x – 1) m. its area 817 m2.(i) Express area in algebraic form.(ii) Covert algebraic expression asequation.(iii) Calculate actual length and breadth of the land.Solution: (i) Length of the land (l) = (2x + 3) mbreadth of the land (b) = (x – 1) marea of land (A) = 817 m2Now, Area of land = length × breadth= (2x + 3) × (x – 1)= (2x2 – 2x + 3x – 3) m2(2x + 3) m(x – 1) m
Acme Mathematics 8 177(ii) Equation from the algebraic expression:Area of land = 817 m2So, 2x2 + x – 3 = 817or, 2x2 + x – 820 = 0(iii) Solving:2x2 + x – 820 = 0or, 2x2 + 41x – 40x – 820 = 0or, x(2x + 41) – 20(2x + 41) = 0or, (2x + 41) (x – 20) = 0Either, 2x + 41 = 0 .............. (i)Or, x – 20 = 0 ................... (ii)From (i) 2x = – 41or, x = – 20.5From (ii) x – 20 = 0or, x = 20Here, x = – 20.5 is not possible.So, x = 20.Hence,Length of the land = (2x + 3) m= (2 × 20 + 3) m = 43 mBreadth of the land = (x – 1) m= (20 – 1) m = 19 m.Classwork1. Solve the following: (a) x(x – 1) = 0 (b) y(y + 2) = 0 (c) (x + 1) (x + 2) = 0(d) (y – 2) (y + 3) = 0 (e) (2x + 3) (x – 2) = 0 (f) (2x + 1) (3x – 1) = 02. Solve for x: (a) x2 = 9 (b) x2 – 36 = 0 (c) 4x2 – 100 = 03. Solve each of the following quadratic equations: (a) x2 + x – 6 = 0 (b) x2 – x – 6 = 0 (c) x2 + 6x + 5 = 0
178 Acme Mathematics 8Exercise 4.121. Solve for x: (a) x2 – 49 = 0 (b) x2 – 3x = 0 (c) x2 + x = 0 (d) x2 + 5x = 0(e) x2 – 3x = 0 (f) 3x2 – 75 = 0 (g) 1 – x2 = 0 2. Solve each of the following quadratic equations: (a) x2 – x – 2 = 0 (b) x2 + 3x – 28 = 0 (c) x2 + 6x + 9 = 0 (d) x2 + 100 = 20x (e) x2 + 25 = 10x (f) x2 – 4x = 32(g) x2 + 12x = – 27 3. Solve each of the given quadratic equations: (a) 3x2 + 8x = 0 (b) 2x2 – 7x = 0 (c) 2x2 + 3x + 1 = 0 (d) 8x2 – 6x + 1 = 0 (e) 9y2 – 12y – 5 = 0 (f) 5y2 + 4y – 9 = 0 (g) 9y2 + 15y + 4 = 0 (h) 6y2 – 11y = 10 (i) 12z2 – 5z – 28 = 0 (j) 21z2 – 10 = z 4. Solve for x: (a) (x + 2)2 = 9 (b) (x – 3)2 + x2 = 9 (c) (x + 3)2 – 9 = 0(d) (3 – x)2 – 16 = 0 (e) 13 – x2 = 4 (f) x + 232 = 1625(g) 1x – 1 + 1x = 32 (h) 3x – 2x + 2 = 125. Study the following figures and answer the questions(a) (b)i. Express area in algebraic form.ii. Covert algebraic expression as equation.iii. Calculate actual length and breadth of the land.6. In the figure, a triangular garden is given. Two sides of the land are (x + 5) m and (x – 10) m. Its total area is 125 m2. How long is third side?(x + 4) m84 m2 (x – 4) m(2x + 5) m(x – 3) m765 m2(x + 5) m(x – 10) m
Acme Mathematics 8 1791. Two equations are given as : 2x + y = 7 and x + y = 4(a) What are the system of equations called?(b) Solve the above equations by using graph.2. Two equations are : 2x – y = 5 and x – y = 1(c) Solve the above equation by using graph.(d) What is the reciprocal of x?3. (a) Solve: x2 – 4 = 0(b) Solve graphically: x + y = 5 and x – y = 14. Do the followings.(a) Solve graphically: 2x – y = 3 and 3x + y = 17(b) At what value of x, the value of 3x2 + 4x + 1 becomes zero.5. Two equations are : 2x + y = 13 and x – y = 5.(a) Write the type of the equation?(b) Solve the above equation by graph.6. Simultaneous linear equation 2x + y = 8 and x + 2y = 7 are given.(a) Write the degrees of the given equations.(b) Solve the above equation by using graph.7. Study the given conditions.The cost of 4 pens and 3 pencils is Rs. 230 and the cost of 2 pens is 70 more than three times of cost of pencils.(a) write the equation.(b) calculate the cost of a pen and pencil graphically.8. (a) What are the roots of x2–25=0.(b) Solve : x2–2x–3=0Mixed Exercise
180 Acme Mathematics 8Evaluation EvaluationTime: 53 minutes Full Marks: 221. The sum of two numbers is 8 and their difference is 2. (a) Express the given statement in the mathematical form. [1] (b) Find the two numbers graphically.) [2]2. (a) State the type of equation x2 + 3x + 2 = 0 [1] (b) Solve the above equation by factorization method. [2]3. Two equations are given below. [1]x + y = 5 and 2x + y = 8 (a) What is simultaneous equations? [1] (b) Solve graphically the given equations. [2]4. Two equations are given below. 2x – y = 5 and x – y = 1 (a) Solve the above equation by using graph. [2](b) What is the reciprocal of x ? [1]5. (a) Solve the following equations using the graphical method. [2]x + 2y = 8 and x + y = 5(b) What type of equations are they ? [1]6. (a) Factorize: 2x2 – 50 [1](b) Solve by graphical method: x + y = 13 and 2x = y + 8 [2]7. (a) What is an equation ? [1](b) Solve the given equation graphically.x + y = 8, x – y = 4 [2]
Acme Mathematics 8 1815UNIT Geometry5.1 Angle and LinesA. Angles made by two linesREVISIONThere are many types of angles. Types of angles are based on their size or pairs. (a) Acute angle An angle whose measure is more than 0° and less than 90° is called an acute angle. In the figure ∠AOB is a acute angle as it is 68°.(b) Right angle An angle whose measure is exactly 90° is called right angle. In the figure ∠AOB is a right angle as it is 90°.(c) Obtuse angle An angle whose measure is more than 90°but less than 180° is called obtuse angle. In the figure ∠AOD is a obtuse angle as it is 105°. (d) Straight angle An angle whose measure is exactly 180° is called a straight angle. In the figure ∠AOD is a straight angle, as it is 180°. (e) Reflex angle An angle whose measure is more than 180° but less than 360° is called reflex angle. In the figure ∠AOB is a reflex angle as it is 240°. (f) Complete angle An angle whose measure is exactly 360° is called a complete angle. In the figure a complete angle is shown.(g) Zero angle An angle whose measure is 0° called zero angle. It has only a line segment. 68°O BA O BA90°O D105°AO D180°A240°O BA360°AOO A
182 Acme Mathematics 8(h) Complementary angle If the sum of two angles is 90°, the angles are called the complementary angle.In the figure, ∠AOC and ∠COB are complementary angles, as their sum is 90°.∠ABC and ∠PQR are complementary angles.(i) Supplementary angles If sum of the two angles is 180°, the angles are called supplementary angles. In the figure ∠AOC and ∠COB are supplementary angle as their sum is 180°. (j) Adjacent angles In the figure OB is common arm of ∠AOB and ∠BOC. Such angles are called adjacent angles. (k) Vertically opposite angles (VOA) When two lines intersect, the angles formed opposite to each other as shown in figure alongside. These opposite angles are called vertically opposite angles. ∠AOD and ∠COB are VOA. ∠AOC and ∠BOD are VOA.Experimental Verification (a) Vertically Opposite angles are equal to each other.Look at the given figure. AB and CD are intersecting lines. O is the point of intersection. ∠AOC, ∠BOD and ∠AOD, ∠COB are vertically opposite angles. Now, Measure the angles using protractor and complete the table given below.∠AOC ∠BOD ∠AOD ∠COB Result80° 80° 100° 100° ∠AOC = ∠BOD and ∠COB = ∠AODHence, vertically opposite angles (V.O.A) are equal to each other.O BCA60°30°AB C70°RPQ20°110°CA O B70°CABODCAOBCO BAD
Acme Mathematics 8 183(b) Sum of Adjacent angles in a straight line is 180°.In the figure alongside ∠AOC and ∠COB are adjacent angles on the straight line AB. Now, Measure the angles and complete the table given below.∠AOC ∠COB Result105° 75° ∠AOC + ∠COB = 180°Hence, the sum of adjacent angles in a straight line is 180°.(c) Sum of parts of straight angle is 180°. In the figure alongside ∠AOD, ∠DOC and ∠COB are the parts of straight angle AOB. Now, measure the angles and complete the table given below.∠AOD ∠DOC ∠COB Resut60° 90° 30° ∠AOD + ∠DOC + ∠COB = 180°(d) The sum of Complementary angle is 90°. (Do your Self) (e) The sum of Supplementary angle is 180°. (Do your Self)Classwork1. Study the given figure, AB and CD are two straight lines inter-secting at O. a, b, c and d are four angles. Fill in the blanks.(a) Two pairs of equal angles are .....................(b) Two pairs of adjacent angles are .................(c) Four pairs of supplementary angles are ...................(d) Two pairs of V.O.A are ......................2. Match the followingsTwo acute angles are supplementaryTwo right angles are complementaryTwo straight angles equalTwo angles 40° and 50° add up less than 180°VOA are add up 360°A O BCADCO BAOcdabC BD
184 Acme Mathematics 83. Write 'True' or 'False'.(a) Two obtuse angles can be complementary.(b) Two adjacent angles can be supplementary.(c) VOA can be 360°.(d) One obtuse angle and one right angle add up 180°.(e) Two straight lines intersect at two points.(f) The complement of an acute angle is an obtuse angle.Exercise 5.11. Name the pairs of vertically opposite angles in the following figures.(a) EFHGA I BC D(b) ACEDBFO2. Find the value of x, y, z in the following figures. (a)CDOABy 110°xz(b)RP SQz (7x + 3)° (4x + 24)°y(c)yO A BCDx + 506° x – 502° (d)A BCO3x° 3x°(e)A O B60°Dy°Cy°(f)CO A5z° Bz°
Acme Mathematics 8 185(g)9z°2ACO Bz°(h)O2x°2x° x°ADCB(i)BA4x°5x°3x°3x°DCO(j)A x 100°20°z ECDy B O(k)BAO CDx°y° (3 x – 21)3. Calculate the value of a, b, c, x, y and z.(a)x°x°b°a°3x°2(b)90° 40°z°2a° a°b°(c)(3x– 6)°90°z°z°(d) (e)70° a110° xz y 110° 110°(f)100° a 55° bc cC+154. Calculate the value of 'x' in each complementary angles of the following angles.(a) 63° and x° (b) (4x + 2)° and 20° (c) 9x° and 6x° (d) (3x – 25)° and (100 – x)° 5. Calculate the value of each supplementary angles.(a) 102° and y° (b) 4x° and 14x° (c) (x + 20)° and (2x – 20)° (d) (5x + 10)° and (148 – 4x)° 6. Study the picture. If points A, B and C lie on the same straight line, find the value of 'x'.(b+5)°x° a°y°b° 3b°(b – 3)°(2x + 4)° (4x – 12)°A B CD
186 Acme Mathematics 87. Angle AOB and COD are complementary to the same angle BOC.(a) Find the value of angle AOB. (b) Calculate angle BOC. 8. (a) From the figure write the relation between a, b and c.(b) If a + b = c, find the value of c. 9. Study the following figure.(a) Define complementary angles.(b) If ∠AOB and ∠COD both are supplementary angles for ∠BOC, find the value of ∠BOA. 10. Study the given figure then write the answer of the following questions:(a) Write the relation between the angles y° and (3x)°. (b) What is vertically opposite angle? Write. (c) Find the value of angles x and y. 11. In the given figure, ∠AOC = x° and ∠BOC = x4°, study the figure and answer the following questions.(a) What is a supplementary angle? (b) Find the value of ∠AOC and ∠BOC. 12. From the given figure, find the following:(a) Find the value of 'a'. (b) Find the value of 'b'. (c) If the value of 'a' is increased by 10° then what is the value of 'b'. ODBAC2x–13°x + 27°bca2x + 10y5x3xBCA Ox° x4 °a + 10°b 80°AB CDO (2x – 13)° (x + 27)°
Acme Mathematics 8 187B. Angles Formed by a transversal ActivityA BC DEFGG MADCOH NA BCE FXYZH DFigure (i) Figure (ii) Figure (iii)Study the given figures and discus which are the transversal in the given figure? Why ?In the figure-(i),AB and CD are two straight lines and the line EF is a transversal. Line EF intersects lines AB and CD at points G and H.In the figure-(ii),AB , CD and EF are three straight lines and the line GH is a transversalLine GH intersects lines AB , CD and EF at points X, Y and Z.In the figure-(iii),AB and CD are two straight lines and the line MN is a straight line.Line MN intersect lines AB and CD at points O. But in the figure-(iii) MN is not a transversal. Why?
188 Acme Mathematics 8Transversal: A transversal is a line that cuts two or more lines at different points. For examples;DACE FBAC DFHEGBAC DE FBThe line AB is a transversal.(i) Interior angles: Consider a transversal AB intersecting two lines CD and EF forming eight angles, ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7, and ∠8. The angles enclosed in the interior of the two lines on the same side of the transversal are the interior angles. In the above figure ∠4 and ∠5, ∠3 and ∠6 are pair of interior angles. (ii) Corresponding angle: The angles on the same side of the lines CD and EF and the same side of the transversal AB are corresponding angles. For example, The pair of angles ∠1 & ∠5, ∠4 & ∠8, ∠2 & ∠6, ∠3 & ∠7 are corresponding angles. (iii) Alternate angle: The angles in the interior and the opposite sides of the transversal AB are the alternate angle. For example, ∠4 and ∠6, ∠3 and ∠5 are alternate angles. (iv) Exterior angles: The angles in the exterior and on the same side of transversal are the exterior angles. For example, ∠1 and ∠8, ∠2 and ∠7 are pairs of exterior angles.CA2 14 35 68 7 FDBECA2 14 35 68 7 FDBECA2 14 35 68 7 FDBE
Acme Mathematics 8 189Remember !If a transversal intersect the two parallel lines then the following properties are satisfied. Interior angles on the same side of the transversal are supplementary.∠4 + ∠5 = 180° Corresponding angles are equal. ∠1 = ∠5, ∠2 = ∠6, ∠4 = ∠8 and ∠3 = ∠7 Alternate angles are equal. ∠4 = ∠6, ∠3 = ∠5 Exterior angles on the same side of the transversal are supplementary. ∠1 +∠8 = 180°, ∠2 +∠7 = 180°Experimental Verification (a) Alternate angles are equal. In the figure AB and CD are parallel lines. EF is a transversal. Two pairs of alternate angles are:(i) ∠AGH and ∠GHD(ii) ∠BGH and ∠GHC Now, Measure the angles ∠AGH, ∠ GHD, ∠BGH and ∠GHC in the figure and fill in the following table.∠AGH ∠GHD ∠BGH ∠GHC Result60° 68° 112° 112° ∠AGH = ∠GHD and = ∠BGH = ∠GHC Hence, the pair of alternate angles formed in parallel lines are equal to each other. (b) Corresponding angles are equal. In the figure AB and CD are parallel lines. EF is a transversal. Two pairs of corresponding angles are: (i) ∠1 and ∠5 (ii) ∠2 and ∠6 (iii) ∠4 and ∠8 (iv) ∠3 and ∠7 Now, Measure the angle in the figure and fill in the following table. ∠1 ∠2 ∠3 ∠4 ∠5 ∠6 ∠7 ∠8 Result70° 110° 70° 110° 70° 110° 70° 110° ∠1 = ∠5, ∠4 = ∠8, ∠2 = ∠6, ∠3 = ∠7Hence, The pair of corresponding angles formed in parallel lines are equal to each other.A58 74 3 BC DB62E1FC DBEGAHFC DBE4 1A 3 28 57 6
190 Acme Mathematics 8(c) The sum of Co-interior angles is 180°. In the figure AB and CD are parallel lines. EF is a transversal. Two pairs of co-interior angles are: (i) ∠1 and ∠2 (ii) ∠4 and ∠3 Now, measure the angles in the figure and fill in the following table.∠1 ∠2 ∠3 ∠4 ∠1 + ∠2 ∠3 + ∠4 Result110° 70° 110° 70° 180° 180° ∠1 + ∠2 = 180° and ∠3 + ∠4 = 180°Hence, the sum of pair of co-interior angles formed in the parallel lines are 180°.Remember 1. Two lines are said to be parallel if alternate angles are equal. 2. Two lines are said to be parallel if corresponding angles are equal. 3. Two lines are said to be parallel if sum of co-interne angle is 180°.Solved ExamplesExample 1 : In the adjoining figure AB||CD and EF is a transversal. If ∠2 = 70°, find the measure of remaining angles. Solution : Here, ∠2 = 70° Now,(i) ∠2 + ∠1 = 180° → a straight angle.or, 70° + ∠1 = 180° or, ∠1 = 180° – 70° ∴ ∠1 = 110° (ii) ∠1 = ∠3 → V.O.Aor, 110° = ∠3 ∴ ∠3 = 110° (iii) ∠4 = ∠2 → V.O.A.or, ∠4 = 70° (iv) ∠5 = ∠1 → Corresponding anglesor, ∠5 = 110° (v) ∠6 = ∠2 → Corresponding angles.or, ∠6 = 70° (vi) ∠7 = ∠5 → V.O.Aor, ∠7 = 110° FC DBE4 1A3 2CFDBE1 24 3A5 68 7
Acme Mathematics 8 191(vii) ∠8 = ∠6 → V.O.Aor, ∠8 = 70° Hence, the measure of remaining angles are:110°,110°,70°,110°,70°, 110° and 70°.Example 2 : In the figure AB||CD, ∠BAC = 60°, and ∠DCE = 50°. Find ∠ACB. Solution : Here, ∠BAC = 60° and ∠DCE = 50°. Now, (i) ∠ACD = ∠BAC → Alternate angle, AB||CD.or, ∠ACD = 60° (ii) ∠ACB + ∠ACD + ∠DCE = 180° → Straight angle.or, ∠ACB + 60° + 50° = 180° or, ∠ACB = 180° – 110° or, ∠ACB = 70° Hence, the measure of ∠ACB is 70°.Classwork1. Write 'True' or 'false'.(a) If a transversal cuts two parallel line, then the interior angles on the same side of the transversal are equal.(b) Two parallel lines can have two parallel transversal.(c) If a transversal cuts two parallel lines the corresponding angles are complementary.(d) If a transversal cuts two parallel lines, interior angles are equal.(e) If a transversal cuts two parallel lines, alternate angles are not equal.(f) If a transversal cuts two parallel lines, corresponding angles are equal.2. Fill in the blanks. If two parallel lines are intersected by a transversal, then(a) Interior angles are .......................(b) Exterior angles are ........................(c) .................... angles are equal.(d) Alternate angles are ......................(e) Sum of exterior angles is .......................(f) Sum of interior angles is .......................(g) Transversal is a line that cuts two or more lines at ................. points.B C EA D60°50
192 Acme Mathematics 8Exercise 5.21. In each figure line AB and CD are parallel. Find the value of 'x' and 'm'.(a)FCxDBEA 110°(b)C 2x4xDA B(c)C DA Bx – 30°120°(d)B2m + 10m + 60DA C (e)x + 30°x D A CB (f)120°EAx 70°BDFC2. Check whether line AB and CD are parallel or not in the following figures.(a)118°62° D A CB (b)105°85°AB DC (c)BCDA139°41°(d) BCDA65°95°(e)92°88°A BC D(f)B CA D75°75°3. Find the measures of angles x, y, z, a and b in the given figures if AB||CD||EF.(a)HGPQACEBDF R a bx yz60°(b)A65°70°axyC EB D(c)BDx + 30°2x + 10°ACXPQYX P XQ YP PX XYYQ QYQEPQFPXYPQ
Acme Mathematics 8 193(d)C D 60°yx z A B70°(e)(CE||BF)(f)60°50°CA Bb xy aD(AC||BD)(g) (h) (i)(j) (k) (l)A 20° a Bbx15° C D(m) (n)A Byz 4x–80C D105° x+12a b(o) A BC130° z5b60°xa yD4. (a) In the adjoining figure lines and angles are given. Show that AB||EF||CD.(b) In the figure, AB||CD and BE||DF. Find the value of x, y and z.70°BAFbaDCC DAxB30°20°105°40°y70° 86°A CBDxza 30°90°E x°z°A Cy°BD40°70°DA C EBy zb axAa P125° 60°yRS xb Q BC DC D45°E FA B25°20°155°BA EDC F40°xzyEEEEFGOH F GEFa90°60°yCFDA BEx 150°GH
194 Acme Mathematics 85. In the figure AB//CD and EF is the transversal.(a) Write one pair of VOA angle. (b) What are angles c and g called? (c) If b = 60°, find angle 'e'. 6. From the adjoining figure answer the following questions:(a) Write a pair of alternate angle.(b) On the basis of angles, name the triangle ABC. (c) Find the value of x.7. In the figure, PQ||RS, ∠QPO = 40° and ∠POR = 85°.(a) If a and b are co-interiors angles in parallel lines, what is the value of a + b? (b) What is alternate angle of ∠QRS?(c) Find the value of x and y.8. In the given figure AB\\\\CD,(a) Which is the corresponding angle of ∠ASR?(b) What is the sum of ∠ASR and ∠CRS? 9. AB and CD are parallel lines in given diagram.(a) Write down a set of adjacent angle.(b) Find the value of 'a' and 'x'. 10. PQ & RS are parallel lines in the given diagram and MP is transversal line.(a) Write a set of co-interior angle. (b) Measure each interior angle and find their sum. (c) Write a conclusion on this basis. ACBFEDbc de fg haE C BD Ax 55°70°85°QSRPO40°xyACBQPD RSACBEHD G2xa140°FPRQMPO SN
Acme Mathematics 8 19511. Write the answer of following question on the basis of given diagram.(a) What does the symbol represents?(b) Find the value of x. (c) Find the value of ∠ABC and ∠BCD. 12. Answer the following questions on the basis of given diagram.(a) What is the relation between corresponding angles between two parallel line. (b) Find the value of x. 13. Study the given figure then write the answer of the following questions:(a) Write the vertically opposite angle and alternate angle of ∠AGH.(b) Are ∠BGE and ∠GHD are equal? Give reason. (c) Find the value of angles x and y. 14. From the given figure, find the following:(a) What is the relation between x° and y°? (b) Find the value of x° and y°. (c) What is the value of 'z'? 15. Study the given figure and answer the following questions.(a) Write the corresponding angle of ∠AHE.(b) Find the value of x. ACB3x –10°x +10° D2x 30°40°CDABOACBEFD H30°Gxy70°x°z° y°ACBEFDx + 10°G 3x – 50°H
196 Acme Mathematics 816. Observe the given adjoining figure and answer the following questions.(a) What is the alternate angle of ∠AGH?(b) Find the value of 'x'. (c) What is the value of ∠QRS? 17. (a) Find the value of x and y from given figure.x°48° y°72°P Q AM N B C(b) In given figure AB||CD, if ∠BAC = 50° and ∠DCE = 60°, study the figure and find the sizes of unknown angles.Project WorkDraw two parallel lines and a transversal on a card board. Name the lines and angles. List the pair of angles and write their relations also. Present it to the 'Wall Magazine'.A GEBD2x80°3xFIH C120°130°P QRS Taz y50°60°A DE B C
Acme Mathematics 8 197EvaluationAB CFEx° y° 150°Dz°AB C D3aax 130°EC FA BG Dy x60° 70°(3x+13)°EA(2x-23)°BC DFGPCQA BD(2x+10)° RS (x+40)°P SQ 3x−5 Rx+7(3y − 5)°A(2y+10)°BC DTime: 58 minutes Full Marks: 241. In the given figure: AB//FE, ∡ACD=150°, ∡BAC=90°, ∡???=?°, ∡???=?° and ∡???=?° are given. (a) Write the corresponding angle of ∡???. [1] (b) Find the value: x & y . [2] (c) The measurement of which angle is greater if we compare the angle ∡?????? ∡???. Find out. [1] 2. (a) Find the value of x from the given figure. [2] (b) Find the value of a from the adjoining figure. [2]3. In the figure, AB//CD(a) What is the sum of ∠CEF, ∠ECF and ∠EFC ?Write with reason. [1](b) What is the value of ∠CEF ?Find it. [1](c) Find the value of x and y. [2]4. In the adjoining figure, CD is parallel to AB. EA and FG are two straight lines. (a) Write the corresponding angle of ∠DCE. [1](b) Find the value of x from the figure. [2](c) For what value of ∠FGD the straight line segments FG and EA will become parallel ? [1]5. In the adjoining figure, AB||CD. If ∠PRB = (2x + 10)o and∠RSD = (x + 40)o ,then answer the following questions.(a) Write a pair of alternate Angle. [1](b) Are angle ∠PRB and angle ∠RSD equal ? Give reason. [1](c) Find the value of x. [2]6. Quadrilateral PQRS is parallelogram .(a) Find the value of x. [1](b) What is the name of quadrilateral where adjacent sides are equal and each angle is 90o. [1](c) Find the value of y. [2]
198 Acme Mathematics 8A. Triangles a. Verification of Properties of sides and angles of Triangles. (i) The sum of the three interior angles of a triangle is 180°. Draw three triangles ABC of different sizes. Using protractor, measure the angles and complete the table given below.AB (i) C B (ii) CAB (iii) CAFig. No. ∠A ∠B ∠C ∠A + ∠B + ∠C Result(i) 64° 68° 48° 180°(ii) 50° 90° 40° 180° The sum is 180°(iii) 38° 110° 32° 180°Hence, The sum of three interior angles of a triangle is 180° (ii) All interior angles of an equilateral triangle are equal. Each angle is 60°Draw three equilateral triangles of different sizes.B CA(i)BAC(ii)A CB(iii)In each ΔABC, sides AB, BC and AC are equal. ∠BAC, ∠ABC and ∠BCA are interior angles of the triangle ABC.Measure ∠BAC, ∠ABC and ∠BCA and tabulate them as follows:Figure ∠BAC ∠ABC ∠BCA Result(i) 60° 60° 60°(ii) 60° 60° 60° ∠BAC = ∠ABC = ∠BCA = 60°(iii) 60° 60° 60°Hence, all interior angles of an equilateral triangle are equal. Each angle is 60°.5.2 Plane Figure
Acme Mathematics 8 199(iii) Each base angle of a right angled isosceles triangle is 45°. Draw three right angled isosceles triangles ABC of different sizes using protractor, scale and compass. CB (i) A (ii) ABC(iii)ABCIn the above figures ∠B = 90°, sides AB and BC are equal and ∠A and ∠C are base angles. Now, measure the angles A and B and complete the table.Figure ∠A ∠C Result(i) 45° 45°(ii) 45° 45° ∠A = ∠C = 45°(iii) 45° 45°Hence, Each base angle of right angled isosceles triangle is 45°. (iv) The exterior angle of a triangle is equal to the sum of its opposite interior angles. Draw three triangle of different sizes. Produce the side AB to X in each triangle. ∠CBX is exterior angle. (i)CA B XCA (ii) B X A (iii) B XCFigure ∠A ∠C ∠A+ ∠C ∠CBX Result(i) 61° 57° 61°+ 57° 118°(ii) 90° 45° 90°+ 45° 135° ∠A + ∠C = ∠CBX(iii) 125° 27° 125°+ 27° 152°Hence, the exterior angle of a triangle is equal to the sum of its opposite interior angles.
200 Acme Mathematics 8(v) Base angles of an isosceles triangle are equal. Draw three isosceles triangles ABC of different sizes, using scale and compass.AB (i) CAB (ii) CCA (iii) BIn the above triangles side AB and AC are equal ∠B and ∠C are base angles. Now, measure the angles B and C and complete the table. Figure ∠B ∠C Result(i) 58° 58°(ii) 60° 60° ∠B = ∠C(iii) 45° 45°Hence, base angle of an isosceles triangle are equal. Solved ExamplesExample 1 : Find the value of x, y and a in the adjoining figure. Solution : Here, ∠CAB = 40°, AC = BC Now, (i) ∠ABC = ∠CAB → base angle of isosceles ∆ABC. y = 40° → given ∠CAB = 40°∴ y = 40°(ii) ∠CAB + ∠ABC + ∠BCA= 180° → Sum of interior angles of ∆ABC. or, 40° + y + x = 180° or, 40° + 40° + x = 180° → y = 40°or, 80° + x = 180°or, x = 180 – 80°or, x = 100° (iii) x + a = 180° → Straight angle or, 100° + a = 180° → x = 100°or, a = 80° A BCx40° yaDE