The Leading Maths - 9

Algebra Algebra Example-5 Find the LCM of a2 + 6a + 5, a2 + 8a + 15, and a2 + 4a + 3. Solution: Here, 1st Exp. = a2 + 6a + 5 = a2 + 3a + 2a + 5 = a(a + 1) +5(a + 1) = (a + 5)(a + 1) 2nd Exp. = a2 + 8a + 15 = a2 + 5a + 3a + 15 = a(a + 3) + 5(a + 3) = (a + 5)(a + 3) 3rd Exp. = a2 + 4a + 3 = (a + 3) (a + 1) ∴ LCM = (a + 5)(a + 3)(a + 1). Example-6 Find the LCM of (x – y) (y – z), (y – x) (y – z) and (z – x) (z – y). Solution: Here, 1st expression = (x – y) (y –z) 2nd expression = (y – x) (y – z) = – ( x – y) (y – z) 3rd expression = (z – x) (z – y) = –(z – x)(x – z) = – (x – y) (y – z) (z – x) ∴ LCM = – (x – y) (y – z) (z – x). Example-7 Find the LCM of x2 – (y + z)2 , y2 – (z + x)2 and z2 – (x + y)2 . Solution: Here, 1st expression = x2 – (y + z)2 = (x – y – z) (x + y + z) 2nd expression = y2 – (z +x)2 = (y – z – x) (y + z + x) 3rd expression = z2 – (x + y)2 = (z – x – y) (z + x + y) ∴ LCM = (x – y – z) (x + y + z)(y – z – x) (z – x – y). PRACTICE 10.2 Read Think Understand Do Keeping Skill Sharp 1. (a) What is the LCM of algebraic expressions ? (b) If there is no common factor in the given algebraic expressions, what is the LCM ? (c) Write the relation between HCF, LCM, first expression and second expression. (d) What is the LCM of (2b + a) (3a – 1) and (2b + a) (3a – 1) ? Check Your Performance 2. Find the Lowest Common Multiples (LCM) of: (a) x – 4 and (x – 4) (x + 4) (b) a2 – 4 and a2 + 2a (c) x2 – 4x + 4 and x2 – 4 (d) x2 – y2 and (x – y)2 (e) (x – 2)2 and x3 – 8 (f) 3ab(a – b)3 and 2a2 b(a –b)2 250 Allied The Leading Mathematics-9 Algebra 251

Algebra Algebra 3. Find the lowest common multiples (LCM) of: (a) 6x2 + 19x + 15 and 9x2 + 9x – 10 (b) a2 – a – 2 and a2 + a – 6 (c) x4 + 4 and 2x3 – 4x2 +4x (d) x4 + 2px3 + p2 x2 and x3 – p2 x 4. Find the lowest common multiples (LCM) of: (a) 24(y3 – x3 ), 32x2 (y2 – x2 ) and 48(xy3 – x3 y2 ) (b) x2 + x – 6, x2 + 4x + 3 and x2 – x – 2 (c) a3 – b3 , a2 – b2 and a4 – b4 (c) x2 + x + 1, x6 – 1 and x3 – 1 5. (a) Find the LCM of the expressions a2 – 12a + 35 and a2 – 8a + 7 with the help of their HCF. (b) m2 – 5m – 14 is an expression. Find out anothersimilar expression such that their HCF is (m – 7) and LCM is m3 – 10m2 + 11m + 70. 6. Find the HCF and LCM of: (a) 4(a2 – 4), 6(a2 – a – 2) and 12(a2 + 3a – 10) (b) 2x2 – 3xy – 2y2 , 6x2 + xy – y2 and 3x2 – 7xy + 2y2 (c) a(c + a) – b(b + c), b(a + b) – c(c + a) and c(b + c) – a(a + b) (d) p2 + q2 + 2pq – 1, q2 – p2 + 2q + 1 and p2 – q2 + 2p + 1 (e) 6x2 – 5x – 4, 8x2 + 2x – 15 and 12x2 – 43x + 35 2. (a) (x – 4) (x + 4) (b) a(a + 2) (a – 2) (c) (x + 2) (x – 2)2 (d) (x + y)(x – y)2 (e) (x – 2)2 (x2 + 2x + 4) (f) 6a2 b(a – b)3 3. (a) (3x + 5) (3x – 2) (2x + 3) (b) (a + 1) (a – 2) (a + 3) (c) 2x(x2 + 2x + 2) (x2 – 2x + 2) (d) x2 (x + p)2 (x – p) 4. (a) 96xy2 (y + x) (y – x) (y2 + yx + x2 ) (b) (x + 1) (x – 1) (x + 3) (c) (a – b) (a + b) (a2 + b2 ) (a2 + ab + b2 ) (d) (x – 1) (x + 1) (x2 + x + 1) (x2 – x + 1) 5. (a) (a – 1) (a – 5) (a – 7) (b) m2 – 12m + 35 6. (a) 2(a – 2), 12(a + 1) (a + 2) (a – 2) (a + 5) (b) 1, (2x + y) (3x – y) (x – 2y) (c) a + b + c, (a + b + c) (a – b) (b – c) (c – a) (d) p + q + 1, (p + q + 1) (p + q – 1) (q + 1 – p) (p – q + 1) (e) 1, (2x + 1) (2x + 3) (3x – 4) (3x – 7) (4x – 5) Answers Project Work Draw a chart finding the HCF and LCM, and their relations and properties in a chart paper. 252 Allied The Leading Mathematics-9

Algebra Algebra 11.1 Solving Simultaneous Equations by Substitution Method At the end of this topic, the students will be able to: ¾ solve the simultaneous linear equations of two variables by substitution method. Learning Objectives Linear Equation An equation or inequality in one variable (i.e., condition in one variable) is said to be linear if the highest power of the variable is one. The general form of linear equation is ax + b = 0, which gives a straight line perpendicular to the x-axis or parallel to the x-axis in the graph. For example, x – 1 =0, 3x + 4 = 0, x = 4, etc. are the linear equations of one variable. An equation involving two variables is called the linear equation in two variables. Its general form is ax + by + c = 0. Each linear equation in two variables defines a straight line. x + 2y = 4, 3x – y + 1 = 0, etc. are the examples of the linear equations in two variables. Simultaneous Equations A set of linear equations, which has common solution, is called simultaneous equations. The general forms of the simultaneous equations of two variables x and y are a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. For example, 2x + 3y = 2 and x – y = 1 are the simultaneous equations. Solution of Simultaneous Linear Equations We have seen that a pair of linear equations in two variables may have one solution, a set of many solutions and no solution. We shall henceforth assume that a system of two linear equations in two variables has a unique solution. Various methods of solving systems of linear equations are known. We shall consider the following three methods: a) Graphical method b) Substitution method c) Elimination method Solving Simultaneous Equations by Substitution Method In this method, one of the two variables x and y is first expressed in terms of the other variable by considering the simpler of the two equations; and then the expression so obtained is substituted in the other equation. This procedure reduces the second equation to a linear equation in one variable. On solving this equation, we get the value of one of the variables. This value, when substituted in any one of the previous equations, yields the value of the other variable. The two values taken together constitute the required solution. CHAPTER 11 LINEAR EQUATIONS PB Allied The Leading Mathematics-9 Linear Equations 253

Algebra Algebra Example-1 Solve each of the following pairs of linear equations by substitution method: (a) y = 2x and x + y = 3 (b) x = 3y and x + y = 4 Solution: (a) Put y = 2x in x + y = 3, we get x + 2x = 3 or, 3x = 3, Dividing both sides by 3, we get x = 1. Putting x = 1 in y = 2x, we get y = 2. So, the required solution is x = 1 and y = 2. Example-2 Solve the following system of linear equations by substitution method: x + y = 2 and 2x + 3y = 5. Solution: Here, we begin with the first equation. Transpose x to the right to obtain y = 2 – x … … … … (i) Putting this value of y in the second equation 2x + 3y = 5 … … … … (ii) Now, 2x + 3 (2 – x ) = 5 or, 6 – x = 5 Example-3 Solve the following equations by substitution method: 2x + 3y = 5 and 5x + 6y = 11 Solution: Consider the first equation 2x + 3y = 5 … … … … (i) Transpose 2x to the right and then divide both sides by 3 to get or, 3y = 5 – 2x or, y = 5 – 2x 3 … … … … (ii) Substituting the value of y from (ii) in the second equation 5x + 6y = 11, Now, 5x + 6 × 5 – 2x 3 = 11 or, 5x + 10 – 4x = 11 or, x + 10 = 11 or, x = 11 – 1 So, x = 1. Again, putting this value x = 1 in eqn (ii), we get y = 5 – 2 × 1 3 = 5 – 2 3 = 3 3 = 1 Hence, the required solution is x = 1 and y = 1. (b) Put x = 3y in x + y = 4, we get 3y + y = 4, or, 4y = 4, Dividing both sides by 4, we get y = 1. Putting y = 1 in x = 3y, we get x = 3. So, the required solution is x = 3 and y = 1. or, –x = –1; i.e., x = 1. Substituting the value x = 1 in (i), We get y = 1. Hence, the required solution is; x = 1 and y = 1. 254 Allied The Leading Mathematics-9 Linear Equations 255

Algebra Algebra Example-4 Solve and test: 5 x + 3 y = 1 and 1 x – 3 y = 2 Solution: Here, 5 x + 3 y = 1 .............. (i) 1 x – 3 y = 2 ..............(ii) Put 1 x = a and 1 y = b, then 5a + 3b = 1 ........... (iii) a – 3b = 2 ............. (iv) From eqn (iv), a = 2 + 3b ........... (v) Substituting the value of a in eqn (iii), we get 5(2 + 3b) + 3b = 1 or, 10 + 15b + 3b = 1 or, 18b = 1 – 10 or, b = – 9 18 = – 1 2 Substituting the value of b in eqn (v), we get a = 2 + 3 – 1 2 = 4 – 3 2 = 1 2 Hence, a = 1 2 ∴ 1 x = 1 2 ⇒ x = 2 b = – 1 2 ∴ 1 y = – 1 2 ⇒ y = – 2 "Alternatively" From eqn (i), 5 x = 1 – 3 y or, 5 x = y – 3 y or, x = 5y y – 3 ............. (iii) From eqn (ii), 1 5y y – 3 – 3 y = 2 or, y – 3 5y – 3 y = 2 or, y – 3 – 15 5y = 2 or, y – 18 = 10y or, – 9y = 18 or, y = 18 – 9 = – 2 Putting the value of y in eqn (iii), we get x = 5y y – 3 = 5(– 2) – 2 – 3 = – 10 – 5 = 2 ∴ x = 2 and y = – 2 Test Put x = 2 and y = – 2 in eqn (i) and (ii), we get 5 2 + 3 – 2 = 1 or, 5 – 3 2 = 1 or, 2 2 = 1 or, 1 = 1, true. And 1 2 – 3 – 2 = 2 or, 1 + 3 2 = 2 or, 4 2 = 2 or, 2 = 2, true. 254 Allied The Leading Mathematics-9 Linear Equations 255

Algebra Algebra Example-5 Solve: 15 x + y – 2 x – y = 1 and 10 x + y + 3 x – y = 5 Solution: Here, 15 x + y – 2 x – y = 1 .................. (i) or, 10 x + y + 3 x – y = 5 .................. (ii) Put 1 x + y = a and 1 x – y = b, then or, 15a – 2b = 1 ............. (iii) or, 10a + 3b = 5 ............ (iv) From eqn (iii), 15a – 1 = 2b or, 15a – 1 2 = b .......... (v) Putting the value of b in eqn (iv), we get 10a + 3 15a – 1 2 = 5 or, 20a + 45a – 3 2 = 5 or, 65a – 3 = 10 or, 65a = 10 + 3 or, a = 13 65 = 1 5 ∴ 1 x + y = 1 5 ⇒ x + y = 5 .......... (vi) Putting the value of a in eqn (v), we get b = 15 × 1 5 – 1 2 = 3 – 1 2 = 2 2 = 1 ∴ 1 x – y = 1 ⇒ x – y = 1 .......... (vii) From the eqn (vi), y = 5 – x ......... (viii) From eqn (vii), x – (5 – x) = 1 or, x – 5 + x = 1 or, 2x = 1 + 5 or, x = 6 2 = 3 From eqn (viii), y = 5 – 3 = 2 ∴ x = 3 and y = 2. 256 Allied The Leading Mathematics-9 Linear Equations 257

Algebra Algebra PRACTICE 11.1 1. (a) What is the value of x if y = 0 and 2x + y = 3? (b) What is the value of x if y = 2 and x + y = 5? 2. Solve the following equations by substitution method: (a) y = x and y = 2x + 1 (b) y = 5x and y + 3x = 16 (c) y – 2x = 0 and y + x – 9 = 0 (d) 2y = x + 3 and 3x + y + 2 = 0 3. Solve the following equations by substitution method: (a) x = 1 and x + y = 1 (b) x = 5y and 7y = 2x + 3 (c) 2x = y and 2y + 5 = 3x (d) 5x = y + 6 and 2x + y = 8 4. Use substitution method to solve the following equations: (a) 2x + y = 8 and 3x – y = – 5 (b) x + 2y = 8 and 3x + y = 15 (c) 2x + 3y = 13 and x + 2y = 7 2 3 (d) 2x + 3y = 6 and y = – 1 3 x – 2 (e) x + y = 3 and 2x – 2y = 10 (f) x + y = 1 and x – 4y = – 5 5. Use substitution method to solve the following equations and test: (a) x + y = 9 (x – y) and 2x + y = 3 (2x – y) – 2 (b) 3x – 2y 5 = 2x – y 3 = 1 (c) x 2 + y 3 = 11 3 and x 3 – y 2 = 1 (d) 7x 3 + y 3 + 1 = 0 and x + y 2 + 3 2 = 0 (e) x – 1 y + 1 = 3 4 and y – 2 x + 2 = 3 4 6. Solve by using substituting method and test: (a) x 3 – 2 y = 14 15 and x 4 + 3 y = 8 5 (b) 11 x – 7y = 18 and 2 x – 3y = 5 (c) 8 x – 9 y = 1 and 10 x + 6 y = 7 (d) 4 x – 3 y = 3 and 8 x – 7 y = 1 (e) x + y xy = 5 6 and y – x xy = 1 6 (f) xy x + y = 1 2 and xy x – y = 1 6 (g) 9 x + y – 4 x – y = 1 and 3 x + y + 2 x – y = 2 (h) 1 x + y + 1 x – y = 7 and 2 x + y + 4 x – y = 17 2. (a) – 1, – 1 (b) 2, 10 (c) 3, 6 (d) – 1, 1 3. (a) 1, 0 (b) – 5, – 1 (c) – 5, – 10 (d) 2, 4 4. (a) 3 5, 34 5 (b) 22 5 , 9 5 (c) 3, 7 3 (d) 12, – 6 (e) 4, – 1 (f) –1 5 , 6 5 5. (a) 5 2, 2 (b) 1, – 1 (c) 6, 2 (d) 0, – 3 (e) 10, 11 6. (a) 4, 5 (b) 1, – 1 (c) 2, 3 (d) 2 9, 1 5 (e) 2, 3 (f) – 1 2 , 1 4 (g) 5 2 , 1 2 (h) 14 33 , – 8 33 Answers 256 Allied The Leading Mathematics-9 Linear Equations 257

Algebra Algebra 11.2 Solving Simultaneous Equations by Elimination Method At the end of this topic, the students will be able to: ¾ solve the simultaneous linear equations of two variables by elimination method. Learning Objectives Given a pair of linear equations in two variables, it is possible to combine them into a single equation in which one of the variables is absent or eliminated. In practice, this can be achieved in the following way: Consider the following pair of linear equations in two variables 2x + 3y = 2 .......................(i) 3x + 4y = 1 .......................(ii) Multiply both sides of equation (i) by 3, and those of equation (ii) by 2 to find 6x + 9y = 6 .........,,,,,,.......(iii) 6x + 8y = 2 .......................(iv) Subtract (iv) from (iii) to get (9 – 8)y = 6 – 2 ∴ y = 4 ............... (v) From equation (i), 2x + 3 × 4 = 2 or, 2x = 2 – 12 or, x = – 10 2 = – 5 Proceeding in the same manner, we may eliminate y and get the value of x. But thisis not necessary. We may put the value of y in any one of the above equations to find the value of x. For ready reference, we simply record the value of x. So, the required solution is; x = – 5 and y = 4 Example-1 Solve the following equations by the method of elimination: x + y = 5 and x – y = 3. Solution: Here, we eliminate y by adding the two equations x + y = 5 … …… … (i) x – y = 3 … … … … (ii) Thus, 2x = 8, i.e., x = 4. Putting this value x = 4 in (i), we get y = 1 So, the required solution set is (4, 1) or, x = 4 and y = 1. LCM of the coefficient 2 and 3 of x is 6. 258 Allied The Leading Mathematics-9 Linear Equations 259

Algebra Algebra Example-2 Solve the following equations by eliminating x, 7x – 2y = 5 and x + 3y = 4. Solution: Multiply the first equation by 1 and the second equation by 7. Then the two equations may be written as 7x – 2y = 5 ..........................(i) 7x + 21y = 28 ....................(ii) Subtract (i) from (ii) to find 23y = 23 or, y = 1 ................................(iii) Putting y = 1 in (i), we get 7x – 2 = 5, 7x = 7 or, x = 1. That is, x = 1, y = 1. Example-3 Solve the following equations by elimination method: 3x – 4y = 10 and 2x + 3y = 1. Solution: Multiply the first equation by 3 and the second equation by 4, and add the resulting equations: 9x – 12y = 30 … …… (i) 8x + 12y = 4 … ... … (ii) That is, 17x = 34 Hence, x = 2 … … … … … (iii) Putting x = 2 in 2x + 3y = 1, we get 4 + 3y = 1 or, 3y = – 3 , or, y = – 1. Hence, x = 2 and y = –1. Example-4 Solve: 14 x + y – 3 x – y = – 1 and 21 x + y – 1 x – y = 2 Solution: Here, 14 x + y – 3 x – y = – 1 ............. (i) 21 x + y – 3 x – y = 2 ................ (ii) Multiplying eqn . (ii) by 3 and subtracting with eqn . (i), we have, 63 x + y – 3 x – y = 6 (–) – 14 x + y – 3 x – y = – 1 49 x + y = 7 or, x + y = 49 7 or, x + y = 7 ............ (iii) + + 258 Allied The Leading Mathematics-9 Linear Equations 259

Algebra Algebra Substituting the value of x + y in eqn (i), we get 14 7 – 3 x – y = – 1 or, 2 + 1 = 3 x – y or, 3 = 3 x – y or, x – y = 3 3 or, x – y = 1.............. (iv) PRACTICE 11.2 1. (a) What is the value of y in the equations x = 2 and x + y = 3? (b) Write the value of x in the equations y = 1 and x – y = 2. 2. Solve the following equations by elimination method: (a) x + y = 3 and x – y = 1 (b) x + 2y = 3 and – x + y = 3 3. Solve the following equations by elimination method: (a) x + 2y = 8 and 2x + 3y = 13 (b) 3x – y = 4 and x + 2y = 13 (c) 3x – 2y = 7 and 5x + y = 3 (d) 3x + 4y = 1 and 3x + 2y = 1 4. Solve the following equations by elimination method and test: (a) 4x – 3y = 3 and 8x – 7y = 1 (b) 4 = 2x - y and y = – 1 2 x + 1 (c) 2x – 3y = 0 and 6x + 6y = 5 (d) 3x – y = 1 and x – 2y = – 8 5. Solve the following equations by elimination method: (a) 1 x + 1 y = 2 and 2 x – 1 y = 1 (b) 1 x + 1 y = 3 and 2 x + 3 y = 8 (c) 2x + 1 y = 4 and x – 1 y = 5 (d) 4x + 9y – 5xy = 0 and x + 2y xy = 2 (e) x + y = xy and x + 9y + 5xy (f) 2 x + y + 1 x – y = 5 6 and 1 x + y + 4 x – y = 1 2. (a) 2, 1 (b) – 1, 2 3. (a) 2, 3 (b) 3, 5 (c) 1, – 2 (d) 1 3, 0 4. (a) 9 2, 5 (b) 2, 0 (c) 1 2, 1 3 (d) 2, 5 5. (a) 1, 1 (b) 1, 1 2 (c) 3, – 1 2 (d) – 1 3, 1 8 (e) 2, 2 (f) 9 2, – 3 2 Answers Adding eqn (iii) and (iv), we get 2x = 8 ⇒ x = 4 From eqn (iii), 4 + y = 7 or, y = 7 – 4 ⇒ y = 3 ∴ x = 4 and y = 3. 260 Allied The Leading Mathematics-9 Linear Equations 261

Algebra Algebra 11.3 Verbal Problems on Simultaneous Equations in Two Variables At the end of this topic, the students will be able to: ¾ solve the verbal (word) problems in the case of simultaneous equations in two variables. Learning Objectives In everyday life, we often find one quantity related to another. For instance, the age difference between a father and his son, the amount of gasoline used by two cars, the sum of the principal and interest and so on. One of the best ways to represent mathematically how a quantity is related to another is to use equations. Depending upon the number of quantities, we arrive at equations involving one, two or more variables. We illustrate with concrete examples how we come across equations while dealing with word problems and then indicate how such problems can be solved. Consider the word problem on the simultaneous equations in two variables as; The price of 2 kg of carrots and 3 kg of cucumbers is Rs.200 and the difference between the rates per kg of them is Rs.10. Find their. At first, we suppose that x and y are the rates per kg of carrot and cucumber. Then by the given problem; 2x + 3y = 195 …...… (i) and x – y = 10 …...... (ii) Both equation (i) and (ii) represent straight line graph. Solving these equations is to find the coordinates of the point of intersection between the two lines represented by the equations (i) and (ii), which are the rates per kg of carrot and cucumber by x and y respectively. Therefore, the basic idea on the elementary step to solve simultaneous equations in two or more variables is to try by graphical method. If the two lines in the graph intersect, there exists a common solution between the two equations and if the two lines are parallel, the equations are not solvable simultaneously. Graphing straight line graph needs a table of several points lying on that line. Let's make table of values for the equations given above. 2x + 3y = 195 x – y = 10 x 15 30 45 x 10 20 30 y 55 45 35 y 0 10 20 Graphing these equations, we obtain; The point P (45, 35) is the point of intersection between the two lines. Hence, x = 45 and y = 35 is the solution of the simultaneous equations. Here, the values x = 45 and y = 35 satisfy both equations (i) and (ii). Therefore, the simultaneous equations in two variables have typical importance in solving problems that involve two variables. Also, they can solve the substitution or putting method and cancellation or elimination method. Consider the following example. O 20 20 x – y = 10 x – y = 10 (45, 35) 40 60 40 60 260 Allied The Leading Mathematics-9 Linear Equations 261

Algebra Algebra Example-1 A number is twice of the other and their sum is 30. (a) Find the numbers. (b) Find the difference between them. Solution: (a) Let x be a number, the other number 2x. By given condition, Then, x + 2x = 30 or, 3x = 30 ∴ x = 10 Thus, the required two numbers are 10 and 2 × 10 = 20. (b) The difference between 20 and 10 is 20 – 10 = 10. Example-2 The age of father at present is double of the age of his son. 16 years ago father’s age was three times to that of his son’s age. (a) How old are the father and son now? (b) What was the age of the father when his son was born? Solution: (a) At present, let the father’s age be x and his son’s age be y. By question, x = 2y …………….. (i) Sixteen years ago the father's and son's ages were respectively, x – 16 and y – 16 Then, x – 16 = 3(y – 16) …..… (ii) (b) Hence, the age of the father was 64 – 32 = 32 years when his son was born. Example-3 Kranti’s present age is three-fifths of her older brother’s present age. After 14 years from now, the difference between their ages is again 14. Find their ages. Solution: To translate the word problem into mathematical equation, we denote the present ages of Kranti and her older brother by x years and y years respectively. Then, by the given conditions, we arrive at the following equations: x = 3 5 y or, 5x – 3y = 0 … ..... … (i) and (y +14) – (x + 14) = 14 or, y – x = 14 …... … … (ii) × 3 Multiplying the eqn (ii) by 3 and adding with the eqn (i), we get Hence, Kranti’s age is 21 years and her older brother’s age is 35 years. Also, we can suppose that; x = 2y ........... (i) x + y = 30 ..... (ii) Now, substituting x = 2y in equn (ii), 2y – 16 = 3(y – 16) or, 2y – 16 = 3y – 48 or, 3y – 2y = 48 – 16 or, y = 32 …....… (iii) By substitution, x = 2 × 32 = 64 Hence, the father’s age is 64 years and the son’s age 32 years.. 3y – 3x = 42 (+) 5x – 3y = 0 2x = 42 ∴ x = 21. Putting the value of x in the eqn (ii), we get y – 21 = 14 or, y = 14 + 21 = 35. 262 Allied The Leading Mathematics-9 Linear Equations 263

Algebra Algebra Example-4 A two-digit number is such that the sum of its digits is 5. If their place is changed and 9 is added to that, you get the original number. (a) Find the number. (b) When its digits are reversed, what changes will be in the original number? Solution: (a) Let the two-digit number be 10x + y, where x is in tens place and y is in ones place. By question; x + y = 5 ……….. (i) and 10x + y + 9 = 10y + x or, 9x – 9y = – 9 or, y – x = 1 ……… (ii) Hence, the number is 10y + x = 10 × 3 + 2 = 32. (b) When its digits are reversed, it will be 23, which is less than 32 by 9. Example-5 If 3 is added to the numerator of a fraction, the value of the fraction becomes 1. When 3 is added to the denominator of the fraction, the value of the fraction becomes 1 4 . What is the value of the fraction? Find it. Solution: Let x y be a fraction then by the question, x + 3 y = 1 or, y = x + 3 ......... (i) and x y + 3 = 1 4 or, y + 3 = 4x or, y = 4x – 3 ......... (ii) From the equations (i) and (ii), we get x + 3 = 4x – 3 or, 4x – x = 3 + 3 or, 3x = 6 ∴ x = 2 Substituting the value of x in the equation (i), we get y = 2 + 3 = 5. Hence, the required fraction is 2 5 . Example-6 In 2070 BS, Shova’s age was 4 times her daughter Aarisha’s age and in 2088 BS, her age will be only 2 times older than her daughter. (a) What are the ages of Shova and her daughter? (b) What is the year of the birth of the daughter? Solution: (a) Let x and y be the ages of Shova and her daughterAarisha in 2070BS respectively. Then by the question, in 2070 BS; x = 4y .......... (i) and in 2088 BS after 18 years; x + 18 = 2(y + 18) or, 4y + 18 = 2y + 36 [ Putting the value of x.] or, 4y – 2y = 36 – 18 or, 2y = 18 ∴ y = 9 and x = 4 × 9 = 36 Hence, the ages of Shova is 36 years and her daughter, 9 years. (b) Thus, the birth year of her daughter Aarisha is 2070 – 9 = 2061 BS. Adding eqn (i) and (ii), we get x + y = 5 –x + y = –1 2y = 6 or, y = 3 …....... (iii) From (iii) and (i) by substitution, x + 3 = 5 or, x = 2. 262 Allied The Leading Mathematics-9 Linear Equations 263

Algebra Algebra Example-7 8 years hence a teacher will be 1 1 5 times as old as his student. 6 years ago, the teacher’ age was as same as his student's will be 8 years hence. Find their ages. Solution: Let x and y be the ages of the teacher teacher and his student student respectively. respectively. Then by the given; x + 8 = 1 1 5 ( y + 8) or, x + 8 = 6 5 ( y + 8) or, 5x + 40 = 6y + 48 or, 5x – 6y = 48 – 40 or, 5x – 6y = 8 ............ ............ (i) Multiplying Multiplying the eqn (ii) by 5 and subtracting subtracting from the eqn (i), we get 5x – 6y = 8 ± 6x 6y = ± 84 –x = – 76 ∴ x = 76. Substituting Substituting the value of x in the equation equation (ii), we get 76 – y = 14 ∴ y = 76 – 14 = 62. Thus, the ages of the teacher teacher and his student student are 76 years and 62 years respectively. respectively. Example-8 Hira said to Ashok “I was 3 times more than 6 years exceed as old as you were when I was as old as you are.” Also said that if the sum of our ages is 98 years, find what our ages now would be. Solution: Let the ages of Hira and Ashok be x years and y years now respectively. Then, x – (x – y) = 3[ y – (x – y)] + 6 or, x – x + y = 3(y – x + y) + 6 or, y = 3(2y – x) + 6 Multiplying the equation (ii) by 5 and adding with the equation (i), we get 5x + 5y = 490 + 3x – 5y = 6 8x = 496 or, x = 496 8 = 62. From the eqn (ii), 62 + y = 98 or, y = 98 – 62 = 36. Hence, now the ages of Hira and Ashok are 62 years and 36 years respectively. Example-9 Anisha started her scooter for the journey from Pokhara to Bairahawa at 10 AM with an average speed of 35km/hr. After an hour, Anish started his motorbike on the same journey with average speed of 45km/hr. At what time would Anish meet Anisha? Solution: Suppose Anish meets Anisha after x hours by motorbike. Then Anisha travelled by the scooter (x + 1) hours. ∴ The distance travelled by Anish = 45x km and x – 6 = y + 8 or, x – y = 8 + 6 or, x – y = 14 ...... (ii) × 6 or, y = 6y – 3x + 6 or, 3x – 5y = 6 ........ (i) and x + y = 98 ..... (ii) × 5 264 Allied The Leading Mathematics-9 Linear Equations 265

Algebra Algebra and distance travelled by Anisha = 35(x + 1) km. Since they meet together on the same way, they travelled equal distances. ∴ 45x = 35(x + 1) or, 45x = 35x + 35 or, 45x – 35x = 35 Example-10 Two buses are coming from two places Kathmandu and Rasuwagadhi of a distance of 150 km on the same road. The speed of one bus is faster than the other by 10 km/hr. After 3 hours, both buses meet together. If both buses were departed at the same time, find the speed of each bus. Solution: Let the speed of the faster bus be x km/hr and the speed of slower bus be y km/hr. Then, x – y = 10 .......... (i) and 3x + 3y = 150 or, 3(x + y) = 150 or, x + y = 50 ......... (ii) Adding the equations (i) and (ii), we get 2x = 60 ∴ x = 30 Putting the value of x in the equation (i), we get 30 + y = 50 ∴ y = 50 – 30 = 20 Thus, the speed of the faster bus is 30 km/he and that of slower bus is 20 km/hr. PRACTICE 11.3 Keeping Skill Sharp 1. Define the following terms: (a) Simultaneous equation of two variables (b) Solution of simultaneous equation of two variables 2. Solve the following simultaneous equations: (a) x = 2 + y and x + y = 8 (b) y = 3 – x and x + 2y = 5 (c) 2x + y = 10 and x – y = 5 (d) x – 2y = 7 an x + 2y = 9 Check Your Performance Answer the given questions for each situation. 3. (i) A number is twice the other, if their sum is 45. (ii) Among two numbers, one is 6 more than other and their sum is 38. (a) Find the numbers. (b) How much more percent is the bigger number than the smaller number. 4. (i) Seven times of a number increased by 8 is equal to 57. Find the number. (ii) A number is multiplied by 5 and then subtracted by 7 becomes 58. Find the number. 264 Allied The Leading Mathematics-9 Linear Equations 265

Algebra Algebra 5. (i) The length of a rectangle is twice as long as its breadth and its perimeter is 30 m. (ii) The length of equal sides of an isosceles triangle is thrice as long as its base and its perimeter is 56 m. (a) Find the length of its sides. (b) Find its area. 6. (i) Find two consecutive numbers whose sum is 25. (ii) Find two consecutive even numbers whose sum is 6. 7. (i) One-sixth age of a mother is the age of her daughter and the sum of their ages is 49. (ii) The ratio of ages of mother and his son is 7:3 now and the difference of their ages is 40. (a) Find their ages. (b) What was the age of the mother when his child was born? 8. (i) If twice the son’s age added to the father’s age, the sum is 70 and twice the father’s age added to son’s age, the sum is 95. (ii) 3 years ago, a father’s age was 4 times the age of his daughter. Now, the sum of their ages is 46. What are their present ages? (a) Find their present ages. (b) How much older is the father than his child? 9. (i) The present age of the father is three times the age of her daughter. After 10 years, the age of the daughter is equal to the age of the father before 20 years. (ii) A year hence, a father will be 4 times as old as his son. Three years ago, the father was 3 times as old as his son will be 4 years hence. (a) Find their ages. (b) How much years younger is the his child than father? 10. (i) The sum of two numbers is 5 and their difference is 3. Find the numbers. (ii) The sum of ages of two brothers is 45 and the difference of their ages is 3. Find their ages. 11. (i) The sum of digits in a two-digit number is 11. The number formed by interchanging the digits of that number will be 45 more than the original number. (ii) A number consists of two digits. The quotient of these two digits being 3 and if 18 is subtracted to the number, then the digits of the number are reversed. (a) Find the original number. (b) Find the more or less percent of the original number than its reverse. 12. (i) A two-digit number is such that the sum of its digits is 3. Seven times the number is equal to four times the number obtained by reversing the digits. What is that number? (ii) The sum of the digits of a two-digit number is 17 and the number is 9 more than the other with these digits in the reverse order. Find the number. (a) Find the number. (b) Find the difference between the number and its reverse. 13. (i) The sum of two digits of a two-digit number and the number obtained by reversing the order of its digits is 84 and they differ by 3. (ii) In a two-digit number, the digit of the unit’s place is greater than that of ten’s place by 2. If 3 times the sum of digits is added to the number, the digits are reversed. (a) Find the number. (b) Find the increased or decreased percent of the original number than its reverse. 266 Allied The Leading Mathematics-9 Linear Equations 267

Algebra Algebra 14. (i) 5 kg of rice and 3 kg of flour cost Rs. 145; and 3 kg of rice and 5 kg of flour cost Rs.151. (ii) 5 kg of apples and 15 kg of mangoes cost Rs. 2350 and 3 kg of apples and 7 kg of mangoes cost Rs. 1190. (a) Find the cost of each item. (b) What is the increased or decreased percent of the first item than the second item? 15. (i) If the numerator of a fraction is multiplied by 4 and a denominator is reduced by 2, the result is 2. If the numerator of the fraction is increased by 15 and 2 is subtracted from the double of the denominator, the result is 9 7 . Find the fraction. (ii) If 1 is added to both numerator and denominator of a fraction; the fraction becomes 4 5 . If 1 is subtracted from numerator and denominator both, the new fraction becomes 3 4 . What is that fraction ? 16. (i) A mother says to her daughter “5 years ago, I was 5 times as old as you were but 10 years hence I shall be only two times as old as you will.” Find their present ages. (ii) Dharmendra said to hisson Sanni “I wasthrice as old as you were when I was as old as you are.” Also, said that if the sum of our ages is 120 years, find what our ages now would be. 17. (i) Katrina started a car for the journey of 70 km from Beni to Jomsom at 7 AM with an average speed of 20km per hour. After one hour, Ranvir started his motorbike on the same journey with average speed of 30 km/hr. At what time would they meet? (ii) A bus departed at 9 PM for the journey of 200km from Kathmandu to Pokhara with an uniform speed of 40km per hour. After one and half hour, Sansar started his motorbike on the same journey with uniform speed of 60km/hr. At what time would they meet? 3. (i) 30, 15 (ii) 16, 22; 37.50% 4. (i) 7 (ii) 13 5. (i) 10 m, 5 m; 50 m2 (ii) 24 m, 24 m, 8 m; 94.66 m2 6. (i) 12, 13 (ii) 2, 4 7. (i) 42 yrs, 7 yrs.; 36 yrs. (ii) 70 yrs., 30 yrs; 40 yrs. 8. (i) 40 yrs,15 yrs; 25 yrs. (ii) 35 yrs., 11 yrs.; 24 yrs. 9. (i) 45 yrs., 15 yrs.; 30 yrs. (ii) 51yrs., 12 yrs.; 39 yrs. 10. (i) 4, 1 (ii) 24 yrs., 21yrs. 11. (i) 38, 54.22% less (ii) 31; 138.46% more 12. (i) 12; 9 (ii) 98; 9 13. (i) 96; 39.13% more (ii) 24; 42.86% less 14. (i) Rs. 17, Rs. 20; 15% (ii) Rs. 140, Rs. 110; 27.27% 15. (i) 3 8 (ii) 7 9 16. (i) 30 yrs., 10 yrs. (ii) 75 yrs., 45 yrs. 17. (i) 10 AM (ii) 1:30 AM Answers Project Work You bought 3 kg of tomatoes and 4 kg of potatoes and paid the total amount from a shop in a day. Again, you bought 2 kg of tomatoes and 3 kg of potatoes and paid the total amount from the same shop in next day. If the shopkeeper did not write the bill, how can you find their rates. Solve them. Prepare a report and present it in your classroom. 266 Allied The Leading Mathematics-9 Linear Equations 267

Algebra Algebra 12.1 Simplification of Indices By Using Their Laws At the end of this topic, the students will be able to: ¾ simplify the given exponential expressions by using their laws. Learning Objectives When different powers of numbers, variable and expressions appear in combination, they are handled with the help of certain rules or laws known as the Laws of Indices or Exponents. We now consider them together with short sketches of proofs based on the associative and distributive laws of multiplication of numbers. Our proofs will be for numbers only. Similar laws apply in the case of algebraic expressions of various forms. Law I : For any number a and for any two positive integers m and n, am × an = am + n Consider, a4 × a5 = a × a × a × a × a × a × a × a × a = a9 = a4 + 5 ∴ am × an = am + n Law II : for any number a and for any two positive integers m and n, am ÷ an = am – n if m > n; = 1 if m = n; = 1 an – m if n > m Consider, a5 a3 = a × a × a × a × a a × a × a = a × a = a2 = a5 – 3 ∴ am an = am ÷ an = am – n (m > n) a3 a5 = a × a × a a × a × a × a × a = 1 a × a = 1 a2 = 1 a5 – 3 ∴ am an = 1 an – m (m < n) 4 times a 5 times a Proof : Since am = a × a × a × ... × a (m factors) and an = a × a × a .... × a (n factors) m factors n factors am × an = (a × a a × .... × a) (a × a a × .... × a) (m + n) factors = a × a × a .... × a = am + n Proof : Since am = a × a × a × .... × a (m factors) and an = a × a × a .... × a (n factors) am an = a.a.a.... m factors a.a.a.... n factors = a.a.a.... (m – n) factors = am-n The proof of the second part is left as an exercise. Again, (for m < n) am ÷ an = a.a.a.... m factors a.a.a.... n factors = 1 a.a.a.... (n – m) factors = 1 an – m CHAPTER 12 INDICES 268 Allied The Leading Mathematics-9 Indices 269

Algebra Algebra Law III : For any number a and b and for any two positive integers m and n, (ab)m = ambm a b m = am bm , (b ≠ 0) (am) n = amn Consider, (2 × 3)2 = (2 × 3) × (2 × 3) = (2 × 2) × (3 × 3) = 22 × 32 ∴ (ab)m = ambm and so on. We shall now extend the above Laws of Indices to those cases where m and n may be negative integers or fractions. i. We have taken the value of a0 to be equal to 1 or a0 = 1 for any non-zero number a. This is justified because a0 = an – n = an an = 1. ii. We shall take a–m as the reciprocal of am for any non - zero number a. i.e, a– m = 1 am This is justified because am – m = a0 = 1. or, am.a– m = 1 or, a–m = 1 am as also am = 1 a – m iii. The nth root of a, i.e, a n is that quantity which multiplied with itself n times gives a. So, we agree to write a n as a1 n because a 1 n.a 1 n.a 1 n ... n times = (a 1 n) n = a n n = a1 = a Hence, we can define a quantity like a m n by a m n = (a 1 n) m Furthermore, we have a m n = (am) 1 n = a n m iv. We have, (ab) m = am bm and am = 1 a– m ∴ a b – m = a– m b– m = bm am = b a m Proof: Clearly, (ab) m = (ab).(ab) .... (ab) (m factors) = (a × a × a × .... m factors) (b × b × b ... m factors) = am bm. The proofs of the rest are left as exercises. 268 Allied The Leading Mathematics-9 Indices 269

Algebra Algebra Thus, the laws of indices stated above are valid when the indices are any rational numbers. i.e., positive or negative, integer or fraction. In addition to the above laws we shall also use the following facts as axioms. 1. If am = an , then m = n, if a ≠ 0. 2. If xn = yn , then x = may or may not be equal to y, if n ≠ 0. We shall now apply the laws of indices to simplify various complicated expressions and solve given problems. Example-1 Simplify: (a) 8 343 2 3 (b) x3 5 x 5 – 2 (c) 25 . 272 . 364 32 . 543 . 45 (d) xa xb 1 a.b. xb xc 1 b.c. xc xa 1 c.a Solutions : (a) 8 343 2 3 = 23 73 2 3 = (23 ) 2 3 (73 ) 2 3 = 22 72 = 4 49 (b) x3 5 x 5 – 2 = (x3 ) 1 5 (x– 2) 1 5 = x 3 5 x– 2 5 = x 3 5.x 2 5 = x 3 5 + 2 5 = x 3 + 2 5 = x 5 5 = x (c) 25 . 272 . 364 32 . 543 . 45 = 25 .(3.3.3)2 .(2.2.3.3)4 32 .(2.3.3.3)3 .(2.2)5 = 25 .(33 ) 2 .(22 .32 ) 4 32 .(2.33 ) 3 .(22 ) 5 = 25 .36 .(22 ) 4 .(32 ) 4 32 .23 .(33 ) 3 .210 = 25 .36 .28 .38 32 .23 .39 .210 = 25 + 8 .36 + 8 23 + 10.32 + 9 = 213.314 213.311 = 314 – 11 = 33 = 27. (d) xa xb 1 a.b. xb xc 1 b.c. xc xa 1 c.a = (xa – b ) 1 a.b (xb – c ) 1 b.c (xc – a ) 1 c.a = x a – b ab . x b – c bc . x c – a ca Example-2 Simplify M xa2 – b a+b 2 . xb2 – c b+c 2 . xc2 – a c+a 2 Solution: Here, xa2 – b a+b 2 . xb2 – c b+c 2 . xc2 – a c+a 2 = x(a + b)(a – b) a + b . x (b + c)(b – c) b + c . x (c + a)(c – a) c + a = xa – b + b – c + c – a = x0 = 1 = x a – b ab + b – c bc + c – a ca = x 1 b + 1 a + 1 c – 1 b – 1 a – 1 c = x0 = 1. 270 Allied The Leading Mathematics-9 Indices 271

Algebra Algebra Example-3 Simplify M (1 – am – n) –1 + (1 – an – m) –1 Solution: Here, (1 – am – n) –1 + (1 – an – m) –1 = 1 1 – am – n + 1 1 – an – m x–1 = 1 x = 1 1 – am an + 1 1 – an am xa – b = xa xb = 1 an – am an + 1 am – an am Example-4 Simplify M p2 (p – y)y – 2p (p – y)y – 1 + 1 (p – y)y – 2 Solution: Here, p2 (p – y)y – 2p (p – y)y – 1 + 1 (p – y)y – 2 = p2 (p – y)y – 2p(p – y) (p – y)y + (p – y)2 (p – y)y 1 am – n = an bm = p2 – 2p(p – y) + (p – y)2 (p – y)y = p2 – 2p2 + 2py + p2 – 2py + y2 (p – y)y = 2p2 – 2p2 + y2 (p – y)y = y2 (p – y)y Example-5 Prove that : xa/b xb/c ab × xb/c xc/b bc × xc/a xa/c ca Solution: Here, xa/b xb/a ab × xb/c xc/b bc × xc/a xa/c ca = x a b × 1 ab x b a × 1 ab × x b c × 1 bc x c b × 1 bc × x c a × 1 ca x a c × 1 ca = x 1 b2 x 1 a2 × x 1 c2 x 1 b2 × x 1 a2 x 1 c2 = x 1 b2 + 1 c2 + 1 a2 – 1 a2 – 1 b2 – 1 c2 = x0 = 1 Example-6 Simplify : 1 1 + xa – b + xa – c + 1 a + xb – c + xb – a + 1 1 + xc – a + xc – b Solution: Here, 1 1 + xa – b + xa – c + 1 a + xb – c + xb – a + 1 1 + xc – a + xc – b = 1 1 + xa xb + xa xc + 1 1 + xb xc + xb xa + 1 1 + xc xa + xc xb = 1 xb + c + xc + a + xa + b xb + c + 1 xc + a + xa + b + xb + c xc + a + 1 xa + b + xb + c + xc + a xa + b = xb + c xb + c + xc + a + x a + b + xc + a xb + c + xc + a + xa + b + xa + b xb + c + xc + a + x a + b = xb + c + xc + a + xa + b xb + c + xc + a + x a + b = 1 = an an – am + am am – an = an an – am – an an – am [ b – a = –(a – b)] = an – am an – am = 1 270 Allied The Leading Mathematics-9 Indices 271

Algebra Algebra Example-7 Simplify : (an ÷ am)l2 + m2 + lm × (am ÷ an)m2 + n2 + mn × (an ÷ al)n2 + l2 + nl Solution: Here, (an ÷ am)l2 + m2 + lm × (am ÷ an )m2 + n2 + mn × (an ÷ al )n2 + l2 + nl = a(l – m) l2 + m2 + lm × a(m – n) m2 + n2 + mn × a(n – l) n2 + l2 + nl = al3 – m3 × am3 – n3 × an3 – l3 = al3 – m3 + m3 – n3 + n3 – l3 = a° = 1. Example-8 Simplify M a + (ab2 ) 1 3 + (a2 b) 1 3 a – b ×   1 – b 1 3 a 1 3   Solution: Here, a + (ab2 ) 1 3 + (a2 b) 1 3 a – b ×   1 – b 1 3 a 1 3   = (a1 3) 3 + a1 3 . b2 3 + a2 3 b 1 3 (a 1 3) 3 – (b1 3) 3 × a 1 3 – b1 3 a 1 3 = a 1 3(a 2 3 + b2 3 + a1 3 b 1 3) (a 1 3 – b1 3)(a 2 3 + a1 3 b 1 3 + b2 3) × a 1 3 – b1 3 a 1 3 = 1 Example-9 Simplify M a2 – 1 b2 a a – 1 b b – a b2 – 1 a2 b b + 1 a a – b Solution: Here, a2 – 1 b2 a a – 1 b b – a b2 – 1 a2 b b + 1 a a – b = a + 1 b a a – 1 b a a – 1 b b – a b + 1 a b b – 1 a b b + 1 a a – b [ a2 – b2 = (a + b) (a – b) and (ab)n = an bn ] = a + 1 b a a – 1 b a + b – a b + 1 a b + a– b b – 1 a b = ab + 1 b a ab – 1 b b ab + 1 a a ab – 1 a b = (ab + 1)a ba × (ab – 1)b bb × aa (ab + 1)a × ab (ab – 1)b = aa + b ba + b =   a b  a + b PRACTICE 12.1 Keeping Skill Sharp 1. (a) What is an index or exponent ? (b) Express a m into exponential form. (c) What is the index of x such that its value becomes 1? 2. Write the following in the exponential form (i.e., by using index notation). (a) 3 × 3 × 3 × 3 (b) 81 (c) a × a × a × a (d) 32 3. Write the following in the product form: (a) 8a3 (b) 64 (c) x2 y3 (d) 16 272 Allied The Leading Mathematics-9 Indices 273

Algebra Algebra 4. What is the value of: (a) 8 – 2 3 (b) 27– 2 3 (c) 8 27 – 2 3 (d) 64– 2 3 (e) 33 × 32 (f) 23 .22 .24 (g) 37 ÷ 24 × 22 (h) 3 2 2 5. Write in simplest exponential form: (a) (i) 22 .23 (ii) (x + y) 2 (x + y) -3 (iii) a2x .a3. ax .a3 (iv) (2x + 3y) 2m (2x + 3y) 3n (b) (ii) (3a) 2 (ii) (xy) 3 (iii) (ab) 2x (iv) [(x + y)(x – y)]4 (c) (i) 0 5 3 2 (ii) 0 a b 3 (iii) 0 a b c (iv) 0 a + y x – y 3 (d) (i) x4 y– 2 (ii) x4 (iii) xy 3 2 × x2 y 3 (iv) (x– 2 3 ÷ y – 1 2) 6 Check Your Performance 6. Simplify: (a) (81)1 4 ×   1 27  2 3 (b)   625 81   – 3 4 (c) 154 × 65 105 × 92 (d) 284 × 64 213 × 182 (e)   81 16   3 4 ÷   32 243   – 3 5 of   27 8   – 2 3 (f)   2 3  – 1 5 10 3 – 3 2 7. Simplify: (a) 1 1 + xa – b + 1 1 + xb – a (b) (1 – xm – n) –1 + (1 – xn – m) –1 (c) a a + b–1 + 1 ab + 1 (d) {(a – b)–2}–m × {(a – b)m}2 8. Simplify: (a) 27x3 ÷ 243x5 (b) 8xy3 × 2x3 y (c) 625x8 y 4 12 (d) –216a6 b9 c 3 0 (e) –8x3 y 3 –9 × 81x4 y 4 4 (f) 121a4 b–8c8 ÷ 1331a–6b9 c 3 3 (g) 8x–9y12z 3 6 ÷ 1296x8 y 4 –4 (h) 81a12b–8c 4 4 ÷ 125a–6b6 c 3 9 (i) 256x–8y 4 16 × 27x6 y 3 –6 × x4 y 4 – 8 (j) 16x4 y6 z 5 2 × 2x6 y4 z 5 –7 ÷ 16x8 y12 9. Simplify: (a) 3n + 1 × 32n – 3 33n – 4 (b) 2n + 2 × {2(n – 1)}(n + 1) 2n(n – 1) ÷ 4n + 1 (c) 5x + 5x + 1 5 . 5x (d) 4a + 22a – 1 3 × 22a – 1 (e) 52x + 1 + 25 × 52x – 1 10 × 52x (f) 4m + 2 – 4m + 1 4m + 1 + 2 . 4m + 1 10. Simplify: (a) 1 1 + xa – b + xc – b + 1 1 + xb – c + x a – c + 1 1 + xc – a + xb – a (b) 1 1 + yr – p + yr – q + 1 1 + yq – r + yq – p + 1 1 + yp – q + yp – r 11. Simplify: (a)   xa xb   c ×   xb xc   a ×   xc xa   b (b)   xp xq   r ×   x–r x–q   p ×   x–p x–r   q 272 Allied The Leading Mathematics-9 Indices 273

Algebra Algebra 12. Simplify: (a) ap/q aq/p pq × aq/r ar/q qr × ar/p ap/r rp (b) xc 1/c xa 1/a × xa 1/a xb 1/b × xb 1/b xc 1/c (c)   xa + b xc   a – b ×   xb + c xa   b – c ×   xc + a xb   c – a (d)   xa xb   a + b – c ×   xb xc   b + c – a ×   xc xa   c + a – b 13. Prove that: (a) x– 1y y– 1z z– 1x = 1 (b) ab 3 2 a2 b 3 4 a– 3b 3 – 6 = 1 (c) xa xb 2 xb xc 2 xc xa 2 = 1 14. Show that: (a) a(x – y)z a(y – z)x a(z – x)y = 1 (b) (xa – b) a + b (xb – c) b + c (xc – a) c + a = 1 (c) (xa – b ) a2 + ab + b2 .(xb – c ) b2 + bc + c2 .(xc – a ) c2 + ca + a2 = 1 15. Simplify: (a) ax ay z ay az x az ax y (b) xa xb a + b xb xc b + c xc xa c + a (c) m + (mn2 ) 1/3 + (m2 n)1/3 m – n ×   1 – n1/3 m1/3   16. Show that: (a) xb – c bc . xc – a ca . xa – b ab = 1 (b)   x– (a2 + b2 ) x ab   b – a ×   xbc x – (b2 + c2 )   b – c ×   x(c2 + a2 ) x – ca   c – a = 1 17. Simplify: (a) x + 1 y m x – 1 y n y + 1 x m y – 1 x n (b) 1 – n m n m – n 1 + m n m m – n m n – 1 n m – n n m + 1 m m – n 6. (a) 1 3 (b) 27 125 (c) 483 5 (d) 26513 27 (e) 4 9 (f) 2 3 7. (a) 1 (b) 1 (c) 1 (d) (a – b)4m 8. (a) 1 3x (b) 4x2 y2 (c) 5x2 y3 (d) – 6a2 b3 (e) – 6x2 y2 (f) a4 c3 b7 (g) 1 3 y x 5 z2 (h) 3a5 5b4 c2 (i) 12xy4 (j) 1 2x2 y4 z 9. (a) 9 (b) 1 2 (c) 11 5 (d) 1 (e) 1 (f) 1 10. (a) 1 (b) 1 11. (a) 1 (b) 1 12. (a) 1 (b) 1 (c) 1 (d) 1 15. (a) 1 (b) 1 (c) 1 17. (a) x y m + n (b) m n Answers Project Work List out the LAWS of INDICES on the full page of a chart paper. 274 Allied The Leading Mathematics-9 Indices 275

Algebra Algebra CONFIDENCE LEVEL TEST - IV Unit IV : Algebra Class: 9, The Leading Maths Time: 45 mins. FM: 22 Attempt all the questions. 1. Observe the pattern of the sequence of the given figures. , , , , ..... (a) Add the next one figure on the same pattern. [1] (b) Find its general term. [2] (c) Find the number of dots of the 20th figure. [1] (d) Find the value of 6 n=3 ∑(2n2 – n + 1) . [2] 2. (a) Factorize: x8 + 9x4 + 81 [2] (b) Find the Highest Common Factor (HCF) of; [3] a2 – b2 – 2b – 1 and a2 – 1 + 2ab + b2 [2] 3. (a) Solve: 7x + 4y = 1 and 5x + 2y = – 1 [2] (b) In 2074 BS, the mother’s age was 4 times her son is age and in 2078 BS, her age will be only 3 times older than her son. What is the year of the birth of the mother? [3] 4. (a) Simplify: 2432n 5 × 33n + 1 9n + 1 × 32(n – 2) [2] (b) Prove that: 1 1+xa–b+xc–b + 1 1+xb–c+xa–c + 1 1+xc–a+xb–a = 1 [2] Best of Luck 274 Allied The Leading Mathematics-9 Indices 275

Algebra Algebra Additional Practice – IV 1. An arithmetic sequence is given: 5, 9, 13, 17, … (a) Find the common differences of the given sequence. (b) Write the general term of the given arithmetic sequence. (c) Which term of the given sequence is 49? Find it. 2. The first five terms of the geometric sequence are given: 5, 10, 20, 40, 80…………… (a) Write the formula to find the general term of a geometric sequence. (b) Find the common ratio of the given sequence. (c) Find the 10th term of the given sequence. (d) Which term of the given sequence is 5120? Find it. 3. The second term and sixth term of a geometric sequence are 192 and 12 respectively. (a) Find the common ratio of the given sequence. (b) Which term of the given sequence is 3? Find it. (c) Find the 8th term of the sequence. (d) If the first three terms make an arithmetic sequence, what should be the value of the third term? Justify. 4. The first four terms of the arithmetic sequence are given: 5, 11, 17, 23, … (a) Find the common difference of the given sequence. (b) Find the 8th term of the sequence. (c) Which term of the given sequence is 131? Find it. (d) If the first three terms make a geometric sequence, what should be the value of the third term? Justify. 5. (a) Find the HCF of : x3 + 27 and x4 + 9x2 + 81 (b) Simplify: xa xb a2 + ab + b2 × xb xc b2 + bc + c2 × xc xa c2 + ca + a2 (c) Simplify: 273n + 1 × (243)–4n/5 9n + 5 × 33n – 7 6. (a) Simplify: xa2 xb2 a + b × xb2 xc2 b + c × xc2 xa2 c + a (b) Factorize: p3 q3 – q3 p3 (c) Factorize: y4 x4 – 21 y2 x2 + 100 7. (a) If xyz = 1, prove that: 1 1 + x + y–1 + 1 1 + y + z–1 + 1 1 + z + x–1 = 1 (b) Find the LCM of: x2 – x + 1 and x4 + x 276 Allied The Leading Mathematics-9 Indices 277

Algebra Algebra 8. There is a two-digit number between 10 and 100. The number is 8 times the sum of its digits. If 45 is subtracted from the number, the places of the digits are interchanged. (a) Write a two-digit number in terms of x and y. (b) Write two equations in terms of x and y. (c) Find the two-digit number. 9. According to the price list of Nepal Oil Corporation dated 2080/01/20, the price of one-liter petrol was Rs 10 more than one-liter diesel. On that day, Ramesh had bought 5-liter petrol and 7-liter diesel on Rs. 2030. (a) Present the given statement in the form of linear equation. (b) What is the price of petrol and diesel per liter? Find it. (c) If the prices of petrol and diesel are decreased by 10%, how much equal quantity of petrol and diesel can be bought by Ramesh in Rs 1224? Calculate it. 10. In a Dangol Furniture Store, the cost of 6 chairs and 5 tables is Rs. 15000 and the cost of 9 tables and 7 chairs is Rs. 21300. (a) Consider the price of chair to be x and table to be y and form two equations. (b) Find the cost of single chair and table. (c) If the cost of chair is increased by 20% and cost of table is decrease by 25%, how many equal number of chairs and tables can be bought in Rs. 32400? 11. Shusil started her car for the journey from Kathmandu to Pokhara at 11 AM with an average speed of 30 km/hr. After two hours, Ashok started his car on the same journey with average speed of 40 km/hr. (a) After what time will they meet? Write in terms of x. (b) After what time would they meet? Calculate it. (c) At what time would they meet? Calculate it. 1. (a) 4 (b) 4n + 1 (c) 12 2. (b) 2 (c) 2560 (d) 11 3. (a) 1 2 (b) 8 (c) 3 (d) 0 4. (a) 6 (b) 47 (c) 22 (d) 24 1 5 5. (a) x2 – 3x + 9 (b) 1 (c) 1 6. (a) 1 (b) p q – q p p2 q2 + 1 + q2 p2 (c) y2 x2 + y x – 10 y2 x2 – y x – 10 7. (b) x4 + x 8. (b) 2x – 7y = 0; x – y = 5 (c) 72 9. (a) x – y = 10; 5x + 7y = 2030 (b) Rs. 175, Rs. 165 (c) 4 l 10. (a) 6x + 56 = 15000; 9x + 7y = 21300 (b) Rs. 500, Rs. 2400 (c) 9 11. (a) x and x + 2 hrs (b) 6 hrs (c) 5 PM Answers 276 Allied The Leading Mathematics-9 Indices 277

278 Allied The Leading Mathematics-9 Triangles 279 Geometry Geometry Analysis of the geometric facts, develop the presentation and ability of the problem solving. Estimated Working Hours : Competency Learning Outcomes verify the relation of facts related to sides and angles of triangle experimentally and theoretically. solve the problems related to sides and angles of triangle. show the relation between corresponding angles and sides of similar triangles. solve the problems related to similar triangles. verify the relations of corresponding sides, angles and diagonals of parallelogram theoretically. solve the problems related to quadrilaterals. construct scalene triangle and trapezium. verify the relation of the chord with the perpendicular drawn from the centre of a circle experimentally and theoretically. verify the relation of the distance up to equal chords drawn from the centre of a circle to the equal chords experimentally and theoretically. solve the problems related to centre and chord of a circle. 13 Triangle 13.1 Angles of Triangle 13.2 Sides and Angles of a Triangle 13.3 Properties of Isosceles Triangle 13.4 Similar Triangles 14. Parallelogram 14.1 Properties of Parallelogram 14.2 Mid Point Theorems 15. Construction 15.1 Construction of Triangles 15.2 Construction of Quadrilaterals 15.3 Construction of Parallelograms 15.4 Construction of Rectangles 15.5 Construction of Squares 15.6 Construction of Rhombuses 15.7 Construction of Trapeziums 16. Similarity 17. Circle Chapters / Lessons GEOMETRY UNIT 28 (Th. + Pr.) HSS Way: Mathematics begins at Home, grows in the Surroundings and takes shape in School. V Specification Grid Unit Areas Total working hour Knowledge Understanding Application Higher ability Total number of items Total number of questions Total Marks No. of items Marks No. of items Marks No. of item Marks No. of item Marks V Geometry 28 2 2 2 3 2 5 2 3 8 3 13

13.1 Angles of Triangle At the end of this topic, the students will be able to: ¾ prove the theorems related on the sum of interior angles of triangle by experimental and theoretical. Learning Objectives Review on Angles and Triangles Angles and triangles are basic forms of geometric figures formed by rays, line-segments or lines. We shall now make a brief review of basic terminology related to these figures. Angle An angle is said to be formed when two rays have the same end point (or initial point). The common point is called the vertex and the two rays its arms. Any two angles are said to be adjacent angles if they have (i) The same vertex, and (ii) A common arm An angle formed by two rays AB and AC is shown alongside. It is denoted by ∠BAC or ∠CAB. For simplicity, it is denoted by ∠A. An angle can be measured in different ways. We follow one particular standard unit called degree. The main types of angles are null, acute, right, obtuse, reflex and complete angles. Now we give two important definitions: 1. Two angles whose sum is one right angle (90o ) are said to be complementary to each other. Thus, 60o and 30o are complementary to each other. 2. Two angles whose sum is one hundred and eighty degrees(180o ) are said to be supplementary to each other. Thus, 60o and 120o are supplementary to each other. Triangle The word ‘triangle’ immediately leads us to think of three angles. But a common definition does convey nothing about angles. It runs like this: ‘A triangle is a plane figure formed or bounded by three straight lines’. If three points A, B and C are not collinear, then the line segments AB, BC and CA form the triangle ABC. The points A, B and C are called the vertices A B C CHAPTER 13 TRIANGLES 278 Allied The Leading Mathematics-9 Triangles 279 Geometry Geometry

of the triangle ABC and the segments AB, BC and CA are called its sides. The sides AB, BC, CA are respectively denoted by the small letters: c, a and b respectively. The triangle itself is denoted by ∆ABC. Note that AB = c is opposite to the angle ACB (or ∠C) of the ∆ABC, BC = a is opposite to the angle CAB (or ∠A) and CA = b is opposite to the angle ABC (or ∠B). Triangles are classified in two ways, either on the basis of the sides or on the basis of angles. Classification of triangle according to sides 1. Equilateral Triangle : If all the sides of a triangle are equal, the triangle is called an equilateral triangle. For instance, ∆ABC is equilateral with AB = BC = CA. If we measure its angles, we shall find that ∠A = ∠B = ∠C = 60o . 2. Isosceles Triangle : If any two sides of a triangle are equal, the triangle is called an isosceles triangle. For instance, ∆PQR is isosceles with PQ = PR. If we measure the base angles, we shall find that ∠Q = ∠R. 3. Scalene Triangle : If all the sides of a triangle are unequal, the triangle is called a scalene triangle. For instance, ∆XYZ is scalene with XY ≠ YZ ≠ ZX. If we measure the angles, we shall find that ∠X ≠ ∠Y ≠ ∠Z. Classification of triangle according to angles 1. Right-angled Triangle: If one angle of a triangle is a right angle, the triangle is called a right-angled triangle. For instance, in the ∆ABC, ∠B = 90o (i.e., a right angle). 2. Obtuse-angled Triangle: If one angle of a triangle is obtuse, the triangle is called an obtuse-angled triangle. For instance, ∆PQR is obtuse-angled with ∠Q > 90o but < 180o . 3. Acute-angled Triangle: If all the angles of a triangle are acute angles, the triangle is called an acute-angled triangle. In the above figure, ∆XYZ is an acute-angled triangle with each angle less than 90o . Before we go ahead, we give below a shortsummary ofterminology and facts that we may need in our proofs. Note: 1. When the two rays forming an angle are opposite to each other, it is said to form a straight angle. Its measure is taken to be 180o . B C A Q R P Y Z X B C A Q R P Y Z X B C A 280 Allied The Leading Mathematics-9 Triangles 281 Geometry Geometry

2. If two parallel lines are cut by a transversal, then, (a) alternate (internal) angles of each pair are equal and conversely. (b) consecutive interior angles of each pair are supplementary and conversely. (c) corresponding angles of each pair are equal and conversely. The first basic theorem on triangles is about the sum of the angles of a triangle. Statement: The sum of the three angles of a triangle is two right angles. Observation Method We can easily show the sum of the interior angles of a triangle is two right angles or 180° by folding paper as follows: Step 1: Cut down a triangle from a paper sheet and write the name ABC on both sides. Step 2: Fold the triangular paper vertically through its vertex A and unfold. Name the perpendicular AD. Step 3: Fold it vertically along EF as shown in the figure such that its vertex A is overlapped with D. Step 4: Again, fold it at E and F vertically such that E and F meet with D. We see that the angles A, B and C of ∆ ABC form a straight line. The sum of their measure is 180°. Hence, the sum of the interior angles of a triangle is two right angles or 180°. A B C D A B D C A B D C A B D C E F E F As a direct consequence of the above theorem, we have the following corollaries: 1. Each angle of an equiangular triangle is 60o . If each angle is x, 3x = 180o . So, x = 60o . 2. In a right-angled triangle, the two remaining angles are complementary. For, if x and y are the remaining two angles, then x + y + 90o = 180o . That is, x + y = 90o . Hence, the angles x and y are complementary. 3. In a right-angled triangle, each of the remaining angles is acute. (Why?) 4. In a right-angled triangle, the right angle is the greatest angle. (Why?) 5. In a right-angled isosceles triangle, the measure of each acute angle is 45o . (Why?) 6. If an angle of a triangle is obtuse, each of the remaining two angles is acute. (Why?) P1 P2 T h b c f g a d e 280 Allied The Leading Mathematics-9 Triangles 281 Geometry Geometry

We shall now see how an angle formed by producing a side of a triangle is related to the two inside angles of a triangle and then derive a theorem known as the Exterior Angle Theorem. Note 1: If the side BC of the triangle ABC is produced to D, the angle ACD is called the exterior angle of the ∆ABC at C. It is denoted by ext. ∠ACD. 2: The anglesAand B are then said to be remote interior angles or interior opposite angles with respect to the ext. ∠ACD of the ∆ABC. 3: Since at each vertex of a triangle two exterior angles can be constructed, we can, in all, six exterior angles can be constructed. Statement: If a side of a triangle is produced, the exterior angle so formed is equal to the sum of its two non-adjacent or opposite interior angles of the triangle. THEOREM 13.1 A. Observation Method Step 1: Draw a triangle ABC and produce its one side BC up to D as shown in the figure alongside. Step 2: Cut down the angles of A and B and arrange on the exterior angle at C of ∆ABC. Both angles A and B overleap with ∠ACD. Hence, ∠BAC + ∠ABC = ∠ACD. B. Experimental Verification Step 1: Two triangles ABC are drawn in different shape and size. Step 2: Produce the side BC up to D, where ∠ACD is an exterior angle of ∆ABC. B C D A Fig. (i) B C D A Fig. (ii) Step 3: The three angles BAC, ABC, ACD are measured, tabulated and the sum of BAC and ABC is then calculated in each case. Fig. ∠BAC ∠ABC ∠ACD ∠BAC + ∠ABC Result (i) 69° 53° 122° 69° + 53° = 122° ∠BAC + ∠ABC = (ii) 40° 105° 145° 40° + 105° = 145° ∠ACD From the above measurement and calculation, we arrive at the following conclusion: If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite (non-adjacent) angles of the triangle. Proved. B C D A B C D A A B 282 Allied The Leading Mathematics-9 Triangles 283 Geometry Geometry

C. Theoretical Proof Given: In ∆ABC, the side BC is produced up to D, where ∠ACD is its exterior angle. To prove: ∠BAC + ∠ABC = ∠ACD Construction: Through C, draw CE parallel to BA. Proof: Statements Reasons 1. ∠BAC = ∠ACE 1. Alternate angles on BA//CE 2. ∠ABC = ∠ECD 2. Corresponding angles on AB//CE 3. ∠ACE + ∠ECD = ∠ACD 3. Whole parts axiom 4. ∠BAC + ∠ABC = ∠ACE + ∠ECD 4. Adding statements (1) and (2) 5. ∠BAC + ∠ABC = ∠ACD 5. From statements (3) and (4) Proved. “Alternative Method” Given: In the ∆ABC, BC is produced to D forming the exterior ∠ACD. To prove: ∠BAC + ∠ABC = ∠ACD Proof: Statements Reasons 1. ∠BAC + ∠ABC + ∠ACB = 180o 1. Sum of the angles of ∆ABC 2. ∠ACB + ∠ACD = 180o 2. BCD being a straight line 3. ∠BAC + ∠ABC + ∠ACB = ∠ACB + ∠ACD 3. From statements (1) and (2) 4. ∠BAC + ∠ABC = ∠ACD 4. Subtracting the same ∠ACB from both sides i.e. by subtraction axiom Proved. An immediate consequence of Theorem 13.1 is the following result known as the Exterior Angle Theorem or Property of Triangle: “An exterior angle of a triangle is greater than either of the interior opposite (non-adjacent) angles.” B C D A E B C D A 282 Allied The Leading Mathematics-9 Triangles 283 Geometry Geometry

Example-1 Observe the adjoining figure. (a) Find the values of x, y and z. (b) Write the relation between x, y and z. Solution: (a) From the given figure, (i) ∠XYZ = ∠APY [ Alternate angles] or, x = 35° (ii) ∠YXZ + ∠XYZ = ∠XZD [ Exterior angle property of ∆XYZ] or, y + 35° = 120° or, y = 120° – 35° = 85° (iii) ∠BQZ + ∠XZD = 180° [ Co-interior angles] or, ∠BQZ + 120° = 180° or, ∠BQZ = 180° – 120° = 60° (iv) ∠XQA = ∠BQZ [ Vertically opposite angles] or, z = 60° (b) x + y + z = 35° + 85° + 60° = 180°. Hence, the sum of the angles x, y and z is 180°. Example-2 Observe the given figure in which AB//CD. If ∠BAO = 35° and ∠DCO = 50°. Find the value of a. Solution: In the given figure, produce AO up to E of CD. (i) ∠CEA = ∠EAB = 35° [Alternate angles] (ii) ∠AOC = ∠CEA + ∠OCD [ Exterior angle property of triangle] or, a = 35° + 50° = 85° Example-3 Study the given figure in which PO and RO are the bisectors of ∠TPQ and ∠PRS respectively. (a) Prove that ∠POR is a right angle. (b) What is the sum of ∠SRO and ∠RSO? Find it. Solution: (a) In the given figure, (i) ∠TPO = ∠QPO [PO is bisector of ∠TPQ] (ii) ∠PRO = ∠SRO [ RO is bisector of ∠PRS] (iii) ∠TPQ + ∠PRS = 180° [Co-interior angles] (iv) ∠TPO + ∠QPO + ∠PRO + ∠SRO = 180° [From the eqn (i), (ii) and (iv)] or, 2∠TPO + 2∠PRO = 180° or, 2(∠TPO + ∠PRO) = 180° ∠TPO + ∠PRO = 90° (vi) ∠POR + ∠TPO + ∠PRO = 180° [ Sum of angles of triangle] or, ∠POR + 90° = 180° ∠POR = 180° – 90° = 90°. Hence, it is proved. (b) ∠SRO + ∠RSO = ∠ROP = 90° [ Exterior angle property of triangle] C A B D Z 35° Y P X Q 120° z x y O C E A 35° 50° a D B T R S N O P Q 284 Allied The Leading Mathematics-9 Triangles 285 Geometry Geometry

PRACTICE 13.1 Keeping Skill Sharp 1. Observe the given figure and answer the following questions. (a) What is the supplementary angle of ∠4 in the given figure? (b) Which is the co-interior angle of ∠5 in the given figure? (c) What is the alternate angle of ∠2 in the given figure? (d) What is the sum of ∠1, ∠2 and ∠3 in ∆ABC? (e) What is the sum of ∠1, ∠5 and ∠6 in the given figure? (f) Which angle is equal to the sum ∠1 and ∠2 in the figure? 2. Observe the given figure and answer the following questions. (a) What is the value of (a + b + c) in the given figure? (b) Which angle is equal to a + b in the given figure? 3. (a) What is the value of 'e' in the adjoining triangle? (b) What is the measure of the angle 'x' in the given figure? Check Your Performance 4. Observe the following figures and answer the given questions. (i) State the exterior angle property of a triangle. (ii) Find the unknown angles (denoted by letters such as a, b, c, x, y, z, etc.) by using exterior angle property of the triangle. (a) 60° 55° x B C A D y (b) R S Q T P 80° 50° c a b (c) X A Y D B C 60° 35° x y (d) T Q R S D P A B C 120° 110° x y z (e) N M K L 2a 3a a x (f) L C N B A M120° 130° x (g) D C E F 55° 130° x (h) E B D F A C P Q 110° x 30° 5. (a) Verify experimentally that if a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles of the triangle. (b) Prove that the sum of cyclic exterior angles of a triangle is four right angles. 4. (a) 65°, 115° (b) 50°, 130°, 130° (c) 85°, 120° (d) 60°, 70°, 50° (e) 30°, 150° (f) 110° (g) 75° (h) 40° Answers D A E B C 4 2 3 1 6 5 b a d c Fig. 3(a) Fig. 3(b) 60° 50° e 50° 120° x 284 Allied The Leading Mathematics-9 Triangles 285 Geometry Geometry

13.2 Properties of Isosceles Triangle At the end of this topic, the students will be able to: ¾ prove the theorems related on the properties of isosceles triangle by experimental and theoretical. Learning Objectives Congruent Triangles In high school geometry, we are mostly concerned with plane figures such as triangles, quadrilaterals, circles, etc. It is easy to visualize how a circle can be fitted or placed exactly over another if both have the same radius. Similarly, it is not difficult to see how a square can be placed exactly over another square having equal side. But, in the case of triangles, fitting one triangle with another or superposing of triangles can be done by drawing triangles on paper, cutting them out and fitting them together. Such activities or experiences of daily life lead us to what is known as congruence of geometrical figures. Fig. E ≅ Fig. F Fig. G ∼ Fig. H Now onwards, we shall agree with the following definition: If two geometrical figures fit exactly on each other, they are said to be congruent or in congruence. Testing the congruency of two figures or superposing of one figure onto another will be done only in the following three ways: 1. sliding or translating in the plane without rotation, 2. rotating about a point in the plane, or 3. inverting upside down (without stretching and tearing). In each case, it is better to start with a fixed point such as end-point of a line-segment, vertex of an angle, centre of a circle. Our problem is to establish the congruency of triangles and then use them for various purposes. We know a triangle has three sides and three angles called six elements of a triangle. But these six elements are determined by the three vertices. So, if three vertices of one triangle are fitted (or matched exactly onto three vertices of another, then they will be congruent. It is known that there are six such ways. The six ways give rise to six conditions. But, fortunately, we do not require all of them. It is found that out of the three sides and three angles, three suitably chosen conditions (or parts) are sufficient to establish the congruency of two triangles. The sides and angles may be chosen in any one of the forms: 1. SAS (Side-angle-side) 2. ASA (Angle-side-angle) 3. AAS (Angle-angle-side) or SAA 4. SSS (Side-side-side) 5. RHS (Right angle-hypotenuse-side) or SHR 286 Allied The Leading Mathematics-9 Triangles 287 Geometry Geometry

1. SAS (Side-Angle-Side) Condition Two triangles are congruent if two sides and the included angle of the one are respectively equal to the corresponding sides and angle of the other. 2. ASA (Angle-Side-Angle) Condition Two triangles are congruent if two angles of the one and the side between them are respectively equal to the corresponding angles and side of the other. 3. AAS (Angle-Angle-Side) Condition Two triangles are congruent if two angles of the one and any one of the sides are respectively equal to the corresponding angles and side of the other. 4. SSS (Side-Side-Side) Condition Two triangles are congruent if three sides of the one are respectively equal to the three sides of the other. 5. RHS (Right Angle-Hypotenuse-Side) Condition Two right-angled triangles are congruent if the hypotenuse and one side of the one are respectively equal to the hypotenuse and one side of the other. Statement: In an isosceles triangle, the angles opposite to the equal sides are equal. OR Base angles of an isosceles triangle are equal. THEOREM 13.2 (Isosceles Triangle Property) A. Observation Method Step-1: Cut down an isosceles triangle ABC with AB = AC. Mark the base angles ∠ABC and ∠ACB on both parts. Step-2: Fold ∆ABC vertically through A as shown in the figure. Then the base angles ∠ABC and ∠ACB are exactly overlapped. Hence, the base angles of an isosceles triangle are equal. B A C B A C A B = C S S S S A A S S A A A A S S A A A A S S S S S S S S H H R R 286 Allied The Leading Mathematics-9 Triangles 287 Geometry Geometry

B. Experimental Verification Step 1: Draw two different isosceles triangles PQR to make PQ = PR. Fig. (i) Q P R Fig. (ii) P Q R Step 2: The base angles (opposite to equalsides) are measured by using protractor and the results are tabulated below: Fig. ∠PQR ∠PRQ Remarks (i) 52° 52° ∠PQR = ∠PRQ (ii) 45° 45° ∠PQR = ∠PRQ Conclusion: From the above table, we conclude that, ‘If two sides of a triangle are equal, then the angles opposite to them are equal.’ OR ‘The base angles of an isosceles triangle are equal.’ C. Theoretical Proof: Given: ABC is an isosceles triangle in which AB = AC. To prove: ∠ABC = ∠ACB Construction: Draw AD bisecting the angle BAC and meeting BC at D. Proof: Statements Reasons 1. In ∆ ADB and ∆ADD (i) AB = AC (S) (ii) ∠BAD = ∠CAD (A) (iii) AD = AD (S) 1. (i) By given (ii) By construction (iii) Common side 2. ∆ADB ≅ ∆ADC 2. By SAS triangles congruent condition from stat. 1. 3. ∠ABC = ∠ACB 3. Corresponding angles of congruent triangles (c.a.c.t.) Hence, it is proved. Corollary: The bisector AD of the vertical angle BAC bisects the base BC perpendicularly. For ∆ABD ≅ ∆ACD implies (i) BD = CD and (ii) ∠ADB = ∠ADC which implies AD ⊥ BC. The above theorem can be proved by dropping a perpendicular from the vertex to the base and then applying the RHS congruent condition or drawing the median AD on BC in ∆ABC. D A B C 288 Allied The Leading Mathematics-9 Triangles 289 Geometry Geometry

“Alternate proof ” Given: ∆ABC is an isosceles triangle in which AB = AC. To prove: ∠ABC = ∠ACB Construction: Draw AD ⊥ BC Statements Reasons 1. In ∆ADB and ∆ADC, (i) ∠BDA = ∠CDA (R) (ii) AB = AC (H) (iii) AD = AD (S) 1. (i) By construction, AD ⊥ BC (ii) By given (iii) Common side 2. ∆ADB ≅ ∆ADC 2. RHS triangles congruent condition 3. ∠ABC = ∠ACB 3. Corresponding angles of congruent triangles (c.a.c.t) Hence, it is proved. Note: A third proof can be given by joining the vertex A to the midpoint D of the base BC and then applying SSS Axiom. Statement: If two angles of a triangle are equal, then the sides opposite to the equal angles are also equal. OR If two angles of a triangle are equal then it is an isosceles triangle. THEOREM 13.3 (Converse of Theorem 13.2) A. Experimental Verification Step 1: Two different line-segments are drawn and each is labelled as BC. B B C A Fig. (ii) C A Fig. (i) Step 2: At each of the end-points B and C, the equal angles are drawn with the help of a compass (or protractor) such that the two arms intersect at A. Step 3: The sides AB and AC are measured in cm and the results are tabulated below: Fig AB AC Remarks (i) 3.0 cm 3.0 cm AB = AC (ii) 2.9 cm 2.9 cm AB = AC Conclusion: ‘If two angles of a triangle are equal, then it is an isosceles triangle.’ D A B C 288 Allied The Leading Mathematics-9 Triangles 289 Geometry Geometry

B. Theoretical Proof Given: ABC is a triangle in which ∠ABC = ∠ACB. To prove: AB = AC or ∆ABC is an isosceles. Construction: Draw AD ⊥ BC. Proof: Statements Reasons 1. In ∆ADB and ∆ADC, (i) ∠ABD = ∠ACD (A) (ii) ∠ADB = ∠ADC (A) (iii) AD = AD (S) 1. (i) By given (ii) AD ⊥ BC (By construction) (iii) Common side 2. ∆ADB ≅ ∆ADC 2. AAS triangles congruent condition 3. AB = AC 3. Corresponding sides of congruent triangles (c.s.c.t.) “Alternative proof” Given: ABC is a triangle in which ∠ABC = ∠ACB. To prove: AB = AC or ∆ABC is an isosceles. Construction: Draw the bisector AD of ∠BAC that meets AB at D. Proof: Statements Reasons 1. In the ∆ADB and ∆ADC, (i) ∠ABD = ∠ACD (A) (ii) ∠BAD = ∠CAD (A) (iii) AD = AD (S) 1. (i) By given (ii) AD bisects of ∠BAC (By construction) (iii) Common side 2. ∆ADB ≅ ∆ADC 2. AAS Axiom 3. AB = AC 3. Corresponding sides of congruent triangles Proved Statement: The straight line drawn perpendicular from the vertex to the base of an isosceles triangle bisects the vertical angle and the base. THEOREM 13.4 A. Experimental Verification Step 1: Two different isosceles triangles ABC are drawn in which AB = AC. D A B C D A B C 290 Allied The Leading Mathematics-9 Triangles 291 Geometry Geometry

D A B C Fig. (ii) D A B C Fig. (i) Step 2: Using set squares (or protractor or otherwise) AD is drawn perpendicular from A to BC. Step 3: BD and DC are measured in cm; and ∠BAD and ∠CAD in degrees. The results are tabulated as shown below: Fig. BD DC Remarks ∠BAD ∠CAD Remarks (i) 2.5 cm 2.5 cm BD = DC 46° 46° ∠BAD = ∠CAD (ii) 1.7 cm 1.7 cm BD = DC 38° 38° ∠BAD = ∠CAD Step 4: Conclusion: ‘The straight line drawn perpendicular from the vertex to the base of an isosceles triangle bisects the vertical angle and the base.’ B. Theoretical Proof Given: ABC is an isosceles triangle in which AB = AC and AD is perpendicular to BC. To Prove: (i) ∠BAD = ∠CAD (ii) BD = DC Proof: Statements Reasons 1. In ∆ADB and ∆ADC, (i) ∠ADB = ∠ADC (R) (ii) AB = AC (H) (iii) AD = AD (S) 1. (i) Both angles being right angles (AD⊥BC) (ii) By given (iii) Common side 2. ∆ADB ≅ ∆ADC 2. RHS triangles congruent condition 3. BD = DC 3. Corresponding sides of congruent triangles 4. ∠BAD = ∠CAD 4. Corresponding angles of congruent triangles Hence, it is proved. D A B C 290 Allied The Leading Mathematics-9 Triangles 291 Geometry Geometry

Statement: In an isosceles triangle, the bisector of the vertical angle is the perpendicular bisector of its base. THEOREM 13.5 A Experimental Verification Step 1: Two different isosceles triangles are drawn and each is labeled ABC in which AB = AC. D A B C Fig. ( i) D A B C Fig. (ii) Step 2: Through the vertex A, the bisector AD of the vertical angle BAC is drawn by using the protractor or otherwise. Step 3: BD and DC are measured in cm; and the adjacent angles ∠BDA and ∠CDA in degrees. The results are tabulated as shown below: Fig. BD DC Remarks ∠BDA ∠CDA Remarks (i) 1.6 cm 1.6 cm BD = DC 90° 90° BD = DC, ∠BDA = ∠CDA, So AD⊥BC (ii) 2.4 cm 2.4 cm BD = DC 90° 90° BD = DC, ∠BDA = ∠CDA, So AD⊥BC Conclusion: ‘In an isosceles triangle, the bisector of the vertical angle is the perpendicular bisector of its base.’ Statement: The line joining the vertex and the mid-point of the base of an isosceles triangle is perpendicular to the base and bisects the vertical angle. *THEOREM 13.6 Experimental Verification Step 1: Two different isosceles triangles are drawn and each is labeled PQR in such a way that PQ = PR. Step 2: The vertex P is joined to the middle point S of the base QR. Step 3: ∠QPS, ∠RPS, ∠PSQ, and ∠PSR are measured in degrees. The results are tabulated as shown below: Fig. ∠QPS ∠RPS Remarks ∠PSQ ∠PSR Remarks (i) 35° 35° ∠QPS = ∠RPS 90° 90° PS⊥PR (ii) 40° 40° ∠QPS = ∠RPS 90° 90° PS⊥QR S P Q R Fig. (i) S P Q R Fig. (ii) 292 Allied The Leading Mathematics-9 Triangles 293 Geometry Geometry

Conclusion: ‘The line joining the vertex and the mid-point of the base of an isosceles triangle is perpendicular to the base and bisects the vertical angle.’ To prove the above theorem, we have to note that the ∆PSQ and ∆PSR are congruent by SSS Axiom, since (i) PQ = PR, (ii) PS = PS, (iii) QS = RS. Then, ∆PQS ≅ ∆PRS implies ∠QPS = ∠QPS and ∠PSQ = ∠PSR. Since the adjacent angles being equal, so PS ⊥ QR. Example-1 In ∆BAC, ∠A = 90o , AB = AC. Find the measure of the base angles. Solution: Here, In ∆BAC, since AB = AC then ∠C = ∠B. [ Base angles] Also, ∠A + ∠B + ∠C = 180o [ Sum of angles of ∆ABC] or, 90o + ∠B + ∠B = 180o or, 2∠B = 180o – 90o = 90° ⇒ ∠B = 45°. Hence, ∠B = ∠C = 45o . Example-2 Observe the given ∆PQR. (a) Find the measures of ∠PQS and ∠PRQ. (b) Is ∆PQR isosceles or equilateral? Justify it. Solution: (a) Here, in ∆PQR, ∠PQS +∠QPR = 120o , [ Exterior angle property on ∆PQR] So, ∠PQS = 120o – ∠QPR = 120o – 60o = 60o . Also, ∠R = 180o – 120o = 60o . (Why?) (b) Since, ∠P = ∠Q = ∠R, the triangle is both isosceles and equiangular. Example-3 Observe the given figure in which AD = CD = BC and ∠ACD = 25°. (a) Find the value of x. (b) Write the relation between ∠ABC and ∠BAC. Solution: (a) In the given figure, AD = CD = BC, ∠ACD = 25°, x = ? Now, we have (i) ∠BAC = ∠ACD = 25° [ Base angles of isosceles ∆ACD] (ii) ∠BDC = ∠BAC + ∠ACD [ Exterior angle property in ∆ACD] = 25° + 25° = 50° (iii) ∠BDC = ∠CBD = 50° [ Base angles of isosceles ∆ACD] (iv) ∠BCD + ∠BDC + ∠CBD = 180° [ Sum of angles of ∆BCD] or, x + 50° + 50° = 180° or, x = 180° – 100° = 80°. (b) Since ∠ABC = 50° and ∠BAC = 25°, so ∠ABC = 2∠BAC. A C B Q 60° 120° R S P B C x 25° D A 292 Allied The Leading Mathematics-9 Triangles 293 Geometry Geometry

Example-4 In the figure, RT is the bisector of ∠PRS and RT//QP. Prove that ∆PQR is an isosceles triangle. Solution: Given: In the given figure, RT is the bisector of ∠PRS, i.e, ∠PRT = ∠SRT and RT//QP. To Prove: ∆PQR is an isosceles triangle. Proof: Statements Reasons 1. ∠PRT = ∠SRT 1. By given 2. ∠PRT = ∠QPR 2. Alternate angles 3. ∠SRT = ∠PQR 3. Corresponding angles 4. ∠QPR = ∠PQR 4. From statements (1), (2) and (3). 5. ∆PQR is an isosceles triangle. 5. Equal base angles in stat. (4). Hence, it is proved. Example-5 In the given isosceles triangle ABC, AB = AC, D and E are the mid-points of AB and AC respectively. Prove that CD = BE. Solution: Given: In the given isosceles triangle ABC, AB = AC, D and E are the mid-points of AB and AC respectively. To Prove: CD = BE Proof: Statements Reasons 1. AB = AC 1. By given 2. AD = BD and AE = CE 2. By given 3. In ∆BCD and ∆BCE, (i) BD = CE (S) (ii) ∠CBD = ∠BCE (A) (iii) BC = BC (S) 3. (i) From statements (1) and (2). (ii) Base angles (iii) Common side 4. ∆BCD ≅ ∆BCE 4. By SAS triangles congruent condition 5. CD = BE 5. Corresponding sides of congruent triangles Hence, it is proved. P R S T Q B C D E A 294 Allied The Leading Mathematics-9 Triangles 295 Geometry Geometry

Example-6 In the given figure, ∆ABC and ∆BDE are the equilateral triangles. Prove that AD = CE. Solution: Given: In the given figure, ∆ABC and ∆BDE are the equilateral triangles. To Prove: AD = CE Proof: Statements Reasons 1. ∠ABC = ∠DBE 1. Angle of the equilateral triangles ABC and BDE 2. ∠ABC + ∠CBD = ∠DBE + ∠CBD 2. Addition axiom 3. In ∆ABD and ∆BCE, (i) AB = BC (S) (ii) ∠ABD = ∠CBE (A) (iii) BD = BE (S) 3. (i) Sides of equilateral triangle ABC (ii) From statement (2) (iii) Sides of equilateral triangle BDE 4. ∆ABD ≅ ∆BCE 4. By SAS congruent condition 5. AD = CE 5. Corresponding sides of congruent triangles Hence, it is proved. PRACTICE 13.2 Keeping Skill Sharp Circle the correct answer. 1. (a) Which angles are equal in ∆ABC alongside? (i) a and b (ii) a and c (iii) b and c (iv) None (b) If ∠Q = ∠R in ∆PQR, which sides are equal? (i) PQ and PR (ii) PQ and QR (iii) PR and QR (iv) None (c) In ∆LMN, LM = LN and LK ⊥ MN. What is the relation between MK and NK? (i) MN = MK (ii) MK = KN (iii) MK > KN (iv) MK < MN (d) In ∆ABC,AB =AC and ∠BDA= ∠CDA. What isthe relation between ∠BAD = ∠CAD? (i) ∠BAD > ∠CAD (ii) ∠BAD < ∠CAD (iii) ∠BAD = ∠CAD (iv) ∠BAD = ∠CAD C D B A E B b c a C A Q R P Fig. 1(a) Fig. 1(b) M K N L B D C A 294 Allied The Leading Mathematics-9 Triangles 295 Geometry Geometry

2. (a) Write down two properties of an isosceles triangle. (b) What is the measure of each acute angle of an isosceles right-angled triangle? (c) What is the relation between the bisector of the vertical angle of an isosceles triangle and its base? (d) Write the relation between the perpendiculars draw from the base points to the equal sides of an isosceles triangle. 3. (a) In BAC, ∠A = 90o , AB = AC. Find the base angles. (b) In ∆PQR, ∠Q = 120o , PQ = QR. Find the base angles. Check Your Performance 4. In the ∆PQR, the measures of some angles are shown. (a) Which angles are equal in an isosceles triangle? (b) Find the values of ∠PQR and ∠PRQ. (c) Is ∆PQR isosceles? Why? 5. Observe the equal sides marked in the following figures and answer the given questions. (i) (ii) (iii) (iv) (a) Find the values of a, b, c, d, e, wherever applicable. (b) Measure the remaining angles. 6. (i) Find the angles of an isosceles triangle in which each of the base angles is half of the vertical angle. (ii) Find the angles of an isosceles triangle in which the vertical angle is half of the base angle. 7. (i) (a) What is an isosceles triangle? Define it. (b) In the adjoining triangle, LM = LN. Show that ∠LMN = ∠MNL. (c) If ∠MLN = 34o , find the measure of ∠LMN and ∠MNL. (ii) Prove that if two angles are equal in a triangle, the sides opposite to the equal angles are also equal. 8. (i) Prove that the perpendicular drawn from the vertex of an isosceles triangle to the base bisects the base. (ii) In the isosceles triangle PR = PQ and QS = RS. (a) Prove that: ∠QPS = ∠RPS (b) Prove that: PS⊥QR (c) If ∠QPS = 30o , find the value of ∠PQR. Q R 150° 110° S P Fig. 4(ii) Q 115° 50° R S P Fig. 4(i) 58° a 4c 3c a b 48° 120° a N L M Q S P R 296 Allied The Leading Mathematics-9 Triangles 297 Geometry Geometry

9. (i) Prove that the base angles of an isosceles right triangle are equal and each measures half of right angle. (ii) If a triangle is an equilateral, show that it is equiangular triangle and each angle measures 60o . (iii) If a triangle is an equiangular, show that it is equilateral triangle. 10. (i) ABC is an isosceles triangle in whichAB =AC andAD bisects ∠CAE. (a) Prove that AD//BC. (b) If ∠EAD = 48o , ∠ACB. (ii) In a triangle ABC, AB = AC and O is an interior point of ∆ABC. OB and OC are the bisectors of angles B and C. Prove that OBC is an isosceles triangle. (iii) PQR is an isosceles triangle in which PQ = PR. M and N are the mid-points of PR and PQ. Prove that QM = NR. 11. (i) In the given figure, ∆PQR is an isosceles triangle, QS = TR and ∠SPT = 40°. (a) Prove that ∆PST is an isosceles triangle. (b) Find the measure of ∠PST. (ii) In the adjoining figure, AB = AC, AD = AE and ∠ADC = 75°. (a) Prove that BE = CD. (b) Find the measure of ∠DAE. 12. (i) In the adjoining figure, ∆ABC and ∆CDE are both isosceles triangles and ∠ACB = ∠DCE. Prove that: ∠ADC = ∠BEC. (ii) In the given figure, ∆PQR and ∆RST are both equilateral triangles. Prove that: PS = QT (iii) In the given figure, D, E and F are the mid-points of the sides AB, BC and CA of the equilateral triangle ABC respectively. Prove that ∆DEF is also an equilateral triangle. 4. (i) 65°, 65°, Yes (ii) 40°, 30°, No 5. (a) 61° (b) 18° (ii) 66°, 33° (iv) 60° 6. (i) 45°, 45°, 90° (ii) 72°, 72°, 36° 7. (i) 73° 8. (ii) 60° 10. (i) 48° 11. (i) 70° (ii) 30° Answers Project Work Show the properties of an isosceles triangle by cutting and folding the paper in the classroom and draw them separately in the full of A4 paper. A D E B C Q S T P R 40° B D E A C 75° Fig. 11(i) Fig. 11(ii) P Q S R T A B D C E Fig. 12(i) Fig. 12(ii) B E C F A D 296 Allied The Leading Mathematics-9 Triangles 297 Geometry Geometry

13.3 Sides and Angles of a Triangle At the end of this topic, the students will be able to: ¾ prove the theorems related on the relation among the sides and angles of triangle by experimental. Learning Objectives Sides and angles are closely related to themselves and to each other. Interesting cases are those in which unequal angles and sides are related to each other. We can prove many of them theoretically. We, however, restrict ourselves to experimental verifications and simple numerical computation of some such relations. Such relations will be stated as theorems. Statement: The sum of any two sides of a triangle is greater than its third side. *THEOREM 13.7 Experimental Verification Step 1: Two triangles ABC are drawn in different shape and size. B C A Fig. (i) B C A Fig. (ii) Step 2: Measure the sides AB, BC and CA in cm and add any two sides. The results are tabulated below: Fig. AB BC CA AB + BC BC + CA AB + CA Remarks (i) 3.5 cm 3.1 cm 3.7 cm 6.6 cm 6.8 cm 7.2 cm AB + BC > CA, BC + CA > AB, AB + CA > BC (ii) 3.6 cm 4.6 cm 5.8 cm 8.2 cm 10.4 cm 11.8 cm From the above measurement and calculation, we arrive at the following conclusion: The sum of any two sides of a triangle is greater than the third side. Proved. 298 Allied The Leading Mathematics-9 Triangles 299 Geometry Geometry

Statement: If two sides of a triangle are unequal then the angle opposite to the longer side is greater than the angle opposite to the shorter side. *THEOREM 13.8 Experimental Verification Step 1: Two triangles ABC are drawn in different shape and size in such a way that AC > BC > AB. B C A Fig. (i) B C A Fig. (ii) Step 2: The angles ∠B and ∠C are measured and tabulated: Fig ∠B opposite to the longer side AC ∠C opposite to the shorter side AB Remarks (i) 104° 26° ∠B > ∠C (ii) 85° 28° ∠B > ∠C From the above measurement and calculation, we arrive at the following conclusion: ‘The angle opposite to the greater side of a triangle is greater than the angle opposite to the smaller.’ Proved. Statement: If two angles of a triangle are unequal, then the side opposite to the greater angle is longer than the side opposite to the smaller. *THEOREM 13.9: (Converse of Theorem 13.8) Experimental Verification Step 1: Two triangles of different shape and size are drawn and each is labeled ABC so as to make ∠B > ∠C. B C A B C Fig. (ii) A Fig. (i) Step 2: The sides AC and AB opposite to the ∠B and ∠C are measured in cm and tabulated. Fig Side AC opposite to greater ∠B Side AB opposite to smaller ∠C Remarks (i) 4.3 cm 2.1 cm AC > BC (ii) 4.6 cm 2.1 cm AC > BC From the above measurement and calculation, we arrive at the following conclusion: ‘The side opposite to the greater angle of a triangle is greater than the side opposite to smaller angle.’ Proved. 298 Allied The Leading Mathematics-9 Triangles 299 Geometry Geometry

Statement: Of all the line segments that can be drawn to a given straight line from a given point outside it, the perpendicular is the shortest. *THEOREM 13.10 Experimental Verification Step 1: Three different line-segments PM, PB and PN are drawn on the line XY from the point P outside XY and drawn PA perpendicular to XY. X M B A N P Y Fig. (i) X M A B N P Y Fig. (ii) Step 2: The lengths of the line segments PM, PB, PN and PA are measured in cm and tabulated below: Fig. PM PB PN PA Remarks (i) 2.5 cm 2 cm 2 cm 1.8 cm PA is the shortest. (ii) 2 cm 2.3 cm 2.9 cm 1.9 cm PA is the shortest. Conclusion: Of all the straight lines drawn to a given straight line from a given point outside it, the perpendicular is the shortest. Proved. Statement: Of all the straight lines drawn to a given straight line from a given point outside of it, the shortest one is the perpendicular to the given line. *THEOREM 13.11: (Converse of Theorem 13.10) Note: Since the shortest distance of a straight line from a given point outside it, is defined as the length of the perpendicular from the point to the given line, no question of experimental verification arise. Moreover, there is no other criterion that tells us how to draw the shortest linesegment from a point to a line that extends both sides endlessly. Example-1 In a right-angled triangle, the side adjacent to the right angle is shorter than the hypotenuse. Give a short proof. Solution: If ABC is a triangle with right angle at B, then hypotenuse is AC and sides adjacent to the right angle B are AB and BC. Since each of the angles ∠A and ∠C is acute, i.e., ∠A or ∠C < ∠B Since side opposite to the smaller angle < side opposite to greater angle, we have BC or AB < AC. B C A 300 Allied The Leading Mathematics-9 Triangles 301 Geometry Geometry