Vedanta Excel in Mathematics Book 9 Final (2080)_compressed

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 201 Vedanta Excel in Mathematics - Book 9 Indices OBJECTIVE QUESTIONS Let’s tick (√) the correct alternative. 1. How many times has x been multiplied in 4x3 ? (A) 4 (B) 3 (C) 7 (D) 12 2. For any two positive integers m and n, am × an is equal to (A) a2mn (B) am + n (C) am – n (D) amn 3. What is the value of (9x) o, x≠0? (A) 1 (B) 9 (C) 9x (D) 0 4. The simplified value of 24 × 42 ÷83 is (A) 2 (B) 4 (C) 1 2 (D) 1 4 5. If m + n = m, what is the value of mn? (A) 0 (B) 1 (C) m (D) n 6. What is the value of ab when a = –3 and b = –2? (A) 9 (B) –9 (C) 1 9 (D) – 1 9 7. What is the value of a° 64 – 2 3 ? (A) 4 (B) 16 (C) 64 (D) 512 8. What is the simplified form of 1 1 + xa – b + 1 1 + xb – a ? (A) 1 (B) x (C) a (D) b 9. The value of 3 9 9 is (A) 3 (B) 9 (C) 27 (D) 81 10. The simplified form of 5m+2 – 5m 5m+1 – 5m is (A) 2 (B) 4 (C) 6 (D) 8 https://www.geogebra.org/m/xffnt9b3 Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Vedanta Excel in Mathematics - Book 9 202 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 12.1 Simultaneous equations - review Let’s take a linear equation, y = x + 4. It is a first degree equation with two variables x and y because each variable is raised to the first power, i.e. 1. The standard form of linear equations in two variables is ax + by = c, where a, b, and c are constants, and a or b both are not zero. The equation, y = x + 4 has as many pairs of solutions as we wish to find. The table given below shows a few pairs of solutions. x 0 1 2 3 –1 –2 –4 y 4 5 6 7 3 2 0 Thus, (0, 4), (1, 5) (2, 6), (3, 7), (–1, 3), (–2, 2), (–4, 0), … are a few pairs of solutions that satisfy the equation y = x + 4. Again, let’s consider another linear equation y = 2x + 1 A few pairs of solutions of this equations are shown in the table below: x 0 1 3 4 –1 –2 y 1 3 7 9 –1 –3 Thus, (0, 1), (1, 3), (3, 7), (4, 9), (–1, –1), (–2, –3) are a few pairs of solutions that satisfy y = 2x + 1. Here, (3, 7) is the common pair of solution that satisfies both the equations simultaneously. Such pair of equations that have only one pair of solution which satisfies both the equations simultaneously at a time are called simultaneous equations. 12.2 Method of solving simultaneous equations There are various methods of solving simultaneous equations. Here, we discuss only two methods: (i) Elimination method (ii) Substitution method (i) Elimination method In this method, we add or subtract the given equations to eliminate one of the two variables by making their coefficients equal. Then, a single equation with one variable so obtained is solved to find the value of the variable. The value of the variable is substituted to any one equation to find the value of the eliminated variable. Unit 12 Simultaneous Equations

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 203 Vedanta Excel in Mathematics - Book 9 Example 3: Solve: 3 x + 2 y = 1 and 4 x + 3 y = 17 12 Solution: 3 x + 2 y = 1 ........................ (i) 4 x + 3 y = 17 12 ....................... (ii) Checking: When x = 3 and y = 4, From equation (i): LHS = x + 2y = 3 + 2 × 4 = 11 = RHS From equation (ii): LHS = 3 + 4 = 7 = RHS Here we are eliminating y. So, to make the coefficient of y same, equation (ii) is multiplied by 5. Example 2: Solve: 2x + 5y = 9 and x – y = 1 Solution: 2x + 5y = 9 ....................... (i) x – y = 1 .......................... (ii) Multiplying equation (ii) by 5 and adding to (i), 2x + 5y = 9 5x – 5y = 5 7 = 14 or, x = 2 Now, substituting the value of x in equation (ii), we get, 2 – y = 1 or, y = 1 \ x = 2 and y = 1 Checking: When x = 2 and y = 1, From equation (i): LHS = 2x+5y = 2×2 + 5×1 = 9 = RHS From equation (ii): LHS =x – y = 2 – 1 = 1 = RHS Worked-out examples Example 1: Solve: x + 2y = 11 and x + y = 7. Solution: x + 2y = 11 ....................... (i) x + y = 7 .......................... (ii) Subtracting equation (ii) from (i) x + 2y = 11 + x + y =+ 7 y = 4 Now, substituting the value of y in equation (ii), we get, x + 4 = 7 or, x = 3 \ x = 3 and y = 4 – – – The coefficients of x are the same in both equations. So, we subtract one equation from the other to eliminate x. Simultaneous Equations

Vedanta Excel in Mathematics - Book 9 204 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Multiplying equation (i) by 3 and (ii) by 2 and subtracting (ii) from (i), 9 x + 6 y = 3 ........................ (i) ± 8 x ± 6 y = – 34 12 ....................... (ii) 1 x = 1 6 or, x = 6 Now, substituting the value of x in equation (i), we get, 3 6 + 2 y = 1 or, y = 4 \ x = 6 and y = 4 Checking: When x = 6 and y = 4, From equation (i); LHS = 3 x +2 y = 3 6 + 2 4 = 1 2+1 2 = 1 = RHS From equation (ii); LHS =4 x +3 y = 4 6 + 3 4 = 2 3+3 4 = 17 12 = RHS (ii) Substitution method In this method, a variable is expressed in terms of another variable from one equation and it is substituted in the remaining equations. Example 5: Solve x + 3y = 1700 and 7x – y = 900 Solution: x + 3y = 1700 ... (i) and 7x – y = 900 ... (ii) From equation (i), x = 1700 – 3y ... (iii) Substituting for x from equation (iii) in equation (ii), we get, 7(1700 – 3y) – y = 900 or, 11900 – 21y – y = 900 or, 11000 = 22y ∴ y = 500 Now, substituting the value of y in equation (iii), we get, x = 1700 – 3 × 500 = 1700 – 1500 = 200 \ x = 200 and y = 500. Example 6: Solve the given system of equation by substitution method. x – 1 y + 1 = 3 4 and x + 2 y – 2 = 4 3 Solution: x – 1 y + 1 = 3 4 or, 4x – 4 = 3y + 3 or, 4x – 3y = 7 or, 4x = 7 + 3y ∴ x = 7 + 3y 4 ... (i) Simultaneous Equations

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 205 Vedanta Excel in Mathematics - Book 9 Simultaneous Equations Also, x + 2 y – 2 = 4 3 or, 3x + 6 = 4y – 8 or, 3x – 4y = –14 ... (ii) Now, substituting the value of x in equation (ii), we get, 3 7 + 3y 4 – 4y = –14 or, 21 + 9y – 16y 4 = –14 or, 21 – 7y = –56 or, 77 = 7y ∴ y = 11 Again, substituting the value of y in equation (i), we get, x= 7 + 3 × 11 4 = 10 Hence, x = 10 and y = 11. General section 1. Find the value of the variables as indicated. a) If x = 2 in 2x + y = 5, find the value of y. b) If x = –1 in x – y = 3, find the value of y. c) If x = –y in x + 2y = –2, find the values of y and x. d) If y = 3 in 2x – y = 5, find the value of x. e) If y = x 2 in x + y = 9, find the value of x. 2. Select the correct pair of solutions of the given pair of equations. a) x + y = 7 and x – y = 3 (i) (4, 3) (ii) (6, 3) (iii) (5, 2) b) 2x – y = –2 and x + 2y = 9 (i) (1, 4) (ii) (4, 6) (iii) (3, 3) c) y – 3x = 10 and 3y – x = 6 (i) (7, –1) (ii) (–3, 1) (iii) (–1, 1) d) y = x + 2 and y = 3y – x = 6 (i) (5, 3) (ii) (4, 6) (iii) (3, 5) e) y = x – 1 2 and y = x + 1 3 (i) (5, 2) (ii) (2, 5) (iii) (2, 1) Creative section - A 3. Solve each pair of simultaneous equations by elimination method. b) 3x + y = 13 x – y = 3 c) x + 2y = 7 x + y = 4 a) x + y = 5 x – y = 1 Checking: When x = 10 and y = 11, From equation (i); LHS = x–1 y+1= 10–1 11+1= 9 12= 3 4 RHS From equation (ii); LHS = x+2 y–2 =10+2 11–2 =12 9 = 4 3 RHS EXERCISE 12.1 e) 2x + 3y = 3 x + 2y = 1 f) 6x + 7y = 5 7x + 8y = 6 d) 3x + 5y = 11 4x – y = 7

Vedanta Excel in Mathematics - Book 9 206 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur h) 2x – 3y – 12 = 0 3x – 2y – 13 = 0 i) 3x + 2y – 15 = 0 5x – 3y – 25 = 0 g) 2x + 3y – 1 = 0 5x + 2y + 3 = 0 j) 4(x – 1) + 5(y + 2) = 10 5(x + 1) – 2(y – 2) = 14 k) 2(x + 2) – 3(y – 1) = 6 7(x – 1) + 4(y + 2) = 12 l) 2x + 5 = 4(y + 1) – 1 3x + 4 = 5(y + 1) – 3 m) 2 x + 6 y = 3 10 x – 9 y = 2 n) 9 x – 4 y = 5 3 x + 8 y = – 3 o) x 5 + 6 y = 4 x 2 – 9 y = 2 4. Solve each pair of simultaneous equations by substitution method. a) y = 2x x + y = 9 d) 3x + 2y = 26 x = y + 2 g) x = 3 – y 2x – y = 3 b) y = 5x x + y = 12 e) y = x – 3 3x + 4y = 2 h) x + 2y = 9 x = 2y + 1 c) y = –4x x + 3y = – 11 f) y = 3x + 1 2x + y = 6 j) 2x + y = 4 2x – y = 8 k) x + 3y = 7 3x – y = 11 l) 2x – 3y = 6 x + 3y = 3 i) 2x + y = 8 x – y = 1 m) x – 1 y + 1 = 1 2 x – 2 y + 2 = 1 3 n) x – 2 y – 2 = 1 2 x + 2 y + 2 = 3 4 o) x + y – 2 3 = 6 y + x + 1 2 = 8 12.3 Application of simultaneous equations Simultaneous equations are broadly used to find two unknown quantities. Considering two unknown quantities which are asked to find in a word problem by any two variables such as x and y, we can make a pair of simultaneous equations under the two given conditions. By solving the equations, we can find the unknown quantities. Worked-out examples Example 1: The sum of two numbers is 45. The greater number is 9 more than the smaller one. (a) Express the above statements in the form of linear equations. (b) Solve the equations and find the numbers. Solution: (a) Let the greater number be x and the smaller one be y. From the first given condition, x + y = 45 or, x = 45 – y ... (i) From the second given condition, x = y + 9 … (ii) The sum of the number is 45. ∴ x + y = 45 The greater number is 9 more than smaller one. ∴ x = y + 9 Simultaneous Equations

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 207 Vedanta Excel in Mathematics - Book 9 Simultaneous Equations (b) Substituting the value of x from equation (i) in (ii), we get, 45 – y = y + 9 or, – y – y = 9 – 45 or, – 2y = – 36 or, y = 18 Now, substituting the value of y in equation (i), we get, x = 45 – 18 = 27 So, the required greater number is 27 and the smaller one is 18. Example 2: Last week, the total cost of 3 kg of apples and 5 kg of oranges was Rs 1,155. Likewise, the cost of 1 kg of apples was the same as the cost of 2 kg of orange. (a) Express the above statements in the form of linear equations. (b) Find the rate of cost of these fruits. (c) If the rate of cost of apples is decreased by 15% and the rate of cost of oranges is increased by 10% this week, how many kilograms of apples and oranges of same quantity can be bought for Rs 588? Solution: (a) Let the rate of cost of apples be Rs x per kg. And, the rate of cost of oranges be Rs y per kg. From the first given condition, 3x + 5y = 1155 … (i) From the second given condition, x = 2y … (ii) (b) Substituting the value of x from equation (ii) in equation (i), we get, 3 × 2y+ 5y = 1155 or, 6y + 5y = 1155 or, 11y = 1155 or, y = 1155 11 = 105 Now, substituting the value of y in equation (ii), we get, x = 2y = 2 × 105 = 210 So, the required rate of cost of apples is Rs 210 per kg and that of oranges is Rs 105 per kg. (c) This week, the rate of cost of 1 kg apples = Rs 210 – 15% of Rs 210 = Rs 178.50 The rate of cost of 1 kg oranges = Rs 105 + 10% of Rs 105 = Rs 115.50 Let the quantity of apples = the quantity of oranges = w kg Now, total cost of w kg apples and w kg oranges = Rs 588 or, 178.5w + 115.5w = 588 or, 294w = 588 or, w = 588 294 = 2 So, this week 2 kg of apples and 2 kg of oranges can be bought for Rs 588. Answer checking The cost of 3 kg apples and 2 kg of oranges = 3 × Rs 210 + 2 × Rs 105 = Rs 1,155 which is given in the question.

Vedanta Excel in Mathematics - Book 9 208 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 3: Two years ago, father was 6 times as old as his son was. Three years hence he will be 5 years older than 3 times the age of his son. (a) Express the above statements in the form of linear equations. (b) Find their present ages. (c) After how many years, will the father be only twice as old as his son? Solution: (a) Let the present age of the father be x years. And the present age of his son be y years. From the first given condition, x – 2 = 6 (y – 2) or, x = 6y – 12 + 2 or, x = 6y – 10 … (i) From the second given condition, x + 3 = 5 + 3 (y + 3) or, x = 5 + 3y + 9 – 3 or, x = 3y + 11 … (ii) (b) Substituting the value of x from equation (i) in equation (ii), we get, 6y – 10 = 3y + 11 or, 6y – 3y = 11 + 10 or, 3y = 21 or, y = 7 Now, substituting the value of y in equation (i), we get, x = 6 × 7 – 10 or, x = 32 Hence, the present age of the father is 32 years and that of the son is 7 years. (c) Let after t years, the father will be only twice as old as his son. So, 32 + t = 2 (7 + t) or, 32 + t = 14 + 2t or, t = 18 Thus, the father will be only twice as old as his son after 18 years. Example 4: A year hence a father will be 5 times as old as his son. Two years ago, he was three times as old as his son will be four years hence. (a) Express the above statements in the form of linear equations. (b) Find their present ages. (c) What will be the age of father when both of them live until the son becomes as old as his father at present? Solution: (a) Let the present age of the father be x years and the present age of the son be y years. From the first condition, x + 1 = 5 (y + 1) or, x = 5y + 5 – 1 or, x = 5y + 4 …. (i) From the second condition, x – 2 = 3 (y + 4) or, x – 2 = 3y + 12 or, x = 3y + 14 …. (ii) Two years ago, the age of the father was (x – 2) years and that of son was (y – 2) years. Father was 6 times as old as his son means x – 2 = 6 (y – 2) 3 years hence, father will be (x + 3) years and the son will be (y + 3) years. 5 years older than 3 times the age of the son means x + 3 = 5 + 3 (y + 3) Simultaneous Equations

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 209 Vedanta Excel in Mathematics - Book 9 Simultaneous Equations (b) Substituting the value of x from equation (i) in equation (ii), we get 5y + 4 = 3y + 14 or, 5y – 3y = 14 – 4 or, 2y = 10 or, y = 5 Now, substituting the value of y in equation (i), we get, x = 5 × 5 + 4 = 29 Hence, the present age of the father is 29 years and that of the son is 5 years. (c) The difference between the ages of the father and his son = 29 years – 5 years = 24 years. So, after 24 years, the age of son will be 5 year + 24 years = 29 years, which is the present age of father. At the same time, the age of father will be 29 years + 24 years = 53 years. Hence, the father will be 53 years old when son’s becomes as old as his father at present. Example 5: The sum of the digits of a two–digit number is 8. If 18 is subtracted from the number, the places of the digits are reversed. (a) Express the above statements in the form of linear equations. (b) Solve the equations and find the number. (c) By how much more or less is the two digit number having same digits at both one’s and ten’s places and the nearest to original number? Solution: (a) Let the digit at tens place of the number be x and the digit at ones place of the number be y. ∴ The number so formed is 10x + y. From the first given condition, x + y = 8 ∴ x = 8 – y … (i) From the second given condition, 10x + y – 18 = 10y + x or, 10x + y – 10y – x = 18 or, 9x – 9y = 18 or, 9 (x – y) = 18 or, x – y = 2 ... (ii) (b) Substituting the value of x from equation (i) in equation (ii), we get, 8 – y – y = 2 or, – 2y = 2 – 8 or, y = 3 Now, substituting the value of y in equation (i), we get, x = 8 – 3 = 5 Hence, the required number is 10x + y = 10 × 5 + 3 = 53. (c) Since, the two digits numbers having same digits at both the places are 11, 22, 33, 44, 55, … etc. The number nearest to 53 is 55. So, 55 is 2 more than the original number 53. Let's consider a number 26. In 26, 2 is in tens place and 6 is in ones place. ∴ 26 = 10 × 2 + 6 Similarly, if x is in tens place and y in ones place, the number is 10x + y. The original number is 10x + y. When the places of the digits are reversed, the new number so formed is 10y + x.

Vedanta Excel in Mathematics - Book 9 210 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 6 : A two digit number is 3 times the sum of its digits. The sum of the number formed by reversing its digits and 9 is equal to 3 times the original number. Find the number. Solution: Let the digit at tens place of the number be x and the digit at ones place of the number be y. Then, the number so formed is 10x + y. Again, the number formed by reversing the digits is 10y + x. From the first given condition, 10x + y = 3(x + y) or, 10x + y= 3x + 3y or, y = 7x 2 ... (i) Substituting the value of y from equation (i) in equation (ii), we get, 7 × 7x 2 – 29x = – 9 or, 49x – 58x = –18 or, –9x = – 18 or, x = 2 Now, putting the value of x in equation (i), we get, y = 7x 2 = 7 × 2 2 = 7 Hence, the required number is 27. Example 7: In a fraction, the numerator is 1 less than the denominator. If 1 is added to the numerator and 5 is added to the denominator, the fraction becomes 1 2 . Find the original fraction. Solution: Let the numerator and the denominator of the original fraction be x and y respectively. Then, the required fraction is x y . From the first given condition, x = y – 1 ... (i) From the second given condition, x + 1 y + 5 = 1 2 or, 2x + 2 = y + 5 or, 2x – y = 3 ... (ii) Now, substituting the value of x from equation (i) in equation (ii), we get, 2(y – 1) – y = 3 or, 2y – 2 – y = 3 or, y = 5 Again, putting the value of y in equation (i) we get, x = y – 1 = 5 – 1 = 4 So, the required fraction is 4 5 . From the second given condition, (10y + x) + 9 = 3(10x + y) or, 10y + x + 9 = 30x + 3y or, 7y – 29x = – 9 ... (ii) Simultaneous Equations

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 211 Vedanta Excel in Mathematics - Book 9 Simultaneous Equations Example 8: The monthly income of Ram and Hari are in the ratio of 3 : 2 and their expenses are in the ratio of 5 : 3. If each of them saves Rs 4000 in a month, find their monthly incomes. Solution: Let the monthly income of Ram be Rs x. And the monthly income of Hari be Rs y. ∴ The monthly expense of Ram = Rs (x – 4,000) The monthly expense of Hari = Rs (y – 4,000) From the first given condition, x y = 3 2 or, x = 3y 2 ... (i) From the second given condition, x – 4,000 y – 4,000 = 5 3 or, 3x – 12,000 = 5y – 20,000 or, 3x = 5y – 8,000 ... (ii) Substituting the value of x from equation (i) in equation (ii), we get, 3 × 3y 2 = 5y – 8,000 or, 9y = 10y – 16,000 or, y = 16,000 Now, substituting the value of y in equation (i), we get. x = 3 × 16000 2 = 24,000 Hence, the monthly income of Ram is Rs 24,000 and that of Hari is Rs 16,000. Example 9: Upendra started his motorbike journey from Damak to Lahan at 7 a.m. with an average speed of 40 km/hr. An hour later Tashi also started his journey from Damak to Lahan with a uniform speed of 60 km/hr. At what time would they meet each other? Solution: Suppose Upendra meets Tashi after x hours. But, Tashi meets Upendra after (x – 1) hour. Now, the distance travelled by Upendra in x hours = 40x km. Also, the distance travelled by Tashi in (x – 1) hours = 60(x – 1) km. When they travelled the equal distance, they meet each other. ∴ 40x = 60(x – 1) or, 2x = 3x – 3 or, x = 3 ∴ They meet after 3 hours. Since, Upendra started his journey at 7 a.m., they meet at 10 a.m.

Vedanta Excel in Mathematics - Book 9 212 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 10: Two buses were coming from two villages situated just in the opposite direction. The uniform speed of one bus is 10 km/hr more than that of another one and they had started their travelling in the same time. If the distance between the villages is 450 km and they meet after 5 hours, find their speeds. Solution: Let the speed of the faster bus be x km/hr and the speed of slower bus be y km/hr. From the first given condition, x – y = 10 ... (i) From the second given condition, 5x + 5y = 450 or, 5(x + y) = 450 or, x + y = 90 ... (ii) Adding equations (i) and (ii), x – y + x + y = 10 + 90 or, 2x = 100 or, x = 50 Now, substituting the value of x in equation (ii), we get, 50 + y = 90 or, y = 40 So, the speed of the faster bus is 50 km/hr and the slower one is 40 km/hr. Example 11: A few number of students are made to stand in rows. If 5 more students were kept in each row, there would be 3 rows less. If 5 less students were kept in each row, there would be 5 rows more. Find the number of students. Solution: Let, the number of rows be x and the number of students in each row be y. Then, the total number of students = xy From the first condition, From the second condition, (x – 3) (y + 5) = xy (x + 5) (y – 5) = xy or, xy + 5x – 3y – 15 = xy or, xy – 5x + 5y – 25 = xy or, 5x – 3y = 15 ... (i) or, –5x + 5y = 25 ... (ii) Adding equation (i) and (ii), we get, 5x – 3y – 5x + 5y = 15 + 25 or, 2y = 40 or, y = 20 Now, putting the value of y in equation (i), we get, 5x – 3 × 20 = 15 or, x = 15 Then, the number of students = 15 × 20 = 300 Hence, there are 300 students. Distance travelled by faster bus = 5x Distance travelled by slower bus = 5y When they meet, they travelled 450 km. travelling from opposite direction. So, 5x + 5y = 450 450 km Simultaneous Equations

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 213 Vedanta Excel in Mathematics - Book 9 Simultaneous Equations EXERCISE 12.2 General section 1. a) The sum of two numbers is 60 and their difference is 10. (i) Express the above statements in the form of linear equations. (ii) Solve the equations and find the numbers. b) The sum of two numbers is 77 and their difference is 55. (i) Express the above statements in the form of linear equations. (ii) Find the numbers by solving the equations. c) A number is twice the other number and their sum is 30. (i) Express the above statements in the form of linear equations. (ii) Find the numbers. d) A number is thrice the other number and their sum is 80. (i) Express the above statements in the form of linear equations. (ii) Find the numbers. e) The sum of two numbers is 59 and the smaller number is less than the bigger one by 7. (i) Express the above statements in the form of linear equations. (ii) Find the numbers. f) The difference of two numbers is 12. The greater number is three times the smaller one. (i) Express the above statements in the form of linear equations. (ii) Find the numbers. g) Three times the sum of two numbers is 42 and five times their difference is 20. (i) Express the above statements in the form of linear equations. (ii) Find the numbers. 2. a) The total cost of a bag and an umbrella is Rs 1,000 and the umbrella is cheaper than the bag by Rs 200. (i) Express the above statements in the form of linear equations. (ii) Find the cost of the bag and the umbrella. b) The total cost of a watch and a radio is Rs 5,000 and the watch is cheaper than the radio by Rs 1,500. (i) Express the above statements in the form of linear equations. (ii) Find the cost of the watch and the radio. c) The sum of age of father and his son is 38 years and the father is 22 years older than his son. (i) Express the above statements in the form of linear equations. (ii) Find the present age of the son and the father.

Vedanta Excel in Mathematics - Book 9 214 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur d) The age of a mother is 6 times the age of her daughter and the difference between their ages is 35 years. (i) Express the above statements in the form of linear equations. (ii) Find their present ages. Creative section - A 3. a) In a fruit shop, the cost of 1 kg of apples is the same as the cost of 2 kg of oranges. The total cost of 4 kg of apples and 6 kg of oranges is Rs 1,540. (i) Express the above statements in the form of linear equations. (ii) Find the rate of cost of apples and oranges. b) In a meat shop, the cost of 1 kg mutton is four times the cost of 1 kg chicken. The total cost of 2 kg mutton and 3 kg chicken is Rs 3,300. (i) Express the above statements in the form of linear equations. (ii) Find the rates of cost of mutton and chicken. c) The cost of tickets of a comedian show of ‘Gaijatra’ is Rs 150 for an adult and Rs 50 for a child. Mr. Shah paid Rs 550 for 5 tickets of the show. (i) Express the above statements in the form of linear equations. (ii) How many tickets were purchased in each category? d) A man buys a few numbers of books at Rs 200 each and a few number of pens at Rs 25 each. If he buys 7 articles and pays Rs 700 altogether, based on this information, answer the following questions. (i) Express the above statements in the form of linear equations. (ii) Find the number of each article bought by him. 4. a) If twice the son's age in years is added to the father's age, the sum is 70. But twice the father's age is added to the son's age, the sum is 95. (i) Express the above statements in the form of linear equations. (ii) Find the present ages of the father and his son. b) The sum of the ages of the father and son is 44 years. After 8 years, the age of the father will be twice the age of the son. (i) Express the above statements in the form of linear equations. (ii) Find their present ages. c) Three years ago the sum of the ages of a father and his son was 48 years and three years hence father's age will be three times that of his son's age. Find the present ages of the father and his son. Simultaneous Equations

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 215 Vedanta Excel in Mathematics - Book 9 d) Eight years ago, the daughter's age was thrice the son's age. Now the daughter's age is 4 years more than the son's age. Find their present ages. e) 2 years ago, father's age was nine times the son's age but 3 years later it will be 5 times only. Find the present ages of the father and the son. f) 5 years ago a man's age was 5 times the age of his daughter's. 3 years hence, twice his age will be equal to 6 times his daughter's age. Find their present ages. g) A year hence a father will be 5 times as old as his son. Two years ago the father was 3 times as old his son will be 4 years hence. Find their present ages. h) 14 years ago the age of the mother was 4 times the age of her daughter. The present age of the mother is 2 times the age of her daughter will be 4 years hence. What are their present ages? i) 3 years later a mother will be 4 times as old as her son. 3 years ago, the mother's age was two times her son's age will be 8 years hence. What are their present ages? j) The present age of the father is three times the age of his son. If the age of the son after 10 years is equal to the age of the father before 20 years, find the present ages of the father and the son. k) The ages of two girls are in the ratio of 5: 7. Eight years ago their ages were in the ratio 7 : 13. Find their present ages. l) Three years ago, the ratio of the ages of A and B was 4: 3. Three years hence, the ratio of their ages will be 11: 9. Find the present ages of A and B. 5. a) A number consists of two digits. The sum of its digits is 16. if 18 is subtracted from the number, the digits interchange their place. (i) What is the two-digit number in which y is at one’s place and x is at ten’s place? (ii) Express the above statements in the form of linear equations. (iii) Solve the equations and find the number. b) The sum of the digits of a two-digit number is 10. When 18 is added to the number, its digits are reversed. (i) Express the above statements in the form of linear equations. (ii) Find the number by solving the equations. c) The digit at tens place of a two digit number is two times the digit at ones place. When 27 is subtracted from the number, its digit are reversed. Find the number. d) The digit at ones place of a two digit number is three times the digit at tens place. When 54 is added to the number, its digits are reversed. Find the number. e) A number consisting of two digits is three times the sum of its digits. If 45 is added to the number, the digits will be interchanged. Find the number. Simultaneous Equations

Vedanta Excel in Mathematics - Book 9 216 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur f) A certain number of two digits is seven times the sum of the digits. If 36 is subtracted from the number, the digits will be reversed. Find the number. g) The sum of the digits in a two-digit number is 11. The number formed by interchanging the digits of the number will be 45 more than the original number. Find the original number. h) A number of two-digits exceeds four times the sum of its digits by 3. If 36 is added to the number, the digits are reversed. Find the number. 6. a) If 4 is subtracted from the numerator of a fraction, its value becomes 1 3. If 5 is added to the denominator of the original fraction, its value becomes 1 2 . What is the original fraction? b) The numerator of a fraction is 5 less than its denominator. If 1 is added to each, its value becomes 1 2 . Find the original fraction. c) If the numerator of a fraction is multiplied by 4 and the 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. d) The ratio of the monthly income and expenditure of Sunayana is 5 : 3. If she increases her income by Rs 5,000 and decreases the expenditure by Rs 3,000, the new ratio becomes 5 : 2. Find her income and expenditure. e) The monthly income of Rita and Bishwant are in the ratio of 4 : 3 and their expenses are in the ratio 3:2. If each of them saves Rs 2000 in a month, find their monthly income. Creative section – B 7. a) In a city, the taxi charge consists of two types of charges: a fixed charge together with the charge for the distance covered. If a person travels 7 km, he pays Rs 390 and for travelling 12 km, he pays Rs 640. (i) Find the fixed charges and the rate of charge per km. (ii) If the fixed charge were increased by 25% and the rate of charge per km remained same, how long would a passenger travel for Rs 300? b) A lending library has a fixed charge for the first 4 days and an additional charge for each day thereafter. Dorje paid Rs 50 for a book kept for a 9 days, while Kazol paid Rs 35 for the book she kept for 6 days. (i) Find the fixed charge and the charge for each extra day. (ii) If the fixed charge were decreased by 20% and the charge for each extra day were increased by 10%, how long would a reader keep a book with the payment of Rs 75. 8. a) The sum of the ages of a father and his son is 40 years. If they both live on till the son becomes as old as the father is now, the sum of their ages will be 96 years. (i) Find the present ages of the son and his father. (ii) After how many years, will the father be only thrice as old as his son? Simultaneous Equations

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 217 Vedanta Excel in Mathematics - Book 9 b) The sum of present ages of a father and his son is 80 years. When the father's age was equal to the present age of the son, the sum of their ages was 40 years. (i) Find their present ages. (ii) How many years ago, was the father six times as old as his son? 9. a) If Prakash gives one of the marbles from what he possesses to Kamala, then they will have equal number of marbles. If Kamala gives one of the marbles from what she possesses to Prakash, Prakash will have double number of marbles. Find the number of marbles possessed by them initially. b) When Gopal gives Rs 10 from his money to Laxmi her money becomes double than that of the remaining money of Gopal. When Laxmi gives Rs 10 to Gopal his money is still Rs 10 less than the remaining money of Laxmi. Find their original sum of money. 10. a) Janak started his bicycle journey from Kohalpur to Dhangadhi at 6:00 a.m. with an average speed of 20 km/hr. Two hours later Ganesh also started his journey from Kohalpur to Dhangadhi with an average speed of 30 km/hr. At what time would they meet each other if both of them maintain non-stop journey. b) Two buses were coming from two different places situated just in the opposite direction. The average speed of one bus is 5 km/hr more than that of another one and they had started their journey in the same time. If the distance between the places is 500 km and they meet after 4 hours, find their speed. 11. a) When the length of a rectangular field is reduced by 5 m and breadth is increased by 3 m, its area gets reduced by 9 sq. m. If the length is increased by 3 m and breadth by 2 m the area increases by 67 sq. m. Find the length and breadth of the room. b) A few number of students are made to stand in a certain number of rows equally. If 3 more students were kept in each row, there would be 2 rows less. If 3 less students were kept in each row, there would be 3 rows more. Find the number of students. Project work and activity section 12. a) Make a pair of simultaneous equations and solve them to get the following values of the variables. (i) (ii) x = 5 y = 7 (iii) x = 4 y = –1 (iv) x = –2 y = 5 (v) x = –3 y = –2 b) Think the values of x and y of your own choice. Then, complete the each pair of equations and solve them. (i) x + y = ....... x – y = ....... (ii) 2x+y = ....... x +2y = ....... (iii) 3x–2y = ....... 2x – y = ....... (iv) x 2 + y 3 = ....... x 3 + y 2 = ....... x = 3 y = 2 Simultaneous Equations

Vedanta Excel in Mathematics - Book 9 218 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur c) Make word problems reflecting to the real life situations as far as possible from the given pairs of simultaneous equations, then solve them. (i) x + y = 30 x – y = 10 (ii) x+y = 36 x = 2y (iii) 3x = y y – x = 20 (iv) x + 2y = 70 x : y = 3 : 2 OBJECTIVE QUESTIONS Let’s tick (√) the correct alternative. 1. Which of the following is a solution of the equation 3x – y = 7? (A) (2, 1) (B) (5, 8) (C) (–1, 4) (D) (3, –2) 2. 5 years ago, Mr. Rai was x years, what will be his age after 5 years? (A) x + 5 (B) x + 10 (C) x – 5 (D) x – 10 3. The present ages of a father and his daughter are 40 years and 10 years respectively, what will be their ages x years hence? (A) 40 + x, 10 + x (B) 40 – x, 10 + x (C) 40+x, 10–x (D) 40 – x, 10 – x 4. If the cost of a dozen of bananas is Rs 45 more than that of 7 bananas, what is the cost of a banana? (A) Rs 5 (B) Rs 7 (C) Rs 9 (D) Rs 10 5. Mrs. Sharma has only one son. Last year, her son was 11 years old. Now, she is thrice as old as her son. What was the age of Mrs. Sharma when her son was born? (A) 23 years (B) 24 years (C) 25 years (D) 36 years 6. The sum of two numbers is 30. If twice the bigger number is thrice the smaller, the smaller is (A) 6 (B) 8 (C) 12 (D) 18 7. The sum of three consecutive numbers is 15, what is the greatest consecutive number? (A) 4 (B) 5 (C) 6 (D) 7 8. Which of the two–digit number has sum of digits 9 and the digit at ten’s place is twice the digit at one’s place? (A) 24 (B) 36 (C) 63 (D) 48 9. If x and y are the digits at ten’s and one’s places of a number, what is the number formed by reversing its digits? (A) 10x + y (B) 10 (x + y) (C) 10y + x (D) 10xy 10. The corresponding linear equation for “the sum of digits of a two–digit number 10x + y is 11” is (A) x + y = 11 (B) 10x + y = 11 (C) 10y + x = 11 (D) xy = 11 Simultaneous Equations

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 219 Vedanta Excel in Mathematics - Book 9 Assessment -IV 1. Consider a sequence 21, 18, 15, … , –33. (a) What is the common difference of this sequence? (b) How many terms are there in the sequence? (c) Which term of this series is zero? 2. Mr. Gurung went for a 4 days hiking trip. Each day he walked 10% more than the distance he walked on the day before. He walked 6655 m in the last day. (a) By what common ratio did he increase the distance on each day after the first day? (b) Find the distance he walked on the first day of the trip? 3. After a knee surgery, the trainer told to Mrs. Chhetri to return to her jogging program slowly. He suggested jogging for 12 minute each day for the first week. Each week thereafter, he suggested that she has to increase that time by 6 minute per day, answer the following questions. (a) Form a sequence of time of each day for the successive weeks. (b) What type of sequence is it? (c) How much weeks will it be before she is up to jogging 60 minute per day? 4. The given expressions are 8a3 + x3 and 16a4 + 4a2 x2 + x4 . (a) Find the algebraic expression with the highest degree and the greatest coefficient which divides the given expressions exactly. (b) Find the algebraic expression with the least degree and the smallest coefficient which is exactly divisible by the given expressions. 5. Simplify: (a) xa xb a2 + ab + b2 × xb xc b2 + bc + c2 × xc xa c2 + ca + a2 (b) 1 1 + mp – q + mr – q + 1 1 + mq – r + mp – r + 1 1 + mr – p + mq – p 6. In a dairy, the rate of cow milk is Rs 120 per litre and the rate of buffalo milk is Rs 130 per litre. Ajita paid Rs 510 for 5 litres of milk. (a) Express the above statements in the form of linear equations. (b) How many litres of cow milk and buffalo milk did she purchase? (c) If the rates of cost of both types of milk were increased by 10%, how many litres of cow milk and buffalo milk of equal quantity would be bought for Rs 1100? 7. Ten years ago, the son’s age was twice the daughter’s age. Now, the son’s age is 3 years more than the daughter’s age. (a) Write the above statements in the form of linear equations. (b) Find their present ages. (c) What was the age of the brother when his sister was just born? 8. In 6 hours Harka walks 6 km more than Dorje walks in 3 hours. In 8 hours Dorje walks 12 km more than Harka walks in 9 hours. (a) Write the above statements in the form of linear equations. (b) Find their speed in km per hour.

Vedanta Excel in Mathematics - Book 9 220 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 13. 1 Triangle –Looking back 1. Let’s fill in the blanks with correct answers as quickly as possible. a) A triangle in which all three sides are equal is called an ……….. b) Every triangle has at least …………. acute angles. c) The sum of interior angles of a triangle is always ………. right angles. d) An obtuse angled triangle has exactly one …………. angle. e) The base angels of an isosceles triangle are ……… 2. Let’s match the following as quickly as possible. a) Each angle measures 60o (i) Isosceles triangle b) One right angle and other angles are equal (ii) Acute angled triangle c) Sum of angles is not 180o (iii) Isosceles right-angled triangle d) Base angles are equal (iv) Equilateral triangle e) No sides are equal (v) No triangle exists f) Each angle is less than a right angle (vi) Scalene triangle 3. Let’s write ‘True’ for true statements and ‘False’ for false statements. a) Every equilateral triangle is an isosceles triangle. b) Every isosceles triangle is an equilateral triangle. c) Every equilateral angle is an acute angled triangle. d) Every acute angled triangle is an equilateral triangle. e) A right-angled triangle may be an isosceles triangle. 13.2 Types of triangle - review Types of triangles by angles On the basis of the sizes of angles, there are three types of triangles: (i) Acute angled triangle (ii) Obtuse angled triangle (iii) Right angled triangle It has three acute angles. It has one obtuse angle. It has one right angle. A B C A B C A B C Unit 13 Geometry - Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 221 Vedanta Excel in Mathematics - Book 9 (i) Scalene triangle (ii) Isosceles triangle (iii) Equilateral triangle Types of triangles by sides On the basis of the length of the sides of triangles, there are three types of triangles. It has none of the equal sides. It has two equal sides. It has all three equal sides. A B C A B C A B C 13.3 Median and altitude of a triangle A median of a triangle is a straight line that joins a vertex of a triangle to the mid-point of its opposite side. A perpendicular drawn from a vertex to the opposite side of a triangle is known as the altitude (or height) of the triangle. Following are the important facts about the medians and altitudes of different types of triangles. (i) In an equilateral triangle, its medians are also the altitudes of the triangle or vice versa. (ii) In an isosceles triangle, the median drawn from its vertex to the base is also its altitude. (iii) In a right-angled triangle, the median drawn to its hypotenuse is half of the hypotenuse in length. (iv) Each median of a triangle divides the triangle in two triangles having equal area. A F E D G B C AD, BE, and CF are the medians of ∆ ABC. All three medians of a triangle are concurrent and the meeting point of medians is called the centroid of the triangle. In the figure, G is the centroid of ∆ ABC. The centroid of a triangle divides each median in the ratio of 2:1. AD, BE, and CF are the altitudes of ∆ABC. The altitudes of a triangle are also concurrent and the meeting point of altitudes of a triangle is called Orthocentre of the triangle. In the figure, O is the orthocentre of the triangle ABC. A F E D O B C 13.4 Properties of triangles Following are a few important properties of triangles. We can verify these properties experimentally as well as theoretically. (i) The sum of the angles of a triangle is equal to two right angles (180°). (ii) The exterior angle of a triangle is equal to the sum of two opposite interior angles. Geometry - Triangle

Vedanta Excel in Mathematics - Book 9 222 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur (iii) The sum of any two sides of a triangle is greater than the third side. (iv) In any triangle, the angle opposite to longer side is greater than the angle opposite to shorter side. (v) The base angles of an isosceles triangle are equal. (vi) In an isosceles triangle, perpendicular drawn from the vertical angle to the base bisects the base. (vii) A straight line that joins the mid-points of any two sides of a triangle is parallel to the third side of the triangle, ... and so on. Theorem 1 The exterior angle of a triangle is equal to the sum of the two opposite interior angles. Experimental verification Step 1: Three triangles ABC with different shapes and sizes are drawn. A side BC is produced to D in each triangle. Step 2: The exterior angle ACD and the opposite interior angles ABC and BAC are measured in each triangle. The measurements are tabulated in the table. Figure ∠ABC ∠BAC ∠ABC + ∠BAC ∠ACD Result (i) ∠ABC + ∠BAC = ∠ACD (ii) ∠ABC + ∠BAC = ∠ACD (iii) ∠ABC + ∠ BAC = ∠ACD Conclusion:The exterior angle of a triangle is equal to the sum of the two opposite interior angles. Worked-out examples Example 1: From the figure given alongside, find the size of ∠ABC. Solution: Here, ∠BAC = (2x – 5o ), ∠ABC = (3x + 10o ) and ∠ACD = (7x –35o ) Now,∠BAC + ∠ABC = ∠ACD [The exterior angle of triangle is equal to the sum of two opposite interior angles.] or, (2x – 5o ) + (3x + 10o ) = 7x – 35o or, 5x + 5o = 7x – 35o or, -2x = - 40o or, x = 20o Hence, ∠ABC = 3x + 10o = 3×20o + 10o = 70o https://www.geogebra.org/m/v3vngznk Vedanta ICT Corner Please! Scan this QR code or browse the link given below: https://www.geogebra.org/m/ethbh2t9 Vedanta ICT Corner Please! Scan this QR code or browse the link given below: A(2x–5°) (3x+10°) (7x–35°) B C D Geometry - Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 223 Vedanta Excel in Mathematics - Book 9 Geometry - Triangle Example 2: In a triangle, the exterior angle formed by extending a side is 100o . If the difference between two interior angles opposite to the exterior angle is 40o , find the interior angles of the triangle. Solution: Let, x and y be two interior angles opposite to the given exterior angle of a triangle. Then, x + y = 100o …equation (i) Also, x – y = 40o or, x = y + 40o … equation (ii) Now, putting the value of x from equation (ii) in equation (i), we get (y + 40o ) + y = 100o or, 2y + 40o = 100o or, 2y = 60o or, y = 30o Again, putting the value of y in equation (ii), we get x = 30 o + 40o = 70o Hence, the first interior angle (x) = 70o , the second interior angle (y) = 30o and the third interior angle = 180o – (x + y) = 180o – (70o + 30o ) = 80o Example 3: In the figure BD and CD are the bisectors of ∠ABC and ∠ACE respectively. If ∠BAC = 56°, find the value of x. Solution: (i) Here, suppose ∠ABD = ∠DBC = a and ∠ACD = ∠DCE = b. In DABC, ∠BAC + ∠ABC = ∠ACE or, 56° + (a + a) = (b + b) [Being exterior angles of triangle equal to the sum of opposite interior angles] or, 56° + 2a = 2b or, 2(28° + a) = 2b ∴ b = 28° + a .......... (i) (ii) In DBCD, ∠DBC + ∠BDC = ∠DCE a + x = b or, a + x = 28° + a [From (i), b = 28° + a] ∴ x = 28° Hence, x = 28° Example 4: In ∆ABC, BN is perpendicular to AC and CM is perpendicular to AB. BN and CM intersect at O. Prove that ∠ BOC = 180°– ∠A [The exterior angle of triangle is equal to the sum of two opposite interior angles.] x 56° A B C E D x 56° A B a a b b C E D A C B N M O

Vedanta Excel in Mathematics - Book 9 224 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Solution: Given: In ∆ABC, BN ⊥ AC and CM ⊥AB. BN and CM intersect at O. To prove: ∠BOC = 180°–∠A Proof: S.N. Statements S.N. Reasons 1. ∠BAC + ∠ABN = ∠BNC i.e., ∠A + ∠ABN = 90o 1. The exterior angle of triangle is equal to the sum of two opposite interior angles. 2. ABN+ ∠BMO = ∠BOC i.e., ∠ABN+ 90o = ∠BOC 2. Same as (1) 3. ∠A + ∠ABN – (∠ABN + 90o ) = 90o – ∠BOC i.e.,∠BOC = 180°–∠A 3. Subtracting statement (2) from statement (1) Proved Example 5: In the adjoining ∆ABC, AY is the bisector of ∠BAC and AX ⊥ BC. Prove that: ∠XAY = 1 2 (∠B – ∠C). Solution: Given: In ∆ABC, AY is the bisector of ∠BAC and AX ⊥ BC. To prove: ∠XAY = 1 2 (∠B – ∠C). Proof: S.N. Statements S.N. Reasons 1. ∠BAC = 2∠BAY 1. AY is the bisector of ∠BAC. 2. ∠BAC+ ∠ABC +∠ACB = 180o 2. The sum of angles of triangle is 180o . 3. 2∠BAY + ∠B +∠C= 180o i.e., ∠BAY = 1 2 (180o – ∠B –∠C) 3. From statements (1) and (2). 4. ∠BAX+ ∠ABX = ∠AXC i.e., ∠BAX = 90o – ∠B 4. The exterior angle of triangle is equal to the sum of two opposite interior angles. 5. ∠BAY - ∠BAX = 1 2 (180o – ∠B –∠C) – (90o – ∠B) i.e., ∠XAY = 1 2 (∠B – ∠C) 5. Subtracting statement (4) from statement (3) and using ∠BAY – ∠BAX = ∠XAY Proved A B X Y C Geometry - Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 225 Vedanta Excel in Mathematics - Book 9 Geometry - Triangle EXERCISE 13.1 General section 1. a) In DABC, M is the mid-point of side BC and AP ⊥ BC. Then, name the altitude and median of the triangle. b) In DXYZ, A, B and C are the mid-points of sides XY, YZ and XZ respectively. Then, name the medians of the triangle. What is the point G called? c) In DPQR. PX ⊥ QR, QY ⊥ PR and RZ ⊥ PQ. Then, name the altitudes of the triangle. What is the point O called? 2. a) From the figure given alongside, write down the relation of ∠BAC, ∠ABC and ∠ACD. b) In ∆PQR, the side QR is extended to the point S. If ∠PRS = 58o , what is the value of ∠P + ∠Q? 3. Answer the following questions a) In a triangle, if an exterior angle is 69o and its opposite interior angles are xo and 2xo , find the sizes of each interior angle of the triangle. b) When a side of a triangle is extended to a point, the exterior angle so formed is 100o . If the two opposite interior angles are in the ratio 2:3, find the sizes of all three interior angles of triangle. c) An exterior angle of a triangle is (4x – 15)° and two opposite interior angles are (x + 10)o and (2x – 5)o . Find the measure of each interior angle of the triangle. d) In a triangle, an exterior angle is (153 – 3x)° and two opposite interior angles are (5x – 3)o and (2x + 6)o . Find the measure of all interior angles of the triangle. 4. Find the unknown sizes of angles: a) b) c) A B C P M X Y Z A G C B P Q R Z Y O X A B D C S R Q P 58° A B C D 120° a 70° b A B C 50° 2x 3x+5° D R 140° 2p S Q P p 2

Vedanta Excel in Mathematics - Book 9 226 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur d) e) f) g) h) i) Creative section 5. a) When a side BC of a triangle ABC is produced to an exterior point D, answer the following questions. (i) Write down the relation between ∠ACD and ∠ABC + ∠BAC. (ii) Verify experimentally that the relationship between ∠ACD and ∠ABC + ∠BAC. (Two triangles of different shapes and sizes are necessary) (iii) If the triangle ABC is an isosceles right angled triangle right-angled at B, find the ratio of measurements of ∠ABC and ∠ACD. b) In triangle PQR, a side is produced. (i) Verify experimentally that the relationship of exterior angle of the triangle with some of the two non-adjacent interior angles of the triangle PQR by drawing two triangles of different shape and size. (ii) If triangle PQR were an equilateral triangle and side QR is produced to a point S, calculate ∠PRS. Also find the ratio of ∠PQR and ∠PRS. 6. a) In ∆PQR, RX ⊥ PQ and QY ⊥ PR, RX and QY intersect at O. Prove that ∠QOR = 180° – ∠P. b) In the given figure ABCD, prove that ∠BCD = ∠BAD + ∠ABC + ∠ADC. [Hint: Join A and C then extended AC to the point E.] c) In the adjoining star shaped figure, prove that ∠A + ∠B + ∠C + ∠D + ∠E = 180° A B O C D 120° A B C Q P R 3x 2x 4x P T Q R 30° 80° y S x A E P Q R S T B C D 30° 35° 50° x 45° B E F G D C x 110° 145° A y 25° x A B C E D P Q R X Y O D C A B A E P Q R S T B C D Geometry - Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 227 Vedanta Excel in Mathematics - Book 9 Geometry - Triangle 7. a) In the figure alongside, BE and CE are the angular bisectors of ∠ABC and ∠ACD respectively. Prove that ∠BAC = 2∠BEC. b) In the given figure, the bisector of ∠ACU meets AU at O. Prove that ∠COT = 1 2 (∠CAT + ∠CUT). c) In the adjoining DABC, AY is the bisector of ∠BAC and AX ⊥ BC. Prove that ∠XAY = 1 2 (∠B – ∠C). 13.5 Triangle inequality property Theorem 2 Let’s study the following activities. The mathematics teacher asked his/her three students Archana, Rahul and Pemba to form triangles with the sticks of given measurements. Student’s name Lengths of sticks Archana 5 cm, 7 cm and 10 cm Rahul 4 cm, 12 cm and 6 cm Pemba 11 cm, 6 cm and 5 cm All of them tried to form a triangle. Archana formed the triangle successfully. When Rahul and Pemba tried to join the ends of the two smaller sticks, they were unable to form triangles, why? The sum of any two sides of a triangle is greater than the third side. Experimental verification Step 1: Three triangles ABC with different shapes and sizes are drawn. A B C D E T C A O U A B X Y C Rahul's work Pemba's work Archana's work https://www.geogebra.org/m/jyyx5kkd Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Vedanta Excel in Mathematics - Book 9 228 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Step 2: All three sides of each of ∆ABC are measured and the lengths are tabulated in the table. Figure AB BC CA AB + BC BC + CA AB + CA Result (i) AB + BC > CA, BC + CA > AB, AB + CA > BC (ii) AB + BC > CA, BC + CA > AB, AB + CA > BC (iii) AB + BC > CA, BC + CA > AB, AB + CA > BC Conclusion: The sum of any two sides of a triangle is greater than the third side. Theorem 3 In any triangle, the angle opposite to the longer side is greater than the angle opposite to the shorter side. Experimental verification Step 1: Three triangles ABC with different shapes and sizes are drawn in such a way that BC is the longest and CA is shortest sides in each triangle. Step 2: The angle opposite to the greatest side BC and the angle opposite to the shortest side CA are measured and tabulated in the table. Figure Angle opposite to BC: ∠A Angle opposite to CA: ∠B Result (i) ∠A > ∠B (ii) ∠A > ∠B (iii) ∠A > ∠B Conclusion:The angle opposite to the longer side of any triangle is greater than the angle opposite to the shorter side. Converse of Theorem 3 In any triangle, the side opposite to the greater angle is longer than the side opposite to the smaller angle. Experimental verification Step 1: Three triangles ABC with different shapes and sizes are drawn in such a way that ∠A is the greatest and ∠B is the smallest angles in each triangle. Geometry - Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 229 Vedanta Excel in Mathematics - Book 9 Geometry - Triangle Step 2: The side opposite to the greatest angle A and the side opposite to the smallest angle B are measured and tabulated in the table. Figure Side opposite to ∠A: BC Side opposite to ∠B: CA Result. (i) BC > CA (ii) BC > CA (iii) BC > CA Conclusion:The side opposite to the greater angle of any triangle is longer than the side opposite to the smaller angle. Theorem 4 Of all straight line segments drawn to a given line from a point outside it, the perpendicular is the shortest one. Experimental verification Step 1: Three straight line segments XY with different lengths and position are drawn. A point P is marked outside the each line segment. Three line segments PA, PB, PC and a perpendicular PQ are drawn from P to XY. Step 2: The lengths of the line segments PA, PB, PC, and PQ are measured and tabulated in the table. Figure Length of the line segments Result PA PB PC PQ (i) Perpendicular PQ is the shortest one. (ii) Perpendicular PQ is the shortest one. (iii) Perpendicular PQ is the shortest one. Conclusion:Of all line segments drawn to a given line from a point outside it, the perpendicular is the shortest one. Worked-out examples Example 1: Can you construct a triangle with sides 3.5 cm, 9 cm and 4 cm? Why? (i) (ii) (iii)

Vedanta Excel in Mathematics - Book 9 230 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Solution: Here, 3.5 cm + 9 cm = 12.5 cm > 4 cm, 9 cm + 4 cm = 13 cm > 3.5 cm 4 cm + 3.5 cm = 7 cm which is not greater than 9 cm So, we cannot construct the triangle with sides 3.5 cm, 9 cm and 4 cm. Example 2: In ∆ABC, if AB = 4 cm and BC = 7 cm, between what two measures should the third side AC fall? Solution: Here, in ∆ABC, AB = 4 cm and BC = 7 cm Since, for any triangle; the sum of any two sides should be greater than the third side and their difference should be smaller than the third side. So, 4 cm + 7 cm = 11 cm and 7 cm – 4 cm = 3 cm Hence, the length of the third side AC lies between 3 cm and 11 cm i.e., 3 cm < AC < 11 cm. Example 3: In the given quadrilateral ABCD, prove that AB + BC + CD > AD. Solution: Given: ABCD is a quadrilateral. To prove: AB + BC + CD > AD Construction: B and C are joined. Proof: S.N. Statements S.N. Reasons 1. AB + BC > AC 1. In ∆ABC, the sum of two sides is greater than the third side. 2. AC + CD > AD 2. In ∆ACD, the sum of two sides is greater than the third side. 3. AB + BC + CD > AD 3. From statements (1) and (2). 13.6 Congruent triangles Two or more triangles are said to congruent if they have the same shape and size. In this case, when one triangle is placed over another triangle, their corresponding parts exactly coincide to each other. The congruent triangles are always similar and they have equal area. 13.7 Conditions of congruency of triangles There are 3 sides and 3 angles in a triangle. Two triangles can be congruent if 3 parts (out of 6 parts) of one triangle are equal to 3 corresponding parts of another triangle as per the following conditions. We use these conditions as axioms. A B C D A B C D Geometry - Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 231 Vedanta Excel in Mathematics - Book 9 Geometry - Triangle (i) S.S.S. axiom When three sides of a triangle are respectively equal to three corresponding sides of another triangle, they are said to be congruent. In ∆s ABC and PQR, a) AB = PQ (S) b) BC = QR (S) c) CA = RP (S) d) ∴ ∆ ABC ≅ ∆ PQR [S.S.S. axiom] Corresponding parts of congruent triangles are also equal. ∴ ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R. Note: The parts opposite to the equal parts of triangles are called the corresponding parts. (ii) S.A.S. axiom When two sides of one triangle and angle made by them are respectively equal to the corresponding sides and angle of another triangle they are said to be congruent triangles. In ∆s ABC and PQR a) AB = PQ (S) b) ∠B = ∠Q (A) c) BC = QR (S) d) ∴∆ABC ≅ ∆PQR [S.A.S. axiom] Now, ∠C = ∠R and ∠A = ∠P [Corresponding angles of congruent triangles] CA = RP [Corresponding sides of congruent triangles] (iii) A.S.A. axiom When two angles and their adjacent side of one triangle are respectively equal to the corresponding angles and sides of another triangle, they are said to be congruent triangles. In ∆s ABC and PQR, a) ∠B = ∠Q (A) b) BC = QR (S) c) ∠C = ∠R (A) d) ∴ ∆ABC ≅ ∆PQR [A.S.A. axiom] Now, AC = PR and AB = PQ [Corresponding sides of congruent triangles] ∠A = ∠P [Corresponding angles of congruent triangles] (iv) R.H.S. axiom In two right angled triangles, when the hypotenuse and one of the two remaining sides are respectively equal, they are said to congruent triangles.

Vedanta Excel in Mathematics - Book 9 232 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur In right angled ∆s ABC and PQR, a) ∠B = ∠Q (R) b) AC = PR (H) c) BC = QR (S) d) ∴ ∆ABC ≅ ∆PQR [R.H.S. axiom] Now, ∠C = ∠R and ∠A = ∠P [Corresponding angles of congruent triangles] AB = PQ [Corresponding sides of congruent triangles] (v) A.A.S. axiom When two angles and a side of one triangle are respectively equal to the corresponding angles and sides of another triangle, they are said to be congruent triangles. This axiom can be verified by using A.S.A. axiom. Here, a) ∠A = ∠P [Given] b) ∠B = ∠Q [Given] c) ∠A + ∠B + ∠C = ∠P + ∠Q + ∠R [Sum of the angles of any triangle is 180°] d) ∴ ∠C = ∠R [From (a), (b) and (c)] Now, in ∆s ABC and PQR, e) ∠B = ∠Q (A) [Given] f) BC = QR (S) [Given] g) ∠C = ∠R (A) [from (d)] h) ∴ ∆ABC ≅ ∆PQR [A.S.A axiom] Now, AB = PQ and AC = PR [Corresponding sides of congruent triangles] EXERCISE 13.2 General section 1. a) Identify and name the longest and the shortest sides of the following triangles. 60° R P Q F 35° 85° D E Geometry - Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 233 Vedanta Excel in Mathematics - Book 9 Geometry - Triangle b) Identify and name the greatest and the smallest angles of the following triangles. X Y Z B E F G C A c) In ∆ABC, if ∠A = 85°, ∠B = 65° and ∠C = 30°, name the longest and the shortest sides of the triangle. d) In ∆PQR, if ∠Q = 75°, and ∠R = 25°, name the longest and the shortest sides of the triangle. e) In ∆XYZ, if XY = 5.6 cm, YZ = 4.7 cm and ZX = 6.5, name the greatest and the smallest angles of the triangle. 2. a) Can you construct a triangle with sides 5 cm, 7 cm and 10 cm? Why? b) Can you construct a triangle with sides 6 cm, 6 cm and 6 cm? Give reason. c) Can you construct a triangle with sides 4 cm, 7 cm and 11 cm? Why? d) Can a triangle be constructed with the sides 5.2 cm, 3.7 cm and 8.5 cm? Why? Creative section - A 3. a) In ∆ABC, if AB = 5 cm and BC = 8 cm, between what two measures should the third side AC fall? b) If the measures of two sides of a triangle are 6 cm and 4 cm, between what two measures should the third side lie? 4. Use the necessary axioms to verify that the following pairs of triangles are congruent. Also, write the equal corresponding parts of the triangles. A P Q B R C a) b) D E F Y X Z c) C d) A B R Q P G E F R S T 5. a) In the adjoining figure, show that ∆ABC ≅ ∆PQR. Also find the unknown sizes of angles of ∆ABC and ∆PQR. 70°

Vedanta Excel in Mathematics - Book 9 234 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur b) In the figure alongside, show that ∆RST ≅ ∆XYZ. Also find the unknown sizes of x cm, y cm and a°. Creative section - B 6. DABC is a given triangle. Answer the following questions. a) Write down the relation between the sum of any two sides of the triangle and its third side. b) Verify the relationship between the sum of any two sides and the third side of triangle experimentally. (Two triangles ABC of different shapes and sizes are necessary) c) Can the triangle ABC be constructed with the sides 7.2 cm, 3.5 cm and 2.9 cm? Give reason. 8. a) In the given figure, AB // PQ and AB = PQ. Prove that ∆ABO ≅ ∆POQ. Also show that AO = OQ and BO = OP 7. a) In ∆PQR, O is the interior point. Prove that OP + OQ + OR > 1 2 (PQ + QR + RP) . P Q R O b) In the figure alongside, O is the interior point of DABC. If AB = 7 cm, BC = 11 cm and CA = 8 cm, show that OA + OB + OC > 13 cm. A B C O 11 cm 7 cm 8 cm b) In the given figure, PQ // SR and PS // QR. Prove that ∆PQR ≅ ∆PRS. Also show that PQ = RS and PS = QR. Project work and activity section 8. a) Take three sticks of different lengths such that the total length of two shorter sticks is greater than longer stick. Now, make a triangle by joining these sticks. b) Take three sticks of different lengths such that the total length of two shorter sticks is less than the longer stick. Can you make a triangle by joining these sticks? Discuss in your class. 9. a) Fold a rectangular sheet of paper diagonally and cut into two halves. (i) What type of triangles did you get? (ii) Are these triangles congruent? Discuss in the class. (iii) Which are the corresponding sides and angles in these two triangles? Discuss in the class. Geometry - Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 235 Vedanta Excel in Mathematics - Book 9 Geometry - Triangle Theorem 5 The bisector of the vertical angle of an isosceles triangle is the perpendicular bisector of the base. Experimental verification Step 1: Three isosceles triangles ABC with different shapes and sizes are drawn in such a way that AB = AC in each triangle. Step 2: The bisector of ∠A is drawn in each triangle. Suppose it is AD and it meets the opposite side BC at D. Step 3: The lengths of BD and DC, and the angles ADB and ADC are measured and tabulated in the table. Figure BD DC Result ∠ADB ∠ADC Result (i) BD = DC ∠ADB = ∠ADC, AD ⊥ BC (ii) BD = DC ∠ADB = ∠ADC, AD ⊥ BC (iii) BD = DC ∠ADB = ∠ADC, AD ⊥ BC Conclusion: The angular bisector of the vertical angle of an isosceles triangle is the perpendicular bisector of its base. Converse of Theorem 5 The straight 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. Experimental verification Step 1: Three isosceles triangles ABC with different shapes and sizes are drawn in such a way that AB = AC in each triangle. Step 2: The mid–point of BC is marked as D. A and D are joined in each triangle.

Vedanta Excel in Mathematics - Book 9 236 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Step 3: ∠ADB, ∠ADC, ∠BAD, and ∠DAC are measured and tabulated in the table. Figure ∠ADB ∠ADC Result ∠BAD ∠DAC Result (i) ∠ADB = ∠ADC, AD ⊥ BC ∠BAD = ∠DAC (ii) ∠ADB = ∠ADC, AD ⊥ BC ∠BAD = ∠DAC (iii) ∠ADB = ∠ADC, AD ⊥ BC ∠BAD = ∠DAC Conclusion: The straight 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. Worked-out examples Example 1: Find the values of x and y from the figure given alongside. Solution: Here, ∠BAC = 90° and AB = AC = CD. (i) ∠ABC = ∠ACB = x [Base angle of isosceles ∆ ABC] (ii) ∠ABC + ∠ACB + ∠BAC = 180° [ Sum of angles of DABC] or, x + x + 90° = 180° or, 2x = 90° ∴ x = 45° (iii) ∠CAD + ∠ADC = y [Base angles of isosceles ∆ ACD] (iv) ∠CAD + ∠ADC = ∠ACB [Being exterior angle of ∆ACD equal to the sum of two non-adjacent interior angles] or, y + y = 45° or, 2y = 45° ∴ y = 22.5° Example 2: In the adjoining figure, find the values of a and b. Solution: Here, (i) ∠SPQ = ∠PSQ = a [ PQ = QS] (ii) ∠SQR = ∠SPQ + ∠PSQ = a + a = 2a [Exterior angle of DPQS] (iii) ∠SQR = ∠SRQ = 2a [ SQ = SR] (iv) ∠SPR + ∠PRS = ∠TSR or, a + 2a = 90° ∴ a = 30° (v) ∠PSQ + ∠QSR + ∠RST = 180° [Being parts of straight angle] or, a + b + 90° = 180° or, 30° + b + 90° = 180° ∴ b = 60° Hence, a = 30° and b = 60° A y x B D C P Q R S a b T Geometry - Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 237 Vedanta Excel in Mathematics - Book 9 Geometry - Triangle Example 3: In DABC, AB = AC and AP ⊥ BC. If AB = (2x + 3) cm, AC = (3y – 1) cm, BP = (y + 1) cm and PC = (x + 2) cm, find the values of x and y. Solution: (i) AB = AC or, 2x + 3 = 3y – 1 2x – 3y = –4 ... (i) (ii) BP = PC [ AB = AC and AP ⊥ BC] or, y + 1 = x + 2 y = x + 1 ... (i) Now, putting the value of y in equation (i) from equation (ii), we get, 2x – 3(x + 1) = –4 or, 2x – 3x – 3 = –4 ∴ x = 1 Again, putting the value of x in equation (ii), we get, y = 1 + 1 = 2 Hence, x = 1 and y = 2 Example 4: In the given figure, AB = AC and BD = EC. Prove that ∆ ADE is an isosceles triangle. Solution: Given: AB = AC and BD = EC To prove: (i) ∆ADE is an isosceles triangle. Proof Statements Reasons 1. (i) (ii) (iii) In ∆ABD and ∆ACE AB = AC (S) ∠ABD = ∠ACE (A) BD = EC (S) 1. (i) (ii) (iii) Given Base angles of an isosceles ∆ABC. Given 2. ∴ ∆ABD ≅ ∆ACE 2. By S.A.S. axiom 3. AD = AE 3. Corresponding sides of congruent triangles 4. DADE is an isosceles triangle 4. From statement (3) Proved Example 5: If the perpendiculars drawn from any two vertices to their opposite sides of a triangle are equal, prove that the triangle is an isosceles triangle. Solution: Given: In ∆ABC, CP ⊥ AB, BQ ⊥ AC and CP = BQ To prove: (i) ∆ABC is an isosceles triangle. Proof Statements Reasons 1. (i) (ii) (iii) In ∆PBC and ∆QBC ∠BPC = ∠BQC (R) BC = BC (A) CP = BQ (S) 1. (i) (ii) (iii) Both are right angles Common sides Given 2. ∴ ∆PBC ≅ ∆BQC 2. By R.H.S. axiom 3. ∠PBC = ∠QCB i.e. ∠ABC = ∠ACB 3. Corresponding angles of congruent triangles 4. DABC is an isosceles triangle 4. Base angles are equal. Proved B (y+1)cm P (x+2)cm (2x + 3)cm (3y – 1)cm C A A D B E C

Vedanta Excel in Mathematics - Book 9 238 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur EXERCISE 13.3 General section 1. a) In the given figure, AB = AC, write down the relation between ∠ABC and ∠ACB. b) In the figure alongside, ∠PQR = ∠PRQ, what is the relation between PQ and PR? c) In the adjoining figure, if XY = XZ and YA = AZ, write down the relation between XA and YZ. d) In the given figure, if LM = LN and LP ⊥ MN, write down the relation between MP and PN, and ∠LMP and ∠LNP. e) In the figure alongside, ON = OE and ∠NOT = ∠TOE, write down the relation between, NT and TE, OT and NE. 2. Find the unknown sizes of angles in the following figures i) j) k) D B x y A C E A l) x 69° x y B D C T P x 30° y Q S R X 40° 30° 30° x U y Y Z V P W A E C K A B C D L M N P D B A F G H B C D X Z Q R S Y Q P Q R X Y A Z O N T E A B C L M P N Geometry - Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 239 Vedanta Excel in Mathematics - Book 9 Geometry - Triangle Creative section - A 3. a) In the given figure, find the sizes of ∠x, ∠y, and ∠z. b) In the adjoining figure, calculate the sizes of ∠p, ∠q, ∠r, and ∠s. 4. a) In the given figure, AB = AC, ∠BAC = 44° and ∠ACD = 24°, show that BC = CD. b) In the figure alongside, SR = QR, ∠QPR = 48° and ∠PRS = 18°, show that PQ = PR. 5. a) In the adjoining figure, find the values of x and y. b) In DABC, AD bisects ∠A and it is perpendicular to base BC. If AB = (3x + 1) cm, AC = (5y – 2) cm, BD = (x + 1) cm and DC = (y + 2) cm, find the values of x and y. c) In the given figure, AB = AC and ∠ACD = 110°, what will be the measure of ∠BAC + ∠ACB? 6. In a given isosceles triangle ABC; bisector of vertical ∠BAC is drawn to a point D on its base BC. Answer the following questions. a) What is the relation between AD and BC? b) What is the relation between BD and CD? c) Construct two triangles ABC and verify experimentally that the bisector of the vertical angle of an isosceles triangle is perpendicular bisector of the base. 50 A C 24° 44° B D P R 18° 48° Q S P (x+8)cm (3y+1)cm (x+4)cm (y+3)cm Q R T A (3x+1)cm (5y–2)cm (x+1)cm (y+2)cm B C D

Vedanta Excel in Mathematics - Book 9 240 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Creative section - B 7. a) In the given figure, AB = AC and BD = EC. Prove that AD = AE. b) In the given figure, AB = AC, BD = EC and ∠DAE = 30°. Prove that ∆ADE is an isosceles triangle. Also calculate the size of ∠ADE. 8. a) In the given figure, PQ = PR, QO bisects ∠PQR and RO bisects ∠PRQ. Prove that OQR is an isosceles triangle. b) In the figure alongside, BD and CD are bisectors of ∠ABC and ∠ACB respectively. If BD = CD, prove that DABC is an isosceles triangle. 9. a) In the given figure, X is the mid-point of QR, XA ⊥ PQ, XB ⊥ PR and XA = XB. Prove that DPQR is an isosceles triangle. b) In the given figure, BN ⊥ AC, CM ⊥ AB and BN = CM. Prove that DABC is an isosceles triangle. c) In the given triangle ABC, AB = AC, BP ⊥ AC and CQ ⊥ AB. Prove that (i) BP = CQ (ii) OP = OQ. 10. a) In the figure alongside, APB and AQC are equilateral triangles. Prove that PC = BQ. (Hint: ∆ APC ≅ ∆ AQB, then PC = BQ) b) In the figure alongside PABQ and ACYX are squares. Prove that PC = BX. 30° P Q R O A B C D A B C M N X Y A B P Q Geometry - Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 241 Vedanta Excel in Mathematics - Book 9 Geometry - Triangle c) In the adjoining figure ABC is an equilateral triangle and BCDE is a square. Prove that AE = AD. d) In the figure alongside PQRS is a square in which PA and SB intersect at O. If PA = SB, prove that PA and SB are perpendicular to each other at O. (Hint: ∆ PBS ≅ ∆ PQA by RHS, then show that ∠SOA = ∠SPO + ∠OPB = 90°) Project work and activity section 11. a) Take a square sheet of paper and fold it through its one diagonal. Cut out it through the diagonal and get two right-angled triangles. Now, fold each right-angled triangle bisecting the vertical angle. (i) Is the folding perpendicular to the base in each triangle? (ii) Does the folding bisect the base in each triangle? (iii) What conclusion can you make from these activities? Discuss in your class. b) Draw a triangle of your own. Produce its all sides as shown in the diagram and get three exterior angle of triangle. Show that sum of three exterior angles of a triangle is 360°. 13.8 Similar triangles - review In the figure given alongside, each angle of DABC is respectively equal to the corresponding angle of DPQR. Therefore, DABC and DPQR have the same shape and they are said to be the similar triangle. We write it as DABC ~ DPQR. However, DABC and DPQR do not have the same size. On the other hand, DDEF and DXYZ are not the similar triangles. Because, none of the angles of these triangles are equal. Following are the required conditions for the similarity between triangles. These are also called the properties of similar triangles. (i) When all angles of one triangle are respectively equal to the corresponding angles of another triangle, the triangles are said to be similar. For example In ∆s ABC and PQR, ∠A = ∠P ∠B = ∠Q ∠C = ∠R ∴ ∆ BC ∼ ∆PQR E D A B C S P Q R O A B C A 30° 115° 35° B R P 30° 115° 35° Q Z X 40° 110° 30° Y 85° F D E 50° 45° A B C P Q R

Vedanta Excel in Mathematics - Book 9 242 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur ‘∼' is the symbol of similarity between the similar triangles. Note: Of course, if two angles of a triangle are respectively equal to two corresponding angles of another triangle, the remaining angles must be equal. So the triangles are said to be similar. (ii) When the ratios of the corresponding sides of triangles are equal, i.e. the corresponding sides of triangles are proportional, the triangles are said to be similar. For example In ∆s KLM and XYZ, KL XY = LM YZ = MK ZX ∴ ∆ KLM ∼ ∆ XYZ Here, a) KL is the opposite side of ∠M and ∠M = ∠Z; the opposite side of ∠Z is XY. So, KL and XY are corresponding sides. b) LM is the opposite side of ∠K and ∠K = ∠X; the opposite side of ∠X is YZ. So, LM and YZ are corresponding sides. c) MK is the opposite side of ∠L and ∠L = ∠Y; the opposite side of ∠Y is ZX. So, MK and ZX are the corresponding sides. Theorem 6 Two equiangular triangles are similar. Experimental verification Step 1: Three pairs of triangles ABC and PQR with two equal corresponding angles are drawn. Here, in ∆ABC and ∆PQR, ∠A = ∠P and ∠B = ∠Q. Step 2: The remaining angles and all the corresponding sides of each pair of triangles are measured. The ratios of corresponding sides are found. K L M X Y Z Geometry - Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 243 Vedanta Excel in Mathematics - Book 9 Geometry - Triangle Fig. ∠C ∠R Result ∆ABC ∆PQR Ratio of corresponding sides AB BC CA PQ QR RP Result AB PQ BC QR CA RP (i) ∠C = ∠R AB PQ = BC QR = CA RP (ii) ∠C = ∠R AB PQ = BC QR = CA RP (iii) ∠C = ∠R AB PQ = BC QR = CA RP Conclusion: As the corresponding sides of equiangular triangles are proportional, each pair of triangles are similar. Theorem 7 If any two corresponding sides of triangles are proportional and the angles included by them are equal, the triangles are similar. Experimental verification Step 1: Three pairs of triangles ABC and PQR are drawn in such a way that in each pair AB PQ = BC QR and ∠B = ∠Q. Step 2: The remaining angles and the remaining sides of each pair of triangles are measured. Then the ratios of the remaining corresponding sides are found. Fig Corresponding angles Corresponding sides AC PR Results ∠A ∠P Result ∠C ∠R Result AC PR (i) ∠A = ∠P ∠C = ∠R AB PQ = BC QR = AC PR (ii) ∠A = ∠P ∠C = ∠R AB PQ = BC QR = AC PR (iii) ∠A = ∠P ∠C = ∠R AB PQ = BC QR = AC PR Conclusion: If two corresponding sides of triangles are proportional and the angle included by them are equal, the triangles are similar.

Vedanta Excel in Mathematics - Book 9 244 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Theorem 8 If two triangles have their corresponding sides proportional, the triangles are similar. Experimental verification Step 1: Three pairs of triangles ABC and PQR are drawn in such a way that in each pair AB PQ = BC QR = AC PR . Step 2: The corresponding angles of each pair of triangles are measured. Fig. ∠A ∠P Result ∠B ∠Q Result ∠C ∠R Result (i) ∠A = ∠P ∠B = ∠Q ∠C = ∠R (ii) ∠A = ∠P ∠B = ∠Q ∠C = ∠R (iii) ∠A = ∠P ∠B = ∠Q ∠C = ∠R Conclusion: If two triangles have their corresponding sides proportional, the triangles are similar. Worked-out examples Example 1: In the adjoining figure, show that ∆ABC ∼ ∆PQC. Then, find the lengths of sides marked with x and y. Solution: 1. In ∆ ABC and ∆ PQC, (i) ∠ACB = ∠PCQ [Common angle] (ii) ∠ABC = ∠PQC [AB // PQ and corresponding angles] (iii) ∠BAC = ∠QPC [Remaining angles of triangles] (iv) ∆ABC ∼ ∆PQC [A.A.A. axiom] 2. AB PQ = BC QC = AC PC [Corresponding sides of similar triangles] or, 10 6 = 2 + x x = 9 + y 9 or, 5 3 = 2 + x x Also, 5 3 = 9 + y 9 or, 5x = 6 + 3x or, 27 + 3y= 45 or, 2x = 6 or, 3y = 18 ∴ x = 3 ∴ y = 6 Hence, x = 3 cm and y = 6 cm. A C B y cm P 9 cm 6 cm 10 cm 2 cm x cm Q Geometry - Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 245 Vedanta Excel in Mathematics - Book 9 Geometry - Triangle Example 2: In the adjoining figure, AC // DE. Prove that AB.DE = AC.BE. Also, find the value of x. Solution: 1. In ∆ ABC and ∆BDE, (i) ∠ACB = ∠BDE [AC // DE and alternate angles] (ii) ∠ABC = ∠DBE [Vertically opposite angles] (iii) ∠BAC = ∠BED [Remaining angles of the triangles] (iv) ∴ ∆ABC ∼ ∆BDE [A.A.A. axiom] 2. AB BE = AC DE = BC BD [Corresponding sides of similar triangle are proportional] or, AB.DE = AC.BE proved. Again, AB BE = BC BD or, 5.6 3.6 = x 4.5 or, x = 7 cm. Example 3: In the given figure, show that DPQR ~ DXQR. If PX = 5 cm and QX = 4 cm, find the length of QR. Solution: 1. In ∆ PQR and ∆ XQR, (i) ∠PQR = ∠XQR [Common angle] (ii) ∠QPR = ∠XRQ [Given] (iii) ∠PRQ = ∠QXR [Remaining angles of the triangles] (iv) ∴ ∆PQR ∼ ∆XQR [A.A.A. axiom] 2. PQ QR = QR XQ [Corresponding sides of similar triangles] or, 9 cm QR = QR 4 cm or, QR2 = 36 cm2 ∴ QR = 6 cm Hence, the length of QR is 6 cm. Example 4: In the figure alongside, ABCD is a parallelogram, in which M is the mid-point of AD, prove that: OB = 2OD. Solution: Given: In parallelogram ABCD, M is the mid-point of AD. To prove: OB = 2OD P R 4 cm 5 cm X Q https://www.geogebra.org/m/jkh6f87x Vedanta ICT Corner Please! Scan this QR code or browse the link given below: A B C M D O

Vedanta Excel in Mathematics - Book 9 246 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Proof: Statements Reasons 1. (i) (ii) (iii) (iv) In DMOD and ∆ BOC ∠OMD = ∠OCB (A) ∠ODM = ∠OBC (A) ∠MOD = ∠BOC (A) ∆MOD ~ ∆BOC 1. (i) (ii) (iii) (iv) AD // BC and alternate angles Same as (i) Vertically opposite angles. By A.A.A axiom 2. OM OC = OD OB = MD BC 2. Corresponding sides of similar triangles 3. BC = AD = 2MD 3. Opposite sides of parallelogram and given. 4. OD OB = MD 2MD ∴ OB = 2OD 4. From statements (3) and (4) Proved EXERCISE 13.4 General section 1. a) In the given figure, ∆ABC ~ ∆APQ. If AB = 15 cm, AP = 10 cm and BC = 12 cm, find the length of PQ. b) In the adjoining figure, PQ // XZ, PX = 6 cm, XY = 15 cm and YQ = 3 cm, find the length of QZ. c) In the adjoining figure, ∆PQR ~ ∆ABR, QR = 20 cm, QB = 8 cm, AB = 6 cm and PR = 15 cm. Find the length of PQ and AR. d) In the adjoining figure, ∆PEM ~ ∆REM, ∠EPM = ∠RME. Find the length of EM. e) In the given figure, ∆ABC ~ ∆ABD, ∠BAD = ∠ACB, find the length of BD. A B C P Q P X Q Z Y P Q B A R P E M R 8 cm 10 cm A 4 cm B D8 cm C Geometry - Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 247 Vedanta Excel in Mathematics - Book 9 Geometry - Triangle Creative section - A 2. a) In the figure alongside AB // DE. Show that ∆ABC ~ ∆DCE and find the value of x. b) In the adjoining figure, ∠BAD = ∠ACB. Prove that ∆ABC ~ ∆ABD and find the lengths of AD and BD. c) In the figure alongside, ∠ABC = ∠ACP. Prove that ∆ABC ~ ∆APC and find the lengths of AB and PB. d) In the given figure, AX ⊥ BC. Prove that ∆ABC ~ ∆AXC and find the lengths of CX and AX. e) In the adjoining figure, AB // QC, PR = 2RQ and QC = 3 cm. Find the length of AP with the suitable reasons. 3. a) A boy 1.8 m tall casts the shadow of length 3 m at 2:30 p.m., what is the height of the tree which casts the shadow of length 30 m at the same time? b) The width of a LED TV of screen size 32 inch is 27.9 inch. What is the screen size of the TV which is 37.5 inch wide? c) A man standing in front of a tower observes the top of the tower in a mirror on the ground. The mirror is placed at a distance of 36 ft. from the foot of the tower. If the man is 2.5 ft. away from the mirror, and the distance of his eye level from the ground is 5 ft, what is the height of the tower? D A B 10 cm 5 cm x cm 9 cm E C A C 8 cm 10 cm 6 cm D B P A B C X 3 cm 4 cm A C B P R Q 3 cm 30 m 1.8 m 3 m 27.9 in 37.5 in 32 in 5 ft. A B2.5ft.C 36 ft. E D

Vedanta Excel in Mathematics - Book 9 248 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Creative section - B 4. a) In the given figure, AB // CD. Prove that (i) ∆AOB ~ ∆COD (ii) AB.DO = CD.AO b) In the adjoining figure, ∠P = ∠S. Prove that (i) ∆POR ~ ∆QOS (ii) PR.OQ = QS.OR (iii) PO.OQ = OR.OS P S Q R c) In the given right angled triangle ABC, ∠ABC = 90° and NM ⊥ AC. Prove that (i) ∆ABC ~ ∆AMN (ii) BC.AM = AB.MN d) In the adjoining figure, ∠PQS = ∠PRQ. Prove that (i) ∆PQS ~ ∆PQR (ii) PQ2 = PS.PR A B N M C f) In the trapezium ABCD, AB // DC and the diagonals AC and BD intersect each other at O. Prove that AB.DO = CD.BO g) In the adjoining figure, ∠ADE = ∠ ACB, and ∠DAC = ∠BAE. Prove that AD.BC = AC.DE A B C D E e) In the given right-angled triangle ABC right-angled at B, BD ⊥ AC and ∠CBD = ∠BAC. Prove that (i) ∆ABC ~ ∆BCD (ii) ∆BCD ~ ∆ABD (iii) BC2 = AC.CD (iv) BD2 = AD.CD (v) BC2 BD2 = AC AD Geometry - Triangle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 249 Vedanta Excel in Mathematics - Book 9 Geometry - Triangle 5. a) In the given figure, AB = DC, AB // DC, M is the mid-point of AB. Prove that (i) DAOM ~ DCOD (ii) CO = 2AO b) In the adjoining diagram, AD // BC and AD = BC. Prove that (i) DEBC ~ DEMN (ii) BE.MN = AD.ME A B C M N D E OBJECTIVE QUESTIONS Let’s tick (√) the correct alternative. 1. The sum of interior angles of any triangle is (A)1 right angle. (B) 2 right angles (C) 3 right angles (D) 4 right angles 2. Which of the following measurements cannot be the side lengths of a triangle? (A) 1 cm, 2 cm, 3 cm (B) 3 cm, 4 cm, 5 cm (C) 7 cm, 4.5 cm, 3.8 cm (D) 6.2 cm, 10.5 cm, 8.4 cm 3. When a side of a triangle is produced to a point, then (A)The exterior angle so formed is equal to the sum of interior angles. (B)The exterior angle so formed is equal to the sum of any two interior angles. (C)The exterior angle so formed is equal to the sum of two opposite interior angles. (D)The exterior angle so formed is equal to adjacent interior angle. 4. An exterior angle of a triangle is 100° and two opposite interior angles are equal. Then measure of each of these angle will be (A) 100° (B) 50° (C) 80° (D) 70° 5. In a ∆ABC, if AB = 3.6 cm, BC = 5.2 cm and AC = 4.5 cm, then the smallest angle is (A) ∠A (B) ∠B (C) ∠C (D) None 6. In ABC, if ∠A = 60o and ∠B = 70o then which of the following statements is correct? (A) AB <BC < AC (B) BC <AC < AB (C) AC < BC < AB (D) AB <AC < AB 7. For any ∆ABC, which of the following is not the triangle inequality property? (A) AB + BC < AC (B) BC + AC < AB (C) AC – AB > BC (D) AB – AC < AB 8. In an isosceles triangle ABC, AB = AC and AM is the median drawn from vertex A upon base BC then which of the following statement is true? (A) AM ⊥ BC (B) ∆ABM ≅∆ACM (C)∠BAM = ∠CAM (D) All of the above 9. Which of the following is NOT the criterion for congruency of triangles? (A) S.S.S. (B) S.A.S. (C)R.H.S. (D) A.A.A. 10. In ∆PQR and ∆XYZ, PQ XY = QR YZ then they will be similar if (A) ∠P = ∠X (B) ∠Q = ∠ X (C) ∠R = ∠Z (D) ∠Q = ∠Y

Vedanta Excel in Mathematics - Book 9 250 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 14.1 Special types of Quadrilaterals The quadrilateral ABCD given alongside is a closed plane figure bounded by four line segments called the sides of the quadrilateral. AC is a diagonal of the quadrilateral. DP ⊥ AC and BQ ⊥ AC are drawn. Therefore, DP = h1 , and BQ = h2 are the heights of DACD and DABC respectively. Here, area of quadrilateral = Area of (DACD + DABC) = 1 2 AC × h1 + 1 2 AC × h2 = 1 2 AC (h1 + h2 ) Thus, area of quadrilateral = 1 2 diagonal × sum of perpendiculars drawn from opposite vertices to the diagonal Trapezium, parallelogram, rectangle, square, rhombus, and kite are the special types of quadrilaterals. Trapezium The quadrilateral in which any two opposite sides are parallel is called a trapezium. In the adjoining figure, PQRS is a trapezium in which PQ // SR. Properties of trapezium (i) The parallel sides of a trapezium are called its bases. In the figure, PQ and SR are the bases. (ii) The non-parallel sides of a trapezium are called its legs. In the figure, PS and QR are the legs. (iii) The straight line segment that joins the mid-points of legs is called the median of the trapezium. XY is the median. (iv) The perpendicular distance between the parallel sides (bases) of a trapezium is called its altitude (height). h is the height of the trapezium. (v) Area of trapezium = 1 2 Height × Sum of the bases = 1 2 h (PQ + SR) Parallelogram The quadrilateral in which opposite sides are parallel is called a parallelogram. In the figure, ABCD is a parallelogram. Q P Unit 14 Geometry: Parallelogram