The Leading Maths - 10

200 Allied The Leading Mathematics-10 Quadratic Equation 201 Algebra (b) Here, mass of the body (m) = 10 gm and kinetic energy (E) = 500 units. Now, E = 1 2 mv2 or, 500 = 1 2 (10) v2 or, v2 = 100 = 102 => v = 10 or – 10 units. (c) Here both values are admissible. The positive value 10 refers to the velocity of the body while going upwards and the negative value –10 corresponds to the velocity of the body while falling downwards. VI Applications of Quadratic Equations on Financial Problems The quadratic equations are also used in compound interest, compound growth, compound depreciation, business fields, etc, So, we remember some relation related to above mentioned topics. Example-10 A sum of money is invested at a compound interest in a company by a man. In 2 years, the amount is four times the money invested. (a) Write the given information in mathematical notation form. (b) What is the rate of compound interest provided by the company? Find it. (c) How many times more money than the invested money would he get after 3 years? Calculate. Solution: (a) Let the sum invested be x. i.e., Principal (P) = x. Then the amount in time (T) = 2 years is 4x. i.e., Compound amount (CA) = 4x (b) Now, we have, CA = P 1 + R 100 T or, 4x = x 1 + r 100 2 or, 4 = 100 + r 100 2 [∴ x ≠ 0] or, 40000 = 10000 + 200r + r2 or, r2 + 200r – 30000 = 0 or, r2 + 300r – 100r – 30000 = 0 or, r(r + 300) – 100(r + 300) = 0 or, (r – 100) (r + 300) = 0 So, r = 100 or – 300 Since the rate of interest cannot be negative, the required rate of interest is 100%. (c) Again, Total amount after 3 years, CA = P 1 + R 100 T = x 1 + 100 100 3 = x × 8 = 8x.

202 Allied The Leading Mathematics-10 Quadratic Equation 203 Algebra ∴ Increased ratio = 8x x = 8 : 1 Hence, he gets 8 times more amount than the invested amount after 3 years. Example-11 A class 10 student plans to go picnic and fixes the budget of Rs. 63000 for all students. Although 10 students can't go to the picnic due to their personal problems, Rs. 210 is to be collected more amount to each student. (a) How many students participate in the picnic ? (b) How much amount does each student pay for the picnic? Find it. (c) What percentage of more amount does each student pay for going to the picnic? Solution: (a) Let the total number of students of a class 10 be x. Since the fixed budget for a picnic is Rs. 63000, so the amount to be paid by each student is Rs. 63000 x . Here, 10 students can't go to the picnic. So, the number of students to be participated in the picnic is x – 10. So, the amount to be paid by each student who wants to go to the picnic is Rs.63000 x – 10 . ................. (i) Then, by question, x – 10 students pay Rs. 210 more. So, the paid amount by x – 10 students will be Rs. 63000 x + Rs. 210 = Rs. 63000 + Rs. 210x x . ........ (ii) From the relations (i) and (ii), we get 63000 x – 10 = 63000 + 210x x or, 63000x = 63000x + 210x2 – 630000 – 2100x or, x2 – 10x – 3000 = 0 or, x2 – 60x + 50x – 3000 = 0 or, x(x – 60) + 50(x – 60) = 0 or, (x – 60) (x + 50) = 0 => x = 60 or x = – 50. Hence, 60 students participated in the picnic. (b) The number of students who participated in the picnic = 60 – 10 = 50. ∴ The paid amount by each student who went to the picnic = Rs. 63000 50 = Rs. 1260. (c) The percentage of more amount paid by each student who went to the picnic is, 210 1260 – 210 × 100% = 210 1050 × 100% = 20%.

202 Allied The Leading Mathematics-10 Quadratic Equation 203 Algebra PRACTICE 7.2 Read Think Understand Do Keeping Skill Sharp 1. (a) What are the roots in px2 + qx = 0? (b) What are the non-negative root in x2 – a = 0? (c) Write the positive root in 2x2 – 18 = 0. 2. (a) When the sum of the square of a positive number and 2 is 38, which is the number ? (i) 6 (ii) – 6 (iii) 40 (iv) 40 (b) If 2 is subtracted from the square of a number, the result is 7. The number is ...... . (i) 3 (ii) 2 (iii) ± 3 (iv) – 3 (c) When the square of a natural number is multiplied by 3, the product is 12. What is it ? (i) ± 2 (ii) 2 (iii) ± 4 (iv) – 2 (d) If 5 is added to 2 times the square of a whole number, the result is 13. What is it ? (i) ± 2 (ii) 0 (iii) 1 (iv) 2 (e) When 3 times of the square of an integer is divided by 2, the quotient is 24. What is it ? (i) 3 (ii) – 4 (iii) ± 4 (iv) 4 Check Your Performance Answer the following questions for each of the problem below. 3. (i) The sum of two numbers is 15 and their product is 56. (ii) The difference of two positive numbers is 1 and their product is 132. (iii) The product of two parts of 19 is 90. (a) Express the given problem in the mathematical notations. (b) Convert them into a single variable equation. Name the type of the equation. (c) Solve this equation and test in the given conditions. (d) When 2 is added to each number, what percentage will the new product be increased than the original product ? 4. (i) If 25 is added to the square of a positive number, the sum is 125. (ii) If 5 is subtracted from the square of a non-negative number, the result is 139. (iii) When the square of a natural number is multiplied by 5, the product is 125. (iv) When 7 times of the square of an integer is divided by 5, the quotient is 35. (v) If 10 is added to 7 times the square of a whole number, the result is 353. (vi) If 17 is subtracted from 18 times the square of a rational number, the sum is 111. (vii) The sum of a positive number and its square is 56.

204 Allied The Leading Mathematics-10 Quadratic Equation 205 Algebra (a) Express the given problem in the mathematical notations. (b) Convert them into a single variable equation. Name the type of the equation. (c) Solve this equation and test in the given conditions. 5. (i) The product of two consecutive integers is 20. (ii) The product of two consecutive even numbers is 143. (iii) One number is 3 more than other and the larger number is equal to their product. (iv) There are three consecutive even integers such that twice their sum equals the product of the first and the second numbers. (v) The sum of a number and its reciprocal is 2. (vi) The sum of the reciprocal of x and x + 3 is 2 3 , x ∈ R. (vii) The sum of the reciprocal of x + 2 and x + 3 is 3 5 , x ∈ R+ . (a) Express the given problem in the mathematical statement. (b) Write the mathematical statement in standard form. (c) Solve this equation and write the correct number. 6. (i) In a number between 9 and 100, the product of the digits is 20 and their sum is 9. (ii) The difference of digits of two-digit number is 1 and their product is 12. (iii) The product of digits in a two-digit number is 18. The number formed by interchanging the digits of that number will be 27 more than the original number. (iv) The product of digits in a two-digit number is 14. If 45 is subtracted from the original number, the number will be reversed. (a) Write the given conditions in the equation forms and combine them. (b) Solve the combined equation and find the number. (c) What percentage is increased when it is reversed? [Note: For (iii)] (c) What percentage is decreased when it is reversed? [Note: For (iv)] 7. (i) The present ages of people are 13 years and 7 years, and x years ago, the product of their ages was 55. (ii) The present ages of two people are 30 years and 14 years respectively, x years ago, the product of their ages was 192. (a) What would be the ages of these people x years ago ? (b) Write the given condition in a notation form. (c) Find the value of x and test it in the given condition. (d) What were their ages x years ago in actuality? Write. (iii) The sum and product of the present ages of two sisters are 34 and 288 respectively. (iv) The difference and product of the present ages of a father and his son are 27 and 810 respectively. (v) The difference of the ages of two brothers is 6 years and the product of their ages will be 280 after 2 years.

204 Allied The Leading Mathematics-10 Quadratic Equation 205 Algebra (a) Write the given conditions in the equation forms and combine them. (b) Solve the single equation and write their present ages. (c) In how many years of the elder sister had the younger sister been born ? [Note: For (iii)] (c) In how many years of the father had his son been born ? Find it. [Note: For (iv)] 8. (i) Raj's present age is twice that of Anju and the product of their ages is 98 years. (ii) The product of the ages of a man 10 years before and 10 years hence is 800. (iii) The present age of father is thrice as old as the age of his daughter 5 years ago and after 5 years, his age will be double of his daughter. (iv) Seven years ago Pratima's age was five times the square of Samjhana's age. Three years hence, Samjhana's age will be two-fifths of Pratima's age. (a) What type of equation do you find from the given problem? (b) Write the given problem in a single equation in standard form. (c) Solve the equation. Write the present ages. (d) By how many years is Raj older than Anju ? Find it. [For No. (i)] (d) By how many years is father older than his daughter ? Find it. [For No. (iii)] (d) How many times older is the age of Pratima than the age of Samjhana? [For No. (iv)] 9. (i) The area of a squared pond is 576 sq. meters. (ii) The length of a pond is 1 m greater than its breadth and its area is 20 m2 . (iii) The area of a rectangular pond is 63 m2 and the perimeter is 32 m. (iv) The semi-perimeter of a rectangular pond is 20 m and its area is 91 m2 . (v) The length of a certain rectangle of area 27 m2 is 6 meters more than its width. (a) Which formula do you use to find the length of the squared pond? [For No. (i)] (a) Which formula do you use of find the length or breath of the rectangle? (b) How long and broad is the pond? (c) How much will he expense for fencing around the pond six times at the rate of Rs. 315.50 per meter ? (vi) The area of a circular pond is 616 sq. meters. (a) Find its radius. (b) Calculate the cost of fence around the pond five times at Rs. 250 per m. (vii) The length of a rectangle is twice of its width. If the rectangle was only 2 meters wider, its area would be 240 sq. meters. Find the area of the given rectangle. (viii)The length of a rectangle is one centimeter more than twice its width. If the width is decreased by 3 cm then the area of the resulting rectangle will be 60 sq. cm. What are the dimensions of the original rectangle?

206 Allied The Leading Mathematics-10 Quadratic Equation 207 Algebra 10. (i) The length of hypotenuse of a right triangular garden exceeds the length of the base by 2 m and exceeds twice the length of perpendicular by 1 m. (ii) The sides of a right angled triangular pond of the garden containing the right angle are less than its hypotenuse by 5 m and 10 m respectively. (iii) The hypotenuse of a right triangular plot of the land is 20 m and the ratio of the remaining sides is 1:3. (a) State the Pythagoras' theorem. (b) Write the relation between the hypotenuse, base and perpendicular of the right triangle in a single variable. (c) Solve the equation obtained in (ii) and find the length of its each side of the right triangle. (d) Establish the Pythagoras' theorem. 11. (i) The total surface area of a right circular tank of height 7 ft is 1848 sq. ft. (ii) The total surface area of a cone of slant height 7 ft is 2451 7 ft2 . (iii) The curved surface area of a hemi-spherical hole in the ground is 1232 m2 . (iv) The total surface area of a pyramid with slant height 25 m is 2400 m2 . (a) Which formula is used to calculate the radius/side of the base of the given solid ? (b) What is the circumference of the circular base of the solid ? Find it. (c) Calculate the amount to be filled with water at Rs. 9.75 per litre. 12. (i) A sum of money is invested at a compound interest in a company by a girl. In 2 years, the amount is 2 times the money invested. (ii) In 2 years, the deposited money is 3 times in a bank in compounded annual interest. (iii) The population of a city increased 6/5 times in 2 years. (iv) A boy bought motorbike and its price reduced 4 time in 2 years. (a) Write the given information in mathematical notation form. (b) What is the rate of compound interest provided by the company? Find it. (c) How many times more money than the invested money would she get after 3 years? Calculate. [For No. 13. (i)] (c) By how much more money on the invested money would he get after 4 years? Calculate. [For No. 13. (ii)] (c) How many times will the population of the city be after 8 years? Calculate. [For No. 13. (iii)] (c) How many times has the cost of the bike been reduced in 4 years? Calculate. [For No. 13. (iv)] 13. (i) A group of girls plan to go picnic and fixes the budget of Rs. 48000 for all girls in equal sharing. 8 girls couldn't go to the picnic due to their personal problems. So, Rs. 300 more amount has to be paid to each girl who went to the picnic.

206 Allied The Leading Mathematics-10 Quadratic Equation 207 Algebra (a) How many students participate in the picnic ? (b) How much amount does each student pay for the picnic? Find it. (c) What percentage of more amount does each person pay for going to the picnic? (ii) A group of people plans to donate for a school and fixes the amount of Rs. 225000 for all people in equal sharing. 5 people couldn't give to donate to the school due to their personal problems. So, Rs. 500 more amount has to be given to each person who wants to donate to the school. 3. (i) (a) x + y = 15, xy = 50 (b) x2 – 15x + 56 = 0 (c) 7, 8 (d) 60.71% (ii) (a) x – y = 1, y = 132 (b) x2 – x – 132 = 0 (c) 11, 12 (d) 37.88% (iii) (a) x + y = 19, xy = 90 (b) x2 – 19x + 90 = 0 (c) 9, 10 (d) 46.66% 4. (i) (a) x2 + 25 = 125 (b) x2 – 100 = 0, pure quadratic equation, (c) 10 (ii) (c) 12 (iii) (c) 5 (iv) (c) ± 5 (v) (c) 7 (vi) (c) ± 8 3 (vii) (c) 7 5. (i) (c) 4, 5 or –5, –4 (ii) 11, 13 (iii) –3, 0 or 1, 4 (iv) (c) 6, 8, 10 (v) (c) 1 (vi) (c) ± 3 2 (vii) (c) – 5 + 109 6 6. (i) (b) 45 or 54 (ii)(b) 43 (iii) (b) 36 (c) 75% (iv) (b) 72 (c) 62.5% 7. (i) (b) 2 (d) 11, 5 (ii) (c) 6 (d) 24, 8 (iii) (b) 18, 16 (c) 2 (iv) (b) 45, 18 (c) 27 (v) (b) 18, 12 8. (i) (c) 14, 7 (d) 7 (ii) (c) 30 (iii) (c) 35, 15 (d) 20 (iv) (c) 27, 9 (d) 3 times. 9. (i) (b) 24 m (c) Rs. 181728 (ii) (b) 5 m, 4 m (c) Rs. 34074 (iii) (b) 9 m, 7 m or 7 m, 9, Rs. 60576 (iv) (b) 13 m, 7 m or 7 m, 13 m, (c) Rs. 75720 (v) (b) 9 m, 3 m (c) Rs. 45432 (vi) (a) 14 m (b) Rs. 110000 (vii) 200 m2 (viii) 15 cm, 7 cm 10. (i) (c) 17, 15, 8 (ii) (c) 25, 20, 15 (iii) (c) 2 10, 6 10 11. (i) (b) 88 ft (c) Rs. 91647.05 (ii) 37.41 (c) Rs. 2890.13 (iii) (b) 88 m (c) Rs. 4312000 (iv) (b) 120 (c) Rs. 4500000 12. (i) (b) 41.42%, 2.83 times (ii) (b) 73.21%, 9 times (iii) (b) 9.54%, 36 25 times (iv) 300%, 16 times 13. (i) (a) 32 (b) Rs. 1500 (c) 25% (ii) (a) 45 (b) Rs. 500 (c) 11.11% Answers Project Work Estimate the perimeter and area of your rectangular room and find the its length and breadth. Now, measure the actual length and breadth of the same room and compare your calculation from estimation. If you need to enlarge some of the length and breadth of the room, what percentage do its perimeter and area will increase? Prepare a report and present it in your classroom.

208 Allied The Leading Mathematics-10 Algebraic Fractions 209 Algebra 8.1 Simplification of Algebraic Fractions At the end of this topic, the students will be able to: ¾ simplify the algebraic rational fractions. Learning Objectives I Introduction Activity 1 Consider the areas of two classrooms of Class: 10 "A" and Class: 10 "B" (x2 – 3x + 2) sq. feet and (x2 – 4) sq. feet respectively. The both areas are the algebraic expressions or polynomials. What is the ratio of their areas ? How can we compute them? The ratio of the areas of the Class: 10 "A" and Class: 10 "B" is, x2 – 3x + 2 x2 – 4 ...... (i) The numerator and denominator of the ratio (i) are algebraic expressions in which the power of variable may or may not be non-negative. This is called algebraic fraction. It both expressions may or may not be either positive or negative. Algebraic fractions are subject to the same laws as arithmetic fractions. But, a rational expression or algebraic rational fraction is an algebraic fraction whose numerator and denominator are both polynomials in which the power of variable is only non-negative. For example, 2x2 – 3 x2 – 6 is a rational fraction and 2x2 – 3 x2 – 6 is not rational fraction, but it is an algebraic fraction. So, all rational fractions are the algebraic fractions, but not vice versa. What is their exact ratio for x ∈ R? For what values of x doesn't it exist or define ? How can we find easily ? For this, we factorise the numerator and denominator of the algebraic fraction (i) as' x2 – 3x + 2 x2 – 4 = x2 – 2x – x + 2 x2 – 4 = x(x – 2) – 1(x – 2) x2 – 22 = (x – 2) (x – 1) (x + 2) (x – 2) = x – 1 x + 2 for x ≠ 2. This is the reduction of algebraic fraction. Here, if x = 2 or – 2 in its denominator, the algebraic fraction of the areas of the room doesn't exist or defined. For positive ratio, the numerator and denominator are either positive or negative. When the value of x greater than – 2. i.e., x = (– 2 , ∞) for denominator and x > 1 for numerator, the ratio will be positive and vice versa. Class: 10 "B" Class: 10 "A" Do not confuse fraction, ratio, quotient and rational number ! CHAPTER 8 ALGEBRAIC FRACTIONS

208 Allied The Leading Mathematics-10 Algebraic Fractions 209 Algebra II Simplification of Algebraic Fractions Four fundamental rules of arithmetic can be extended to the case of algebraic fractions. In addition, we often shall have to apply the following fundamental principles of fraction: A fraction remains unchanged if its numerator and denominator are divided by the same non-zero expression. In other words, if a b denotes an algebraic fraction, then a b = a c b c or, ac bc such that b ≠ 0 and c ≠ 0. For instance, 6 9 = 2.3 3.3 = 2 3 ; 6ax2 y 9bxy2 = 2.3.a.x.x.y 3.3.b.x.y.y = 2ax 3by ; ax + ay bx + by = a(x + y) b(x + y) = a b ; x2 – 4 x2 – 3x + 2 = (x – 2) (x + 2) (x – 2) (x – 1) = x + 2 x – 1 In dealing with one or more fractions, it is often advisable to reduce a given fraction to its simplest form as shown above. The next step is to find the algebraic sum of two or more fractions. The basic procedure is indicated below: a d + b d + c d + ..... + m d = a + b + c + ... + m d , provided that d ≠ 0. For instance, 2x 3x + y – y 3x + y = 2x – y 3x + y Thus, we have seen how fractions with the same or common denominators are added. Here the numerator is the algebraic sum of the numerators while the common denominator is retained in the sum of the fractions. To add fractions having different denominators, we first express the given fractions as equivalent fraction having the same denominators and then proceed as explained above. It is desirable, but not essential, to use the least common denominator (LCD) of all the fractions involved. For instance, 1 + a b = b a + a b = b + a b = a + b b a b – c d = ad bd + bc bd = ad + bc bd x ab + y bc + z ca = cx abc + ay abc + bz abc = cx + ay + bz abc ; 1 a + b + 1 a – b = a – b (a + b) (a – b) + a + b (a + b) (a – b) = a – b + a + b (a + b) (a – b) = 2a a2 – b2 Note : To add two or more fractions, find the LCM of the denominators, this will be used as the common denominator; divide it by the denominator of the first fraction, and multiply its numerator by the quotient so obtained; the same process is repeated for the other fractions also. The sum of the numerators so obtained with the common denominator is the required sum of the fractions. Cyclic Order a ± b ab b ± bc c c ± a ca a b c

210 Allied The Leading Mathematics-10 Algebraic Fractions 211 Algebra Example-1 Express as a single fraction : a x2 y + b y2 z – c z2 x . Solution: Here, a x2 y + b y2 z – c z2 x The LCM of x2 y, y2 z and z2 x is x2 y2 z2 . Expressing each fraction as an equivalent fraction with this LCM as the denominator, a x2 y = ayz2 x2 y . yz2 = ayz2 x2 y2 z2 b y2 z = bzx2 y2 z . zxz2 = bzx2 x2 y2 z2 c z2 x = cxy2 z2 x . xy2 = cxy2 x2 y2 z2 Now, a x2 y + b y2 z – c z2 x = ayz2 x2 y2 z2 + bzx2 x2 y2 z2 – cxy2 x2 y2 z2 = ayz2 + bzx2 – cxy2 x2 y2 z2 . In short, Here, a x2 y + b y2 z – c z2 x = ayz2 + bzx2 – cxy2 x2 y2 z2 Example-2 Simplify: x2 + 1 x + 1 + x2 – x + 1 x2 – 1 Solution: Here, x2 + 1 x + 1 + x2 – x + 1 x2 – 1 = x2 + 1 x + 1 + x2 – x + 1 (x + 1) (x – 1) = (x2 + 1) (x – 1) + x2 – x + 1 (x + 1) (x – 1) = x3 + x – x2 – 1 + x2 – x + 1 x2 – 1 = x3 x2 – 1 Example-3 Simplify : 2 a + b – 2 a – b + 4a a2 – b2 Solution: Here, 2 a + b – 2 a – b + 4a a2 – b2 = 2 a + b – 2 a – b + 4a (a + b) (a – b) = 2(a – b) – 2(a + b) + 4a (a + b) (a – b) = –4b + 4a (a + b) (a – b) = 4(a – b) (a + b) (a – b) = 4 a + b LCM of (x + 1) and (x + 1) (x – 1) is (x + 1) (x – 1). x2 + 1 x + 1 + x2 – x + 1 (x + 1) (x – 1) LCM of (a + b), (a – b) and (a + b) (a – b) is (a + b) (a – b). 2 a + b – 2 a – b + 4a (a + b) (a – b)

210 Allied The Leading Mathematics-10 Algebraic Fractions 211 Algebra Example-4 Simplify: x + 1 x – 1 + x – 1 x + 1 – 3x2 x – 1 Solution: Here, x + 1 x – 1 + x – 1 x + 1 – 3x2 x – 1 = x2 + 2x + 1 + x2 – 2x + 1 (x – 1) (x + 1) – 3x2 – 1 x – 1 = 2x2 + 2 – 3x3 – 3x2 (x – 1) (x + 1) = 2 – x2 – 3x3 x2 – 1 Example-5 Simplify : 1 x2 – 8x + 15 + 1 x2 – 4x + 3 – 2 x2 – 6x + 5 . Solution: Here, 1 x2 – 8x + 15 + 1 x2 – 4x + 3 – 2 x2 – 6x + 5 = 1 (x – 3) (x – 5) + 1 (x – 1) (x – 3) – 2 (x – 5) (x – 1) = x – 1 + x – 5 – 2x + 6 (x – 3) (x – 5) (x – 1) = 0 (x – 3) (x – 5) (x – 1) = 0 PRACTICE 8.1 Read Think Understand Do Keeping Skill Sharp 1. (a) What is rational expression? (b) What is simplification? (c) In which condition, a(x) b(x) + c(x) d(x) is defined ? (d) The simplest form of p(x) q(x) + r(x) q(x) is ............ . (d) What is the sum of 1 a + b and 1 a – b ? Write. 2. Circle ( ) the correct answer. (a) In which condition is p(x) q(x) a rational fraction ? (i) p(x) is algebraic expression. (ii) q(x) is algebraic expression. (iii) p(x) and q(x) are both algebraic expressions. (iv) p(x) and q(x) are not both algebraic expressions. (b) a a + b + b a + b = ........ (i) a + b (ii) a – b (iii) (a + b)2 (iv) 1

212 Allied The Leading Mathematics-10 Algebraic Fractions 213 Algebra (c) x x + y + x x – y = ........ (i) 2x2 x2 – y2 (ii) 2x x2 – y2 (iii) 2x2 x2 + y2 (iv) x x – y (d) a – b a2 – ab + b2 + a + b a2 + ab + b2 = ........... . (i) 2ab a4 – a2 b2 + b4 (ii) 2a2 a4 + a2 b2 + b4 (iii) 2b3 a4 + a2 b2 + b4 (iv) 2a3 a4 + a2 b2 + b4 (e) Which is the simplest form of a2 – b2 a3 – b3 ? (i) a + b a2 + ab + b2 (ii) a – b a2 – ab + b2 (iii) a2 + b2 a2 + a2 b2 + b2 (iv) a2 + b2 a2 – a2 b2 + b2 Check Your Performance 3. Simplify the following rational fractions: (a) 1 a – 2 + 1 a – 3 (b) 1 x – 2 – 1 x – 3 (c) x + 2 x + 3 – x – 1 x + 2 (d) x3 x – 1 – x3 x + 1 (e) x + 2 – x – 2 x – 1 (f) 1 a(a – b) + 1 a(a + b) 4. Prove the following relations: (a) 1 x + y + x xy + y2 = 1 y (b) a + b a – b – a – b a + b = 4ab a2 – b2 (c) 1 x – 3y – y x2 – 9y2 = x + 2y x2 – 9y2 (d) 1 (p – 2)2 + 3 p2 – 4 = 4p – 4 (p – 2)2 (p + 2) 5. Simplify : (a) 1 a2 – 6a + 9 – 1 a2 – 5a + 6 (b) 1 x2 – 3x + 2 + 1 x2 + 3x – 10 (c) 1 x + 3 – 1 x + 2 + 1 x2 + 5x + 6 (d) 2x – 6 x2 – 6x + 9 – 2x – 3 x2 – x – 6 6. Simplify by reducing to a single fraction: (a) 1 a – x – 1 a + x – 2x a2 + x2 (b) a a – b – b a + b – 2ab a2 – b2 (c) a x2 – x + a x – x3 – a x2 – 1 (d) x (x – y) (x – z) + y (y – z) (y – x) + z (z – x) (z – y) (e) 1 a2 – 5a + 6 + 2 (a – 1) (3 – a) + 1 a2 – 3a + 2 (f) 2(x – 1) x2 – 4x + 3 + x – 3 x2 – 7x + 12 – 3(x – 4) x2 – 5x + 4

212 Allied The Leading Mathematics-10 Algebraic Fractions 213 Algebra (g) 1 1 – x + x2 + 1 1 + x + x2 – 2x 1 + x2 + x4 (h) a + b a2 + ab + b2 + a – b a2 – ab + b2 + 2b3 a4 + a2 b2 + b4 (i) x – 6 x2 – 6x + 36 + x + 6 x2 + 6x + 36 + 432 x4 + 36x2 + 1296 (j) 1 bc(a – b) (c – a) – 1 ca(b – c) (b – a) – 1 ab(c – a) (c – b) 7. Simplify : (a) x – y x + y + x + y x – y – 2(x2 – y2 ) x2 + y2 – 4(x4 – y4 ) x4 + y4 (b) 1 + a 1 – a + 1 – a 1 + a + 4a 1 + a2 + 8a3 1 – a4 (c) 1 + x 1 – x + 1 – x 1 + x – 1 + x2 1 – x2 – 1 – x2 1 + x2 (d) 8 a8 – 1 + 4 a4 + 1 + 2 a2 + 1 + 1 a + 1 (e) 1 x + y + 2x x2 + y2 + 4x3 x4 + y4 – 8x7 x8 – y8 (f) 1 x + 2 + 4 x2 + 4 + 32 x4 + 16 + 1024 x8 – 256 8. Prove that : (a) p + 1 p – 1 + p + 1 p + 1 – p – 1 p + 2 + p + 1 p – 2 = – 4(2p2 + 1) (p2 – 4) (p2 – 1) (b) x +1 x – 1 – x – 1 x + 1 – 4x x2 + 1 = 8x x4 – 1 (c) a3 (b + c) (a – b) (a – c) + b3 (c + a) (b – c) (b – a) + a3 (a + b) (c – a) (c – b) = ab + bc + ca (d) 2a a + b + 2b b + c + 2c c + a = 3 – (a – b) (b – c) (c – a) (a + b) (b + c) (c + a) (e) If a + b + c = 0, then prove that: (b + c)2 3bc + (c + a)2 3ca + (a + b)2 3ab = 1 3. (a) 2a – 5 (a – 2) (a – 3) (b) 1 (2 – x) (x – 3) (c) 2x + 7 (x + 2) (x + 3) (d) 2x3 x2 – 1 (e) x2 x – 1 (f) 2 a2 – b2 5. (a) 1 (a – 3)2 (a – 2) (b) 2(x + 2) (x – 1) (x – 2) (x + 5) (c) 0 (d) 7 (x – 3) (x – 2) 6. (a) 4x3 a4 – x4 (b) a – b a + b (c) (d) (e) (f) 7x – 25 (x – 3) (x – 4) (x – 1) (g) 2 1 + x + x2 (h) 2(a + b) a2 + ab + b2 (i) 2(x + 6) x2 + 6x + 36 (j) 0 7. (a) 16x4 y4 x8 – y8 (b) 2(1 + a) (1 – a) (c) 4x2 1 – x4 (d) 1 a – 1 (e) 1 b – a (f) 1 x – 2 Answers

214 Allied The Leading Mathematics-10 Indices 215 Algebra 9.1 Solving Exponential Equations At the end of this topic, the students will be able to: ¾ solve the exponential equations. Learning Objectives I Review of Laws of Indices Take a pot of rose flowers and take out two branches. Put in the same pot for the propagation of two new rose plants and grow. How many rose plants are there in the pot? Again, every rose plant is cut down into two-two branches of three rose plants from the same pot. If all cutting rose plants are alive, how many rose plants are there at all? If this process continues at 4 times, how many rose plants are there at all? It gives a sequence of the number of rose plants below. Start 1st time 2nd time 3rd time 4th time ......... Pattern 1 3 9 27 81 ........ ........ 1 31 32 33 34 ........ 3x , x ∈ N If you get 729 new rose plants by the same process mentioned above, how many times are rose plants there propagated? For this, we make a relation 3x = 729. This relation is called an exponential equation. Now, we solve the exponential equation 3x = 729 by using laws of indices below. Here, 3x = 729 or, 3x = 36 Here, the bases of the indices are the same/equal. So, their exponents are also the same/equal. ∴ x = 6. Hence, the rose propagates at 6 times to obtain 729 plants. In other hand, we obtain 625 plants at 4 times propagation of the plants, how many plants are propagated at first ? For this, the number of plants at initial is x. Then, x4 = 625 = 54 => x = 5. Hence, 5 plants are continuously propagated 4 times, we obtain 625 plants. Discuss reality. Similarly, how many plants are propagated to obtain 256 plants at the time as the number of 2nd time 1st time Plant propagation by cutting Oh! It is the process of growth rate. Initial plant, Pi = 1, Growth rate, R = 200%, Propagation times, T = 6 times ∴ Total plants at 6 times, PT = Pi(1 + R/100)T or, P6 = 1(1 + 200/100)6 = 36 = 729 plants. CHAPTER 9 INDICES

214 Allied The Leading Mathematics-10 Indices 215 Algebra initial plants ? We get another rule, xx = 256 => xx = 44 ∴ x = 4. 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. Similar laws are applied in the case of algebraic expressions of various forms. You have come across the following formulae on laws of indices. Recall them. (a) xm × xn = xm + n In multiplication of the same bases, the powers are added. (b) xm ÷ xn = xm xn = xm – n In division of the same bases, the powers are subtracted. (c) (xm) n = (xn ) m = xmn Powers are multiplied while working with power of powers. (d) x n = x1/n Meaning of nth root (Reciprocal). (e) xm n = x n m = xm/n Meaning of nth root and mth power. (f) (xy)m = xmym Distributive property of powers. (g) x0 = 1 Meaning of zero power. (h) 1m = 1 mth power of 1 is 1, where x ∈ R (x ≠ 0), m, n ∈ R. II Exponential Equations Activity 1 An equation in which the exponents are variables is known as exponential equation. For instance; (i) 8 = 2x , (ii) 2x + 4 = 8x , (iii) 2x + 5 = x2x + 1 + 60 are examples of exponential equations. The equation 2x = 8 can be solved by using trail and error. Let x = 1, 21 ≠ 8, When x = 2, 22 ≠ 8 When x = 3, 23 = 8 Therefore, x = 3 But all equations cannot be solved by using trial and error. Some exponential equations can be solved by using the following properties: For all positive real numbers a ≠ 1 and all real numbers x and y. (i) ax = ay ⇔ x = y; (ii) ax = bx ⇔ a = b; (iii) xx = aa ⇔ x = a. Note : 1) If ax = 1, then ax = a0 ⇒ x = 0.2) If ax = bx , then a = b. Using this property, the equation given above can be solved as follows: Here, 8 = 2x or, 23 = 2x ∴ x = 3.

216 Allied The Leading Mathematics-10 Indices 217 Algebra Example-1 Solve and verify: (i) 3x = 1 81 (ii) 4x – 2 = 8 (iii) 2 3 x = 9 4 Solutions: (i) 3 x = 1 81 or, 3x = 1 34 or, 3x = 3–4 ∴ x = – 4 Verification, Put x = – 4 in the eqn (i), 3– 4 = 1 81 or, 1 3– 4 = 1 81 or, 1 3– 4 = 1 81, which is true. Example-2 Solve: (i) 32x – 3 + 32x = 84, (ii) 2x + 2x+2 = 5. Solutions: (i) Here, 32x – 3 + 32x = 84, or, 32x 33 + 32x = 84 or, 32x 27 + 32x = 84 or, 1 + 1 27 32x = 84 or, 28 27 × 32x = 84 or, 32x = 3 × 27 = 34 Since each base is 3 ≠ 0, the exponents are also equal so, 2x = 4 => x = 2. Example-3 Solve: (i) 2x + 3 × 3x+4 = 18, (ii) 3x–1 + 31–x = 2. Solutions: (i) Here, 2x + 3 × 3x+4 = 18 or, 2x . 23 × 3x . 34 = 18 or, (2 × 3)x (8 × 81) = 18 or, 6x = 18 8 × 81 or, 6x = 1 36 or, 6x = 1 62 or, 6x = 6–2 ∴ x = –2 (ii) 4x – 2 = 8 or, (22 ) x–2 = 23 or, 22x – 4 = 23 ∴ 2x – 4 = 3 or, 2x = 7 or, x = 7 2 . Verification, 47/2 – 2 = 8 or, 22(7/2 – 2) = 8 or, 27 – 4 = 8 or, 8= 8, which is true. (iii) 2 3 x = 9 4 or, 2 3 x = 32 22 or, 2 3 x = 2–2 3–2 or, 2 3 x = 2 3 –2 ∴ x = – 2. Verification, 2 3 -2 = 9 4 or, 9 4 = 9 4 , which is true. (ii) Here, 2x + 2x + 2 = 5 or, 2x + 2x .22 = 5 or, 2x (1 + 4) = 5 or, 2x × 5 = 5 or, 2x = 5 5 = 1 or, 2x = 20 Since each base is 2 ≠ 0, the exponents are also equal so, x = 0.

216 Allied The Leading Mathematics-10 Indices 217 Algebra (ii) Here, 3x–1 + 31–x = 2 or, 3x 31 + 31 3x = 2 or, (3x ) 2 + 9 3.3x = 2 or, (3x ) 2 + 9 = 6.3x or, (3x ) 2 – 6.3x + 9 = 0 or, (3x ) 2 – 2.3x .2 + 32 = 0 or, (3x – 3)2 = 0 ∴ 3x – 3 = 0 or, 3x = 31 ∴ x = 1. Example-4 Solve: 4x – 6.2x+1 + 32 = 0. Solution: Here, 4x – 6.2x+1 + 32 = 0 or, 22x – 6.2x .21 + 32 = 0 or, (2x ) 2 – 12.2x + 32 = 0 or, (2x ) 2 – 8.2x – 4.2x + 32 = 0 or, 2x (2x – 8) – 4(2x – 8) = 0 or, (2x – 8) (2x – 4) = 0 Either, 2x – 8 = 0 or, 2x = 23 ∴ x = 3. OR, 2x – 4 = 0 or, 2x = 22 ∴ x = 2 Thus, x = 2 or 3. Example-5 If a = bx , b = cy and c = az then prove that xyz = 1. Solution: Given, a = bx and b = cy ∴ a1/x = b and b1/y = c Now, c = az or, b1/y = az or, a 1 x × 1 y = az ∴ 1 xy = z ⇒ xyz = 1. Proved. Note : You can solve by supposing 2x as a in the equation (2x ) 2 –12.2x+32 = 0 as a2 –12a+ 32=0.

218 Allied The Leading Mathematics-10 Indices 219 Algebra PRACTICE 9.1 Read Think Understand Do Keeping Skill Sharp 1. (a) What is an exponential equation? (b) If ax = ay , express x in term of y. (c) If xp = ap , what is the value of x ? (d) If 2x = 4, what is the value of x ? (e) If 34 = x4 , what is the value of x ? (f) If 4xx = 16, the value of x is ....... . 2. Circle ( ) the correct answer. (a) Which is false ? (i) If ap = aq , then a = p. (ii) If ap = aq , then p = q. (iii) If ap = bp , then a = b. (iv) If aa = bb , then a = b. (b) If x6 = x3p, which is the value of p ? (i) 2 (ii) 3 (iii) 6 (iv) 4 (c) If 42x + 1 = 45 , which is the value of p ? (i) 5 (ii) 2 (iii) 3 (iv) 6 (d) If 62x + 3 = (x + 1)2x + 3, which is the value of p ? (i) 5 (ii) 0 (iii) 6 (iv) 3 (e) If 3x – 81 = 0, then x = .......... . (i) 1 (ii) 2 (iii) 3 (iv) 4 Check Your Performance 3. Solve and verify your solution: (a) 22y + 1 = 8 (b) 95x – 2 = 34 (c) 1 2 x = 0.125 (d) 16–x = 2 (e) 312 = 812x + 3 (f) 1 4 2+x = 1 8 x–1 (g) 3x + 1 = 81 (h) 4x = 1 256 (i) 1 27 x – 2 = 81x + 1 4. Solve and verify your solution: (a) 2x+3 + 2x = 25 + 4 (b) 3x+2 + 3x+1 = 4 3 (c) 2x+2 + 2x+3 2 = 1 (d) 4x+3 + 4x +2 = 1280 (e) 2x+1 – 2x = 8 (f) 42x = 256 – 16x 5. (a) If a = bc , b = ca and c = ab then prove that abc = 1. (b) If a = 10x , b = 10y and ay bx = 100 then prove that xy = 1.

218 Allied The Leading Mathematics-10 Indices 219 Algebra (c) If xy = yx , prove that : x y x = xx–y. (d) If ax = by and bx = ay , prove that a = b. 6. Solve and verify your solution: (a) 22x + 7. 2x – 8 = 0 (b) 22x+1 + 6. 2x – 8 = 0 (c) 4.3x+1 – 9x = 27 (d) 5.4y + 1 – 16y = 64 (e) 9x + 9 = 10.3x (f) 2x+3 + 1 2x = 9 (g) 7x + 343 7x – 56 = 0 (h) 2x + 1 2x = 4 1 4 (i) 3y + 3–y = 9 1 9 (j) 5a + 5–a = 25 1 25 (k) 7x + 1 7x = 7 1 7 (l) 16p + 64 = 5.4p+1 7. (a) If a = xp .y, b = xq .y and c = xr .y then prove that aq–r.br–p.cp–q = 1. (b) If a = xq+r.yp , b = xr+p.yq and c = xp+q.yr then prove that aq–r.br–p.cp–q = 1. (c) If 2x = 3y = 12z then prove that: 1 z = 1 y + 2 x (d) If ax = by = cz then prove that: 1 x + 1 y + 1 z = 1 (e) If x = y2 + z2 , y = z2 + x2 and z = x2 + y2 then prove that: x x + 1 + y y + 1 + z z + 1 = 1 8. (a) Prove that the values of x obtained from the equation 4 × 3x + 1 – 9x = 27 satisfy in the next equation 32x – 4 × 3x+1 + 27 = 0. (b) Prove that the Roots of x obtained from the equation 5x×x × 5x – 15625 = 0 satisfy in the other equation 9x×x × 9x – 81 = 0. 2. (a) 1 (b) 4 3 (c) 3 (d) – 1 4 (e) 0 (f) 7 (g) 3 (h) –4 (i) 1 2 4. (a) 2 (b) –2 (c) –3 (d) 2 (e) 3 (f) 7 4 6. (a) 0 (b) 0 (c) 1, 2 (d) 1, 2 (e) 0, 2 (f) 0, –3 (g) 1, 2 (h) 2, –2 (i) ± 2 (j) 2, –2 (k) ± 1 (l) 2, 1 Answers Project Work Make a chart about the notation of indices and its laws and exponential equation in a chart paper.

220 Allied The Leading Mathematics-10 Indices 221 Algebra CONFIDENCE LEVEL TEST - IV Unit IV : Algebra Unit IV : Algebra Class: 10, The Leading Maths Time: 45 mins. FM: 23 Attempt all questions. 1. Ramesh saves Rs. 200 during the first month, Rs. 300 in the next month and Rs. 400 in the third month. He continues to save money in this sequence up to 12 months. (a) Write the general term of the arithmetic sequence. [1] (b) How much amount does he save in the 7th month? Calculate it. [1] (c) How much amount does he save in 12 months? Calculate it. [2] (d) How much percent is increase in each term in the AS? Find it. [1] 2. The given numbers are in the form of geometric series 2048 + 1024 + 512 + ……. + 1. (a) How many terms are there with the new three terms in the above series? Find it. [2] (b) Find the sum of the given series. [2] 3. The length of land is 5 m longer than its breadth. The area of a given land is 50 m2 and area of Chapari is 2 m2 . (Area of rectangle = l × b) (a) Find the length and breadth of the given land. [2] (b) Calculate the cost of the fence when it fences 5 times around the given land at Rs. 35 per meter. [2] (c) How many maximum number of such types of Chapari can be kept on the given land? Calculate it. [1] 4. (a) Solve: 2x x2 – 1 – 1 x – 1 = 2 [2] (b) Do all the roots of the above quadratic equation satisfy ? Give reason. [1] (c) Find the value of 1 2x – 2y – 1 2x + 2y – y x2 – y2 . [2] 5. (a) Solve: 4x – 1 4x = 16 1 16 [3] (b) Simplify: a – b a2 – ab + b2 + a + b a2 + ab + b2 – 2a3 a4 – a2 b2 + b4 [3] Best of Luck

220 Allied The Leading Mathematics-10 Indices 221 Algebra Additional Practice – IV Sequence and Series 1. (a) What is the arithmetic mean between 2 and 12? (b) What is the geometric mean between 3 and 12? (c) Insert the four arithmetic means between 3 and 18? (d) Insert the four geometric mean between 2 and 162? (e) What is the relation between arithmetic mean (AM) and geometric mean (GM) ? (f) What is the sum of the natural numbers from 1 to 20? (g) Write the sum of the first 10 odd numbers. 2. (a) If 2a + 1, a2 – 3 and 2a + 9 are three successive terms of an AP, find its mean. (b) If 2p + 1, 2p + 3 and 6p + 5 are first three terms of a geometric sequence, find the positive value of p. Also, find its sum of first 10 terms. 3. Calculate the sum of the following series: (a) 9 + 33 3 + 15 2 + 27 4 + ……. to 25 terms. (b) 4 + 9 + 14 + 19 + ………. + 54 4. Find the sum of first 25 terms of the series in AP whose nth terms are given below: (a) un = 5 – 2n (b) an = 20 – 3n 5. (a) Find the common difference of a series in AP from the following information: The sum of the first 21 terms (S21) = 0, the first term (a) = 20. (b) The sum of the first 6 terms of AP is 2166. If its common difference is 104 then find the first term. 6. (a) If AM between the numbers 5 and 3x is 25, find the last number. [Ans: 15 (b) If the GM having positive common ratio between the numbers x – 1 and 3x + 1 is 8, find the numbers. 7. (a) The 7th term and 12th term of an AP are 31 and 61 respectively. Find the sum of first 17 terms. (b) The 6th term of a GP with positive ratio is 9 times of its 4th term. If its 1st term is 2, find the sum of first 8 terms. 8. (a) The sum of first seven terms of an arithmetic series is 161. If the tenth term exceeds double of the fifth term by 1, find the sum of first twelve terms. (b) The sum of the first 4 terms of a geometric progression having positive common ratio is 20 and the sum of first two terms is 2. Find the sum of first ten terms of the GP. 9. (a) The sum of three consecutive numbers of an AP is 36. If the sum of the square of the first number and last number is 296, find the numbers

222 Allied The Leading Mathematics-10 Indices 223 Algebra (b) The product of four successive numbers of a GP is 81. If the sum of the second number and third number is 15 2 , find the numbers. 10. (a) If 1 27 , x, y, 27 are in GP, find the values of x and y. (b) If p, –5, q, r, 52 are in an arithmetic series, find the values of p, q and r. 11. (a) There are n-arithmetic means between 2 and 2048. If the fourth mean is 746, find the value of n. (b) Find the number of geometric means between 1 100 and 10000 when the second mean and fourth mean are in the ratio 1:100. 12. (a) 8 arithmetic means are inserted between two terms a and b such that the fourth mean and seventh mean are 10 and 16 respectively. Find the values of a and b. (b) 4 GM’s are inserted between a and 96. If the third mean is 24, find out the other means and the first term. 13. (a) If AM and GM between two numbers are 257 and 32 respectively, find the numbers. (b) Find the two positive numbers whose sum is 90 and the AM exceeds the GM by 18. 14. (a) How many terms must be taken of the series 2 + 10 + 50 + 250 + ...... to make the sum 7812? (b) How many terms must be taken of the series 2 + 8 + 32 + ….… to make the sum 682? 15. (a) The 2nd and 4th terms of a GP having positive common ratio are – 6 and – 24 respectively. Find the sum of the first seven terms. (b) If the sixth term is 16 times of the second term and the sum of the first seven terms is 127 2 in a geometric series having positive common ratio, find the sum of the first nine terms. 16. (a) The sum of three numbers in AP is 33. If 1 is added to the first two numbers and 2 is subtracted from the double of the third number then resulting numbers are in GP, find the numbers. (b) The product of three consecutive numbers is 13. If 2 and 3 are added to the first two numbers, the resulting numbers with the third number form an AP. Find the numbers. 17. (a) If 1 b + c , 1 c + a , 1 a + b are in AP, then prove that a2 , b2 , c2 are in AP. (b) If a, b, c are in GP and x and y the AM between a, b and b, c respectively, show that: a x + c y = 2 and 1 x + 1 y = 2 b Solution of Quadratic Equation 18. Solve: (a) (x – 1)2 + x2 – 1 = 0 (b) (x + 1)3 – x3 – 1 = 0 19. Solve and test: x + 2 x – 2 + x – 2 x + 2 = 3 2 20. Solve: (a) x2 – 4x + 3 = 0 (b) x2 – 7x + 10 = 0

222 Allied The Leading Mathematics-10 Indices 223 Algebra 21. Solve after factorization of the left member: (a) x2 – 2x – 3 = 0 (b) x2 – 3x – 10 = 0 22. Solve: (a) 2x2 – x – 10 = 0 (b) 4x2 – 6x – 4 = 0 23. Solve and test the following quadratic equations: (a) 4 x – 1 – 5 x + 2 = 3 x (b) 5 x + 1 – 1 x = 6 x + 4 24. Solve and test M (3x – 5) (2x – 5) = x(x + 2) – 3 25. Solve the given equation by completing the square: x2 + 8x – 6 = 0 26. Solve the equation by the method of completing square: 6 + 4t – 2t2 = 0 27. Solve and test: (a) 6x2 + 7x – 3 = 0 (b) x – 2 x – 1 = x + 4 2x + 2 28. Solve the following equations by using the quadratic root formula: (a) x2 – 3x – 8 = 0 (b) x2 – 2x – 8 = 0 (c) x2 + 2x – 8 = 0 29. Solve for x and test: (a) x + 2 x – 2 – x – 2 x + 2 = 5 6 (b) 2x – 1 2x + 1 – 2x + 1 2x – 1 = – 2 1 3 (c) x + 1 = 2x – 7 x + 5 – 5x + 8 x + 5 Verbal Problems on Quadratic Equations 30. (a) If 3 is added to the one-third of the square of a number, the sum is 30. Find the number. (b) If the double of a number is equal to its square, find the number. (c) Divide 23 into two parts so that their product will be 60. 31. (a) If each of the adjacent sides of the right angle in a rightangled triangle is 1cm, find the length of the hypotenuse. (b) The lengths of the sides containing the right angle of a right–angled triangle are (x + 1) cm and (x + 2) cm. If the area of the triangle is 10 cm2 , find x. (c) In a rectangle the length is increased by 2 and breadth is reduced by 2 units, the area is reduced by 28 sq. units. If the breadth is increased by 2 units and length is reduced by 1 unit, the area is increased by 33 square units. Find the length and breadth of the rectangle. (d) A traffic sign is to have the shape of a triangle with the height 5 inches more than the length of the base. How long should be the base and height, if the area must be 250 sq. inches? 32. (a) The product of two consecutive odd integers is 30 more than the smaller integer. What are they? Find. (b) The sum of two numbers is 18 and the sum of their square is 234. Find the numbers. (c) The sum of digits of a two-digit number is 7 and their product is 12. Find the numbers. θ Opposite Hypotenuse Adjacent

224 Allied The Leading Mathematics-10 Indices 225 Algebra 33. (a) The difference of the age of two brothers is 4 years and the product of their ages is 221. Determine the ages of these two brothers. (b) The product of Rama’s age before 5 years with his age 9 years later is 15. Find her present age. (c) Shova was 27 years old when her son was born. If the product of their ages now is 90, find their present ages. Simplifications 34. (a) What is the sum of x x + y and x x + y ? (b) What is the value of a a + b – c + b a + b – c – c a + b – c ? 35. Simplify the following rational fractions: (a) 1 bc + 1 ca + 1 a + b (b) 1 x – y + 1 x + y – 2y x2 – y2 (c) a a – b – b a + b – b2 a2 – b2 36. Prove that the following relations: (a) x2 + xy + y2 x + y + x2 – xy + y2 x – y = 2x3 x2 – y2 (b) x3 + y3 x2 – xy + y2 – x3 – y3 x2 + xy + y2 = 2y 37. Simplify : (a) x2 – x – 6 x2 – 9 + x2 – 2x – 24 x2 + x – 12 (b) x – z (x – y) (z – x) – y – z (y – x) (y – z) 38. Simplify by reducing to a single fraction: (a) 2x – 1 2x2 + 5x – 3 – x – 3 x2 – x – 6 – 2x + 5 x2 – x – 6 (b) x + 1 1 + x + x2 – x – 1 1 – x + x2 – 2 1 – x2 + x4 (c) x2 – (y – z)2 (x + z)2 – y2 + y2 – (x – z)2 (x + y)2 – z2 + z2 – (x – y)2 (y + z)2 – x2 39. Simplify : (a) 1 x – 1 – 1 x + 1 – 2 x2 + 1 – 4 x4 + 1 (b) 1 a + 1 – a a2 – 1 + a2 a4 – 1 + a4 a8 – 1 (c) 1 p – 2q + 1 p + 2q + 2p p2 + 4q2 + 4p3 p4 + 16q4 (d) 1 + x 1 – x – 1 – x 1 + x + 4x 1 + x2 + 8x3 1 – x4 40. Simplify: (a) a4 a2 + 1 + a4 a2 – 1 – 1 a2 + 1 – 1 a2 – 1 (b) 2 x + 1 – 2 x + 2 + 1 (x + 1)2 (x + 2)2 – 1 (x + 1)2 41. Prove that : 1 1 + x + x2 – 1 1 – x + x2 – 2x 1 – x2 + x4 – 4 1 + x4 + x8 = 4(x –1) (1 – x + x2 ) (1 – x2 + x4 ) Indices 42. What is the value of x in the following exponential equations ? (a) 3x = 81 (b) 1 49 – 2 = 7x (c) 15x = 1

224 Allied The Leading Mathematics-10 Indices 225 Algebra 43. Solve and verify your solution: (a) 25 4 x = 5 2 (b) 49 81 y = 9 7 (c) 81–x = 3 44. Solve and verify your solution: (a) 32x +3 = 32x + 1 + 648 (b) 4 y – 1 + 192 = 4y (c) 2x–8 ×5x–9 = 2 45. Solve and verify your solution: (a) 51–x + 5x–1 = 5 1 5 (b) 3a–3 + 34–a – 4 = 0 (c) 1 5x–3 + 1 52–x = 6 (d) xx + 108 × x–x – 31 = 0 46. (a) If x1/m = y1/n = z1/p then xyz = 1 provided m + n + p = 0. (b) If xa = yb = zc and y2 = zx then prove that : 1 a + 1 c = 2 b . 47. (a) Verify that the values of x obtained from the equation 9y – 2 × 3y – 3 = 0 satisfy in the other equation 32y – 4 × 3y + 3 = 0. (b) Prove that the values of x obtained from the equation 2y + 2–y = 21 2 satisfy in the other equation 7y + 7–y = 71 7 . Sequence and Series 1. (a) 7 (b) 6 (c) 6, 9, 12, 15 (d) 6, 18, 54 (e) AM ≥ GM (f) 210 (g) 100 2. (a) 13 or 1 (b) 1 2 ; 1023 256 3. (a) 0 (b) 319 4. (a) – 525 (b) – 475 5. (a) – 2 (b) 101 6. (a) 15 (b) 4, 16 7. (a) 731 (b) 6560 8. (a) 456 (b) 14762 9. (a) 10, 12, 14 or 41, 12 10 (b) 3 8 , 3 2 , 6, 24 or 24, 6, 3 2 , 3 8 10. (a) 1 3 , 3 (b) – 24, 14, 33 11. (a) 10 (b) 5 12. (a) 2, 20 (b) 6, 12, 48; 3 13. (a) 2, 512 (b) 81, 9 14. (a) 6 (b) 5 15. (a) – 381 (b) 255 16. (a) 3, 11, 19 or 17, 11, 5 (b) 1, 3, 9 or 9, 3, 1 18. (a) 0, 1 (b) 0, – 1 19. ± 2 5 20. (a) 1, 3 (b) 2, 5 21. (a) 3, −1 (b) 5, −2 22.(a) – 2, 5 2 (b) 2, – 1 2 23. (a) 3, – 1 2 (b) 1 2, 4 24. 4, 7 5 Answers

226 Allied The Leading Mathematics-10 Indices PB Algebra 25. – 4 ± 22 26. – 1, 3 27. (a) 1 3 , – 3 2 (b) 0, 5 28. (a) 3 ± 41 2 (b) 4, −2 (c) 2, − 4 29. (a) 10, − 2 5 (b) 6 ± 85 14 (c) 4, 5 Verbal Problems on Quadratic Equations 30. (a) 9 or –9 (b) 0 or 2 (c) 20, 3 31. (a) 2 cm (b) 3 (c) 23, 11 (d) 20 inches, 25 inch 32. (a) 5, 7 (b) 3, 15 (c) 43 or 34 33. (a) 17, 13 yrs. (b) 6 (c) 30, 3yrs. Simplifications 35. (a) a2 + b2 + 2ab + abc abc(a + b) (b) 2 x + y (c) a2 a2 – b2 37. (a) 2(x2 – 4x – 6) x2 – 9 (b) 0 38. (a) –(x2 + 6x + 3) (x + 3) (x + 2) (x –3) (b) – 4x2 1 + x4 + x8 (c) 1 39. (a) 8 x8 – 1 (b) 1 1 – a8 (c) 8p7 p8 – 256q8 (d) 8x 1 – x2 40. (a) 2a2 (b) 1 (x + 2)2 Indices 42. (a) 4 (b) 4 (c) 0 43. (a) 1 2 (b) – 1 2 (c) – 1 4 44. (a) 3 2 (b) 4 (c) 9 45. (a) 0, 2 (b) 3, 4 (c) 2, 3 (d) 2, 3

Geometry PB Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 227 Analysis, presentation and problem solving on geometric theorems and development of ability of constructing triangle and quadrilateral on the basis of given measures. Marks Weightage : 13 Estimated Working Hours : Competency Learning Outcomes 1. To prove theoretically the relationship between areas of parallelograms and triangles standing on the same base and between the same parallel lines. 2. To solve the problems related to the areas of parallelograms and triangles standing on the same base and between the same parallel lines. 3. To construct the triangles and quadrilaterals with equal areas. 4. To search the relation between the central angle, inscribed angle and their corresponding arc of a circle. 5. To prove experimentally and theoretically the relation between the central angle and inscribed angle, and inscribed angles standing on the same arc. 6. To prove experimentally and theoretically the relation between the opposite angles of a cyclic quadrilateral. 7. To solve the problems related to the properties of angle and arc of a circle. 10. Area of Triangle and Quadrilateral 10.1 Theorem on Area of Parallelograms and Triangles 11. Construction 11.1 Construction of Triangles or Quadrilaterals with Equal Area 11.2 Construction of Parallelogram and Triangle with Equal Area 12. Circle 12.1 Theorems on Arc and Its Corresponding Angles of Circle 12.2 Theorem on Inscribed Angle on Semi-circle 12.3 Theorem on Cyclic Quadrilateral Chapters / Lessons GEOMETRY V UNIT 28 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 HSS Way: Mathematics begins at Home, grows in the Surroundings and takes shape in School.

Geometry 228 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 229 R E V I S I T I N G M E N S U R A T I O N WARM-UP Revisiting on Areas of Triangles and Quadrilaterals What is triangle ? How many types of triangles are there? What types of triangles are called an equilateral triangle, isosceles triangle and right triangle ? What types of plane figure is called quadrilateral ? Tell the types of quadrilateral? Why is the parallelogram different from trapezium ? Discuss the properties of the types of quadrilaterals. What is the area of a plane figure ? Tell and write the formulae of the plane figures mentioned in the table below. SN Name Figure Properties Area 1. Rectangle b l ● Opposite sides are parallel and equal. ● Measurement of each angle is a right. ● Diagonals are equal and bisect each other. A = ......... 2. Square l ● All sides and angles are equal. ● Diagonals are equal and bisect each other at right angle. A = ......... 3. Parallelogram b h ● Opposite sides are parallel and equal. ● Opposite angles are equal. ● Diagonals bisect each other. A = ......... 4. Rhombus d d1 2 ● All sides are equal. ● Opposite sides and angles are parallel. ● Diagonals bisect perpendicularly. A = ......... 5. Trapezium h b1 b2 ● One pair of opposite sides is parallel, called bases. ● Legs (non-parallel sides) may or may not be equal. A = ......... 6. Quadrilateral d h1 h2 ● All sides may or may not be equal or parallel. A = .........

Geometry 228 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 229 R E V I S I T I N G M E N S U R A T I O N 7. Triangle h b ● All sides may or may not be equal. A = ......... 8. Right-angled Triangle h b p ● Any one angle is a right. ● h2 = p2 + b2 , where h = hypotenuse, p = perpendicular and b = base. A = ......... 9. Triangle a b c ● All sides are not to be equal. A = ......... 10. Isosceles Triangle a a b ● Any two sides are equal. ● Base angles are equal. ● Perpendicular drawn from vertex to the base bisects the base and vertical angle. A = ......... 11. Equilateral Triangle a ● All sides are equal. ● All angles are equal to 60o . ● Perpendicular drawn from vertices to the sides bisects each side as well as vertical angle and vice versa. A = ......... A Pair of Angles Complementary Angles Supplementary Angles Complete Angle Vertically Opposite Angles Sum of two angles is 90°. Sum of two angles is 180°. Sum of two angles is 360°. Two non-adjacent angles formed by two intersecting lines. 1. Two non-adjacent interior angles on opposite sides of the intersecting transversal line to two parallel lines are called alternate angles. They are Z formed angles. They are equal. 2. Two non-adjacent interior and exterior angles on the same sides of the intersecting transversal line to two parallel lines are called corresponding angles. They are F formed angles. They are equal. 3. Two non-adjacent interior angles on the same sides of the intersecting transversal line to two parallel lines are called corresponding angles. They are C formed angles. Their sum are 180°. Axiom on Area of Plane Figures 1. If two plane figures are congruent, their areas are obviously equal. For example; If ∆ABC ≠ ∆PQR, then Area of ∆ABC = Area of ∆ABC. i.e., Ar(∆ABC) = Ar(∆PQR). Basic Theorem on Area of Parallelogram 1. A diagonal of a parallelogram bisects it in area. 2. A median of an equilateral triangle bisects it in area. P R T S Q D C A B

Geometry 230 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 231 10.1 Theorems on Area of Parallelograms and Triangles At the end of this topic, the students will be able to: ¾ show theoretically that parallelograms and triangles on the same base and between the same parallels are equal in area. ¾ solve the problems related to the theorems on parallelograms and triangles, and its types. Learning Objectives I Plane Figures Between the Same Parallel Lines Activity 1 Look at the following pictures. What do you conclude about them? In each picture, two plane surfaces are joined at one edge,which is also common for each. Make some sketches of these pictures as shown in the adjoining Geoboard (or squared grid or graph), which are engaged under some rules of geometry. From the above discussion, some figures have the same base or equal bases between the parallel lines as shown in the figure alongside, in which ∆XYZ and ∆XYW are standing on the same base XY and between the same parallel lines XB and ZC. But ∆XYZ and ∆ABE are respectively standing on the equal bases XY and AB, and between the same parallel lines XB and ZC and so on. Let us discuss the theorems used in celebrating flags or banners and Buddhist mantra flags. Let us discuss the theorems (a mathematical statement that is proved using rigorous mathematical reasoning) related to the area of triangles and parallelograms standing on the same base and between the same parallel lines as follows: Geoboard Z W E F CD X Y A B l1 l2 The area theorems of triangles and parallelograms is used in the Buddhist Mantra Flags. CHAPTER 10 AREA OF TRIANGLE AND QUADRILATERAL

Geometry 230 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 231 Parallelograms standing on the same base and between the same parallel lines are equal in area. THEOREM 1 (A)* Observation Method Step-1: Cut down the two congruent parallelograms from the chart paper and level it on the both sides as shown in the adjoining figure. Step-2: Arrange these two parallelograms as shown in the figure alongside, where DG//AB or EF. Hence, the parallelograms ABCD and ABGD standing on the same base AB or EF and between the same parallel lines AB or EF and DG are equal in area. Let's make two parallelograms of different size (or a parallelogram and another rectangle) which stand on the same base and between the same parallel lines by using Geoboard or graph. Now, discuss the area of parallelograms and rectangle in the fig. (i) and fig. (ii) of geoboard. For fig. (i), Base length (b) = 3 units, height (h) = 5 units. So, Areas = ..... For fig. (ii), Base length (b) = 3 units, height/breadth (h/b) = 4 units. So, Areas = ..... (B) Theoretical Proof Given: Draw two parallelograms ABCD and ABEF on the same base AB and between the same parallel lines AB and FC. To Prove: Area of ABCD = Area of ABEF Proof: SN Statements SN Reasons 1. In ∆ADF and ∆BCE, (i) AF = BE (S) (ii) ∠AFD = ∠BEC (A) (iii) ∠ADF = ∠BCE (A) 1. (i) Opposite sides of parallelogram (ii) Corresponding angles (AF//BE) (iii) Corresponding angles (AD//BC) 2. ∆ADF ≅ ∆BCE 2. SAA congruent condition 3. Ar. of ∆ADF = Ar. of ∆BCE 3. Area of congruent triangles 4. Ar. of Trap. ABCF + Ar. of ∆ADF =Ar.ofTrap.ABCF + Ar.of ∆BCE 4. By addition axiom 5. Area of ABCD = Area of ABEF 5. By remaining parts of statement (4) D A H E G F C B Chart paper G H D G BF EA Geoboard Fig (ii) Fig (i) A B FD C E

Geometry 232 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 233 Hence, the parallelograms standing on the same base and between the same parallel lines are equal in area. Proved. “Alternatively” Construction: Draw the height PQ for both ABCD and ABEF. Proof: SN Statements SN Reasons 1. Ar. of ABCD = AB×EQ 1. Formula of the area of parallelogram (A = b × h) 2. Ar. of ABEF = AB×DP 2. Same as reason of statement (1). 3. EQ = DP 3. Heights between the same pair of parallel lines 4. Ar. of ABCD = Ar. of ABEF 4. From the statements (1), (2) and (3). Hence, the parallelograms standing on the same base and between the same parallel lines are equal in area. Proved. Converse 1.1 If two parallelograms standing on the same base and between a pair of lines are equal in area, the two lines are parallel. Sub-Theorem 1.1 Parallelogram and rectangle standing on the same base and between the same parallels are equal in area. Sub-Theorem 1.2 The area of a square is equal to the area of parallelogram standing on the same base and between the same parallel lines are equal. Sub-Theorem 1.3 Rhombus and parallelogram standing on the same base and between the same parallels are equal in area. Sub-Theorem 1.4 Rhombus and rectangle standing on the same base and between the same parallels are equal in area. Corollary 1.1 Parallelograms standing on the equal bases and between pair of parallel lines are equal in area. The area of triangle is half of the area of parallelogram on the same base and between the same parallel lines. ‘OR’ The area of parallelogram is twice of the area of triangle on the same base and between the same parallel lines. THEOREM 2 F A Q P B E D C Geoboard

Geometry 232 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 233 Activity 2 (A)* Observation Method Step-1: Cut down the two congruent parallelograms from the chart paper and also cut down diagonally into two equal triangles and level the parallelogram and one triangle on the both sides as shown in the adjoining figure. Step-2: Arrange the parallelogram ABCD and the triangle EFG such that AB and EF are coincide as shown in the figure alongside, where GC//AB or EF. Hence, the ABCD and ∆ABG stand on the same base AB or EF and between the same parallel lines AB or EF and GC. It is clear that the area of ∆ABG is half of the area of the parallelogram ABCD standing on the same base AB or EF and between the same parallel lines AB or EF and GC. Let's make one parallelogram (or rectangle, square or rhombus) and one triangle of different size which stand on the same base and between the same parallel lines by using Geoboard or graph or squared grid. (B) Theoretical Proof Given: ∆BCE and parallelogram ABCD are on the same base BC and between the same parallels AE and BC. To prove: Area of ∆BCE = 1 2 Area of ABCD. Construction: Draw BF parallel to CE to meet AE at F. Proof: SN Statements SN Reasons 1. BCEF is a parallelogram. 1. By construction. 2. Ar. of ∆BCE = 1 2 of Ar. BCEF 2. The diagonal BE divides the BCEF into two triangles of equal area. 3. Ar. of ABCD = Ar. of BCEF 3. Standing on the same base BC and between the same parallels AE & BC. 4. Ar. of ∆BCE = 1 2 Ar. of ABCD 4. From statements (2) and (3). D A E G F C B Chart paper G D C A B E F Geoboard A F D E B C

Geometry 234 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 235 Hence, the area of triangle is half of the area of parallelogram on the same base and between the same parallel lines. Proved. “Alternatively” Construction: Draw the height PQ for ∆ABE and CP for ABCD. Proof: SN Statements SN Reasons 1. Ar. of ABCD = AB × CP 1. By the formula of the area of parallelogram. 2. Ar. of ∆ABE = 1 2 AB × EQ 2. By the formula of the area of triangle. (A = 1 2 b × h) 3 CP = EQ 3. Altitudes between the same pair of parallel lines 4. Ar. of ∆ABE = 1 2Ar. of ABCD 4. From the statements (1) and (2). Hence, the area of triangle is half of the area of parallelogram on the same base and between the same parallel lines. Proved. Converse 2.1 If the area of a triangle is half of a parallelogram standing on the same base and between a pair of straight lines, the pair of straight lines are parallel. Corollary 2.1 Area of a parallelogram (or rectangle, rhombus or square) is equal to twice of a triangle standing on the equal bases and between the same parallel lines. Sub-Theorem 2.1 Area of a triangle is equal to half of a rectangle standing on the same base and between the same parallel lines. Sub-Theorem 2.2 Area of a triangle is equal to half the product of the base and corresponding altitude. That is, Area of a triangle = 1 2 (base) × (altitude). Sub-Theorem 2.3 Area of a triangle is equal to half of a square standing on the same base and between the same parallel lines. Sub-Theorem 2.4 Area of a triangle is equal to half of a rhombus standing on the same base and between the same parallel lines. E D C Q A B P

Geometry 234 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 235 Triangles standing on the same base and between the same parallel lines are equal in area. THEOREM 3 Activity 3 (A)* Observation Method Step-1: Cut down a parallelogram from the chart paper and also cut down diagonally into two equal triangles and level the triangles on the both sides as shown in the adjoining figure. Step-2: Arrange the triangles ABC and DEF such that AB and DE are coincide as shown in the figure alongside, where FC//AB or DE. Hence, ∆ABG and ∆DEF stand on the same base AB or DE and between the same parallel lines AB or DE and FC. It is clear that the areas of ∆ABG and ∆DEF standing on the same base AB or DE and between the same parallel lines AB or DE and FC are equal. Let's make pair of triangles in different shape and size with common height which stand on the same base and between the same parallel lines by using Geoboard. (B) Theoretical Proof Given: Draw triangles ABC and ABD on the same base Area of ∆BCE AB and between the same parallel lines l1 and l2. To prove: Ar. (∆ABC) = Ar. (∆ABD). Construction: Draw BE parallel to AD from B to meet l1 at E. Proof: SN Statements SN Reasons 1. ABED is a parallelogram. 1. Being parallel opposite sides. 2. Ar. (∆ABD) = 1 2 Ar. ( ABED) 2. Being diagonal BD of ABED. 3. Ar. (∆ABC) = 1 2 Ar. ( ABED) 3. Being standing on the same base AB and between the same parallels l1 and l2. 4. Ar. (∆ABC) = Ar. (∆ABD) 4. From the statements (2) and (3). Hence, the triangles standing on the same base and between the same parallel lines are equal in area. Proved. “Alternatively” Construction: Draw the height PQ for ∆ABC and ∆ABD. C E D F Chart paper A B F C A B D E l1 l2 D E C A B l1 l2 D P C Q A B Geoboard

Geometry 236 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 237 Proof: SN Statements SN Reasons 1. Ar. (∆ABC) = 1 2AB × CP 1. By the formula of the area of triangle. 2. Ar. (∆ABD) = 1 2AB × DQ 2. By the formula of the area of triangle. 3. CP = DQ 3. Heights between the same pair of parallel lines. 4. Ar. (∆ABC) = Ar. (∆ABD) 4. From the statements (1) and (2). Hence, the triangles standing on the same base and between the same parallel lines are equal in area. Proved. Converse 3.1 If the area of triangles standing on the same base and between the same pair of straight lines in the same side are equal, the pair of straight lines are parallel. Corollary 3.1 The triangles standing on the equal bases and between the same parallel lines are equal in area. Example-1 Observe the adjoining figure, in which ABCD is a square and ABCF is a parallelogram. (a) Are the area of the square ABCD and the parallelogram ABEF equal ? Given reason. (b) Which formula is used to find the area of the square? (i) l × b (ii) l 2 (iii) 2(l + b) (iv) 4l (c) Find the area of the parallelogram ABEF. Solution: (a) The area of the square ABCD and the parallelogram ABEF equal are equal because they are standing on the same base Ab and between the same parallel lines AB and DE. (b) The formula A = l 2 is used to find the area of the square. (c) Here, the length of side of the square ABCD (l) = 7 cm Now, we have (i) The area of the square ABCD = l 2 = 72 = 49 cm2 (ii) The area of ABEF = The area of the square ABCD = 49 cm2 . Example-2 In the adjoining figure, the area of ∆QRS is 19 cm2 . (a) What is the relation between PQRS and ∆PQS? Given reason. (b) Find the area of TUPQ. (c) Write the relation between TUPQ and ∆QRS. D B C A Ar. (∆ABC) = Ar. (∆BCD) A 7 cm B D F C E U P Q T S R

Geometry 236 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 237 Solution: (a) PQRS = 2 × ∆PQS because the diagonal QS bisects PQRS in area. (b) Here, the area of ∆QRS = 19 cm2 , the area of TUPQ = ? Now, we know that (i) Area of PQRS =2 × Areaof∆QRS [ Diagonal QS of PQRS] = 2 × 19 = 38 cm2 (ii) Area of TUPQ = Areaof PQRS = 38 cm2 (c) TUPQ = 2 × ∆QRS. Example-3 In the given figure, the area of ∆BCD is 70 cm2 . (a) State the relation between ∆ABD and ∆ABE in area. (b) Find the area of ∆ABE. (c) Is the area of ∆BCD and ∆ADE equal ? Give reason. Solution: (a) The area of ∆ABD and ∆ABE standing on the same base AB and between the same parallel lines AB and EC are equal. (b) In the figure, the area of ∆BCD = 70 cm2 . Now, (i) Area of ∆ABD = Area of ∆BCD = 70 cm2 . [ Diagonal BD of ABCD] (ii) Area of ∆ABE = Area of ∆ABD = 70 cm2 . [Being standing on the same base and between the same parallels] “Alternatively” (i) Area of ABCD = 2 × Area of ∆BCD = 2 × 70 cm2 = 140 cm2 . [ Why ?] (ii) Area of ∆ABE = 1 2 × Area of ABCD = 1 2 × 124 = 62 cm2 . [ Why ?] (c) Area of ∆BCD = Area of ∆ADE because both triangles are the half of the area of ABCD and ABDE on the same base AB and between the same parallel lines AB and EC. Example-4 In the adjoining figure, if the area of ABCD is 54 cm2 , find the area of BPQR. Solution: In the given figure, join PC. (i) Area of ABCD = 54 cm2 (ii) Area of ∆BPC = 1 2 × Area of ABCD = 1 2 × 54 = 27 cm2 [ Being standing on the same base BC and between the same parallels BC and AB] (iii) Area of BPQR = 2× Area of ∆BPC = 2 × 27 = 54 cm2 . [ Being ....] [ Standing on the same base PQ and between the same parallel lines PQ and UR] E D C A B R B C Q A P D

Geometry 238 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 239 Example-5 In the given figure, RT = 16 cm and AB = 4 cm. (a) Which formula is used to find the area of the parallelogram QRTB? (b) Find the length of the side of the square PQRS. (c) If the base and height of the parallelogram are increased by 10%, how much percent of side of the square PQRS is increased? Solution: (a) Area of QRTB = Base (RT) × Height (AB). (b) In the given figure, RT = 16 cm and AB = 4 cm. Now, we have (i) Area of QRTB = RT × AB [ A = b × h] = 16 × 4 = 64 cm2 (ii) Area of PQRS = Area of QRTB = 64 cm2 [ Being standing on the same base QR and between the same parallels QR & PT] (iii) ∴ The length of the side of the square PQRS = Area of PQRS [ A = l 2 ] = 64 = 8 cm (c) In the given figure, RT = 16 + 10% of 16 = 17.6 cm AB = 4 + 10% of 4 = 4.4cm. Now, we have (i) Area of QRTB = RT × AB [ A = b × h] = 17.6 × 4.4 = 77,44 cm2 (ii) Area of PQRS = Area of QRTB = 77.44 cm2 [ Being standing on the same base QR and between the same parallels QR & PT] (iii) ∴ The length of the side of the square PQRS = Area of PQRS [ A = l 2 ] = 77.44 = 8.8 cm. (iii) ∴ The increased percent in length of the square PQRS = (8.8 - 8)/8 × 100% = 10% Example-6 In the given figure, CDGH is a rectangle and CDEF is a parallelogram. (a) Write the relation among the area of rectangle and parallelogram in the given figure. (b) Prove that: Area of ∆CDH = 1 2 × Area of CDEF. (c) If the base and height of CDEF are 6 cm and 10 cm, what is the area of ∆CDH? Solution: (a) The area of rectangle and parallelogram are equal because they stand on the same base CD and between the same parallel lines CD and HE. (b) Given: CDGH is a rectangle and CDEF is a parallelogram. To Prove: Area of ∆CDH = 1 2 × Area of CDEF Q R A P S B T 16 cm 4 cm Q R A P S B T 17.6 cm 4.4 cm C D H G F E

Geometry 238 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 239 SN Statements SN Reasons 1. Ar. of ∆CDH = 1 2 × Ar. of CDGH 1. Being diagonal DH of rectangle CDGH 2. Ar. of CDGH = Ar. of CDEF 2. Standing on the same base CD and between the same parallel lines CD andHE 3. Ar. of ∆CDH = 1 2 × Ar. of CDEF 3. From the statement (i) and (ii) Proved. (c) In CDEF, base length (b) = 6 cm, height (h) = 10 cm Now, (i) Area of CDEF = b × h = 6 × 10 = 60 cm2 (ii) Area of ∆CDH = 1 2 × Area of CDEF = 1 2 × 60 = 30 cm2 Example-7 In the given figure, AB = PQ and the area of the parallelogram ABCD is 64 cm2 . (a) State the area of the parallelograms standing on the equal bases and between the same parallel lines. (b) Find the area of PQRS. (c) Write the relation among ABCD and PQRS in area. Solution: (a) The area of the parallelograms standing on the equal bases and between the same parallel lines are equal. (b) In the given figure, AB = PQ, Area of the parallelogram ABCD = 64 cm2 Construction : Join AS and BR Now, we have (i) Area of ABCD = Area of ABRS = 64 cm2 [ Standing on the same base AB and between the same parallel lines AQ and DR] (ii) Area of ABRS = Area of PQRS = 64 cm2 [ Standing on the same base SR and between the same parallel lines DR and AQ] (c) Area of ABCD = Area of PQRS. Example-8 In the adjoining figure, D is the mid-point of the side AB of ∆ABC and ADEC is a parallelogram. Prove that: Ar(∆ABC) = Ar( ADEC) Solution: Given: In the given figure, D is the mid-point of the side AB of ∆ABC and ADEC is a parallelogram. To prove: Ar(∆ABC) = Ar( ADEC) A B P Q D C S R C E A D B

Geometry 240 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 241 Proof: SN Statements SN Reasons 1. Ar. (∆ABC) = 2 × Ar. (∆ACD) 1. The median CD of ∆ABC. 2. Ar. ( ADEC) = 2×Ar. (∆ACD) 2. Median CD of ∆ABC. 3. Ar. (∆ABC) = Ar. ( ADEC) 3. From statements (1) and (2). Example-9 In the given figure, PQRS is a parallelogram. O is a point on QR. PO and SR are produced to meet at T and Q and are joined. Prove that Area of ∆SOR = Area of ∆QOT. Solution: Given: In the given figure, PQRS is a parallelogram. O is a point on QR. PO and SR are produced to meet at T, Q and T are joined. To prove: Area of ∆SOR = Area of ∆QOT Proof: SN Statements SN Reasons 1. Ar. of ∆POS = 1 2 × Area of PQRS 1. Standing on the same base PS and between the same parallels PS and QR. 2. Ar. of ∆PQO + Ar. of ∆SOR = 1 2 × Area of PQRS 2. Remaining triangles in statement (1). 3. Ar. of ∆PQT = 1 2 × Area of PQRS 3. Standing on the same base PQ and between the same parallels PQ and ST. 4. Ar. of ∆PQT = Ar. of ∆PQO + Ar. of ∆QOT 4. By whole parts axiom. 5. Ar. of ∆PQO + Ar. of ∆SOR = Ar. of ∆PQO + Ar. of ∆QOT 5. From statements (2), (3) and (4). 6. Area of ∆SOR = Area of ∆QOT 6. By cancellation axiom of statement (5) Proved. Example-10 In the adjoining figure, PQRS and LQMN are two parallelograms equal in area. Prove that LR//SN. Solution: Given: In the figure, PQRS and LQMN are two parallelograms equal in area. To prove: LR//SN. Construction: Join LS and RN. Proved. P S Q R T O P L N Q R M S

Geometry 240 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 241 Proof: SN Statements SN Reasons 1. Ar. of PQRS = 1 2 × Area of LQMN 1. By given. 2. Ar. of ∆LRS = 1 2 ×Area of PQRS 2. Standing on the same base RS and between the same parallels SR and QP. 3. Ar. of ∆LRN = 1 2 × Area of LQMN 3. Same reason as statement (2). 4. Ar. of ∆LRS = Ar. of ∆LRN 4. From the statements (1), (2) and (3). 5. LR//SN 5. Standing on the same base LR and between the same parallels LR and SN in statement (4). Proved. Example-11 In the adjoining figure, DB//CE. (a) Write two pairs of triangles standing on the same base and between the same parallel lines DB and CE. (b) Prove that the area of the quadrilateral ABCD is equal to the area of the triangle ADE. (c) If the area of ∆ABD and ∆BDE are 20 cm2 and 35 cm2 respectively, find the area of the quadrilateral ABCD. Solution: (a) Two pairs of triangles standing on the same base and between the same parallel lines DB and CE are ∆BCD and ∆BDE, and ∆CDE and ∆BCE. (b) Given: In the given figure, DB//CE. To prove: Ar. (Quad.ABCD) = Ar. (∆ADE) Proof: SN Statements SN Reasons 1. Ar. (∆BCD) = Ar. (∆BDE) 1. Standing on the same base DB and between the same parallels DB and CE. 2. Ar. (∆BCD) + Ar. (∆ADB) = Ar. (∆BDE) + Ar. (∆ADB) 2. By addition axiom. 3. Ar. (Quad.ABCD) = Ar. (∆ADE) 3. By whole parts axiom. Proved. (c) If the area of ∆ABD and ∆BDE are 20 cm2 and 35 cm2 respectively, then the area of the quadrilateral ABCD = Area of ∆ABD + Area of ∆BDE = 20 + 35 = 55 cm2 . D C A B E

Geometry 242 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 243 Example-12 In ∆ABC, XY is parallel to BC. BY and CX are joined which meet at O. Prove that Ar. (∆BOX) = Ar. (∆COY). Solution: Given: In ∆ABC, XY is parallel to BC. BY and CX are joined which meet at O. To prove: Ar. (∆BOX) = Ar. (∆COY) Proof: SN Statements SN Reasons 1. Ar. (∆BXY) = Ar. (∆CXY) 1. Standing on the same base XY and between the same parallels XY and BC. 2. Ar. (∆BXY) – Ar. (∆XOY) = Ar. (∆CXY) – Ar. (∆XOY) 2. By subtraction axiom. 3. Ar. (∆BOX) = Ar. (∆COY) 3. By remaining parts of statement (2). Proved. Example-13 In the figure, ABCD is a trapezium in which AB//EF//DC. Prove that Ar. (∆ADF) = Ar. (∆BCE). Solution: Given: In the figure, ABCD is a trapezium in which AB//EF// DC. To prove: Ar. (∆ADF) = Ar. (∆BCE). Proof: SN Statements SN Reasons 1. Ar. (∆AEF) = Ar. (∆BEF) 1. Standing on the same base EF and between the same parallels EF and AB. 2. Ar. (∆DEF) = Ar. (∆CEF) 2. Standing on the same side EF and between the same parallels EF and DC. 3. Ar. (∆AEF) + Ar. (∆DEF) = Ar. (∆BEF) + Ar. (∆CEF) 3. By adding statements (1) and (2). 4. Ar. (∆ADF) = Ar. (∆BCE) 4. By whole part axiom of statement (3). Proved. Example-14 In the given figure, AB//ED and Ar. (∆ABE) = Ar. (∆ACD). Prove that AD//BC. Solution: Given: In the given figure, AB//ED and Ar. (∆ABE) = Ar. (∆ACD). To prove: AD//BC. Construction: Join BD. Proof: SN Statements SN Reasons 1. Ar.(∆ABE) = Ar.(∆ACD) 1. By given. 2. Ar.(∆ABE) = Ar.(∆ABD) 2. Standing on the same base AB and between the same parallels AB and ED. A X Y O B C D E A B F C E D C A B E D C A B

Geometry 242 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 243 3. Ar.(∆ACD) = Ar.(∆ABD) 3. From statements (1) and (2). 4. AD//BC 4. Equal triangles on the same base AD and between the same lines AD and BC. Example-15 In ∆PQR, S, A and B are the mid-points of QR, PQ and PS respectively. If C is any point in QR, prove that: Area of ∆PQR = 8 × Area of ∆ABC. Solution: Given: In ∆PQR, S, A and B are the mid-points of QR, PQ and PS respectively. Also, C is any point in QR. To prove: Area of ∆PQR = 8 × Area of ∆ABC Construction: Join BQ. Proof: SN Statements SN Reasons 1. AB//QS 1. The mid-points A and B in ∆PQS. 2. Ar.(∆ABQ) = Ar. (∆ABC) 2. Standing on the same base AB and between the same parallels AB and QS. 3. Ar.(∆PQB) = 2 × Ar.(∆ABQ) 3. The median AB of ∆PQB. 4. Ar.(∆PQS) = 2 × Ar.(∆PQB) 4. The median BQ of ∆PQS. 5. Ar.(∆PQR) = 2 × Ar.(∆PQS) 5. The median PS of ∆PQR. 6. Ar.(∆PQR) = 8 × Ar.(∆ABC) 6. From statements (2) to (5) Proved. PRACTICE 10.1 Read Think Understand Do Keeping Skill Sharp 1. (a) What is the area of a parallelogram? (b) What is the relation between parallelograms standing on the same base and between the same parallel in area? (c) Write the relation of the area of the rectangle ABEF and the parallelogram ABCD in the given figure. (d) If ABCD is a square in the given figure, what is the area of the parallelogram ? (e) What is the relation between the triangle and the rectangle standing on the same base and between the same parallel lines? P A B Q C S R A 5 cm B F D E C A B F E D C

Geometry 244 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 245 (f) In the figure, PQRS is a rhombus having area 66 cm2 . Write the area of ∆PTQ. 2. Fill in the blanks. (a) The area of parallelogram is ........... the area of a triangle standing on the same base and between the same parallel lines. (b) In the given figure, if ABCD is a rectangle with AB = 4 cm and AD = 3 cm, the area of ∆AEB .......... . (c) The area of the triangles standing on the same base and between the same parallel lines are ........ . (d) In the figure, the area of ΔBDE is18 cm2 and the area of ΔABD is 12 cm2 . The area of the quadrilateral ABCD is ......... . 3. Circle ( ) the correct answer. (a) In the given figure, the area of ∆KMN is .......... . (i) 96 cm2 (ii) 64 cm2 (iii) 48 cm2 (iv) 144 cm2 (b) Two parallelogram ABCD and ABMN stand on the same base AB and between the same parallel lines AB and DM. If the area of the parallelogram ABMN is 42 sq. cm, which is the area of the parallelogram ? (i) 84 cm2 (ii) 21 cm2 (iii) 42 cm2 (iv) 63 cm2 (c) KMLN and ∆MPL stand on the same base and between the same parallel lines. If the area of KMLN is 20 sq. cm, which is the area of ∆MPL ? (i) 40 cm2 (ii) 20 cm2 (iii) 10 cm2 (iv) 30 cm2 (d) A parallelogram and a triangle stand on the same base and between the same parallel lines. If the area of the triangle is 20 sq. units, which is the area of the parallelogram ? (i) 10 units (ii) 20 units (iii) 30 units (iv) 40 units (e) In the given figure, the area of ∆PRE is ....... . (i) 9 cm2 (ii) 18 cm2 (iii) 36 cm2 (iv) 72 cm2 (f) Which is true relation for the area of triangles standing on the same base and between the same parallel lines ? (i) Double (ii) Half (iii) Equal (iv) Not equal (g) ∆ASK and ∆KST stand on the same base and between the same parallel lines. If the area of ∆KST is 30 sq. units, which is the area of ∆ASK ? (i) 15 units (ii) 30 units (iii) 60 units (iv) 45 units (h) In the given figure, AB = 10 cm and CP = 8 cm, the area of ∆ABD is ....... . (i) 80 cm2 (ii) 60 cm2 (iii) 40 cm2 (iv) 20 cm2 T S R P Q D C E A B D A B E C L M K N 8 cm 12 cm P 9 cm 8 cm R D F C E D P C A B

Geometry 244 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 245 Check Your Performance 4. (i) In the given figure, ABCD is a rectangle. (a) Are the area of the rectangle ABCD and the parallelogram ABRS equal ? Given reason. (b) Which formula is used to find the area of rectangle? Write. (c) Find the area of the parallelogram ABRS. (ii) In the given figure, the area of the PSTU is 62 cm2 . (a) What is the relation between PQRS and ∆PQS? Given reason. (b) Find the area of ∆QRS. (c) Write the relation between PSTU and ∆QRS. 5. (i) In the given diagram, PQRS is a parallelogram, T is any point on QR, SU⊥PQ, SU = 16 cm and PQ = 20 cm. (a) Write the formula to find the area of PQRS in area. (b) Find the area of the triangle ∆SPT. (c) Is the area of PQRS equal to twice of the area of the sum of ∆PQT and ∆RST? Give reason. (ii) In the adjoining figure, PQR is an equilateral triangle and QRST is a parallelogram. If QR = 12 cm, find the area of QRST. 6. (i) In the adjoining figure, the area of ∆PKN is 50 cm2 . (a) Find the area of sq. PKLM. (b) Find the length of the diagonal PL of the square PKLM. (ii) In the figure, the area of the ∆PQR is 36 cm2 and PQRT and PRST are parallelograms. (a) Find the area of PQRT. (b) Find the area of ∆RST. (iii) In the adjoining figure, PQSR is a parallelogram and RS is produced to T such that RS = ST. If the area of parallelogram PQSR is 25 cm2 , find the area of ∆PRT. 7. (i) In the figure, WXYZ is a rhombus in which YZ is produced to the point U. If WY = 12 cm and XZ = 8 cm, find the area of ∆WUX. A 10 cm 6 cm B D S C R Q P S T R U S R P Q T U P Q R T S 12 cm N M L P K Q R S P T R S T P Q W X U Z Y

Geometry 246 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 247 (ii) In the figure, the square ABCD and the ∆ABE are standing on the same base AB and between the same parallel lines AB and DE. If BD = 6 cm, find the area of ∆AEB. (iii) In the given figure, KLMN is a square whose perimeter is 80 cm and KL is produced upto the point J. If E is the mid-point of MN, find the area of ∆JEN. 8. (i) In the figure, the parallelogram PQSR and ∆PQT stand on the same base and between the same parallel lines. The area of the ∆TPQ is 50 cm2 . Find the area of the ∆PRS. (ii) In the given figure, BD is the diagonal of the parallelogram ABCD. If the area of the ∆BCD is 30 cm2 , find the area of ∆ABE. 9. (i) In the figure, ABCD is a trapezium and AEFD is a parallelogram. If E is the mid-point of AB and the area of ∆ABC is 42 cm2 , find the area of ∆ADF. (ii) In the figure, the area of ABCD = 64 cm2 and the area of ∆ABE = 12 cm2 . Find the area of ∆CDE. 10. (i) (a) Write the relation among the area of square PQST and parallelogram PQAB standing on the same base PQ and between the same parallel lines PQ and BS. (b) Prove that: Area of square PQST = Area of PQAB. (c) If the length of a side of the square is 8 cm, what is the area of ∆PQAB? (ii) In the given figure, OPQR and OPST stand on the same base OP and between the same parallel lines OP and TQ. (a) State the relation among OPQR and OPST in area. (b) Prove that: Area of ∆ROT = Area of ∆QPS (c) Prove that: Area of OPQR = Area of OPST (d) If the length of a side of the base of OPQR is 8 cm and its height is 6 cm, what is the area of OPST? (iii) From the given figure, the area of the rectangle ABCD is equal to the area of the parallelogram ABEF. (a) State the relation among ABCD and ABEF in area. (b) Prove that: Area of ∆ADF = Area of ∆BCE (c) Prove that: Area of ABCD = Area of ABEF (d) If AB = 12 cm and AD = 18 cm, what is the area of ABEF? D C E A B L K J M N E R S T P Q E D C A B D F C A E B A D B E C T R S Q O P A B D C F E

Geometry 246 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral 247 11. (i) (a) Draw a parallelogram ABCD and ∆AXB on the same base between the same parallel lines. (b) Prove that, the area of ABCD is two times of the area of ΔAXB. 12. (i) In the adjoining figure, PQRS is a rhombus. (a) Prove that: Area of ΔPQS = Area of ΔPQR (b) Prove that: Area of ΔPQS = Area of ΔPRS. (ii) ABCD is a parallelogram. P is any point on AB and Q is any point on AD. Prove that: Area of ∆ABQ + Area of ∆CDQ = Area of ∆ADP + Area of ∆BCP. (iii) In the parallelogram ABCD, P is any point inside it. Prove that: Ar. (∆APB) + Ar. (∆CPD) = Ar. (∆APD) + Ar. (∆BPC) = 1 2 × Ar. ( ABCD). [Hints: Draw a straight line parallel to AB or BC through P.] (iv) In the figure, O is any point on the diagonal BD of the parallelogram ABCD. EF//AD and GH//DC. Prove that: Area of AEOG = Area of CHOF. 13. (i) In the adjoining figure, if ΔDEF and ΔEFG stand on the same base EF and between the same parallels EF and DG, then prove that their areas are equal. (ii) From the given figure, prove that the areas of the rightangled triangle LMN and ΔKLM standing on the same base LM and between the same parallels LM and NK are equal. 14. (i) In the figure, PQRS is a quadrilateral in which RQ is produced to T such that SQ//PT. Prove that ∆RST is equal in area to the quadrilateral PQRS. (ii) ABCD is a quadrilateral and DE is drawn parallel to AC to intersect BC produced at a point E. Prove that area of ∆ABC = area of quadrilateral ABCD. D X C A B S R P Q A Q D B C P A B P D C A G D E F O B H C D G E F L M N K P S T Q R

Geometry 248 Allied The Leading Mathematics-10 Area of Triangle and Quadrilateral PB (iii) In the adjoining figure, the diagonal RP of a quadrilateral PQRS is parallel to ST. Prove that: Area of quad. PQRS = Area of quad. PQRT (iv) In the figure, ABCD is a trapezium in which DE = AE and CF = BF. Prove that Ar. (∆AFD) = Ar. (∆BEC). 15. (i) In ∆ABC, XY is parallel to BC. BY and CX are intersected meet at O. Prove that Ar. (∆AYB) = Ar. (∆AXC). (ii) In ∆ABC, S and T are the mid-points of PQ and PR respectively. Prove that Ar. (∆BOC) = Ar. (Quad. AXOY). (iii) D and E are points on the side AB and AC of a ∆ABC respectively such that Ar. (∆BCE) = Ar (∆BCD). Show that DE//BC. 16. (i) In the given figure, AB//ED and AD//BC. Prove that Ar. (∆ABE) = Ar. (∆ACD). (ii) In the given figure, ABCD is a parallelogram. Diagonal BD is produced to E. Prove that: Ar. (∆ADE) = Ar. (∆CDE) [Hints: Join AC and use the median theorem.] (iii) In the adjoining figure, PQRS and LQMN are two parallelograms equal in area. Prove that LR//SN. 4. (i)(c) 60 cm2 (ii) (b) 31 cm2 5. (i) (b) 160 cm2 (ii) 36 3cm2 6. (i)(a) 100 cm2 (b) 10 2cm2 (ii) (a) 72 cm2 (b) 36 cm2 (iii) 25 cm2 7. (i) 24 cm2 (ii) 9 cm2 (iii) 100 cm2 8. (i) 50 cm2 (ii) 30 cm2 9. (i) 21 cm2 (ii) 20 cm2 Answers Project Work Draw the possible figures to show that the area of the parallelograms and triangles standing on the same base and between the same parallel lines separately equal. Similarly the area of the parallelogram is twice of the area of triangle standing on the same base and between the same parallel lines separately equal. P T Q S R D E A B F C E A B D C P S L N Q R B

PB Allied The Leading Mathematics-10 Construction 249 Geometry At the end of this topic, the students will be able to: ¾ construct a triangle/quadrilateral equal area to the given triangle/quadrilateral. Learning Objectives I Introduction High school geometry, on one hand, makes us think, collect reasons and then argue systematically or logically; and, on the other hand, it helps in developing enough skill in drawing and constructing geometrical figures. This skill is indispensable in all walks of life. Science, engineering and technology cannot be imagined in the absence of such a process of thinking, arguing, and drawing and reproducing exact copies. Drawing geometrical figures differs basically from the geometrical constructions of the same geometrical figures. Geometrical figures can be drawn freely or with the help of any instrument that is convenient. But a geometrical construction is usually done with the help of a ruler (to be more precise, an ungraduated ruler) and a compass. Geometrical construction begins with the construction (or ‘drawing’ in the ordinary sense) of a line through two given points. This is followed by the construction of a circle having a given point as centre and a given line-segment as its radius. Then, constructions of intersecting line– segments, or a line segment and a circle, or circles under various conditions play the fundamental roles. Standard angles can be constructed with the help of compass and ruler. But all angles cannot be constructed in this way. In such cases, we have to take the help of a protractor to construct the given angles. Graded steps of construction techniques can be found elsewhere. It, however, does not imply that geometrical constructions cannot be done without a systematic training of such a course. We can begin with anything and at any time. What we need is: (a) a rough sketch of the figure we want to construct, (b) appropriate labeling of the rough sketch with given data (or measurement), and (c) basic idea necessary before stepwise construction. To minimize the errors that may occur during the construction of a geometrical figure, the following points should be kept in mind: (a) every point should be observed vertically, (b) observed point or points should be marked by sharply pointed pencil, and (c) accurate instruments should be used. We shall be concerned with geometrical constructions of “triangles and quadrilaterals of equal areas”. CHAPTER 11 CONSTRUCTION

250 Allied The Leading Mathematics-10 Construction 251 Geometry Review on Construction of Triangles and Quadrilaterals Let us recall the construction of different types of triangle and quadrilateral as follows: (a) Triangle: A scalene triangle can be constructed in the following conditions: (i) When three sides are given. (SSS) (ii) When two sides and included angle are given. (SAS) (iii) When two angles and included side are given. (ASA) (iv) When one angle and its two continuous sides are given. (ASS) (v) When right angle and hypotenuse and a side are given. (RHS) (b) Quadrilateral: A quadrilateral can be constructed in the following conditions: (i) When four sides and a diagonal are given. (ii) When four sides and an angle are given. (iii) When three sides and two angles on the middle side among them are given. (iv) When two sides, a diagonal and the angles at the ends of the diagonal are given. (c) Parallelogram: A parallelogram can be constructed in the following conditions: (i) When two sides and included angle are given. (ii) When two sides and a diagonal joining their end points are given. (iii) When a side, a diagonal and the angle between them are given. (iv) When a side and two diagonals are given. (v) When two diagonals and the angle between them are given. (vi) When a side, an angle and a diagonal through the vertex of the angle are given. (d) Rectangle: A rectangle can be constructed in the following conditions: (i) When two sides are given. (ii) When a side and a diagonal are given. (iii) When a side, a diagonal and the angle between the diagonals are given. (iv) When a side, a diagonal and the angle between the diagonals are given. (e) Rhombus: A rhombus can be constructed in the following conditions: (i) When a side and a diagonal are given. (ii) When a side and an angle are given. (iii) When a side an angle between a diagonal and the side are given. (iv) When two diagonals are given. (f) Square: A square can be constructed in the following conditions: (i) When a side is given. (ii) When a diagonal is given. (g) Trapezium: A trapezium can be constructed in the following conditions: (i) When base side, a base angle and non parallel sides are given. (ii) When parallel sides/bases, an angle and the side which made the angle are given.