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Contents, Blueprint, , vi, , Latest Syllabus, , vii, , CBSE Notification, , x, , Exclusive 2019 CBSE Topper Tips, , xiii, , FAQ, , xiv, , Time Management, , xvi, , CBSE Papers, CBSE Sample Paper 20th September 2019 with Marking Scheme, March 2019 Paper Solved by CBSE Topper, March 2018 Paper with CBSE Topper Answer Sheet, , 1, 15, 42, , Sample Question Papers, Sample Paper 1 (Solved), , 58, , Sample Paper 2 (Solved), , 74, , Sample Paper 3 (Solved), , 91, , Sample Paper 4 (Solved), , 110, , Sample Paper 5 (Solved), , 129, , Sample Paper 6 (Solved), , 146, , Sample Paper 7 (Solved), , 164, , Sample Paper 8 (Solved), , 182, , Sample Paper 9 (Solved), , 200, , Sample Paper 10 (Solved), , 218, , Sample Paper 11 (Self-assessment), , 235, , Sample Paper 12 (Self-assessment), , 240, , Sample Paper 13 (Self-assessment), , 246, , Sample Paper 14 (Self-assessment), , 252, , (Self-assessment papers’ solution are on our website)

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(vi), 20Q (20 marks), , 3Q, , 1Q, , 2Q, , 1Q, , 2Q, , 6Q (12 marks), , Note: All questions would be compulsory. However, an internal choice is provided in each section., , Total, , Creating: Compile information together in a different way by, combining elements in a new pattern or proposing alternative, solutions, , Evaluating: Present and defend opinions by making judgments, about information, validity of ideas, or quality of work based on a, set of criteria, , Analysing: Examine and break information into parts by, identifying motives or causes. Make inferences and find evidence, to support generalizations, , 5Q, , 6Q, , Understanding: Demonstrate understanding of facts and, ideas by organizing, comparing, translating, interpreting, giving, descriptions, and stating main ideas, , Applying: Solve problems to new situations by applying acquired, knowledge, facts, techniques and rules in a different way., , 6Q, , 15% marks, , 25% marks, , Remembering: Exhibit memory of previously learned material by, recalling facts, terms, basic concepts, and answers., , Typology of Questions, , Short Answer, (2 marks), , Objective type, (1 mark), , 8Q (24 marks), , 3Q, , 2Q, , 1Q, , 2Q, , 30% marks, , Short Answer, (3 marks), , 6Q (24 marks), , 1Q, , 1Q, , 3Q, , 1Q, , 30% marks, , Long Answer, (4 marks), , Question Paper Design

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2019 CBSE, Topper Tips, Everyone! This page is very important as I have summarised my thoughts below, on key, points to keep in mind whilst attempting the Maths Class 10 Board Exam:, , Units Representation Should Be Correct, The following units should be written correctly always:, Length - cm, mm, m, km (not as cms, mms, ms, kms), Area - sq cm, sq m, sq km (not as cm2, mm2, m2, km2), Volume - cu cm, cu m, cu km (not as cm3, mm3, m3, km3, etc), Speed and Mass - km/h, kg, g (not as km/hr, kgs, gs), , Double-check numerical Values, Very common mistake we make is to not copy the correct values (numbers, equations, etc) from the Question itself as we are in a rush. Ensure you read the questions word by, word with great care., , Use Of Graph Paper, Graph paper should be used when necessary. Diagrams should be neatly drawn to score, full marks., , Final Answer Marks, ½ to 1 Marks are allocated just for the concluding answer. Ensure to mention the final, answer neatly and correctly at the end of the solution. However, remember your working, or method/steps contain majority marks., , Use the first 15 minutes effectively, There is 33% internal choice this time in each section. You get good 15 minutes in the, beginning to read the question paper. Use this time to mark the questions with choice,, the one you are more confident in attempting, to avoid confusion and save time while, writing the exam., , Prioritise your Sections Order, Decide which Section would you want to attempt first and which Section at last. Always, attempt the easy questions first. This way your confidence will grow and you will be, mentally ready to take on the more challenging questions., (xiii)

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SAMPLE PAPER, , 1, , CBSE - Class 10, , MATHE MATICS, (STANDARD), , Time Allowed: 3 Hours, , Maximum Marks: 80, , General Instructions:, (i ) All questions are compulsory., (ii ) The question paper consists of 40 questions divided into four sections A, B, C, & D., (iii ) Section A contains 20 questions of 1 mark each, Section B comprises of 6 questions, of 2 marks each. Section C comprises of 8 questions of 3 marks each. Section D, comprises 6 questions of 4 marks each., (iv ) There is no overall choice. However internal choices have been provided in two, questions of 1 marks each, two questions of 2 marks each, three questions of, 3 marks each and three questions of 4 marks each. You have to attempt only one, of the alternatives in all such questions., (v ) Use of calculators is not permitted., , SECTION - A, Q 1 – 10 are multiple choice questions. Select the most appropriate answer from the, given options., , 1. The HCF of 36 and 54 is:, (A) 2, , (B) 6, , 1, (C) 9, , (D) 18, , 2. Which of the following is not a zero of the polynomial, , 1, , 3, , p(x) = x – 7x + 6, (A) 1, , (B) 2, , (C) –2, , (D) –3, , 3. The discriminant of the quadratic equation x 2 – 4x + 1 = 0 is:, (A) 10, , (B) 11, , (C) 12, , 1, , (D) 14, , 4. Which term of the AP: 4, 9, 14, .............. is 254?, (A) 50, , 58, , th, , th, , (B) 51, , (C) 52, , nd, , 1, rd, , (D) 53, , Mathematics Class X

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5. The centroid of DABC, where A(–4, 6), B(2, –2) and C(2, 5), is:, (A) (0, 2), , (B) (0, 3), , (C) (1, 3), , 1, , (D) (1, 2), , 6. A man goes 15 m due west and then 8 m due north. Now far is he from the starting point?, (A) 15 m, , (B) 8 m, , (C) 17 m, , 1, , (D) 16 m, , 7. 4 tan2 A – 4 sec2 A is equal to:, (A) 2, , 1, , (B) 3, , C) 4, , (D) –4, , 8. If the perimeter of a semi-circular protractor is 36 cm, then its diameter is:, (A) 12 cm, , (B) 13 cm, , (C) 14 cm, , 1, , (D) 15 cm, , 9. A box contains 20 balls bearing number 1, 2, 3, 4, ............., 20. A ball is drawn at random, 1, , from the box. What is the probability that the number on the ball is divisible by 7?, (A), , 1, 10, , (B), , 2, 7, , (C), , 3, 20, , (D), , 1, 5, , 10. Mean of twenty observations is 15. If two observations 3 and 14 are replaced by 8 and 9, 1, , respectively, then the new mean will be:, (A) 14, , (B) 15, , (C) 16, , (D) 17, , (Q 11 – 15) Fill in the blanks:, , 11. The quadratic equation 2x 2 + px + 3 = 0 has two equal roots if p = ...................., , 1, , 12. The sum of the first 20 natural numbers is ................... ., , 1, , 13. If tan q =, , 1, , 3 , then sec q = ................... ., , 14. The probability of an impossible event is ................... ., , 1, , OR, , 1, , An unbiased dice is rolled once. The probability of getting a prime number is ..................., , 15. The mean of first ten multiples of 2 is ..................... ., , 1, , (Q 16 – 20) Answer the following:, , 16. If the mean and mode of a discrete data is 6 and 9, find the median of the data., , 1, , OR, If, , f i = 15,, , f i xi = 3p + 36 and the mean of the distribution is 3, then find the value of p., , 1, , 17. Find the probability of getting doublet in a single throw of a pair of dice., , 1, , 18. Determine the degree of the polynomial (x + 1)(x 2 – x – x4 + 1)., , 1, , 19. If 2x, x + 10, 3x + 2 are in A.P, find the value of x., , 1, , 20. Find the value of cos 15°, using the result cos (A – B) = cos A cos B + sin A sin B., , 1, , Sample Paper-1, , 59

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SECTION - B, Read the following question carefully and answer the questions that follow., 21. Two unbiased coins are tossed. Find the probability of getting: (i) two heads (ii) at least, one head., 22. Find the quotient and the remainder when p(x) = x 3 – 4x is divided by g(x) = x 2 – 2x, , 2, 2, , 23. Usman asked her classmate Mamta to calculate the value of, , 2, , “sin 60 cos 30° + cos 60º sin 30°”., Mamta calculated the value as shown below:, sin 60º cos 30º + cos 60º sin 30º= sin (60º+ 30º) + cos (60º + 30º), = sin 90º + cos 90º, =1+0, =1, i. Examine if Mamta’s calculation is correct or not., ii. If not, point out the inaccuracy and give the correct calculation. If yes, calculate if, the answer will still be “1” if angles 60° and 30° in the equation were changed to, 45°., , 24. Find a relationship between x and y such that the point (x, y) is equidistant from the points, 2, , (3, 6) and (–3, 4)., OR, , 2, , Show that the points (4, 2), (7, 5) and (9, 7) are collinear., , 25. If the circumference of a circle increases from 4p to 8p, then find the percentage increase, 2, , in the area of the circle., , 26. If 0.3528 is expressed in the form, , p, , find the smallest values of m, n and p., 2 5n, OR, m, , 2, 2, , Using prime factorisation, find the LCM of 150 and 210., , SECTION - C, Read the following question carefully and answer the questions that follow., , 27. If the HCF of 657 and 963 is expressible in the form 657 × 22 + 963y, then find the value, 3, , of y., , 28. Anjali places a mirror on level, , A, ground to determine the height, of a tree (see the diagram). She, stands at a certain distance so, that she can see the top of the, tree reflected from the mirror. Tree, Ajnjali’s eye level is 1.8 m above, the ground. The distance of, B, Anjali and the tree from the, mirror are 1.5 m and 2.5 m, respectively., , 60, , Mirror, , C Anjali's, (Eye-level), 1.8 m, 1.5 m D, , 2.5 m, M, , 3, Mathematics Class X

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i., , Name the two similar triangles that are formed in the diagram., , ii. State the criterion of similarity that is applicable to here., iii. Find the height of the tree., , 29. Without using tables, evaluate:, sin(50° + q) – cos(40° – q) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89°, , 3, , OR, Show that:, cot 30 cot 60 1, = cot 90, cot 30 + cot 60, , 3, , 30. In a circle of radius 7 cm, a chord makes an angle of 60° at the, centre of the circle. Find: (a) area of the circle (b) area of sector, AOB (c) area of minor segment APB, , O, , (Take, , 60°, , 3 = 1.73), , A, OR, , B, , P, , 3, , 3, , Draw a line segment of length 8 cm. Divide it into three equal parts., , 31. Solve for x and y:, 3, , 3x + 2y = 11, 2x + 3y = 4, OR, Determine the AP whose 3rd term is 5 and the 7th term is 9., , 3, , 32. In what ratio is the line segment joining the points (–2, –3) and (3, 7) divided by the, 3, , y-axis? Also, find the coordinates of the point of division., , 33. A solid wooden toy is in the shape of a right circular cone mounted an a hemi-sphere, of radius 4.2 cm. The total height of the toy is 10.2 cm. Find the volume of the toy., Take p =, , 3, , 22, 7, , 34. Find the mode from the following:, Age (in years), Number of persons, , 3, , 0-10, , 10-20, , 20-30, , 30-40, , 40-50, , 50-60, , 6, , 11, , 21, , 23, , 14, , 5, , SECTION - D, 35. Construct a DABC in which AB = 6.5 cm, ∠B = 60° and BC = 5.5 cm. Also, construct, a triangle A BC similar to DABC, whose sides are, , 2, the corresponding sides of DABC., 5, , 4, , 36. Given the 1 is a zero of the polynomial 2x 3 + x 2 – 2x – 1, find all the zeros., , 4, , 37. State and prove the converse of Pythagoras Theorem., , 4, , Sample Paper-1, , 61

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OR, In a DABC, ∠B is an acute-angle and AD ⊥ BC., , 4, , Prove that, (i) AC2 = AB2 + BC2 – 2 BC × BD, (ii) AB2 + CD2 = AC2 + BD2, , 38. The shadow of the tower, standing on a level ground is found to be 40 m longer when, the sun's altitude is 30° than what it is 60°. Find the height of the tower., OR, Two pillars of equal height, stand on either side of a roadway which is 150 m wide. From, a point on the roadway between the pillars, the elevations of the top of the pillars are 60°, and 30°. Find the height of the pillars and the position of the point. [Use 3 = 1.73 ], 39. An umbrella has 8 ribs which are equally spaced, as shown in the figure. Assuming the, umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive, ribs of the umbrella., , 4, , 4, 4, , 40. Show graphically that the following system of equations has no solutions:, 4, , x – 2y = 6; 3x – 6y = 0, OR, Using the quadratic formula, solve for x:, 3x 2 + 2 5 x 5 = 0, , 62, , 4, , Mathematics Class X

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1, , SOLUTION, CBSE - Class 10, SECTION - A, 1. (D), Since,, , 1, 2, , 2, , 1, , 3, , 36 = 2 × 3 and 54 = 2 × 3, The HCF (36, 54) = 21 × 32 , i.e. 18, , 2. (C), , 1, , Here,, p(1) = (1)3 – 7(1) + 6 = 0, p(2) = (2)3 – 7(2) + 6 = 0, p(–2) = (–2)3, ( )+ ≠0, 3, p(–3) = (–3) – 7(–3) + 6 = 0, So, –2 is not a zero of p(x)., , 3. (C), , 1, 2, , Discriminant = b – 4ac, = (–4)2 – 4(1)(1) = 16 – 4 = 12, , 4. (B), , 1, , Here, a = 4 and d = 5, Let nth term of the AP be 254. Then,, an = a + (n – 1) d, ⇒, 254 = 4 + (n – 1) (5), ⇒, 5(n – 1) = 250, ⇒, n = 51, , 5. (B), , 1, , 4+2+2 6 2+5, ,, , i.e. (0, 3)., 3, 3, , The centroid of DABC is, , 6. (C), Obriously,, , 1, ON =, , OW + NW, 2, , =, , 152 + 8 2, , =, , 225 + 64, , =, , 289, , = 17 m., Sample Paper-1, , 2, , N, , 8m, , W, , 15 m, , O, , 63

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7. (D), , 1, 2, , 2, , 2, , 2, , 4 tan A – 4 sec A = – 4 (sec A – tan A), =–4×1, =–4, , 8. (C), , 30, , 0, , 10 2, 0, 30, 170 1, 60, 40, 150, 14, 0, , 15, , 170, 16 0, 10, 20, , 180, , 40, , = AB + AB, = 2r + pr, 0, ⇒ (2 + p)r = 36, A, 36 2, 72, 72 7, ⇒, d = 2r =, =, =, = 14 cm, 22, 2+p 2+, 36, 7, 9. (A), , 0, , Perimeter of semi-circular protractor, , 1, 14, , 80 90 10 0 11, 0 1, 70, 20, 0 90 80 7, 60, 0, 10 10, 1, 13, 60, 0, 5 0 1 20, 50, 0, 13, , O, , 0, , 180, , B, , 1, , Out of 20 numbers, multiples of 7 are only two, namely 7 and 14., 2, 1, So,, P(a multiple of 7) =, i.e., 20, 10, 10. (B), sum of 20 observations, Mean =, 20, ⇒ Sum of 20 observations = 15 × 20, = 300, Sum of 20 observations = Sum of 18 observations + 3 + 14, = Sum of 18 observations + 8 + 9, Thus, Mean remains the same., (, , 1, , 3 + 4 = 8 + 9), , 11. 2x 2 + px + 3 = 0 will have equal roots, when p2 – 4(2)(3) = 0, i.e. when p2 – 24 = 0, or, , p=, , 24, , 12. Sum of first 20 natural numbers =, tan q =, , 13., So,, , or, , p=, , 1, , 2 6, , 20 (20 + 1), = 210, 2, , 1, , 3 gives q = 60°, , 1, , sec q = sec 60° = 2, , 14. P (impossible event) = 0, , 1, OR, , Of the six possible outcomes 1, 2, 3, 4, 5 and 6 only, 3, prime numbers. So, the required probability is , or, 6, 2 + 4 + 6 + .... + 20, 15. Mean of first 10 multiplies of 2 is, 10, 16. ince, ode = 3 edian −, ean, 9 = 3 × Median – 2 × 6, 3 × Median = 9 + 12, Median = 21 / 3 = 7, , 64, , 2, 3, 5 are favourable outcomes as, 1, ., 2, i.e. 11., , 1, , 1, Mathematics Class X

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OR, 3 p + 36, =3, 15, 3p = 45 – 36, , Since, , Mean =, , p = 9/3 = 3, , 17. Out of the 36 possible outcomes:, (1, 1), (1, 2) …..(1,6), (2, 1), (2, 2) …..(2,6), (3, 1), (3, 2) …..(3,6), (4, 1), (4, 2) …..(4,6), (5, 1), (5, 2) …..(5,6), (6, 1), (6, 2) …..(6,6), Six are doublets- (1,1), (2,2) , (3,3), (4,4) (5,5), (6,6)., 6, 1, So, required probability is, , or ., 36, 6, 18. Given polynomial in standard form is: –x5 – x4 + x 3 + 1, , 1, 1, , So, its degree is 5, , 19. Since 2x, x + 10, 3x + 2 are in A.P.,, , 1, , Using formula, 2 × Middle number = First number + Third number, 2 (x + 10) = 2x + (3x + 2), 2x + 20 = 5x + 2, 3x = 18 or x = 6, This, on simplification, gives us x = 6 ., , 20. Taking A = 45° and B = 30°, the given result gives:, , 1, , cos (45° – 30°) = cos 45° cos 30° + sin 45° sin 30°, 1, 3, 1, 1, cos 15 =, +, 2, 2, 2 2, , ⇒, , =, , 3 +1, 2 2, , SECTION - B, 21. No. of possible outcomes = 4,, (i) No. of favourable outcomes = 1, namely HH, So, required probability =, , 1, 4, , Sample space, S = {HH, HT, TH, TT}, , 2, , (ii) No. of favourable outcomes = 3, namely HH, HT, TH, So, required probability =, , Sample Paper-1, , 3, 4, , 65

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22., , 2, , x+2, x 2 – 2x x 3 + 0. x 2 – 4x, x 3 – 2x 2, – +, , Thus, quotient = x + 2,, , 2x 2 – 4x, , Remainder = 0, , 2x 2 – 4x, –, +, 0, , 23. (i) Mamta’s solution is not correct., , 2, , (ii) Procedure is incorrect, though the result obtained is correct., The correct procedure is, sin 60° cos 30° + cos + 60° sin 30°, 3 3 1 1, ⋅, + ⋅, 2 2, 2 2, 3 1, = +, 4 4, =1, =, , 24. Let P(x, y), A (3, 6) and B(–3, 4)., , 2, 2, , 2, , As P is equidistant from A and B, PA = PB or PA = PB, i.e., , (x – 3)2 + ( y – 6)2 = (x + 3)2 + ( y – 4)2, , ⇒, , (x 2 – 6x + 9) + ( y2 – 12y + 36) = (x 2 + 6x + 9) + ( y2 – 8y + 16), , ⇒, , 12x + 4y – 20 = 0, or, , 3x + y – 5 = 0, , which is the required relationship between x and y., OR, The given points (4, 2), (7, 5) and (9, 7) will be collinear, if, [4(5 – 7) + 7(7 – 2) + 9(2 – 5)] is equal to zero., Now,, 4(5 – 7) + 7(7 – 2) + 9(2 – 5), = 4(– 2) + 7(5) + 9(–3), = – 8 + 35 – 27, =0, , 2, , Thus, the given points are collinear., , 25. Let r and R be the radius of the initial circle and the increased circle, respectively. Then, , 2, , 2pr = 4p and 2pR = 8p, ⇒, r = 2 and R = 4, ∴ Area of the initial circle = p(2)2 , i.e. 4p; and, Area of the increased circle = p(4)2 , i.e. 16p, Thus, % increase in the area is 400., , 66, , Mathematics Class X

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26., , 3528, 441, 441, =, = 1 4, 1250 2 5, 10000, Thus, smallest values of m, n and p are m = 1, n = 4 and p = 441., , 2, , 0.3528 =, , OR, , 2, , The prime factorisations of 150 and 210 are:, 1, , 1, , 2, , 150 = 2 × 3 × 5, So,, , and, , 1, , 1, , 1, , 1, , 210 = 2 × 3 × 5 × 7, , LCM (150, 210) = 21 × 31 × 52 × 71, i.e. 1050, , SECTION - C, 27. Here,, , 3, , 657 = 3 × 3 × 73, and, 963 = 3 × 3 × 107, So,, HCF(657, 963) = 9, Now,, 9 = 657 × 22 + 963y, ⇒, 963y = –14445, ⇒, y = –15, Thus, the value of y is –15., , 28. (i) DABM and DCDM, , 3, , A, , Mirror, Tree, , C Anjali's, (Eye-level), 1.8 m, , B, , 2.5 m, , 1.5 m D, , M, (ii) Since, angle of incidence and angle of reflection are the same, ∠AMB = ∠CMD, Also, ∠ABM = ∠CDM = 90°, So, by AA similarly criterion, DABM ~ DCDM, AB BM, (iii) As DABM ~ DCDM,, =, CD DM, AB, 2.5, ⇒, =, 1.8, 1.5, ⇒, AB = 3, Thus, the height of the tree is 3 metres., , 29. Given expression is:, , 3, , sin(50° + q) – cos(40° – q) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89°, = cos[90° – (50° + q)] – cos(40° – q) + (tan 1° tan 89°) (tan 10° tan 80°) (tan20° tan70°), = cos(40° – q) – cos(40° – q) + (tan 1° cot 1°) (tan 10° cot 10°) (tan 20° cot 20°), = 0 + (1) (1) (1), =1, Sample Paper-1, , 67

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OR, LHS =, , cot 30 cot 60 1, cot 30 + cot 60, , 3, , 1, 1, 3, 1, 3+, 3, 0, = 0 = cot 90° = RHS, 1, 3+, 3, 3, , =, , =, , 30. (i) Area of the circle = p(7)2 sq cm = 154 sq cm, , 3, , 60, 2, × p(7)2 sq cm = 25 sq cm, 3, 360, (iii) Area of minor segment APB = Area of the sector OAPB – Area of DOAB, (ii) Area of sector of angle 60° =, , = 25, , 2, 3, , 3 7, 2 2, , 2, , sq cm, , 3, .49 sq cm, 4, , = 22.67, , = [25.67 – 21.19] sq cm, = 4.48 sq (approx.), OR, A, , P, , Q, 8 cm, , B, , 3, , A1, A2, A3, X, , 31. Given equations are:, , 3, 3x + 2y = 11, , .....(i), , 2x + 3y = 4, , .....(ii), , Eq. (ii) gives,, , y=, , 4 2x, 3, , .....(iii), , Substituting this value of y in Eq. (1), we have, 3x + 2, ⇒, ⇒, , 68, , 4 2x, 3, , = 11, , 9x + 8 – 4x = 33, 5x = 25 i.e., , x=5, , Mathematics Class X

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Substituting this value of x in Eq. (iii), we have:, y=, Thus, x = 5, y = –2, , 4 2 5, 3, , i.e., , y = –2, , OR, , 3, , Let a be the first term and d be the common difference of AP., Then,, a3 = a + 2d = 5 and a 7 = a + 6d = 9, Solving these simultaneously, we get:, a = 3 and d = 1, Thus, the required AP is 3, 4, 5, 6, ............, , 32. Let P (0, y), a point on y – axis, divide the join of points, (–2, –3) and (3, 7) in the ratio K : 1., Then,, 3K 2 7K 3, ,, P(0, y) = P, K +1 K +1, 3K 2, 2, =0 ⇒ K=, K +1, 3, , ⇒, , y), P(0,, (–, , (3, 7 ), , 1, , 3, , K, ), 2 , –3, , So, the required ratio is 2 : 3., Further,, 7, P 0,, , 2, 3, 3, 2, +1, 3, , i.e., , P(0, 1), , 33. From the figure, we have radius of the base of the cone = 4.2 cm;, and height = 6 cm., So, volume of the toy, = volume of the cone + volume of hemi-sphere, 1, 2, p (4.2)2 (6) + p(4.2)3 cu.cm, =, 3, 3, , A, , 3, 10.2 cm, B, , O, , C, , = (776.16 + 155.232) cu.cm, = 931.392 cu.cm, , 34. Here, the modal class is 30 – 40., , 3, , For this class,, l = 30, h = 10, f = 23, f 0 = 21, f 2 = 14, So,, , Sample Paper-1, , Mode = l +, , f, 2f, , f0, f0, , f2, , ×h, , = 30 +, , 23 21, × 10, 46 21 14, , = 30 +, , 20, 9, = 31, 11, 11, , 69

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SECTION - D, 35., , 4, , A, , 6 .5, , cm, , A, , B, , Here, DA BC, , B1, , C, B2, , C, , 5.5 cm, B3, , DABC, in which BC =, , B4, , B5, , X, , 2, BC, 5, , 36. As 1 is a zero of p(x) = 2x 3 + x 2 – 2x – 1,, , 4, , (x – 1) is a factor of p(x), Now, we divide p(x) by (x – 1) and get the quotient, 2x 2 + 3x + 1, x – 1 2x 3 + x 2 – 2x – 1, 2x 3 – 2x2, – +, 3x 2 – 2x – 1, 3x 2 – 3x, – +, x–1, x–1, – +, 0, Quotient = 2x 2 + 3x + 1 = 2x 2 + 2x + x + 1, = 2x(x + 1) + 1(x + 1), = (x + 1) (2x + 1), 1, Thus, the other two zeros are –1 and – ., 2, , 70, , Mathematics Class X

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37. Statement: In a triangle, if the square of one side is equal to the sum of the squares of, 4, , the other two sides, then the angle opposite to the first side is a right angle., Proof: Let DABC be a triangle in which, 2, , 2, , AB + BC = AC, , 2, , A, , .....(i), , We need to prove that DABC is a right-angled, right-angled at B., Construct a right triangle, PQR, right-angled at Q such that, PQ = AB and QR = BC., , ⇒, , 2, , 2, , AB + BC = PR, , 2, , C, , Q, , R, , P, , In DPQR, since ∠Q = 90°, PR 2 = PQ2 + QR 2, , B, , (By Pythagoras Theorem), , .....(ii) (, , AB = PQ, BC = QR), , From (i) and (2), we have, AC = PR, Therefore, by SSS congruence criterion, DABC, which gives, , ∠B = ∠Q, , But,, , ∠Q = 90°, , ⇒, , DPQR,, , B = 90°, , Hence, DABC is a right triangle, right-angled at B., OR, A, , (i) From DADC, we have, , 4, , AC2 = AD2 + DC2, = AD2 + (BC – BD)2, = AD2 + BC2 + BD2 – 2 BC × BD, 2, , 2, , B, , D, , C, , 2, , = (AD + BD ) + BC – 2 BC × BD, = AB2 + BC2 – 2 BC + BD, (ii) From DABD, we have, AB2 = AD2 + BD2, From DADC, we have, CD2 = AC2 – AD2, Adding these, we get, AB2 + CD2 = AC2 + BD2, , 38. Let h metres be the height of the tower., , Y, , 4, , From DAXY,, h, = tan 60° =, AX, ⇒, From DBXY,, , Sample Paper-1, , 3, , Tower (h), , h, AX =, m, 3, B, , 30°, 40 m, , 60°, A, , X, , 71

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h, 1, = tan 30° =, BX, 3, ⇒, , BX =, , 3hm, , Now,, AX = BX – 40, i.e., , BX = AX + 40, h, h=, + 40, 3, , ⇒, ⇒, , 3 h = h + 40 3, , ⇒, , 2 h = 40 3, , or, , h = 20 3 m, , Thus, the height of the tower is 20 3 metres, OR, Let h metres be the height of each pillar and point P be x metres away from pillar AB., h, B, D, From DPCD,, = tan 60° = 3, PC, h, ⇒, PC =, .....(i), 3, h, h, h, 1, From DPAB,, = tan 30° =, PA, 3, 30°, 60°, A, P, C, ⇒, PA = 3 h, .....(ii), x, 150 m, Now,, PA + PC = 150, ⇒, , 3h +, , h, = 150, 3, , ⇒, , 3h + h = 150 3, , ⇒, , 4h = 150 3, , ⇒, Also,, , 4, , h=, PA =, , 150, 3 = 64.875 m, 4, 3h=, , 150, 3 = 112.5 m, 4, , Thus, the height of the pillar is 64.875 m and the position of the point is at a distance of, 112.5 m from the pillar AB., , 39. Area between the two consecutive ribs of the umbrella, , 72, , =, , 1, × Area of the circle of radius 45 cm., 8, , =, , 1, 22, ×, × (45)2 sq cm, 8, 7, , 4, , Mathematics Class X

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=, , 22275, sq cm, 28, , = 795, , 40., , 15, sq cm., 28, , 4, , Table for x – 2y = 6, , Table for 3x – 6y = 0, , x, , 6, , 8, , 10, , x, , 2, , 4, , –2, , y, , 0, , 1, , 2, , y, , 1, , 2, , –1, , The graph of the two equations is:, , y, 3, 2, 1, x, , –4, , –2, , 2, , 0, –1, , 4, , 6, , 8, , 10, , x, , –2, y, Since the two lines are parallel. The given system of equation have no solutions., OR, , 4, , Using the quadratic formula, we have:, x=, =, , Thus,, , Sample Paper-1, , 2 5, , 2 5, , 2, , 4(3)( 5), , 2 3, 20 + 60, , 2 5, 6, , =, , 2 5 4 5, 6, , =, , 2 5, , or, 6, , =, , 5, , or – 5, 3, , x=, , 5, , or – 5, 3, , 6 5, 6, , 73