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ICSE Class X Board Exam 2023 : Mathematics

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Sample Question Paper-1 (Specimen Paper issued by CISCE dated 12th July, 2022) Mathematics Class-10 SOLVED Time Allowed : 2 hours Maximum Marks : 80 Answers to this Paper must be written on the paper provided separately. You will not be allowed to write during the first 15 minutes. This time is to be spent in reading the Question Paper. The time given at the head of this Paper is the time allowed for writing the answers. Attempt all questions from Section A and any four questions from Section B. All working, including rough work, must be clearly shown, and must be done on the same sheet as the rest of the answer. Omission of essential working will result in loss of marks. The intended marks for questions or parts of question are given in brackets [ ] Mathematical tables are provided. Section-A (Attempt all questions from this Section) Question 1. Choose the Correct answer to the questions from the given options: [15 Marks] (i) The SGST paid by a customer to the shopkeeper for an article which is priced at `500 is `15. The rate of GST charged is: (a) 1.5% (b) 3% (c) 5% (d) 6% (ii) When the roots of a quadratic equation are real and equal then the discriminant of the quadratic equation is: (a) Infinite (b) Positive (c) Zero (d) Negative (iii) If (x 1) is a factor of 2x2 ax 1, then the value of a is: (a) 1 (b) 1 (c) 3 (d) 3 a b p (iv) Given c d X = q . The order of matrix X is: (a) 2 2 (b) 1 2 (c) 2 1 (d) 1 1 (v) 57, 54, 51, 48, ......... are in Arithmetic Progression. The value of the 8th term is: (a) 36 (b) 78 (c) 36 (d) 78 (vi) The point A (p, q) is invariant about x = p under reflection. The coordinates of it s image A is: (a) A (p, q) (b) A ( p, q) (c) A (p, q) (d) A ( p, q) To know about more useful books Click Here 44 OSWAAL ICSE Sample Question Papers, mathematics, Class-X (vii) In the given diagram the ABC is similar to DEF by the axiom: (a) SSS (b) SAS (c) AAA (d) RHS (viii) The volume of a right circular cone with same base radius and height as that of a right circular cylinder, is 120 cm3. The volume of the cylinder is: (a) 240 cm3 (b) 60 cm3 (c) 360 cm3 (d) 480 cm3 (ix) The solution set for the given in equation is: 8 2x < 8, x W (a) { 4, 3, 2, 1, 0, 1,2, 3, 4} (b) { 4, 3, 2, 1} (c) {0, 1, 2, 3} (d) { 8, 7, 6, 5, 4, 3, 2, 1, 0, 1,2, 3, 4, 5,6 7, 8} (x) The probability of the Sun rising from the east is P(S). The value of P(S) is: (a) P(S) = 0 (b) P(S) < 0 (c) P(S) = 1 (d) P(S) > 1 2 x 2 1 8 8 (xi) If 0 1 + 3 4 0 = 12 1 . The value of x is: (a) 2 (b) 3 (c) 4 (d) 5 (xii) The centroid of a ABC is G (6, 7). If the coordinates of the vertices A, Band C are (a, 5), (7, 9) and (5, 7) respectively. The value of a is: (a) 9 (b) 6 (c) 3 (d) 7 (xiii) In the given diagram AC is a diameter of the circle and ADB=35 . The degree measure of x is: (a) 55 (b) 35 (c) 45 (d) 70 (xiv) If the nth term of an Arithmetic Progression (A.P.) is (n + 3), then the first three terms of the (A.P.) are: (a) 1, 2, 3 (b) 2, 4, 6 (c) 4, 5, 6 (d) 7, 8, 9 (xv) The median of a grouped frequency distribution is found graphically by drawing: (a) a linear graph (b) a histogram (c) a frequency polygon (d) a cumulative frequency curve Sample Question Papers 45 Question 2 (i) Salman deposits ` 1200 every month in a recurring deposit account for 2 years. If the rate of interest is 6% per annum, find the amount he will receive on maturity. [4] (ii) 3, 9, m, 81 and n are in continued proportion. Find the values of m and n. [4] cos A 1 + sin A (iii) Prove that: 1 + sin A + cos A = 2 sec A Question 3 (i) The inner circumference of the rim of a circular metal tub is 44 cm. [4] [4] Find: (a) The inner radius of the tub (b) The volume of the material of the tub if it s outer radius is 8 cm. 22 7 Give your answer correct to three significant figures. Use p = (ii) From the given figure: [4] (a) Write down the coordinates of A and B. (b) If P divides AB in the ratio 2:3, find the coordinates of point P. (c) Find the equation of a line parallel to line AB and passing through origin. (iii) Use graph set for this question. Take 2 cm = 1 unit along the axes. [5] Plot the OAB, where O (0, 0), A (3, 2), B (2, 3). (a) Reflect the OAB through the origin and name it as OA B . (b) Reflect the OA B on the y axis and name it as OA B . (c) Reflect the OA B on the x axis and name it as OA B . (d) Join the points AA B B A A B B and give the geometrical name of the closed figure so formed. OSWAAL ICSE Sample Question Papers, mathematics, Class-X 46 Section-B (Attempt any four questions from this Section.) Question 4 (i) The following bill shows the GST rates and the marked price of articles: [3] BILL: COMPUTERS Articles Marked price Rate of GST Graphic Card Rs 15500.00 18% Laptop adapter Rs 1900.00 28% Find the total amount to be paid for the above bill. (ii) Solve the following quadratic equation, [3] 7x2 + 2x 2 = 0 Give your answer correct to two places of decimal (iii) Use graph sheet for this question. Draw a histogram for the daily earnings of 54 medical stores in the following table and hence estimate the mode for the following distribution. Take 2 cm = `500 units along the x-axis and 2 cm = 5 stores along the y-axis. [4] Daily earnings (`) No. of medical stores 4500 5000 5000 5500 5500 6000 6000 6500 6500 7000 20 14 12 5 3 Question 5 3 2 6 4 A= , B = and C = , Evaluate AB 5C 1 4 1 5 (i) [3] (ii) In the given figure, O is the centre of circle. The tangent PT meets the diameter RQ produced at P. [3] (a) Prove PQT ~ PTR (b) If PT = 6 cm, QR = 9 cm. Find the length of PQ (iii) Factorise the given polynomial completely, using Remainder Theorem: [4] 6x3 + 25x2 + 31x + 10 Question 6 (i) ABCD is a square where B (1, 3), D (3, 2) are the end points of the diagonal BD. [3] Find: (a) the coordinates of point of intersection of the diagonals AC and BD (b) the equation of the diagonal AC (ii) Prove that: sec 2 + co sec 2 = sec .co sec [3] (iii) The first, the last term and the common difference of an Arithmetic Progression are 98, 1001 and 7 respectively. Find the following for the given Arithmetic Progression: [4] (a) number of terms n . (b) Sum of the n terms. Sample Question Papers 47 Question 7 (i) A box contains some green, yellow and white tennis balls. The probability of selecting a green ball 1 1 is and yellow ball is . If the box contains 10 white balls, then find: [3] 4 3 (a) total number of balls in the box. (b) probability of selecting a white ball. (ii) A cone and a sphere having the same radius are melted and recast into a cylinder. The radius and height of the cone are 3 cm and 12 cm respectively. If the radius of the cylinder so formed is 2 cm, [3] find the height of the cylinder. (iii) In the given diagram, ABCD is a cyclic quadrilateral and PQ is a tangent to the smaller circle at E. Given AEP = 70 , BOC = 110 . Find: [4] (a) ECB, (c) BFC, Question 8 (i) Solve the following equation: 7 x x 7 4 < , x R 3 2 3 6 (b) BEC, (d) DAB, [3] Represent the solution set on a number line. (ii) The following table gives the petrol prices per litre for a period of 50 days. Price (`) 85 90 90 95 95 100 100 105 [3] 105 110 No. of days 12 10 8 15 5 Find the mean price of petrol per litre to the nearest rupee using step-deviation method. (iii) In the given diagram, ABC is a triangle and BCFD is a parallelogram. [4] AD : DB = 4 : 5 and EF = 15 cm. Find: (a) AE : EC (c) BC (b) DE OSWAAL ICSE Sample Question Papers, mathematics, Class-X 48 Question 9 (i) Amit takes 12 days less than the days taken by Bijoy to complete a certain work. If both, working together, takes 8 days to complete the work, find the number of days taken by Bijoy to complete the work, working alone. [4] (ii) Use a graph sheet for this question. The daily wages of 120 workers working at a site are given below: [6] Wages (`) 250 300 300 350 350 400 400 450 450 500 500 550 550 600 No. of 8 15 20 30 25 15 7 workers Use 2 cm = `50 and 2 cm = 20 workers along x axis and y axis respectively to draw an ogive and hence estimate (a) the median wages (b) the inter quartile range of wages (c) percentage of workers whose daily wage is above `475. Question 10 (i) Solve for x, using the properties of proportion. [3] 2+x + 3 x =3 2 + x 3 x (ii) Using ruler and compasses, construct a regular hexagon of side 4.5 cm. Hence construct a circle circumscribing the hexagon. Measure and write down the length of the circum-radius. [3] (iii) An observer standing on the top of a lighthouse 150 m above the sea level watches a ship sailing away. As he observes, the angle of depression of the ship changes from 50 to 30 . Determine the distance travelled by the ship during the period of observation. Give your answer correct to the nearest meter. (Use Mathematical Table for this question.) [4] SOLUtions Sample Question Paper-1 Mathematics Section-A 1. (i) Option (d) is correct. Explanation: SGST paid = ` 15 Purchase price = ` 500 Total GST = SGST + CGST = 15 + 15 ( SGST = CGST) = 30 Rate = = Amount of GST 100 Total Pr ice 30 100 = 6% 500 (ii) Option (c) is correct. Explanation: We know that, x= b b 2 4 ac 2a Since the roots are real and equal b b x= , 2a 2a b2 4ac = 0 D = 0 (iii) Option (b) is correct. Explanation: Since (x 1) is a factor of 2x2 ax 1 \ x = 1 2(1)2 a(1) 1 = 0 2 a 1 = 0 1 a = 0 1 = a a = 1 (iv) Option (c) is correct. a b Explanation: Here c d is a 2 2 matrix p and q is a 2 1 matrix So, to obtain a 2 1 matrix from a 2 2 matrix, we multiply a 2 2 matrix by 2 1 matrix. A m n matrix is multiplied by n p matrix to obtain a m p matrix. (v) Option (a) is correct. Explanation: AP = 57, 54, 51, 48, ........ Here, a = 57 d = 54 57 d = 3 Here, an = a + (n 1)d a8 = 57 + (8 1)( 3) a8 = 57 +7( 3) a8 = 57 21 a8 = 36 (vi) Option (a) is correct. Explanation: We know that only those points which lie on the line are invariant points when reflected in the line. So, only those points are invariant which are on the line x = p. Hence, line passing through x = p will be considered x axis and thus the coordinates of image are A (p, q) (vii) Option (b) is correct. 4 3 = Explanation: Here 24 18 AB BC = DE EF and ABC = DEF So, by SAS similarity criterion, ABC ~ DEF (viii) Option (c) is correct. Explanation: Volume of cone = 120 cm3 We know that, 1 Volume of cone = pr2h 3 1 2 pr h = 120 3 pr2h = 120 3 Volume of cylinder = 360 cm3 (ix) Option (c) is correct. Explanation: Here 8 2x < 8 8 2 2x 2 < 8 2 4 x < 4 50 OSWAAL ICSE Sample Question Papers, mathematics, Class-X x = 4, 3, 2, 1, 0, 1, 2, 3 But since x W, x = {0, 1, 2, 3} (x) Option (c) is correct. Explanation: Since the sun always rises from east, it is a certain event. Therefore, the probability P(S) = 1 (xi) Option (d) is correct. 2 x 6 3 8 8 Explanation: 0 1 + 12 0 = 12 1 2 + 6 x + 3 8 8 0 + 12 1 + 0 = 12 1 On comparison x + 3 = 8 x = 5 (xii) Option (b) is correct. Explanation: The centroid of triangle x + x 2 + x3 y1 + y 2 + y3 = 1 , 3 3 a+7+5 5+9+7 , (6, 7) = 3 3 a + 12 21 , (6, 7) = 3 3 a + 12 On comparison, 6 = 3 18 = a + 12 a = 6 (xiii) Option (a) is correct. Explanation: Join CD Since AC is diameter ADC = 90 (Angle in a semicircle are equal to 90 ) CDB + ADB = 90 CDB + 35 = 90 CDB = 55 Now CAB = CDB (Angles in the same segment) CAB = 55 (xiv) Option (c) is correct. Explanation: Since nth term = (n + 3) an = (n + 3) a1 = (1 + 3) a1 = 4 a2 = (2 + 3) = 5 and a3 = (3 + 3) = 6 So, the first three terms of the AP are 4, 5 and 6 (xv) Option (d) is correct. Explanation: The median for a grouped frequency distribution is found graphically by drawing a cumulative frequency curve. 2. (i) Monthly deposit = ` 1200, n = 30 months, r = 6% n( n + 1) P r Interest = 2400 = 30( 30 + 1) 1200 6 2400 30 31 1200 6 = 2400 = 30 31 3 = 2790 So, Maturity Value = P n + I = 1200 30 + 2790 = 36000 + 2790 = ` 38790 (ii) Since, 3, 9, m, 81 and n are in continued proportion So, Now, 3 9 m 81 = = = 9 m 81 n 3m = 9 9 m = 81 3 m = 27 27n = 81 81 n = 81 3 n = 243 cos A 1 + sin A (iii) LHS = + 1 + sin A cos A cos2 A + (1 + sin A )2 (1 + sin A ) (cos A ) cos2 A + 1 + sin 2 A + 2 sin A (1 + sin A ) (cos A ) 1 + 1 + 2 sin A (1 + sin A )(cos A ) (cos2A + sin2A = 1) 2 + 2 sin A (1 + sin A )(cos A ) 2 (1 + sin A ) (1 + sin A )(cos A ) 2 cos A 2 sec A = RHS 1 cos A = sec A Hence Proved Solutions 51 3. (i) Inner circumference of circular metal tub = 44 cm 2pr = 44 22 r = 44 2 7 7 1 r = 44 22 2 r = 7 cm Now, Volume of material of tub 2 (R3 r3) = 3 2 22 = (83 73) 3 7 2 22 = (512 343) 3 7 2 22 = 169 3 7 = 354.095 cm3 = 354 cm3 (ii) (a) Coordinates of A = (5, 0), Coordinates of B = (0, 3) m1 x 2 + m2 x1 m1 y 2 + m2 y1 , (b) P = m + m m1 + m2 1 2 2( 0 ) + 3( 5) 2( 3) + 3( 0 ) , = 2+3 2 + 3 SECTION-B 4. (i) MP of Graphic card = `15,500 GST = 18% GST = 0.18 15500 = `2790 So, SP = 15500 + 2790 SP = `18,290 Now, MP of Laptop Adapter = `1900 GST = 28% GST = 0.28 1900 GST = `532 So, SP = 1900 + 532 SP = `2432 Total amount to be paid = 18,290 + 2432 15 6 6 = , = 3, 5 5 5 = `20,722 Therefore, the coordinates of point P, are 6 3, . 5 (c) For parallel lines, m1 = m2 3 0 3 3 = = Slope of line AB = 0 5 5 5 Now, y y1 3 = x x1 5 y 0 3 = x 0 5 y 3 = x 5 3x = 5y 3x + 5y = 0 (ii) x= b b 2 4 ac 2a x= 2 ( 2 )2 4(7 )( 2 ) 2(7 ) x= 2 4 + 56 14 x= 2 60 14 x= 2 7.746 14 x= 2 + 7.746 2 7.746 and x = 14 14 x= 5.746 and x = 9.746 14 14 x = 0.41 and x = 0.696 ~ 0.70 OSWAAL ICSE Sample Question Papers, mathematics, Class-X 52 (iii) 5. (i) 3 2 6 A= and B = 1 4 1 18 2 16 So, AB = = 6 +4 2 4 C= 5 20 So, 5C = 25 Now, 16 20 16 + 20 36 AB 5C = = = 2 25 2 25 27 (ii) (a) Here, in PQT and PTR PQT = PRT (Alternate Segment Theorem; For any circle, the angle formed between the tangent and the chord through the point of contact of the tangent is equal to the angle formed by the chord in the alternate segment) Also, QPT = TPR (common) So, by AA similarity criterion, PQT ~ PTR (b) Now, If a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment. So, PT2 = PR PQ Let PQ be x cm 6 6 = (x + 9)(x) 36 = x2 + 9x 2 x + 9x 36 = 0 2 x + 12x 3x 36 = 0 x(x + 12) 3(x + 12) = 0 (x 3)(x + 12) = 0 x = 3 and x = 12 Since, x represents length of a segment, it can t be negative. Thus PQ = 3 cm (iii) p(x) = 6x3 + 25x2 + 31x + 10 Here, p( 2) = 6( 2)3 + 25( 2)2 + 31( 2) + 10 = 6 8 + 25 4 + 31 2+10 = 48 + 100 62 + 10 = 0 So, x = 2 is the zero of the given polynomial. x + 2 is a factor of the given polynomial. Now, by long division: x + 2) 6x3 + 25x2 + 31x + 10 (6x2 + 13x + 5 6x3 + 12x2 ( ) ( ) 13x2 + 31x + 10 13x2 + 26x ( ) ( ) + 5x + 10 5x + 10 ( ) ( ) 0 p(x) = (x + 2)(6x2 + 13x + 5) = (x + 2)(6x2 + 10x + 3x + 5) = (x + 2)[2x(3x + 5) + 1(3x + 5)] p(x) = (x + 2)(2x + 1)(3x + 5) Solutions 53 6. (i) D (3, 2) C (2, 5/2) (iii) a = 98, an = 1001, d = 7 (a) We know that, an= a + (n 1)d 1001 = 98 + (n 1)(7) 903 = (n 1)(7) 129 = n 1 n = 130 So, the number of terms is 130. B (1, 3) A Sn = n (a + l) 2 (b) Now, x1 + x 2 y1 + y 2 Midpoint of BD = 2 , 2 2 = 65 1099 = 71,435 Therefore, sum of n terms is 71,435. Since ABCD is a square, diagonals bisect each other diagonals AC and BD intersect each other at the midpoint of BD 1+3 3+ 2 = , 2 2 4 5 = , 2 2 5 = 2, 2 Q diagonals of square are at right angles with each other. So m1m2 = 1 = 130 ( 98 + 1001) 7. (i) (a) P (selecting green ball) = P (selecting yellow ball) = 1 4 1 3 Now, P (selecting green ball) + P (selecting yellow ball) + P (selecting white ball) = 1 1 1 + + P (selecting white ball) = 1 4 3 5 y 3 2 2 1 3 x 2 = 1 7 + P (Selecting white ball) = 1 12 P (selecting white ball) = 1 1 2y 5 2 2 x 4 = 1 2y 5 =2 2x 4 2y 5 = 4x 8 4x 2y 3 = 0 2 2 (ii) LHS = sec + co sec 1 1 = + 2 2 cos sin sin 2 + cos2 = 2 2 cos sin 1 = 2 cos sin 2 = sec 2 q c osec 2q = sec cosec = RHS Hence proved 7 12 5 12 10 5 = Total number of balls 12 Total number of balls = 24 Therefore, total number of balls = 24 P (selecting white ball) = (b) P (selecting white ball) = 5 12 (ii) Since, a cone and sphere are melted and recasted into a cylinder Volume of cone + Volume of sphere = Volume of cylinder 1 2 4 3 r h r = pR2H 3 3 4 2 4 (3) (12) + (3)3 = (2)2H 3 3 36 + 36 = 4H 72 = 4H H = 18 Height of cylinder = 18 cm OSWAAL ICSE Sample Question Papers, mathematics, Class-X 54 (iii) x x 5 3 2 3 2 x 3x 5 6 3 5x 5 6 3 And x 2 x 2 7 x < 2 6 x< 7 3 (a) Since AEP = 70 BEP = 70 BCE = ECB = 70 (Alternate segments theorem) (b) Since, BOC = 110 So, BEC = 110 2 = 55 (Degree measure theorem) (c) Here, reflex (ii) Price (`) Class mark 85 90 12 87.5 10 2 24 90 95 10 92.5 5 1 10 95 100 8 97.5=A 0 0 0 BFC = 250 2 = 125 100 105 15 102.5 5 1 15 105 110 5 107.5 10 2 10 (d) Since, BCE = 70 BCD = 180 70 BCD = 110 Now, DAB + BCD = 180 (Opposite angles of a cyclic quadrilateral are supplementary) DAB = 180 110 x 7 x 7 7 7 7 4 3 3 2 3 3 6 3 x 12 7 x 7 3 2 6 x 5 x 7 3 2 6 x 5 x 7 3 3 2 6 x 5 x 3 3 2 Sfiui = 9 Sfi =50 (linearpair) DAB = 70 x x 7 7 8. (i) 4 3 2 3 6 So, fiui BOC = 360 110 = 250 (Degree measure theorem) d di=xi u = i A i h No. of days X = A + fi ui h fi 9 = 97.5 + 5 50 9 = 97.5 + 10 = 97.5 0.9 X = 96.6 So, price of petrol per litre (to nearest rupee) = ` 97 (iii) (a) Solutions 55 Here, DF || BC (Opposite sides of parallelogram are parallel) DE || BC AD AE = DB EC (Basic Proportionality Theorem) 4 AE = 5 EC (b) ADE = ABC (Corresponding angles) And AED = ACB (Corresponding angles) By AA Similarity criterion, DADE ~ DABC AE DE AD = = DB AC BC (By CPST) Now, Since AD 4 = DB 5 DB 5 = AD 4 DB 5 + 1 = +1 AD 4 5+4 DB + AD = 4 AD AB 9 = AD 4 4 AE DE AB = = = AD 9 AC BC 9. (i) Let number of days taken by Bijoy to complete the work alone = x Time taken by Amit to finish the work alone = (x 12) days 1 Work done by Bijoy in one day = x 1 And Work done by Amit in one day = x 12 Time taken when both work together = 8 days. According to question, 1 1 1 = 8 x x 12 x 12 x 1 = ( x )( x 12 ) 8 2 x 12 1 = 2 x 12 x 8 16x 96 = x2 12x x2 12x 16x + 96 = 0 x2 28x + 96 = 0 2 x 24x 4x + 96 = 0 x(x 24) 4(x 24) = 0 (x 24) (x 4) = 0 x 24 = 0 and x 4 = 0 x = 24 and x = 4 but if x = 4, then time taken by Amit will be x 12 = 4 12 = 8 days; which is not possible. Therefore, time taken by Bijoy to complete the work alone is 24 days. (ii) DE 4 = DF 9 Wages (in `) Wages (in `) (BC = DF; opposite sides of parallelogram are equal) No. of workers (f) No. of workers (c.f.) 250-300 8 Less than 300 8 300-350 15 Less than 350 23 350-400 20 Less than 400 43 400-450 30 Less than 450 73 450-500 25 Less than 500 98 500-550 15 Less than 550 113 550-600 7 Less than 600 120 DE 4 = 9 DE+15 9DE = 4DE + 60 5DE = 60 DE = 60 5 DE = 12 cm (c) Also, since DE 4 = BC 9 12 4 = BC 9 BC = 27 cm 56 OSWAAL ICSE Sample Question Papers, mathematics, Class-X Percentage = 34 100 120 = 28.33% 10. (i) Here 2 x 3 x 2 x 3 x = 3 On applying componendo and dividendo: 3 1 3 1 = 2 x 3 x 2 x 3 x 2 x 3 x 2 x 3 x 2 2 x 4 = 2 3 x 2 On squaring both sides 482.5 (a) N = 120, which is even th 120 Median = term 2 = 60th term. From the graph, 60th term is: 430 Therefore, median = 430. th 3N (b) Upper quartile = term 4 th 3 120 = term 4 = 90th term From the graph, 90th term is: 482.5 th N Now, Lower quartile = term 4 = 30th term from the graph, 30th term is: 370 Inter quartile range = Q3 Q1 = 482.5 370 = 112.5 (c) Number of workers above wage rate of ` 475 = 120 86 = 34 (2)2 = 4 = 2+x 3 x 2 2+x 3 x 12 4x = 2 + x 10 = 5x x = 10 5 x = 2 (ii) Step of construction: (a) Draw a line segment AB = 4.5 cm. (b) Taking A and B as centres draw lines AF, BC each of angle 120 and each of length 4.5 cm. Similarly, we draw other segment CD, DE and EF. Hence, we get the regular hexagon ABCDEF. (c) Draw perpendicular bisector of AB and BC which meets at point O. (d) Taking O as centre and OA as radius draw a circle passes through point A, B, C, D, E and F. For a hexagon OA = Radius Solutions 57 Now, (iii) Find CD Here AB BC 150 1.1918 = BC tan 50 = 150 1.1918 BC = 125.86 m AB tan 30 = BD BC = 0.57735 = 150 BD 150 BD = 0 . 57735 BD = 259.80 m So, distance travelled during observation = BD BC = 259.80 125.86 = 133.94 m ~ 134 m

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