Time Allowed : 3 hours<\/strong> General Instructions<\/strong><\/span><\/p>\n SECTION-A<\/strong><\/p>\n Question 1.<\/strong><\/span> Question 2.<\/strong><\/span> Question 3.<\/strong><\/span> Question 4.<\/strong><\/span> Question 5.<\/strong><\/span> Question 6.<\/strong><\/span> SECTION-B<\/strong><\/p>\n Question 7.<\/strong><\/span> Question 8.<\/strong><\/span> Question 9.<\/strong><\/span> Question 10.<\/strong><\/span> Question 11.<\/strong><\/span> Question 12.<\/strong><\/span> SECTION-C<\/strong><\/p>\n Question 13.<\/strong><\/span> OR<\/strong><\/p>\n Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.<\/p>\n Question 14.<\/strong><\/span> Question 15.<\/strong><\/span> Question 16.<\/strong><\/span> Question 17.<\/strong><\/span> Question 18.<\/strong><\/span> OR<\/strong><\/p>\n Find the area of a rhombus if its vertices, taken in order, are (3, 0), (4, 5), (- 1, 4) and (- 2, – 1).<\/p>\n Question 19.<\/strong><\/span> OR<\/strong><\/p>\n If A, B, C are interior angles of a \u2206ABC, then show that : Question 20.<\/strong><\/span> Question 21.<\/strong><\/span> OR<\/strong><\/p>\n The radii of the ends of a frustum of a cone 40 cm high are 20 cm and 11 cm. Determine its total surface area and volume. (Use \u03c0 = 3.14)<\/p>\n Question 22.<\/strong><\/span> SECTION-D<\/strong><\/p>\n Question 23.<\/strong><\/span> Question 24.<\/strong><\/span> Question 25.<\/strong><\/span> OR<\/strong><\/p>\n Construct a triangle ABC with side BC = 7 cm, \u2220B = 45\u00b0, \u2220A= 105\u00b0. Then, construct a triangle whose sides are 3\/4 times the corresponding sides of \u2206ABC.<\/p>\n Question 26.<\/strong><\/span> Question 27.<\/strong><\/span> Question 28.<\/strong><\/span> OR<\/strong><\/p>\n If tan \u03b8 + sin \u03b8 = m and tan \u03b8 – sin \u03b8 = n, show that (m\u00b2 – n\u00b2) = \u221amn.<\/p>\n Question 29.<\/strong><\/span> OR<\/strong><\/p>\n The angle of elevation of the top of a tower from certain point is 30\u00b0. If the observer moves 10 metres towards the tower, the angle of elevation doubles. Find the height of the tower.<\/p>\n Question 30.<\/strong><\/span> SOLUTIONS<\/strong><\/span><\/p>\n SECTION-A<\/strong><\/p>\n Solution 1:<\/strong><\/span> Solution 2:<\/strong><\/span> Solution 3:<\/strong><\/span> Solution 4:<\/strong><\/span> Solution 5:<\/strong><\/span> Solution 6:<\/strong><\/span> SECTION-B<\/strong><\/p>\n Solution 7:<\/strong><\/span> Case II:<\/strong> When r = 1 Case III:<\/strong> When r = 2 Case IV:<\/strong> Solution 8:<\/strong><\/span> Solution 9:<\/strong><\/span> Solution 10:<\/strong><\/span> Solution 11:<\/strong><\/span> Solution 12:<\/strong><\/span> SECTION-C<\/strong><\/p>\n Solution 13:<\/strong><\/span> OR<\/strong><\/p>\n Let the number be a = 3q + r where r = 0,1, 2. Case II:<\/strong> When r = 1 Case III:<\/strong> When r = 2 Solution 14:<\/strong><\/span> Solution 15:<\/strong><\/span> Solution 16:<\/strong><\/span> Solution 17:<\/strong><\/span> Solution 18:<\/strong><\/span> Solution 19:<\/strong><\/span> Solution 20:<\/strong><\/span> Solution 21:<\/strong><\/span> Solution 22:<\/strong><\/span> SECTION-D<\/strong><\/p>\n Solution 23:<\/strong><\/span> Solution 24:<\/strong><\/span> Solution 25:<\/strong><\/span> OR<\/strong><\/p>\n Given : BC = 7 cm
\nMaximum Marks : 80<\/strong><\/p>\n\n
\nFind the smallest number which is divisible by 85 and 119.<\/p>\n
\nWrite a quadratic polynomial, the sum and product of whose zeroes are 3 and – 2 respectively<\/p>\n
\nIs x = – 4 a solution of the equation 2x\u00b2 + 5x – 12 = 0 ?<\/p>\n
\nIn the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD.
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1979\/44217830115_f9004d5cbe_o.png?resize=180%2C113&ssl=1\")
<\/p>\n
\nFind the distance between (a, b) and (- a, – b).<\/p>\n
\nIf tan \u03b8 = cot (30\u00b0 + \u03b8), find the value of \u03b8.<\/p>\n
\nShow that any positive odd integer is of the form 4m + 1 or 4m + 3, where m is some integer.<\/p>\n
\nWhat is the quotient and the remainder, when x\u00b2 + 3x +1 divides 3x4<\/sup> + 5x3<\/sup> – 7x\u00b2 + 2x + 2 ?<\/p>\n
\nFind the 10th term from the end of the A.P. 8, 10, 12, …, 126.<\/p>\n
\nIn what ratio does the point P(2, – 5) divide the line segment joining A(- 3, 5) and B(4, – 9) ?<\/p>\n
\nA bag contains 8 green balls, 5 red balls and 7 white balls. A ball is drawn at random from the bag. Find the probability of getting :
\n(i) neither a green ball nor a red ball.
\n(ii) a white ball or a green ball.<\/p>\n
\nFor the following distribution :
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1914\/43316399770_a91123e99c_o.png?resize=268%2C169&ssl=1\")
\nFind the sum of class marks of the median class and modal class.<\/p>\n
\nOn a morning walk, three boys step off together and their steps measure 45 cm, 40 cm and 42 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps ?<\/p>\n
\nFind the value of k for which 2x\u00b2 – (k – 2)x + 1 = 0 has equal roots.<\/p>\n
\nFor which values of a and b, will the following pair of linear equations have infinitely many solutions ?
\nx + 2y = 1; (a – b)x + (a + b)y = a + b – 2<\/p>\n
\nA circle is inscribed in a quadrilateral ABCD in which \u2220B = 90\u00b0. If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius (r) of the circle.
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1968\/44408957264_ffeecef87c_o.png?resize=122%2C133&ssl=1\")
<\/p>\n
\nThe diagonals of a trapezium ABCD, with AB || DC, intersect each other at the point O. If AB = 2 CD, find the ratio of the area of \u2206AOB to the area of \u2206COD.<\/p>\n
\nIf A(4, – 8), B(3, 6) and C(5, – 4) are the vertices of a \u2206ABC, D is the mid-point of BC and P is a point on AD joined such that \\(\\\\ \\frac { AP }{ PD } \\) = 2, find the coordinates of P.<\/p>\n
\nWithout using the trigonometric tables, evaluate the following :
\n\\(\\frac { 11 }{ 7 } \\frac { \\sin { { 70 }^{ O } } }{ cos{ 20 }^{ O } } -\\frac { 4 }{ 7 } \\frac { cos{ 53 }^{ O }cosec{ 37 }^{ O } }{ { tan15 }^{ O }tan{ 35 }^{ O }tan{ 55 }^{ O }tan{ 75 }^{ O } } \\)<\/p>\n
\n\\(cos\\left( \\frac { B+C }{ 2 } \\right) =sin\\frac { A }{ 2 } \\)<\/p>\n
\nABCD is a rectangle such that AB = 10 cm and AD = 7 cm. A semi-circle is drawn with AD as diameter such that the rectangle and semi-circle don’t overlap. Find the perimeter of this figure. (Use \u03c0 = 22\/7)<\/p>\n
\nA cone, a cylinder and a hemisphere are of equal base and have the same height. What is the ratio of their volumes ?<\/p>\n
\nTwo different dice are rolled together. Find the probability of getting :
\n(i) the sum of numbers on two dice to be 5.
\n(ii) even numbers on both dice.<\/p>\n
\nDetermine, graphically, the vertices of the triangle formed by the lines y = x, 3y = x and x + y = 8.<\/p>\n
\nIf the pth term of an A.R is 1\/q and qth term is 1\/p, prove that the sum of first pq terms of the A.P is \\(\\left( \\frac { pq+1 }{ 2 } \\right) \\)<\/p>\n
\nLet ABC be a right triangle in which AB = 6 cm, BC = 8 cm and \u2220B = 90\u00b0. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.<\/p>\n
\nState and prove the Pythagoras theorem.<\/p>\n
\nDraw “more than ogive” for the following distribution :
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1909\/43316399580_de0ccfb83c_o.png?resize=268%2C245&ssl=1\")
<\/p>\n
\nIf tan \u03b8 + sec \u03b8 = l, then prove that:
\n\\(sec\\theta =\\frac { { l }^{ 2 }+1 }{ 2l } \\)<\/p>\n
\nFrom a balloon vertically above a straight road, the angles of depression of two cars at an instant are found to be 45\u00b0 and 60\u00b0. If the cars are 100 m apart, find the height of the balloon.<\/p>\n
\nThe radii of the circular ends of a bucket of height 15 cm are 14 cm and r cm (r < 14 cm). If the volume of bucket is 5390 cm\u00b3, then find the value of r.<\/p>\n
\nSmallest number which is divisible by 85 and 119 = L.C.M. of numbers 85 and 119
\n\u223485 = 17 x 5
\n119 = 17 x 7
\n\u2234 LCM (85, 119) = 17 x 5 x 7 = 595. Ans.<\/p>\n
\nGiven, Sum of zeroes = 3
\nProduct of zeroes = – 2
\nWe know, quadratic polynomial is given as,
\n= k[x\u00b2 – (Sum of zeroes)x + Product of zeroes]
\n= k[x\u00b2 – 3x – 2] where k \u2260 0.<\/p>\n
\nLet p(x) = 2x\u00b2 + 5x – 12 = 0
\nFor x = – 4
\np(x) = 2( – 4)\u00b2 + 5( – 4) – 12
\n= 32 – 20 – 12 = 0
\np(x = – 4) =0
\nx = – 4 is the solution of equation 2x\u00b2 + 5x – 12 = 0.<\/p>\n
\nGiven : PA = 12 cm
\nand QC = QD = 3 cm
\nWe know, tangents drawn from an external point to a circle are equal.
\n\u2234PA = PB
\nQC =CA
\nand DQ = DB
\nNow,
\nPA + PB = 2 x 12
\nPC + AC + BD + PD = 24 [\u2235AC=QC,BD=QD]
\nPC + QC + QD + PD = 24
\nPC + 3 + 3 + PD = 24
\nPC + PD =24 – 6
\nPC + PD = 18 cm.<\/p>\n
\nGiven points are A(a, b) and B(- a, – b)
\nBy distance formula,
\nAB = \\(\\left| \\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } \\right| \\)
\n= \\(\\left| \\sqrt { { (-a-a) }^{ 2 }+{ (-b-b) }^{ 2 } } \\right| \\)
\n= \\(\\left| \\sqrt { { 4a }^{ 2 }+{ 4b }^{ 2 } } \\right| \\)
\nAB = \\(2\\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \\) units<\/p>\n
\nGiven
\ntan \u03b8 = cot (30\u00b0 + \u03b8)
\ncot (90\u00b0 – \u03b8) = cot (30\u00b0 + \u03b8)
\n90\u00b0 – \u03b8 = 30\u00b0 + \u03b8
\n2\u03b8 = 90\u00b0 – 30\u00b0
\n\u03b8 = \\(\\\\ \\frac { 60 }{ 2 } \\) = 30\u00b0
\n\u03b8 = 30\u00b0.<\/p>\n
\nLet the number be a = 4q + r where r = 0,1, 2, 3.
\nCase I:<\/strong> When r = 0
\na = 4q
\n=> a = 2(2(q)
\nIt is an even number.<\/p>\n
\na =4q + 1
\nIt is an odd number.<\/p>\n
\na =4q + 2
\n=> a = 2(2q + 1)
\nIt is an even number.<\/p>\n
\na = 4q + 3
\n=> a = 2(2q + 1) + 1
\nIt is an odd number.
\nHence, any positive odd number is the form of (4m + 1) or (4m + 3).<\/p>\n
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1906\/45131386811_27eaa06973_o.png?resize=504%2C285&ssl=1\")
<\/p>\n
\nGiven A.P. is 8, 10, 12, …, 126.
\nHere l = 126, d = 10 – 8 = 12 – 10 = 2
\n10th term from the end = l + (n – 1) (- d)
\n= 126 + (10 – 1) (- 2)
\n= 126 – 18 = 108
\nHence 10th term from the end = 108.<\/p>\n
\nWe have A( – 3, 5), B(4, – 9) and P(2, – 5).
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1969\/45082413462_d7f8d59566_o.png?resize=195%2C87&ssl=1\")
\nBy section formula,
\n\\({ p }_{ x }=\\frac { { m }_{ 1 }{ x }_{ 2 }+{ m }_{ 2 }{ x }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \\)
\n\\(2=\\frac { { m }_{ 1 }(4)+{ m }_{ 2 }(-3) }{ { m }_{ 1 }+{ m }_{ 2 } } \\)
\n2m1<\/sub> + 2m2<\/sub> = 4m1<\/sub> – 3m2<\/sub>
\n2m2<\/sub> + 3m2<\/sub> = 4m1<\/sub> – 2m1<\/sub>
\n5m2<\/sub> = 2m1<\/sub>
\n\\(\\frac { { m }_{ 1 } }{ { m }_{ 2 } } =\\frac { 5 }{ 2 } \\)
\n\u2234 m1<\/sub> : m2<\/sub> = 5 : 2
\nSo, point P divides line segment AB in the ratio 5 : 2.<\/p>\n
\nWe have, Green balls = 8, Red balls = 5, White balls = 7
\nTotal balls in the bag = 8 + 5 + 7 = 20
\n(i) Number of balls which are neither a green ball nor a red ball
\n= No. of white balls = 7
\nProbability of neither a green ball nor a red ball = \\(\\frac { No.\\quad of\\quad white\\quad balls }{ total\\quad balls } =\\frac { 7 }{ 20 } \\)
\n(ii) A white ball or a green ball = 7 + 8 = 15
\nProbability of a white ball or a green ball nor a red ball = \\(\\frac { white+green\\quad balls }{ total\\quad balls } =\\frac { 15 }{ 20 } =\\frac { 3 }{ 4 } \\)<\/p>\n
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1959\/45131386601_d686321124_o.png?resize=596%2C169&ssl=1\")
\nMedian = \\(\\\\ \\frac { N }{ 2 } \\) th term = \\(\\\\ \\frac { 76 }{ 2 } \\) = 38th term
\nFrequency just above 38 is 67 corresponding to class 15 – 20.
\nMedian class = 15 – 20
\nClass mark = \\(\\\\ \\frac { 15+20 }{ 2 } \\) = 17.5
\n30 is the highest frequency corresponding to class 15 – 20.
\n\u2234 Modal class = 15 – 20
\n\u2234 Class mark = \\(\\\\ \\frac { 15+20 }{ 2 } \\) = 17.5
\nSum of class marks of median class and modal class = 17.5 + 17.5 = 35.<\/p>\n
\nGiven, three boys step off together and their steps measure 45 cm, 40 cm and 42 cm.
\nMinimum distance each should walk in complete steps = L.C.M. of the numbers 45, 40 and 42
\n\u2235 45 = 3 x 3 x 5
\n40 = 2 x 2 x 2 x 5
\n42 = 2 x 3 x 7
\nL.C.M. = 2 x 2 x 2 x 3 x 3 x 5 x 7 = 2520
\nHence, minimum distance each should walk = 2520 cm = 25.20 m. Ans.<\/p>\n
\nCase I:<\/strong> When r = 0
\na = 3q
\na\u00b2 – 9q\u00b2 = 3(3q\u00b2)
\na\u00b2 = 3m, where 3q\u00b2 = m<\/p>\n
\na =3q + 1
\na\u00b2 = (3q + 1)\u00b2 = 9q\u00b2 + 6q + 1
\n= 3(3q\u00b2 + 2q) + 1
\na\u00b2 = 3m + 1, where m = 3q\u00b2 + 2q<\/p>\n
\na = 3q + 2
\na\u00b2 = (3q + 2)\u00b2
\na\u00b2 = 9q\u00b2 + 12q + 4
\na\u00b2 = 9q\u00b2 + 12q + 3 + 1
\na\u00b2 = 3(3q\u00b2 + 4q + 1) + 1
\na\u00b2 = 3m + 1, where m – 3q\u00b2 + 4q + 1
\nHence, square of any integer is of the form 3m or 3m + 1. Hence Proved.<\/p>\n
\nGiven equation is,
\n2x\u00b2 – (k – 2)x + 1 = 0
\nOn comparing the equation with ax\u00b2 + bx + c = 0, we get
\na = 2, b = – (k – 2) and c = 1
\nFor equal roots
\nb\u00b2 – 4ac = 0
\n{- (k – 2)}\u00b2 – 4 x 2 x 1 = 0
\n(k- 2)\u00b2 – 8 = 0
\n(k – 2)\u00b2 = 8
\nTaking square root both sides
\nk – 2 = \u00b1 \u221a8
\nk = 2 \u00b1 2\u221a2<\/p>\n
\nGiven equations are
\nx + 2y – 1 =0
\nand (a – b)x + (a + b)y – (a + b – 2) = 0
\nOn comparing the given equations with a1<\/sub>x + b1<\/sub>y + c1<\/sub> = 0 and a2<\/sub>x + b2<\/sub>y + c2<\/sub> = 0, respectively, we get
\na1<\/sub> = 1, b1<\/sub> = 2 and c1<\/sub> = – 1
\na2<\/sub> = a – b, b2<\/sub> = a + b and c2<\/sub> = – (a + b – 2)
\nFor infinitely many solutions
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1935\/45082778892_620daa4254_o.png?resize=786%2C441&ssl=1\")
<\/p>\n
\nIn the figure,
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1945\/45082778672_0b6b146761_o.png?resize=199%2C140&ssl=1\")
\nOP = OQ = r cm
\nBP = BQ
\n[Tangents from point B]
\n\u2220B = \u2220P = \u2220Q = 90\u00b0
\n\u2234 OQBP is a square
\nOP = BQ = BP = OP = r cm
\nnow,DR = DS = 5 cm [tangent from D]
\n\u2234 AR = AD – DR = 23 – 5 = 18 cm
\n=> AR = AQ = 18 cm [tangent from A]
\nnow,AB = AQ + BQ
\n=> 29 = 18 + r
\n\u2234 r = 29 – 18 = 11 cm
\nHence radius of circle = 11 cm<\/p>\n
\nGiven : ABCD is a trapezium in which AB || CD and AB = 2CD.
\nIn trapezium ABCD,
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1903\/45082778612_ef027c2a9b_o.png?resize=188%2C158&ssl=1\")
\nAB || CD
\n\u2220DCO = \u2220OAB
\n(Alternative angles)
\n\u2220CDO = \u2220OBA
\n(Alternative angles)
\n\u2206AOB ~ \u2206COD
\n\\(\\frac { AO }{ CO } =\\frac { BO }{ OD } =\\frac { AB }{ CD } \\)
\n\\(\\frac { ar(\\triangle AOB) }{ ar(\\triangle COD) } =\\frac { { AB }^{ 2 } }{ { CD }^{ 2 } } =\\frac { { (2CD) }^{ 2 } }{ { CD }^{ 2 } } =\\frac { { 4CD }^{ 2 } }{ { CD }^{ 2 } } =\\frac { 4 }{ 1 } \\)
\n\u2234 ar (\u2206AOB) : ar (\u2206COD) = 4 : 1.<\/p>\n
\nGiven : In \u2206ABC,
\n\\(\\frac { AP }{ PD } =2\\)
\nAP : PD = 2 : 1
\nLet the coordinates of P be (x, y).
\nSince, D is the mid-point of side BC of \u2206ABC,
\nTherefore, AD is the median and point P is the centroid of \u2206ABC.
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1936\/31258256418_89609ab091_o.png?resize=767%2C377&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1942\/45082778532_f4697b4b7d_o.png?resize=766%2C303&ssl=1\")
<\/p>\n
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1952\/31258256028_2433121b3a_o.png?resize=773%2C636&ssl=1\")
<\/p>\n
\nGiven : ABCD is a rectangle and AED is semi-circle of radius r , drawn on side AD
\nClearly, \\(r=\\frac { AD }{ 2 } =\\frac { 7 }{ 2 } cm\\)
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1957\/31258255748_a079094698_o.png?resize=200%2C113&ssl=1\")
\nPerimeter of the figure = AB + BC + CD + arc DEA
\n= 10 + 7+ 10 + \u03c0r
\n[\u2235 AB = CD, AD = BC]
\n= \\(27+\\frac { 22 }{ 7 } \\times \\frac { 7 }{ 2 } \\)
\n= 27 + 11
\n= 38 cm.<\/p>\n
\nLet the radius of hemisphere be r unit
\nRadius and height of the cylinder = r unit respectively,
\nand Radius and height of the cone = r unit respectively.
\nVolume of the cone : Volume of the cylinder : Volume of hemisphere
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1930\/31258255678_79c3c7e88f_o.png?resize=775%2C747&ssl=1\")
<\/p>\n
\nWhen two dice are rolled together, possible outcomes are
\n(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
\n(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
\n(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
\n(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
\n(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
\n(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
\nn(S) = 36
\n(i) Sample space for sum of numbers on two dice is 5 = (1, 4), (2, 3), (3, 2), (4, 1).
\n\u2234n(E) = 4
\nP(E) = \\(\\frac { n(E) }{ n(S) } =\\frac { 4 }{ 36 } =\\frac { 1 }{ 9 } \\)
\n(ii) Sample space for even number on both dice = (2, 2), (4, 4), (6, 6), (2, 4), (2, 6), (4, 2), (4, 6), (6, 2), (6, 4)
\nn(E) = 9
\nP(E) = \\(\\frac { n(E) }{ n(S) } =\\frac { 9 }{ 36 } =\\frac { 1 }{ 4 } \\)<\/p>\n
\nGiven equations are,
\ny = x…(i)
\n3y = x…(ii)
\nand x + y = 8…(iii)
\nOn solving, we get
\ny = x
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1945\/30194506157_546e7ebf66_o.png?resize=233%2C221&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1913\/30194506237_0d7744c08f_o.png?resize=672%2C604&ssl=1\")
<\/p>\n
\nLet the first term of A.P. be a and common difference be d.
\nNow, ap<\/sub> = \\(\\\\ \\frac { 1 }{ q } \\) (given)
\n=> a+(p-1)d = \\(\\\\ \\frac { 1 }{ q } \\)…(i)
\nAlso aq<\/sub> = \\(\\\\ \\frac { 1 }{ p } \\)
\n=> a + (q – 1)d = \\(\\\\ \\frac { 1 }{ p } \\)…(ii)
\nOn subtracting equation (i) from equation (ii), we get
\n(p – 1)d – (q – 1)d = \\(\\frac { 1 }{ q } -\\frac { 1 }{ p } \\)
\n(p – 1 – q + 1)d = \\(\\\\ \\frac { p-q }{ pq } \\)
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1922\/30194506317_01453a9553_o.png?resize=537%2C734&ssl=1\")
<\/p>\n
\nSteps of construction :
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1949\/30194506417_49b4e1483e_o.png?resize=230%2C285&ssl=1\")
\n(1) Draw BC = 8 cm and make a right angle at point B such that \u2220XBC = 90\u00b0.
\n(2) Cut BA = 6 cm on BX and join AC.
\n(3) With B as centre and any radius, draw arcs intersecting AC at S and Q.
\n(4) With S and Q as centre and same radius, draw two arcs which are intersecting at the point R.
\n(5) Join BR which intersect AC at the point D. Hence BD \u22a5 AC.
\n(6) Draw a perpendicular bisector of side BC which intersect side BC at point O.
\n(7) Taking O as centre and radius equal to side BO = OC, draw a circle which passes through the point B, D and C.
\n(8) Taking A as centre and radius 6 cm (v AB = 6 cm), draw an arc intersecting the circle at the point T.
\n(9) Join AT and extend it to the point P.
\n\u2234 AP is also tangent from the point A.
\nSo, AP and AB are the required tangents.<\/p>\n