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 Show n(n + 1) (n – 1) is divisible by 3 for any positive integer ‘n’.<\/p>\n Question 14.<\/strong><\/span> Question 15.<\/strong><\/span> Question 16.<\/strong><\/span> OR<\/strong><\/p>\n Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that \u2206ABC ~ \u2206PQR.<\/p>\n Question 17.<\/strong><\/span> Question 18.<\/strong><\/span> OR<\/strong><\/p>\n Prove that : \\(\\frac { 1+cosA }{ sinA } +\\frac { sinA }{ 1+cosA } =2cosecA\\)<\/p>\n Question 19.<\/strong><\/span> Question 20.<\/strong><\/span> Question 21.<\/strong><\/span> OR<\/strong><\/p>\n A square field and an equilateral triangular park have equal perimeters. Cost of tilling the field at the rate of \u20b9 5 per m\u00b2 is \u20b9 720. What is the cost of grassing the park at the rate of \u20b9 10 per m\u00b2 correct to nearest rupee ?<\/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 State and prove the Basic Proportionality Theorem.<\/p>\n Question 26.<\/strong><\/span> Question 27.<\/strong><\/span> OR<\/strong><\/p>\n From the top of a building 100 m high, the angles of depression of the top and bottom of a tree are observed to be 45\u00b0 and 60\u00b0 respectively. Find the height of the tree.<\/p>\n Question 28.<\/strong><\/span> Question 29.<\/strong><\/span> OR<\/strong><\/p>\n A solid consisting of a right circular cone of height 120 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. How many litres of water is left in the cylinder, if the radius of cylinder is 60 cm and its height 180 cm ? (Use \u03c0 = \\(\\\\ \\frac { 22 }{ 7 } \\))<\/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> 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> Solution 14:<\/strong><\/span> Solution 15:<\/strong><\/span> Solution 16:<\/strong><\/span> OR<\/strong><\/p>\n Given, \u2206ABC and \u2206PQR in which AD and PM are medians drawn on sides BC and QR respectively. It is also given that 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> Solution 26:<\/strong><\/span> Solution 27:<\/strong><\/span> Solution 28:<\/strong><\/span>
\nMaximum Marks : 80<\/strong><\/p>\n\n
\nUse prime factorization method to find the HCF of 300 and 1000.<\/p>\n
\nFind the value of k for which the quadratic equation 2x\u00b2 + 5x + k = 0 has equal roots.<\/p>\n
\nFind the 101th term of the A.P. – 4, – 2, 0, 2, 4, …<\/p>\n
\n\u2206ABC is similar to \u2206DEF. If AB = 10 cm, DE = 12 cm, perimeter of \u2206ABC = 25 cm, find the perimeter of \u2206DEF.<\/p>\n
\nFind the distance of A(9, – 40) from origin.<\/p>\n
\nGiven that tan A = \\(\\frac { 1 }{ \\sqrt { 5 } } \\) what is the value of \\(\\frac { { cos }^{ 2 }A-2{ sin }^{ 2 }A }{ { cos }^{ 2 }A+2{ sin }^{ 2 }A } \\) ?<\/p>\n
\nUse Euclid’s algorithm to find the HCF of 612 and 408. Also find their LCM using their HCF.<\/p>\n
\nFind the 10th term from the end of the A.P. 60, 130, 200, …, 2160.<\/p>\n
\nSolve for x and y : ax + by = a\u00b2 + b\u00b2; bx + ay = 2ab.<\/p>\n
\nFind a point on the Y-axis which is equidistant from the points A( – 4, 3) and B(6, 5).<\/p>\n
\nMean and median of a slightly asymmetric data are 52 and 54.5 respectively. Find the mode of the data.<\/p>\n
\nA card is drawn from a well-shuffled deck of 52 playing cards. Find the probability of getting :
\n(i) A black king or a red queen
\n(ii) A non-face card.<\/p>\n
\nShow \u221a7 is irrational.<\/p>\n
\nOn dividing x\u00b3 – 3x\u00b2 + x + 2 by a polynomial g(x), the quotient and remainder are x – 2 and – 2x + 4, respectively. Find g(x).<\/p>\n
\nSolve graphically : 3x – 2y = 5, 2x + 3y = 12. Also find area bounded by these lines with X-axis.<\/p>\n
\nABCD is a quadrilateral. Diagonals AC and BD meet at point O. Also AO x DO = BO x CO. Show ABCD is a trapezium.<\/p>\n
\nFind the area of the quadrilateral whose vertices taken in order are P( – 5, – 3), Q( – 4, – 6), R(2, – 1) and S(1, 2).<\/p>\n
\nProve that : \\(\\frac { sinA+cosA }{ sinA-cosA } +\\frac { sinA-cosA }{ sinA+cosA } =\\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } \\)<\/p>\n
\nProve that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre of the circle.<\/p>\n
\nThe area of an equilateral triangle ABC is 17320.5 cm\u00b2. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded region. (Use \u03c0 = 3.14 and \u221a3 = 1.73205)
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1903\/30190258397_458a43601f_o.png?resize=165%2C151&ssl=1\")
<\/p>\n
\nA solid sphere made of copper has a diameter of 6 cm. It is melted and recast into small spherical balls of diameter 2 cm each. Assuming that there is no wastage in the process, find the number of small spherical balls obtained from the given sphere.<\/p>\n
\nThe following table shows the marks obtained by 100 students of Class X in a school during a particular academic session. Find the mode of this distribution.
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1919\/45128047441_701aa5060f_o.png?resize=300%2C251&ssl=1\")
<\/p>\n
\nIn a class test, the sum of the marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in maths and 4 marks less in science, the product of marks obtained in the two subjects would have been 180. Find the marks obtained in the two subjects separately.<\/p>\n
\n180 logs are stacked in the following manner : 19 logs in the bottom row, 18 in the next row, 17 in the row next to it and so on. In how many rows are the 180 logs placed and how many logs are in top row ?<\/p>\n
\nProve that in a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides.<\/p>\n
\nIf tan A = n tan B and sin A = m sin B, prove that \\({ cos }^{ 2 }A=\\frac { { m }^{ 2 }-1 }{ { n }^{ 2 }-1 } \\)<\/p>\n
\nThe angle of elevation of the top of a building at a point on the level ground is 30\u00b0. After walking a distance of 100 m towards the foot of the building along the horizontal line through the foot of the building on the same level ground, the angle of the elevation of the top of the building is 60\u00b0. Find the height of the building.<\/p>\n
\nDraw \u2206ABC with AB = 5 cm, \u2220B = 60\u00b0 and BC = 4 cm. Now construct a triangle whose sides are \\(\\\\ \\frac { 4 }{ 5 } \\) times the corresponding sides of \u2206ABC. Give the steps of construction.<\/p>\n
\nA metallic right circular cone 20 cm high and whose semi-vertical angle is 30\u00b0 is cut into two parts from the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1\/16 cm, find the length of the wire.<\/p>\n
\nThe following table gives the yield per hectare of wheat of 60 farms in a village.
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1917\/30190258127_816b772cfb_o.png?resize=311%2C216&ssl=1\")
\nDraw a more than ogive for the above data.<\/p>\n
\nGiven numbers are 300 and 1000.
\n\u2235 300 = 3 x 2\u00b2 x 5\u00b2
\n1000 = 2\u00b3 x 5\u00b3
\n\u2234 HCF (300, 1000) = 2\u00b2 x 5\u00b2 = 100. Ans.<\/p>\n
\nGiven quadratic equation is,
\n2x\u00b2 + 5x + k = 0
\nHere, A = 2, B = 5, C = k
\nEquation has equal roots, if
\nD = 0
\n=> B\u00b2 – 4AC = 0
\n=> 25 – 4 x 2 x k = 0
\n=> 8k = 25
\n=> k = \\(\\\\ \\frac { 25 }{ 8 } \\)<\/p>\n
\nGiven A.P. is – 4, – 2, 0, 2, 4,…
\nHere, a = – 4, d = 2, n = 101
\na101<\/sub> = a + 100d = – 4 + 100 x 2
\n= 196.<\/p>\n
\nGiven, AB = 10 cm, DE = 12 cm, pr. (\u2206ABC) = 25 cm
\nIt is given that \u2206ABC ~ \u2206DEF
\n=> \\(\\frac { pr.\\left( \\triangle ABC \\right) }{ pr.\\left( \\triangle DEF \\right) } =\\frac { AB }{ DE } \\)
\n=> \\(\\frac { 25 }{ pr.\\left( \\triangle DEF \\right) } =\\frac { 10 }{ 12 } \\)
\n=> pr.(\u2206DEF) = 30 cm<\/p>\n
\nGiven points are A(9, – 40) and O(0, 0).
\nUsing distance formula,
\nOA = \\(\\sqrt { { (9-0) }^{ 2 }+{ (-40-0) }^{ 2 } } \\)
\n= \\(\\sqrt { { 9 }^{ 2 }+{ (-40) }^{ 2 } } \\)
\n= \\(\\sqrt { 81+1600 } \\)
\n= \\(\\sqrt { 1681 } \\)
\n= 41 units.<\/p>\n
\nGiven
\ntan A = \\(\\frac { 1 }{ \\sqrt { 5 } } \\)
\nWe have,
\n\\(\\frac { { cos }^{ 2 }A-2{ sin }^{ 2 }A }{ { cos }^{ 2 }A+{ sin }^{ 2 }A } \\)
\nDividing Nr<\/sub> and Dr<\/sub> by cos\u00b2 A, we get
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1919\/31254763538_6e3590d414_o.png?resize=247%2C301&ssl=1\")
<\/p>\n
\nGiven numbers are 612 and 408.
\nNow, using Euclid’s division algorithm
\n612 = 408 x 1 + 204
\n408 = 204 x 2 + 0
\nHCF = 204
\nWe know that HCF x LCM = a x b
\n=> 204 x LCM = 612 x 408
\n=> LCM = \\(\\\\ \\frac { 612X408 }{ 204 } \\)
\n= 1224<\/p>\n
\nGiven AP is 60, 130, 200,…., 2160.
\nHere, l = 2160, d = 130 – 60 = 70
\n\u2234 nth term from end = l – (n – l)d
\n10th term from end = 2160 – (10 – 1) x 70
\n= 2160 – 630 = 1530. Ans.<\/p>\n
\nGiven equations are,
\nax + by = a\u00b2 + b\u00b2 …(i)
\nand bx + ay = 2ab …(ii)
\nOn multiplying equation (i) by a and equation (ii) by b, we get
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1965\/45079587562_8d5fc99669_o.png?resize=782%2C337&ssl=1\")
<\/p>\n
\nGiven points are A( – 4, 3) and B(6, 5).
\nLet the point on Y-axis be P(0, b). (Given)
\nNow, PA = PB
\n=> PA\u00b2 = PB\u00b2
\n=> (0 + 4)\u00b2 + (b – 3)\u00b2 = (0 – 6)\u00b2 + (b – 5)\u00b2
\n=> 16 + b\u00b2 – 6b + 9 = 36 + b\u00b2 – 10b + 25
\n=> b\u00b2 – 6b + 25 = b\u00b2 – 10b + 61
\n=> 4b = 36
\n=> b = 9
\nThe required point is (0, 9). Ans.<\/p>\n
\nGiven : Mean = 52, Median = 54.5.
\nWe know, Mode = 3 Median – 2 Mean
\n= 3 x 54.5 – 2 x 52
\n= 163.5 – 104
\n=> Mode = 59.5. Ans.<\/p>\n
\nTotal number of cards = 52
\nNo. of black king cards = 2
\nNo. of red queen cards = 2
\nNo. of face cards = 12
\n(i) P(black king or red queen) = P(black king) + P(red queen)
\n\\(=\\frac { 2 }{ 52 } +\\frac { 2 }{ 52 } =\\frac { 4 }{ 52 } =\\frac { 1 }{ 13 } \\)
\n(ii) P(non-face card) = 1 – P(face card)
\n\\(=1-\\frac { 12 }{ 52 } =\\frac { 40 }{ 52 } =\\frac { 10 }{ 13 } \\)<\/p>\n
\nLet \u221a7 is a rational number.
\nThen, \u221a7 can be written in the form
\n\u221a7 = \\(\\\\ \\frac { p }{ q } \\)
\nsuch that q \u2260 0
\nand p and q are co prime integers
\nOn squaring both sides,
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1916\/45128841811_f9bf3ddef7_o.png?resize=792%2C639&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1921\/45128842001_1c9eddfee7_o.png?resize=764%2C269&ssl=1\")
<\/p>\n
\nWe have,
\nDividend = x\u00b3 – 3x\u00b2 + x + 2
\nQuotient = x – 2
\nRemainder = – 2x + 4
\nWe know,
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1901\/45128842151_729d8f3629_o.png?resize=774%2C616&ssl=1\")
<\/p>\n
\nGiven equations are,
\n3x – 2y = 5…(i)
\nand 2x + 3y = 12…(ii)
\nOn solving, we get
\nl1<\/sub>\u00a0: 3x – 2y = 5
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1934\/45128842411_e9dd1026d6_o.png?resize=194%2C67&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1943\/45128842751_fcbd6802a0_o.png?resize=777%2C700&ssl=1\")
<\/p>\n
\nGiven : AO x DO = BO x CO
\nTo prove : ABCD is a trapezium.
\nProof : We have,
\nAO x DO = BO x CO
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1913\/31255267158_7516d1d603_o.png?resize=239%2C172&ssl=1\")
\n=> \\(\\\\ \\frac { AO }{ OC } \\) = \\(\\\\ \\frac { BO }{ DO } \\)
\nAlso \u22201 = \u22202
\n\u2234 By SAS similarity axiom,
\n\u2206AOB ~ \u2206COD
\n\u22203 = \u22204
\n\u22203 and \u22204 from a pair of alterate angles
\nAB || CD
\nHence, ABCD is a trapezium.
\nHence Proved.<\/p>\n
\n\\(\\frac { AB }{ PQ } =\\frac { BC }{ QR } =\\frac { AD }{ PM } \\)
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1920\/31255267658_0ec439e31b_o.png?resize=788%2C471&ssl=1\")
<\/p>\n
\nGiven, vertices of quadrilateral are P(- 5, – 3), Q(- 4, – 6), R(2, – 1) and S(1, 2).
\nConstruction : Join R.
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1960\/31255267778_9dd29315c4_o.png?resize=225%2C194&ssl=1\")
\nNow, quadrilateral PQRS is divided into two triangles, \u2206PSR and \u2206PQR.
\n\u2234ar(\\(\\Box \\) PQRS) = ar (\u2206PQR) + ar (\u2206PSR)
\n= \\(\\\\ \\frac { 1 }{ 2 } \\) |- 5( – 6 + 1) – 4( – 1 + 3) + 2( – 3 + 6) | + \\(\\\\ \\frac { 1 }{ 2 } \\) | – 5(2 + 1) + 1( – 1 + 3) + 2( – 3 – 2) |
\n= \\(\\\\ \\frac { 1 }{ 2 } \\) | 25 – 8 + 6 | + \\(\\\\ \\frac { 1 }{ 2 } \\) | – 15 + 2 – 10 |
\n= \\(\\\\ \\frac { 1 }{ 2 } \\) | 23 | + \\(\\\\ \\frac { 1 }{ 2 } \\) | – 23 |
\n= 23 sq. unit.<\/p>\n
\nGiven
\nL.H.S = \\(\\frac { sinA+cosA }{ sinA-cosA } +\\frac { sinA-cosA }{ sinA+cosA } \\)
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1951\/45128841561_0ea3d3dc98_o.png?resize=623%2C296&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1977\/45128841701_ca760b4260_o.png?resize=513%2C421&ssl=1\")
<\/p>\n
\nGiven : l and m are two parallel tangents, n is the intercept of tangents which intersect circle at point C.
\nTo prove : \u2220AOB = 90\u00b0.
\nConstruction : Join OP and OC.
\nProof : In \u2206OPA and \u2206OAC, we have
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1946\/45128841321_42a308d8b3_o.png?resize=197%2C166&ssl=1\")
\nOP = OC [Radii]
\nAP = AC
\n[Tangents to a circle from an external point are equal in length]
\nAO = AO (common)
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1974\/45128841431_328bb3a019_o.png?resize=791%2C401&ssl=1\")
<\/p>\n
\nGiven ar(\u2206ABC) = 17320.5 cm\u00b2
\nlet radius of circle be r
\nThen,AB = BC = CA = 2r
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1963\/30191057017_0a63f6d5ea_o.png?resize=775%2C530&ssl=1\")
<\/p>\n
\nGiven,
\nDiameter of sphere of copper = 6 cm
\nRadius, R = \\(\\\\ \\frac { 1 }{ 2 } \\) = 3 cm
\nDiameter of small spherical balls = 2 cm
\nRadius, r = \\(\\\\ \\frac { 2 }{ 2 } \\) = 1 cm
\nLet number of small spheres be n.
\nThen,
\nn x Volume of small sphere = Volume of large sphere
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1950\/30191056457_6e2660ea21_o.png?resize=775%2C418&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1957\/30191056787_68de5344f9_o.png?resize=775%2C255&ssl=1\")
<\/p>\n
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1916\/30191056177_83848b0e7c_o.png?resize=571%2C518&ssl=1\")
<\/p>\n
\nLet marks obtained in mathematics be x and marks obtained in science be y.
\nAccording to the question,
\nx + y = 28
\n=> y = 28 – x…(i)
\nAlso (x + 3).(y – 4) = 180 ,
\n=> (x + 3) (28 – x – 4) = 180
\n=> (x + 3) (24 – x) = 180
\n=> 24x – x\u00b2 + 72 – 3x = 180
\n=> x\u00b2 – 21x + 108 = 0
\n=> x\u00b2 – 12x – 9x + 108 = 0
\n=> x(x – 12) – 9(x – 12) = 0
\n=> (x – 12) (x – 9) = 0
\n=> x = 12 or x = 9.
\n\u2234Marks obtained in mathematics = 12 or 9.
\nIf marks obtained in mathematics = 12 then marks obtained in science = 28 – 12 = 16
\nIf marks obtained in mathematics = 9 then marks obtained in science = 28 – 9 = 19<\/p>\n
\nLet number of rows be ‘n’.
\nNo. of logs in bottom row = 19
\nIn next row = 18 and so on Also, total logs = 180
\n=> 19 + 18 + 17 + … (n terms) = 180
\nThis is an A.P.
\nHere, a = 19, d = – 1, n = ?, Sn<\/sub> = 180
\nWe know,
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1971\/30191413327_92ab5aca96_o.png?resize=774%2C641&ssl=1\")
<\/p>\n
\nGiven : A right-angled triangle ABC in which \u2220B = 90\u00b0.
\nTo prove : (Hypotenuse)\u00b2 = (Base)\u00b2 + (Perpendicular)\u00b2
\ni.e\u201e AC\u00b2 = AB\u00b2 + BC\u00b2
\nConstruction : From B, draw BD \u22a5 AC.
\nProof : In triangles ADB and ABC, we have
\n\u2220ADB = \u2220ABC
\nand, \u2220A = \u2220A
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1959\/43315962790_b2a4b1cb7e_o.png?resize=790%2C676&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1979\/30191413507_482b396800_o.png?resize=790%2C519&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1917\/43315962910_906afe9556_o.png?resize=771%2C168&ssl=1\")
<\/p>\n
\nGiven
\nsin A = m sin B,
\ntan A = n tan B
\n=> \\(m=\\frac { sinA }{ sinB } ,n=\\frac { tanA }{ tanB } \\)
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1913\/43315963120_ef0553a760_o.png?resize=783%2C503&ssl=1\")
<\/p>\n
\nLet AB be the building, C and D be the points of observation.
\nGiven : DC = 100 m
\nLet BC = x m and AB = h m
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1963\/43315963160_afffe339e5_o.png?resize=807%2C354&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1958\/43315963250_b937603fb2_o.png?resize=801%2C732&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1929\/43315963320_c82eea3e37_o.png?resize=779%2C296&ssl=1\")
<\/p>\n
\nSteps of construction :
\n(1) Draw a line segment BC = 6 cm.
\n(2) Construct \u2220YBC = 60\u00b0.
\n(3) Taking B as centre and radius 5 cm, draw on arc, intersecting BY at A.
\n(4) Join AC. Thus, \u2206ABC is obtained.
\n(5) Draw a ray BX such that \u2220CBX is an acute angle and BX lies in exterior of \u2206ABC.
\n(6) Mark B1<\/sub> B2<\/sub>, B3<\/sub> and B4<\/sub> on BX such that BB1<\/sub> = B1<\/sub>B2<\/sub> = B2<\/sub>B3<\/sub> = B3<\/sub>B4<\/sub>.
\n(7) Join B3<\/sub> to C.
\n(8) From B4<\/sub>, draw a line parallel to B3<\/sub>C to meet BC produced at C’.
\n(9) From C’ draw a line parallel to CA to meet AB produced at A’.
\nThus, \u2206A\u2019BC’ is the required triangle whose sides are \\(\\\\ \\frac { 4 }{ 3 } \\) of corresponding sides of \u2206ABC.
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1912\/45129252771_97a5c10de8_o.png?resize=179%2C332&ssl=1\")
<\/p>\n
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