{"id":13441,"date":"2024-02-05T11:41:49","date_gmt":"2024-02-05T06:11:49","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=13441"},"modified":"2024-02-06T17:23:41","modified_gmt":"2024-02-06T11:53:41","slug":"tamil-nadu-10th-maths-model-question-paper-4-english-medium","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/tamil-nadu-10th-maths-model-question-paper-4-english-medium\/","title":{"rendered":"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium"},"content":{"rendered":"<p>Students can Download Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium Pdf, <a href=\"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-10th-maths-model-question-papers\/\">Samacheer Kalvi 10th Maths Model Question Papers<\/a> helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.<\/p>\n<h2>Tamilnadu Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium<\/h2>\n<p>Instructions<\/p>\n<ul>\n<li>The question paper comprises of <span style=\"color: #0000ff;\">four<\/span> parts.<\/li>\n<li>You are to attempt all the parts. An internal choice of questions is provided wherever applicable.<\/li>\n<li>All questions of Part I, II, III and IV are to be attempted separately.<\/li>\n<li>Question numbers <span style=\"color: #0000ff;\">1<\/span> to <span style=\"color: #0000ff;\">14<\/span> in <span style=\"color: #0000ff;\">Part I<\/span> are Multiple Choice Qu\u00e8stions of <span style=\"color: #0000ff;\">one-mark<\/span> each. These are to be answered by choosing the most suitable answer from the given four alternatives and.writing the option code and the corresponding answer.<\/li>\n<li>Question numbers <span style=\"color: #0000ff;\">15<\/span> to <span style=\"color: #0000ff;\">28<\/span> in <span style=\"color: #0000ff;\">Part II<\/span> \u00e0re <span style=\"color: #0000ff;\">two-marks<\/span> questions. These are to be answered in about one or two sentences.<\/li>\n<li>Question numbers <span style=\"color: #0000ff;\">29<\/span> to <span style=\"color: #0000ff;\">42<\/span> in <span style=\"color: #0000ff;\">Part III<\/span> are <span style=\"color: #0000ff;\">five-marks<\/span> questions. These are to be answered in about three to five short sentences.<\/li>\n<li>Question numbers <span style=\"color: #0000ff;\">43<\/span> to <span style=\"color: #0000ff;\">44<\/span> in <span style=\"color: #0000ff;\">Part IV<\/span> are <span style=\"color: #0000ff;\">eight-marks<\/span> questions. These are to be answered in detail. Draw diagrams wherever necessary.<\/li>\n<\/ul>\n<p>Time: 3 Hours<br \/>\nMax Marks: 100<\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">PART &#8211; 1<\/span><\/p>\n<p><span style=\"color: #0000ff;\">I. Choose the correct answer. Answer all the questions. [14 \u00d7 1 = 14]<\/span><\/p>\n<p>Question 1.<br \/>\nIf there are 10 relations from a set A = {1, 2, 3,4, 5} to a set B, then the number of elements in B is &#8230;&#8230;&#8230;&#8230;.. .<br \/>\n(1) 3<br \/>\n(2) 2<br \/>\n(3) 4<br \/>\n(4) 8<br \/>\nAnswer:<br \/>\n(2) 2<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 2.<br \/>\nIf g = {(1,1),(2, 3),(3,5),(4,7)} is a function given by g(x) = \u03b1x + \u03b2 then the values of \u03b1 and \u03b2 are &#8230;&#8230;&#8230;&#8230;.. .<br \/>\n(1) (-1,2)<br \/>\n(2) (2,-1)<br \/>\n(3) (-1,-2)<br \/>\n(4) (1,2)<br \/>\nAnswer:<br \/>\n(2) (2,-1)<\/p>\n<p>Question 3.<br \/>\nIf 7<sup>4k<\/sup> = &#8230;&#8230;&#8230;&#8230;.. (mod 100)<br \/>\n(1) 1<br \/>\n(2) 2<br \/>\n(3) 3<br \/>\n(4) 4<br \/>\nAnswer:<br \/>\n(1) 1<\/p>\n<p>Question 4.<br \/>\nThe next term ot the sequence \\(\\frac{3}{16}, \\frac{1}{8}, \\frac{1}{12}, \\frac{1}{18}\\) is &#8230;&#8230;&#8230;&#8230;.. .<br \/>\n(1) \\(\\frac{1}{24}\\)<br \/>\n(2) \\(\\frac{1}{27}\\)<br \/>\n(3) \\(\\frac{2}{3}\\)<br \/>\n(4) \\(\\frac{1}{81}\\)<br \/>\nAnswer:<br \/>\n(2) \\(\\frac{1}{27}\\)<\/p>\n<p>Question 5.<br \/>\nIf (x &#8211; 6) is the HCF of x<sup>2<\/sup> &#8211; 2x &#8211; 24 and x<sup>2<\/sup> &#8211; kx &#8211; 6 then the value of K is &#8230;&#8230;&#8230;&#8230;.. .<br \/>\n(1) 3<br \/>\n(2) 5<br \/>\n(3) 6<br \/>\n(4) 8<br \/>\nAnswer:<br \/>\n(2) 5<\/p>\n<p>Question 6.<br \/>\nFind the matrix X if 2X + \\(\\left[ \\begin{matrix} 1 &amp; 3 \\\\ 5 &amp; 7 \\end{matrix} \\right] =\\left[ \\begin{matrix} 5 &amp; 7 \\\\ 9 &amp; 5 \\end{matrix} \\right]\\) &#8230;&#8230;&#8230;&#8230;.. .<br \/>\n(1) \\(\\left[ \\begin{matrix} -2 &amp; -2 \\\\ 2 &amp; -1 \\end{matrix} \\right] \\)<br \/>\n(2) \\(\\left[ \\begin{matrix} 2 &amp; 2 \\\\ 2 &amp; -1 \\end{matrix} \\right] \\)<br \/>\n(3) \\(\\left[ \\begin{matrix} 1 &amp; 2 \\\\ 2 &amp; 2 \\end{matrix} \\right] \\)<br \/>\n(4) \\(\\left[ \\begin{matrix} 2 &amp; 1 \\\\ 2 &amp; 2 \\end{matrix} \\right] \\)<br \/>\nAnswer:<br \/>\n(2) \\(\\left[ \\begin{matrix} 2 &amp; 2 \\\\ 2 &amp; -1 \\end{matrix} \\right] \\)<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 7.<br \/>\nThe two tangents from an external points P to a circle with centre at O are PA and PB. If \u2220APB = 70\u00b0 then the value of \u2220AOB is &#8230;&#8230;&#8230;&#8230;.. .<br \/>\n(1) 100\u00b0<br \/>\n(2) 110\u00b0<br \/>\n(3) 120\u00b0<br \/>\n(4) 130\u00b0<br \/>\nAnswer:<br \/>\n(2) 110\u00b0<\/p>\n<p>Question 8.<br \/>\nThe equation of a line passing through the origin and perpendicular to the line 7x &#8211; 3y + 4 = 0 is &#8230;&#8230;&#8230;&#8230;.. .<br \/>\n(1)7x &#8211; 3y + 4 = 0<br \/>\n(2) 3x &#8211; 7y + 4 = 0<br \/>\n(3) 3x + 7y = 0<br \/>\n(4) 7x &#8211; 3y = 0<br \/>\nAnswer:<br \/>\n(3) 3x + 7y = 0<\/p>\n<p>Question 9.<br \/>\nIf x = a tan \u03b8 and y = b sec \u03b8 then &#8230;&#8230;&#8230;&#8230;.. .<br \/>\n(1) \\(\\frac{y^{2}}{b^{2}}-\\frac{x^{2}}{a^{2}}=1\\)<br \/>\n(2) \\(\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1\\)<br \/>\n(3) \\(\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=0\\)<br \/>\n(4) \\(\\frac{y^{2}}{b^{2}}-\\frac{x^{2}}{a^{2}}=0\\)<br \/>\nAnswer:<br \/>\n(1) \\(\\frac{y^{2}}{b^{2}}-\\frac{x^{2}}{a^{2}}=1\\)<\/p>\n<p>Question 10.<br \/>\nThe ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is &#8230;&#8230;&#8230;&#8230;.. .<br \/>\n(1) 1:2:3<br \/>\n(2) 2:1:3<br \/>\n(3) 1:3:2<br \/>\n(4) 3:1:2<br \/>\nAnswer:<br \/>\n(4) 3:1:2<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 11.<br \/>\nThe probability of getting a job for a person is \\(\\frac { x }{ 3 }\\). If the probability of not getting the job is \\(\\frac { 2 }{ 3 }\\) then the value of x is &#8230;&#8230;&#8230;&#8230;.. .<br \/>\n(1) 2<br \/>\n(2) 1<br \/>\n(3) 3<br \/>\n(4) 1.5<br \/>\nAnswer:<br \/>\n(2) 1<\/p>\n<p>Question 12.<br \/>\nVariance of the first 11 natural numbers is &#8230;&#8230;&#8230;&#8230;.. .<br \/>\n(1) \u221a5<br \/>\n(2) \u221a10<br \/>\n(3) 5\u221a2<br \/>\n(4) 10<br \/>\nAnswer:<br \/>\n(4) 10<\/p>\n<p>Question 13.<br \/>\nIf \u03b1 and \u03b2 are the roots of the equation ax<sup>2<\/sup> + bx + c = 0 then (\u03b1 + \u03b2)<sup>2<\/sup> is &#8230;&#8230;&#8230;&#8230;.. .<br \/>\n(1) \\(-\\frac{b^{2}}{a^{2}}\\)<br \/>\n(2) \\(\\frac{c^{2}}{a^{2}}\\)<br \/>\n(3) \\(\\frac{b^{2}}{a^{2}}\\)<br \/>\n(4) \\(\\frac{b c}{a}\\)<br \/>\nAnswer:<br \/>\n(3) \\(\\frac{b^{2}}{a^{2}}\\)<\/p>\n<p>Question 14.<br \/>\nIf K(x) = 3x &#8211; 9 and L(x) = 7x &#8211; 10 then Lok is &#8230;&#8230;&#8230;&#8230;.. .<br \/>\n(1) 21x + 73<br \/>\n(2) -21x + 73<br \/>\n(3) 21x &#8211; 73<br \/>\n(4) 22x &#8211; 73<br \/>\nAnswer:<br \/>\n(3) 21x &#8211; 73<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">PART &#8211; II<\/span><\/p>\n<p><span style=\"color: #0000ff;\">II. Answer any ten questions. Question No. 28 is compulsory. [10 \u00d7 2 = 20]<\/span><\/p>\n<p>Question 15.<br \/>\nLet A = {1,2,3,4,&#8230;, 45} and R be the relation defined as \u201cis square of \u201d on A. Write R as a subset of A \u00d7 A. Also, find the domain and range of R.<br \/>\nAnswer:<br \/>\nA = {1,2,3,4. . . .45} .<br \/>\nThe relation is defined as &#8220;is square of&#8217;<br \/>\nR = {(1, 1) (2, 4) (3, 9) (4, 16) (5, 25) (6, 36)} .<br \/>\nDomain of R = {1, 2, 3, 4, 5, 6}<br \/>\nRange of R = {1,4,9,16,25,36}<\/p>\n<p>Question 16.<br \/>\nIf f(x) = 3x &#8211; 2, g(x) = 2x + k and if fog = gof, then find the value of k.<br \/>\nAnswer:<br \/>\nf(x) = 3x &#8211; 2, g(x) = 2x + k<br \/>\nfog(x) = f(g(x)) = f(2x + k) = 3(2x + k) &#8211; 2 = 6x + 3k &#8211; 2<br \/>\nThus, fog(x) = 6x + 3k &#8211; 2.<br \/>\ngof(x) = g(3x &#8211; 2) = 2(3x &#8211; 2) + k<br \/>\nThus, gof(x) = 6x &#8211; 4 + k.<br \/>\nGiven that fog = gof<br \/>\nTherefore, 6x + 3k &#8211; 2 = 6x &#8211; 4 + k<br \/>\n6x &#8211; 6x + 3k &#8211; k = &#8211; 4 + 2 \u21d2 k = &#8211; 1<\/p>\n<p>Question 17.<br \/>\nFind the rational form of the number \\(0 . \\overline{123}\\).<br \/>\nAnswer:<br \/>\nLetx = \\(0 . \\overline{123}\\)<br \/>\n= 0.123123123&#8230;.<br \/>\n= 0.123 + 0.000123 + 000000123 + &#8230;.<br \/>\nThis is an infinite G.P<br \/>\nHere a = 0.123, r = \\(\\frac{0.000123}{0.123}\\) = 0.001<br \/>\nS<sub>n<\/sub> = \\(\\frac{a}{1-r}=\\frac{0.123}{1-0.001}=\\frac{0.123}{0.999}=. \\frac{41}{333}\\)<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 18.<br \/>\nHow many consecutive odd integers beginning with 5 will sum to 480?<br \/>\nAnswer:<br \/>\nFirst term (a) = 5<br \/>\nCommon difference (d) = 2 (consecutive odd integer)<br \/>\nS<sub>n<\/sub> = 480<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39842\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-1.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 1\" width=\"94\" height=\"97\" \/><br \/>\n\\(\\frac { n }{ 2 }\\) [2a + (n &#8211; 1)d] = 480<br \/>\n\\(\\frac { n }{ 2 }\\) [10 + (n &#8211; 1)2] = 480<br \/>\n\\(\\frac { n }{ 2 }\\) [10 + 2n &#8211; 2] = 480<br \/>\n\\(\\frac { n }{ 2 }\\) (8 + 2 n) = 480<br \/>\nn( 4 + n) = 480<br \/>\n4n + n<sup>2<\/sup> &#8211; 480 = 0<br \/>\nn<sup>2<\/sup> + 4n &#8211; 480 = 0<br \/>\n(n + 24) (n &#8211; 20) = 0<br \/>\nn + 24 = 0 or n &#8211; 20 = 0<br \/>\nn = -24 or n = 20 [number of terms cannot be negative]<br \/>\n\u2234 Number of consecutive odd integers is 20<\/p>\n<p>Question 19.<br \/>\nSimplify \\(\\frac{5 t^{3}}{4 t-8} \\times \\frac{6 t-12}{10 t}\\)<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39841\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-2.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 2\" width=\"351\" height=\"160\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-2.png 351w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-2-300x137.png 300w\" sizes=\"(max-width: 351px) 100vw, 351px\" \/><\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 20.<br \/>\nSolve the following quadratic equations by factorization method \u221a2x<sup>2<\/sup> + 7x + 5 \u221a2 = 0<br \/>\nAnswer:<br \/>\n\u221a2x<sup>2<\/sup> + 7x + 5\u221a2 = 0<br \/>\n\u221a2x<sup>2<\/sup> + 2x + 5x + 5\u221a2 = 0<br \/>\n\u221a2x(x + \u221a2) + 5(x + \u221a2) = 0<br \/>\n(x + \u221a2) + 5(x + \u221a2) = 0<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39811\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-32.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 32\" width=\"125\" height=\"75\" \/><br \/>\n(x + \u221a2) or 5(x + \u221a2) = 0 (equate the product of factors to zero)<br \/>\nx = -\u221a2 or \u221a2x = -5 x \u21d2 x = \\(\\frac{-5}{\\sqrt{2}}\\)<br \/>\nThe roots are -\u221a2, \\(\\frac{-5}{\\sqrt{2}}\\)<\/p>\n<p>Question 21.<br \/>\nFind the value of a, b, c, d from the equation \\(\\begin{pmatrix} a-b &amp; 2a+c \\\\ 2a-b &amp; 3c+d \\end{pmatrix}=\\begin{pmatrix} 1 &amp; 5 \\\\ 0 &amp; 2 \\end{pmatrix}\\)<br \/>\nAnswer:<br \/>\nThe given matrices are equal. Thus all corresponding elements are equal.<br \/>\nTherefore, a &#8211; b = 1 &#8230;(1)<br \/>\n2a + c = 5 &#8230;(2)<br \/>\n2a &#8211; b = 0 &#8230;.(3)<br \/>\n3c + d = 2 &#8230;(4)<br \/>\n(3) gives 2a &#8211; b = 0 &#8230;(4)<br \/>\n2 a = b &#8230;(5)<br \/>\nPut 2a = b in equation (1), a &#8211; 2a = 1 gives a = -1<br \/>\nPut a = -1 in equation (5), 2(-1) = b gives b = -2 .<br \/>\nPut a = -1 in equation (2), 2(-1) + c = 5 gives c = 7<br \/>\nPut c = 7 in equation (4), 3(7) + d = 2 gives = -19<br \/>\nTherefore, a = -1, b = -2, c = 7, d = -19<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 22.<br \/>\nIn \u2206ABC, D and E are points on the sides AB and AC respectively such that DE || BC If AD = 8x -7 , DB = 5x &#8211; 3 , AE = 4x &#8211; 3 and EC = 3x &#8211; 1, find the value of x.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39840\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-3.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 3\" width=\"297\" height=\"179\" \/><br \/>\nGiven AD = 8x &#8211; 7; BD = 5x &#8211; 3; AE = 4x &#8211; 3; EC = 3x &#8211; 1<br \/>\nIn \u2206ABC we have DE || BC<br \/>\nBy Basic proportionality theorem<br \/>\n\\(\\frac{A D}{D B}=\\frac{A E}{E C}\\)<br \/>\n\\(\\frac{8 x-7}{5 x-3}=\\frac{4 x-3}{3 x-1}\\)<br \/>\n(8x &#8211; 7) (3x -1) = (4x -3) (5x -3)<br \/>\n24x<sup>2<\/sup> &#8211; 8x &#8211; 21x + 7 = 20x<sup>2<\/sup> &#8211; 12x &#8211; 15x + 9<br \/>\n24x<sup>2<\/sup> &#8211; 20x<sup>2<\/sup> &#8211; 29x + 27x + 7 &#8211; 9 = 0<br \/>\n4x<sup>2<\/sup> &#8211; 2x &#8211; 2 = 0 .<br \/>\n2x<sup>2<\/sup> &#8211; x &#8211; 1 = 0 (Divided by 2)<br \/>\n2x<sup>2<\/sup> &#8211; 2x + x &#8211; 1 = 0<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39839\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-4.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 4\" width=\"89\" height=\"74\" \/><br \/>\n2x(x &#8211; 1) + 1 (x &#8211; 1) = 0<br \/>\n(x &#8211; 1) (2x + 1) = 0<br \/>\nx &#8211; 1 = 0 or 2x + 1 = 0<br \/>\nx = 1 or 2x = -1 \u21d2 x = \\(\\frac { 1 }{ 2 }\\) (Negative value will be omitted)<br \/>\nThe value of x = 1<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 23.<br \/>\nThe hill in the form of a right triangle has its foot at (19, 3) . The inclination of the hill to the ground is 45\u00b0. Find the equation of the hill joining the foot and top.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39838\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-5.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 5\" width=\"256\" height=\"214\" \/><br \/>\nSlope of AB (m) = tan 45\u00b0 = 1<br \/>\nEquation of the hill joining the foot and the top is 45\u00b0<br \/>\ny &#8211; y<sub>1<\/sub> = m(x &#8211; x<sub>1<\/sub><br \/>\ny &#8211; 3 = 1(x &#8211; 19)<br \/>\ny &#8211; 3 = x &#8211; 19<br \/>\n&#8211; x + y &#8211; 3 + 19 = 0<br \/>\n&#8211; x + y + 16 = 0<br \/>\nx &#8211; y &#8211; 16=0<br \/>\nThe required equation is x &#8211; y &#8211; 16 = 0<\/p>\n<p>Question 24.<br \/>\nProve that \\(\\sqrt{\\frac{1+\\sin \\theta}{1-\\sin \\theta}}\\) = sec \u03b8 + tan \u03b8<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39837\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-6.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 6\" width=\"424\" height=\"301\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-6.png 424w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-6-300x213.png 300w\" sizes=\"(max-width: 424px) 100vw, 424px\" \/><\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 25.<br \/>\nIf the total surface area of a cone of radius 7cm is 704 cm<sup>2<\/sup>, then find its slant height.<br \/>\nAnswer:<br \/>\nGiven that, radius r = 7 cm<br \/>\nNow, total surface area of the cone = \u03c0r(l + r)sq. units<br \/>\nT.S.A = 704 cm<sup>2<\/sup><br \/>\n704 = \\(\\frac { 22 }{ 7 }\\) \u00d7 7(l + 7)<br \/>\n32 = l + 7 implies l = 25 cm<br \/>\nTherefore, slant height of the cone is 25 cm.<\/p>\n<p>Question 26.<br \/>\nThe first term of an A.P is 6 and the common difference is 5. Find the A.P and its general term.<br \/>\nAnswer:<br \/>\nGiven a = 6, d = 5<br \/>\nGeneral term t<sub>n<\/sub> = a + (n &#8211; 1) d<br \/>\n= 6 + (n &#8211; 1)5<br \/>\n= 6 + 5n &#8211; 5<br \/>\n= 5n + 1<br \/>\nThe general form of the A.P is a, a + d, a + 2d &#8230;&#8230;&#8230;<br \/>\nThe A.P. is 6, 11, 16, 21 &#8230;. 5n + 1<\/p>\n<p>Question 27.<br \/>\nIf \u03b8 is an acute angle and tan \u03b8 + cot \u03b8 = 2 find the value of tan<sup>7<\/sup>\u03b8 + cot<sup>7<\/sup>\u03b8<br \/>\nAnswer:<br \/>\nGiven tan \u03b8 + cot \u03b8 = 2<br \/>\ntan \u03b8 + \\(\\frac{1}{\\tan \\theta}\\) = 2<br \/>\n\\(\\frac{\\tan ^{2} \\theta+1}{\\tan \\theta}\\) = 2<br \/>\ntan<sup>2<\/sup>\u03b8 + 1 = 2 tan \u03b8<br \/>\ntan<sup>2<\/sup>\u03b8 &#8211; 2 tan \u03b8 + 1 = 0<br \/>\n(tan \u03b8 &#8211; 1)<sup>2<\/sup> = 0<br \/>\n\u2234 tan\u03b8 &#8211; 1 = 0<br \/>\ntan\u03b8 = 1<br \/>\ntan\u03b8 = tan45 \u21d2 \u03b8 = 45\u00b0<br \/>\ntan<sup>7<\/sup>\u03b8 + cot<sup>7<\/sup>\u03b8 = tan<sup>7<\/sup>45\u00b0 + cot <sup>7<\/sup>45\u00b0<br \/>\n= (1)<sup>7<\/sup> + (1)<sup>7<br \/>\n= 2<\/sup><\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 28.<br \/>\nCards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly one card is drawn from this box. Find the probability that the number on the card is a number which is a perfect square.<br \/>\nAnswer:<br \/>\nSample space = {2, 3, 4,&#8230; 101}<br \/>\nn(s) = 100<br \/>\nLet A be the event of getting perfect square numbers<br \/>\nA= {4, 9, 16, 25, 36, 49, 64, 81, 100}<br \/>\nn(A) = 9<br \/>\nP(A) = \\(\\frac{n(\\mathrm{A})}{n(\\mathrm{S})}=\\frac{9}{100}\\)<br \/>\nProbability of getting a card marked with a number which is a perfect square is \\(\\frac{9}{100}\\)<\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">PART &#8211; III<\/span><\/p>\n<p><span style=\"color: #0000ff;\">III. Answer any ten questions. Question No. 42 is compulsory. [10 \u00d7 5 = 50]<\/span><\/p>\n<p>Question 29.<br \/>\nConsider the functions f(x) = x<sup>2<\/sup>, g(x) = 2x and h(x) = x + 4 Show that (fog)oh = fo(goh)<br \/>\nAnswer:<br \/>\nf(x) = x<sup>2<\/sup> ; g(x) = 2x and h(x) = x + 4<br \/>\n(fog) x = f[g(x)]<br \/>\n= f( 2X)<br \/>\n= (2x)<sup>2<\/sup><br \/>\n= 4x<sup>2<\/sup><br \/>\n(fog) oh (x) = fog [h(x)]<br \/>\n= fog(x + 4)<br \/>\n= 4(x + 4)<sup>2<\/sup><br \/>\n= 4[x<sup>2<\/sup> + 8x + 16]<br \/>\n= 4x<sup>2<\/sup> + 32x + 64 &#8230;.(1)<br \/>\ngoh (x) = g[h(x)]<br \/>\n= g(x + 4)<br \/>\n= 2(x + 4)<br \/>\n= 2x + 8<br \/>\nfo(goh)x = fo[goh(x)}<br \/>\n= f[2x + 8}<br \/>\n= (2x + 8)<sup>2<\/sup><br \/>\n= 4x<sup>2<\/sup> + 32x + 64 &#8230;&#8230;. (2)<br \/>\nFrom (1) and (2) we get (fog) oh = fo(goh)<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 30.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39836\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-7.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 7\" width=\"680\" height=\"92\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-7.png 680w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-7-300x41.png 300w\" sizes=\"(max-width: 680px) 100vw, 680px\" \/><br \/>\n(i) f(4)<br \/>\n(ii) f(-2)<br \/>\n(iii) f(4) + 2f(1)<br \/>\n(iv) \\(\\frac{f(1)-3 f(4)}{f(-3)}\\)<br \/>\nAnswer:<br \/>\nThe function f is defined by three values in intervals I, II, III as shown by the side<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39835\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-8.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 8\" width=\"420\" height=\"72\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-8.png 420w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-8-300x51.png 300w\" sizes=\"(max-width: 420px) 100vw, 420px\" \/><br \/>\nFor a given value of x = a, find out the interval at which the point a is located, there after find f(a) using the particular value defined in that interval.<br \/>\n(i) First, we see that, x = 4 lie in the third interval.<br \/>\nTherefore, f(x) = 3x &#8211; 2 ; f(4) = 3(4) &#8211; 2 = 10<\/p>\n<p>(ii) x = -2 lies in the second interval.<br \/>\nTherefore, f(x) = x<sup>2<\/sup> &#8211; 2 ; f(-2) = (-2)<sup>2<\/sup> &#8211; 2 = 2<\/p>\n<p>(iii) From (i), f(4) = 10.<br \/>\nTo find f(1), first we see that x = 1 lies in the second interval.<br \/>\nTherefore,f(x) = x<sup>2<\/sup> &#8211; 2 =&gt; f(1) = 1<sup>2<\/sup> &#8211; 2 = -1<br \/>\nSo, f(4) + 2f(1) = 10 + 2 (-1) = 8<\/p>\n<p>(iv) We know that f(1) = -1 and f(4) = 10.<br \/>\nFor finding f(-3), we see that x = -3 , lies in the first interval.<br \/>\nTherefore, f(x) = 2x + 7; thus, f(-3) = 2(-3) + 7 = 1<br \/>\nHence, \\(\\frac{f(1)-3 f(4)}{f(-3)}=\\frac{-1-3(10)}{1}=-31\\)<\/p>\n<p>Question 31.<br \/>\nIf (m + 1)<sup>th<\/sup> term of an A.P. is twice the (n + 1)<sup>th&gt;<\/sup> term, then prove that (3m + 1)<sup>th<\/sup> term is twice the (m + n + 1)<sup>th<\/sup> term.<br \/>\nAnswer:<br \/>\nt<sub>n<\/sub> = a + (n &#8211; 1)d<br \/>\nGiven t<sub>m+1<\/sub> = 2 t<sub>n+1<\/sub><br \/>\na + (m + 1 &#8211; 1)d = 2[a + (n + 1 &#8211; 1)d]<br \/>\na + md = 2(a + nd) \u21d2 a + md = 2a + 2nd<br \/>\nmd &#8211; 2nd = a<br \/>\nd(m &#8211; 2n) = a &#8230;&#8230;. (1)<br \/>\nTo prove that t<sub>3m + 1<\/sub> = 2(t<sub>3m + n + 1<\/sub>)<br \/>\nL.H.S. = t<sub>3m + 1<\/sub><br \/>\n= a + (3m + 1 &#8211; 1 )d<br \/>\n= a + 3md<br \/>\n= d(m &#8211; 2n) + 3md (from 1)<br \/>\n= md &#8211; 2nd + 3md<br \/>\n= 4md &#8211; 2nd<br \/>\n= 2d (2m &#8211; n)<\/p>\n<p>R.H.S. = 2 (t<sub>m + n + 1<\/sub>)<br \/>\n= 2[a + (m + n + 1 &#8211; 1) d]<br \/>\n= 2 [a + (m + n)d]<br \/>\n= 2 [d (m &#8211; 2n) + md + nd)] (from 1)<br \/>\n= 2 [dm &#8211; 2nd + md + nd]<br \/>\n= 2 [2md &#8211; nd] = 2d (2m &#8211; n)<br \/>\nR.H.S = L.H.S<br \/>\n\u2234 t<sub>(3m + 1)<\/sub>= 2t<sub>(m + n + 1)<\/sub><\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 32.<br \/>\nFind the sum to n terms of the series 5 + 55 + 555 + &#8230;.<br \/>\nAnswer:<br \/>\nThe series is neither Arithmetic nor Geometric series. So it can be split into two series and then find the sum.<br \/>\n5 + 55 + 555 + &#8230;. + n terms = 5[1 + 11 + 111 + &#8230;. + n terms]<br \/>\n= \\(\\frac { 5 }{ 9 }\\)[9 + 99 + 999 + &#8230;. + \u00ab terms]<br \/>\n= \\(\\frac { 5 }{ 9 }\\)[(10 &#8211; 1) + (100 &#8211; 1) + (1000 &#8211; 1) + &#8230;. + n terms)]<br \/>\n= \\(\\frac { 5 }{ 9 }\\)[(10 + 100 + 1000 + &#8230;. +n terms) -n]<br \/>\n= \\(\\frac{5}{9}\\left[\\frac{10\\left(10^{n}-1\\right)}{(10-1)}-n\\right]=\\frac{50\\left(10^{n}-1\\right)}{81}-\\frac{5 n}{9}\\)<\/p>\n<p>Question 33.<br \/>\nVani, her father and her grand father have an average age of 53. One-half of her grand father\u2019s age plus one-third of her father\u2019s age plus one fourth of Vani\u2019s age is 65. If 4 years ago Vani\u2019s grandfather was four times as old as Vani then how old are they all now?<br \/>\nAnswer:<br \/>\nLet the age of Vani be&#8221;x&#8221; years<br \/>\nVani father age = &#8220;y&#8221; years<br \/>\nVani grand father = &#8220;z&#8221; years<br \/>\nBy the given first condition.<br \/>\n\\(\\frac{x+y+z}{3}=53\\)<br \/>\nx + y + z = 159&#8230;.(1)<br \/>\nBy the given 2nd condition.<br \/>\n\\(\\frac{1}{2} z+\\frac{1}{3} y+\\frac{1}{4} x=65\\)<br \/>\nMultiply by 12<br \/>\n6z + 4y +3x = 780<br \/>\n3x + 4y + 6z = 780 &#8230;.(2)<br \/>\nBy the given 3<sup>rd<\/sup> condition<br \/>\nz &#8211; 4 = 4 (x &#8211; 4) \u21d2 z &#8211; 4 = 4x &#8211; 16<br \/>\n-4x + z = -16 + 4<br \/>\n4x &#8211; z = 12 &#8230;.(3)<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39834\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-9.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 9\" width=\"344\" height=\"303\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-9.png 344w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-9-300x264.png 300w\" sizes=\"(max-width: 344px) 100vw, 344px\" \/><br \/>\nSubstitute the value of x = 24 in (3)<br \/>\n4 (24) &#8211; z = 12<br \/>\n96 &#8211; z = 12<br \/>\n-Z = 12 &#8211; 96<br \/>\nz = 84<br \/>\nSubstitute the value of<br \/>\nx = 24 and z = 84 in (1)<br \/>\n24 + y + 84 = 159<br \/>\ny + 108 = 159<br \/>\ny = 159 &#8211; 108<br \/>\n= 51<br \/>\nVani age = 24 years<br \/>\nVani&#8217;s father age = 51 years<br \/>\nVani &#8211; grand father age = 84 years<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 34.<br \/>\nIf A = \\(\\left( \\begin{matrix} 1 &amp; 2 &amp; 1 \\\\ 2 &amp; -1 &amp; 1 \\end{matrix} \\right)\\) and B = \\(\\left( \\begin{matrix} 2 &amp; -1 \\\\ -1 &amp; 4 \\\\ 0 &amp; 2 \\end{matrix} \\right)\\) show that (AB)<sup>T<\/sup> = B<sup>T<\/sup>A<sup>T<\/sup><br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39833\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-10.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 10\" width=\"337\" height=\"303\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-10.png 337w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-10-300x270.png 300w\" sizes=\"(max-width: 337px) 100vw, 337px\" \/><br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39832\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-11.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 11\" width=\"316\" height=\"302\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-11.png 316w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-11-300x287.png 300w\" sizes=\"(max-width: 316px) 100vw, 316px\" \/><\/p>\n<p>Question 35.<br \/>\nIn figure, O is the centre of the circle with radius 5 cm. T is a point such that OT = 13 cm and OT intersects the circle E, if AB is the tangent to the circle at E, find the length of AB.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39831\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-12.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 12\" width=\"244\" height=\"144\" \/><br \/>\nIn the right \u2206 OTP,<br \/>\nPT<sup>2<\/sup> = OT<sup>2<\/sup> &#8211; OP<sup>2<\/sup><br \/>\n= 13<sup>2<\/sup> &#8211; 5<sup>2<\/sup><br \/>\n= 169 &#8211; 25 = 144<br \/>\nPT = \u221a144 = 12 cm<br \/>\nSince lengths of tangent drawn from a point to circle are equal.<br \/>\n\u2234 AP = AE = x .<br \/>\nAT = PT &#8211; AP<br \/>\n= (12 &#8211; x) cm<br \/>\nSince AB is the tangent to the circle E.<br \/>\n\u2234 OE \u22a5 AB .<br \/>\n\u2220OEA= 90\u00b0<br \/>\n\u2220AET = 90\u00b0<br \/>\nIn \u2206AET, AT<sup>2<\/sup> = AE<sup>2<\/sup> + ET<sup>2<\/sup><br \/>\nIn the right triangle AET,<br \/>\nAT<sup>2<\/sup> = AE<sup>2<\/sup> + ET<sup>2<\/sup><br \/>\n(12 &#8211; x)<sup>2<\/sup> = x<sup>2<\/sup> + (13 &#8211; 5)<sup>2<\/sup><br \/>\n144 &#8211; 24x + x<sup>2<\/sup> = x<sup>2<\/sup> + 64<br \/>\n24x = 80 \u21d2 x = \\(\\frac{80}{24}=\\frac{20}{6}=\\frac{10}{3}\\)<br \/>\nBE = \\(\\frac{10}{3}\\) cm<br \/>\nAB = AE + BE<br \/>\n= \\(\\frac{10}{3}+\\frac{10}{3}=\\frac{20}{3}\\)<br \/>\n\u2234 Length of AB = \\(\\frac{20}{3}\\) cm<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 36.<br \/>\nFind the area of the quadrilateral whose vertices are at (-9, 0), (-8, 6), (-1, -2) and (-6,-3)<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39830\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-13.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 13\" width=\"205\" height=\"174\" \/><br \/>\nLet the vertices A(-9, 0), B(-8, 6), C(-1, -2) and D(-6, -3)<br \/>\nPlot the vertices in a graph and take them in counter &#8211; clock wise order.<br \/>\nArea of the Quadrilateral DCB<br \/>\n= \\(\\frac { 1 }{ 2 }\\) [(x<sub>1<\/sub>y<sub>2<\/sub> + x<sub>2<\/sub>y<sub>3<\/sub> + x<sub>3<\/sub>y<sub>4<\/sub> + x<sub>4<\/sub>y<sub>1<\/sub>) &#8211; (x<sub>2<\/sub>y<sub>1<\/sub> + x<sub>3<\/sub>y<sub>2<\/sub> + x<sub>4<\/sub>y<sub>3<\/sub> + x<sub>1<\/sub>y<sub>4<\/sub> )]<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39829\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-14.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 14\" width=\"298\" height=\"58\" \/><br \/>\n= \\(\\frac { 1 }{ 2 }\\) [27 + 12- 6 + 0 -(0 + 3 + 16 &#8211; 54)]<br \/>\n= \\(\\frac { 1 }{ 2 }\\) [33 -(-35)]<br \/>\n= \\(\\frac { 1 }{ 2 }\\) [33 + 35] = \\(\\frac { 1 }{ 2 }\\) \u00d7 68 = 34 sq. units.<br \/>\nArea of the Quadrilateral = 34 sq. units<\/p>\n<p>Question 37.<br \/>\nA pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point \u2018A\u2019 on the ground is 60\u00b0 and the angle of depression to the point \u2018A\u2019 from the top of the tower is 45\u00b0. Find the height of the tower. (\u221a3 = 1.732)<br \/>\nAnswer:<br \/>\nLet BC be the height of the tower and CD be the height of the pole.<br \/>\nLet\u2018A\u2019be the point of observation.<br \/>\nLet BC = x and AB = y.<br \/>\nFrom the diagram,<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39827\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-16.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 16\" width=\"216\" height=\"254\" \/><br \/>\n\u2220BAD = 60\u00b0 and \u2220XCA = 45\u00b0 = \u2220BAC ,<br \/>\nIn right triangle ABC, tan 45\u00b0 = \\(\\frac{B C}{A B}\\)<br \/>\ngives 1 = \\(\\frac { x }{ y }\\) so, x = y &#8230;&#8230; (1)<br \/>\nIn right triangle ABD, tan60\u00b0 = \\(\\frac{B D}{A B}=\\frac{B C+C D}{A B}\\)<br \/>\ngives \u221a3 = \\(\\) so, \u221a3y = x + 5<br \/>\nwe get \u221a3x = x + 5 [From (1)]<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39828\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-15.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 15\" width=\"398\" height=\"57\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-15.png 398w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-15-300x43.png 300w\" sizes=\"(max-width: 398px) 100vw, 398px\" \/><br \/>\nHence, height of the tower is 6.83 m.<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 38.<br \/>\nA shuttle cock used for playing badminton has the shape of a frustum of a cone is mounted on a hemisphere. The diameters of the frustum are 5 cm and 2 cm. The height of the entire shuttle cock is 7 cm. Find its external surface area.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39826\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-17.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 17\" width=\"208\" height=\"193\" \/><br \/>\nRadius of the lower end of the frustum (r) = 1 cm<br \/>\nRadius of the upper end of the frustum (R) = 2.5 cm<br \/>\nHeight of the frustum (h) = 6 cm<br \/>\nLet &#8220;l&#8221; be the slant height of the frustum<br \/>\nl = \\(\\sqrt{h^{2}+(\\mathrm{R}-r)^{2}}\\)<br \/>\n= \\(\\sqrt{6^{2}+(2.5-1)^{2}}\\)<br \/>\n= \\(\\sqrt{36+2.25}\\) = \\(\\sqrt{38.25}\\)<br \/>\n= 6.18 cm<br \/>\nExternal surface area of shuttle cock = C.S.Aof the frustum + C.S.Aof a hemisphere<br \/>\n= \u03c0l(R + r) + 2 \u03c0r<sup>2<\/sup><br \/>\n= \u03c0 [6.18 (2.5 + 1) + 2 \u00d7 1<sup>2<\/sup>] cm<sup>2<\/sup><br \/>\n= \\(\\frac { 22 }{ 7 }\\)[6.18 \u00d7 3.5 + 2]<br \/>\n= \\(\\frac { 22 }{ 7 }\\) \u00d7 (21.63 + 2)<br \/>\n= \\(\\frac { 22 }{ 7 }\\) \u00d7 23.63 cm<sup>2<\/sup> = 74.26 cm<sup>2<\/sup><br \/>\nExternal surface area = 74.26 cm<sup>2<\/sup><\/p>\n<p>Question 39.<br \/>\nThe mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4,10,12 and 14, then find the remaining two observations.<br \/>\nAnswer:<br \/>\nLet the missing two observation be &#8216;a&#8217; and &#8216;b&#8217;<br \/>\nArithmetic mean = 8<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39825\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-18.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 18\" width=\"297\" height=\"241\" \/><br \/>\n\\(\\frac{2+4+10+12+14+a+b}{7}\\) = 8 \u21d2 \\(\\frac{42+a+b}{7}\\) = 8<br \/>\na + b + 42 = 56<br \/>\na + b = 56 &#8211; 42<br \/>\na + b = 14 &#8230;&#8230;&#8230; (1)<br \/>\nVariance = 16<br \/>\nVariance = \\(\\frac{\\Sigma x_{i}^{2}}{n}-\\left(\\frac{\\Sigma x_{i}}{n}\\right)^{2}\\)<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39824\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-19.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 19\" width=\"483\" height=\"114\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-19.png 483w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-19-300x71.png 300w\" sizes=\"(max-width: 483px) 100vw, 483px\" \/><br \/>\n560 &#8211; 460 = a<sup>2<\/sup> + b<sup>2<\/sup><br \/>\na<sup>2<\/sup> + b<sup>2 <\/sup>= 100 \u21d2 (a + b)<sup>2<\/sup> &#8211; 2ab = 100 [a<sup>2<\/sup> + b<sup>2<\/sup> = (a+b)<sup>2<\/sup> &#8211; 2 ab]<br \/>\n14<sup>2<\/sup> &#8211; 2 ab = 100 \u21d2 196 &#8211; 2 ab = 100 [a + b = 14(from (1)]<br \/>\n196 &#8211; 100 = 2ab<br \/>\n96 = 2ab \u21d2 ab = \\(\\frac{96}{2}\\) = 48<br \/>\n\u2234 b = \\(\\frac{48}{a}\\) &#8230;&#8230; (2)<br \/>\nSubstitute thr value of b = \\(\\frac{48}{a}\\) in (1)<br \/>\na + \\(\\frac{48}{a}\\) = 14 \u21d2 a<sup>2<\/sup> + 48 = 14a<br \/>\na<sup>2<\/sup> &#8211; 14a + 48 = 0 \u21d2 (a &#8211; 6) (a &#8211; 8) = 0<br \/>\na = 6 or 8<br \/>\nwhen a = 6<br \/>\nb = \\(\\frac{48}{a}=\\frac{48}{6}=8\\) = 8<\/p>\n<p>when a = 8<br \/>\nb = \\(\\frac{48}{a}=\\frac{48}{6}=8\\) = 6<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39823\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-20.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 20\" width=\"81\" height=\"88\" \/><br \/>\n\u2234 Missing observation is 8 and 6 (or) 6 and 8<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 40.<br \/>\nFrom a solid circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base is removed. Find (i) the volume of the remaining solid also, find (ii) the whole surface area.<br \/>\nAnswer:<br \/>\nCircular cylinder<br \/>\nRadius of the base (r) = 6 cm<br \/>\nHeight of the cylinder (h) = 10 cm<\/p>\n<p>Circular Cone<br \/>\nRadius of the base (R) = 6 cm<br \/>\nHeight of the cone (H) = 10 cm<\/p>\n<p>(i) Volume of the remaining solid = Volume of the cylinder &#8211; Volume of the cone<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39822\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-21.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 21\" width=\"115\" height=\"130\" \/><br \/>\n= \u03c0r<sup>2<\/sup>h &#8211; \\(\\frac { 1 }{ 3 }\\)\u03c0R<sup>2<\/sup>H<br \/>\n= \u03c0(r<sup>2<\/sup>h &#8211; \\(\\frac { 1 }{ 3 }\\)R<sup>2<\/sup>H)<br \/>\n= \\(\\frac { 22 }{ 7 }\\) [6 \u00d7 6 \u00d7 10 &#8211; \\(\\frac { 1 }{ 3 }\\) \u00d7 6 \u00d7 6 \u00d7 10] cm<sup>2<\/sup><br \/>\n= \\(\\frac { 22 }{ 7 }\\) [360 &#8211; 120] cm<sup>3<\/sup><br \/>\n= \\(\\frac { 22 }{ 7 }\\) \u00d7 240 cm<sup>3<\/sup><br \/>\n= \\(\\frac { 5280 }{ 7 }\\) cm<sup>3<\/sup> = 754.29 cm<sup>3<\/sup><\/p>\n<p>(ii) Slant height of a cone = \\(\\sqrt{h^{2}+r^{2}}\\)<br \/>\n= \\(\\sqrt{10^{2}+6^{2}}\\)<br \/>\n= \\(\\sqrt{100+36}\\)<br \/>\n= \\(\\sqrt{136}\\) = 11.66 cm<br \/>\nWhole surface area of the solid = curved surface area of the cylinder + curved surface area of the cone + base area<br \/>\n= 2 \u03c0rh + \u03c0RI + \u03c0r<sup>2<\/sup><br \/>\n= \u03c0[2 \u00d7 6 \u00d7 10 + 6 \u00d7 11.66 + 6 \u00d7 6]<br \/>\n= \\(\\frac { 22 }{ 7 }\\)[120 + 69.96 + 36] cm<sup>2<\/sup><br \/>\n= \\(\\frac { 22 }{ 7 }\\) \u00d7 225.96 cm<sup>2<\/sup><br \/>\n= \\(\\frac { 4971.12 }{ 7 }\\) cm<sup>2<\/sup><br \/>\n= 710.16 cm<sup>2<\/sup><br \/>\n(i) Volume of the remaining solid = 754.29 cm<sup>3<\/sup><br \/>\n(ii) whole surface area = 710.16 cm<sup>2<\/sup><\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 41.<br \/>\nIf a and p are the roots of the equation 3x<sup>2<\/sup> &#8211; 6x + 1=0 form the equation whose roots are 2\u03b1 + \u03b2 and 2\u03b2 + \u03b1<br \/>\nAnswer:<br \/>\n\u03b1 and \u03b2 are the roots of 3x<sup>2<\/sup> &#8211; 6x + 1 = 0<br \/>\n\u03b1 + \u03b2 = \\(\\frac { 6 }{ 3 }\\) = 2<br \/>\n\u03b1\u03b2 = \\(\\frac { 1 }{ 3 }\\)<br \/>\nGiven the roots are 2\u03b1 + \u03b2; 2\u03b2 + \u03b1<br \/>\nSum of the roots = 2\u03b1 + \u03b2 + 2\u03b2 + \u03b1<br \/>\n= 2(\u03b1 + \u03b2) + (\u03b1 + \u03b2)<br \/>\n= 2(2) + 2<br \/>\n= 6<br \/>\nProduct of roots = (2\u03b1 + \u03b2) (2\u03b2 + \u03b1)<br \/>\n= 4\u03b1\u03b2 + 2\u03b1<sup>2<\/sup> + 2\u03b2<sup>2<\/sup> + \u03b1\u03b2<br \/>\n= 5\u03b1\u03b2 + 2(\u03b1<sup>2<\/sup> + \u03b2<sup>2<\/sup>)<br \/>\n5\u03b1\u03b2 + 2[(\u03b1 + \u03b2)<sup>2<\/sup> &#8211; 2\u03b1\u03b2]<br \/>\n= 5 \u00d7 \\(\\frac { 1 }{ 3 }\\) + 2 (4 &#8211; 2 \u00d7 \\(\\frac { 1 }{ 3 }\\))<br \/>\n= \\(\\frac { 5 }{ 3 }\\) + \\(2\\left(\\frac{12-2}{3}\\right)\\)<br \/>\n= \\(\\frac { 5 }{ 3 }\\) + 2 \u00d7 \\(\\frac { 10 }{ 3 }\\)<br \/>\n= \\(\\frac{5}{3}+\\frac{20}{3}\\)<br \/>\n= \\(\\frac { 25 }{ 3 }\\)<br \/>\nThe quadratic polynomial is x<sup>2<\/sup> &#8211; (sum of the roots)x &#8211; product of the roots = 0<br \/>\nx<sup>2<\/sup> &#8211; 6x + \\(\\frac { 25 }{ 3 }\\) = 0<br \/>\n3x<sup>2<\/sup> &#8211; 18x + 25 = 0<\/p>\n<p>Question 42.<br \/>\nTw o dice are rolled simultaneously. Find the probability that that sum of the numbers on the faces is neither divisible by 3 nor by 4.<br \/>\nAnswer:<br \/>\nSample space = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3.1) , (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5.1) , (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}<br \/>\nn(S) = 36<br \/>\nLet A be the event of getting the sum is divisible by 3 &#8216;<br \/>\nA = { (1,2) (2,1) (1,5) (5,1) (2,4) (4,2) (3,3) (3,6) (6,3) (4,5) (5,4) (6,6)}<br \/>\nn (A) = 12<br \/>\nP(A) = \\(\\frac{n(\\mathrm{A})}{n(\\mathrm{S})}=\\frac{12}{36}\\)<br \/>\nLet B be the event of getting a sum is divisible by 4.<br \/>\nB = {(1,3) (2,2) (2,6) (3,1) (3,5) (4,4) (5,3) (6,2) (6,6)} n (B) = 9<br \/>\nn(B) = 9<br \/>\nP(B) = \\(\\frac{n(\\mathrm{B})}{n(\\mathrm{S})}=\\frac{9}{36}\\)<br \/>\nA \u2229 B = {(6,6)}<br \/>\nn(A \u2229 B) = 1<br \/>\nP(A \u2229 B) = \\(\\frac{n(\\mathrm{A} \\cap \\mathrm{B})}{n(\\mathrm{S})}=\\frac{1}{36}\\)<br \/>\nP(A \u222a B) = P(A) + P(B) &#8211; P (A \u2229 B)<br \/>\n\\(=\\frac{12}{36}+\\frac{9}{36}-\\frac{1}{36}\\)<br \/>\n\\(=\\frac{12+9-1}{36}=\\frac{20}{36}\\)<br \/>\nNeither divisible by 3 nor by 4<br \/>\nP(A&#8217; \u2229 B&#8217;) = P(A \u222a B)&#8217;<br \/>\n= 1 &#8211; p(A \u222a B) = 1 &#8211; \\(\\frac{20}{36}=\\frac{36-20}{36}\\)<br \/>\n= \\(\\frac{16}{36}=\\frac{4}{9}\\)<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">PART-IV<\/span><\/p>\n<p><span style=\"color: #0000ff;\">IV. Answer all the questions. [2 \u00d7 8 = 16]<\/span><\/p>\n<p>Question 43.<br \/>\n(a) Draw the two tangents from a point which is 10 cm away from the centre of a circle of radius 5 cm. Also, measure the lengths of the tangents.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39821\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-22.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 22\" width=\"261\" height=\"196\" \/><br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39820\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-23.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 23\" width=\"564\" height=\"470\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-23.png 564w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-23-300x250.png 300w\" sizes=\"(max-width: 564px) 100vw, 564px\" \/><br \/>\nSteps of construction:<\/p>\n<ol>\n<li>With. O as centre, draw a circle of radius 5 cm.<\/li>\n<li>Draw a line OP = 10 cm.<\/li>\n<li>Draw a perpendicular bisector of OP, which cuts OP at M.<\/li>\n<li>With M as centre and MO as radius draw a circle which cuts previous circle at A and B.<\/li>\n<li>Join AP and BP. AP and BP are the required tangents.<\/li>\n<\/ol>\n<p>Verification: In the right \u2206 OAP<br \/>\nPA<sup>2<\/sup> = OP<sup>2<\/sup> &#8211; OA<sup>2<\/sup><br \/>\n= 10<sup>2<\/sup> &#8211; 5<sup>2<\/sup> = \\(\\sqrt{100-25}\\) = \u221a75 = 8.7 cm.<br \/>\nLength of the tangent is 8.7 cm<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>[OR]<\/p>\n<p>(b) Construct a \u2206PQR which the base PQ = 4.5 cm, R = 35\u00b0 and the median from R to RG is 6 cm.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39819\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-24.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 24\" width=\"196\" height=\"209\" \/><br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39818\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-25.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 25\" width=\"424\" height=\"533\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-25.png 424w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-25-239x300.png 239w\" sizes=\"(max-width: 424px) 100vw, 424px\" \/><br \/>\nSteps of construction<\/p>\n<ol>\n<li>Draw a line segment PQ = 4.5 cm<\/li>\n<li>At P, draw PE such that \u2220QPE = 60\u00b0<\/li>\n<li>At P, draw PF such that \u2220EPF = 90\u00b0<\/li>\n<li>Draw the perpendicular bisect to PQ, which intersects PF at O and PQ at G.<\/li>\n<li>With O as centre and OP as radius draw a circle.<\/li>\n<li>From G mark arcs of radius 5.8 cm on the circle. Mark them at R and S<\/li>\n<li>Join PR and RQ. PQR is the required triangle.<\/li>\n<li><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/li>\n<\/ol>\n<p>Question 44.<br \/>\n(a) Draw the graph of y = x<sup>2<\/sup> and hence solve x<sup>2<\/sup> &#8211; 4x &#8211; 5 = 0.<br \/>\nAnswer:<br \/>\nGiven equations are y = x<sup>2<\/sup> and x<sup>2<\/sup> &#8211; 4x &#8211; 5 = 0<br \/>\n(i) Assume the values of x from &#8211; 4 to 5.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39817\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-26.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 26\" width=\"663\" height=\"66\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-26.png 663w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-26-300x30.png 300w\" sizes=\"(max-width: 663px) 100vw, 663px\" \/><br \/>\n(ii) Plot the points (-4, 16), (- 3, 9), (-2,4), (-1, 1), (0, 0), (1, 1), (2,4), (3, 9), (4, 16), (5, 25).<br \/>\n(iii) Join the points by a smooth curve.<br \/>\n(iv) Solve the given equations<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39816\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-27.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 27\" width=\"260\" height=\"117\" \/><br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39815\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-28.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 28\" width=\"509\" height=\"116\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-28.png 509w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-28-300x68.png 300w\" sizes=\"(max-width: 509px) 100vw, 509px\" \/><br \/>\n(v) The points of intersection of the line and the parabola are (-1, 1) and (5, 25).<br \/>\nThe x-coordinates of the points are -1 and 5.<br \/>\nThus solution set is {- 1, 5}.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39814\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-29.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 29\" width=\"454\" height=\"596\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-29.png 454w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-29-229x300.png 229w\" sizes=\"(max-width: 454px) 100vw, 454px\" \/><\/p>\n<p>[OR]<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium\" width=\"224\" height=\"19\" \/><\/p>\n<p>(b) Draw the graph of y &#8211; x<sup>2<\/sup> -5x &#8211; 6 and hence solve x<sup>2<\/sup> &#8211; 5x &#8211; 14 = 0<br \/>\nAnswer:<br \/>\nLet y = x<sup>2<\/sup> &#8211; 5x &#8211; 6<br \/>\n(i) Draw the graph of y = x<sup>2<\/sup> &#8211; 5x &#8211; 6 by preparing the table of values as below.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39813\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-30.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 30\" width=\"674\" height=\"167\" srcset=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-30.png 674w, https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-30-300x74.png 300w\" sizes=\"(max-width: 674px) 100vw, 674px\" \/><br \/>\n(ii) Plot the points (-3, 18), (-2, 8), (-1, 0), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10), (5, -6), (6, 0) and (7, 8).<br \/>\n(iii) Join the points by a free hand to get smooth curve.<br \/>\n(iv) To solve x<sup>2<\/sup> &#8211; 5x &#8211; 14 = 0, subtract x<sup>2<\/sup> &#8211; 5x &#8211; 14 = 0 from y = x<sup>2<\/sup> &#8211; 5x &#8211; 6.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-39812\" src=\"https:\/\/samacheerkalviguru.com\/wp-content\/uploads\/2020\/03\/Tamil-Nadu-10th-Maths-Model-Question-Paper-4-English-Medium-31.png\" alt=\"Tamil Nadu 10th Maths Model Question Paper 4 English Medium - 31\" width=\"215\" height=\"118\" \/><br \/>\nThe equation y = 8 represent a straight line draw a straight line through y = 8 intersect the curve at two places. From the two points draw perpendicular line to the X &#8211; axis it will intersect at -2 and 7.<br \/>\nThe solution is -2 and 7<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Students can Download Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium Pdf, Samacheer Kalvi 10th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium Instructions The question paper comprises &hellip;<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/13441"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=13441"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/13441\/revisions"}],"predecessor-version":[{"id":60497,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/13441\/revisions\/60497"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=13441"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=13441"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=13441"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}