{"id":6041,"date":"2019-08-28T13:53:31","date_gmt":"2019-08-28T13:53:31","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=6041"},"modified":"2021-07-02T15:45:59","modified_gmt":"2021-07-02T10:15:59","slug":"karnataka-sslc-maths-model-question-paper-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/karnataka-sslc-maths-model-question-paper-3\/","title":{"rendered":"Karnataka SSLC Maths Model Question Paper 3 With Answers"},"content":{"rendered":"

Students can Download Karnataka SSLC Maths Model Question Paper 3 with Answers, Karnataka SSLC Maths Model Question Papers with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/span><\/p>\n

Karnataka State Syllabus SSLC Maths Model Question Paper 3 with Answers<\/h2>\n

Time: 3 Hours
\nMax Marks: 80<\/p>\n

I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 \u00d7 8 = 8 )<\/span><\/p>\n

Question 1.
\nIn the following numbers, irrational number is
\na) 0.232332333……
\nb) 0.23233
\nc) 0.232323
\nd) 0.2323
\nAnswer:
\na) 0.232332333……<\/p>\n

Question 2.
\n10 sec2<\/sup>A – 10tan2<\/sup>A =
\na) sec2<\/sup>A
\nb) 10
\nc) 1
\nd) 0
\nAnswer:
\nb) 10
\nSolution:
\n10 sec2<\/sup>A – 10tan2<\/sup>A
\n10(sec2<\/sup>A – tan2<\/sup>A) = 10(1)= 10<\/p>\n

Question 3.
\nThe length of the tangent drawn to a circle of radius 3cm from 5cm away from the centre is
\na) 4cm
\nb) 5cm
\nc) 3cm
\nd) 2cm 4
\nAnswer:
\na) 4cm
\nSolution:
\nd2<\/sup> = r2<\/sup> + t2<\/sup>
\nt2<\/sup> = d2<\/sup> – r2<\/sup>
\n52<\/sup> – 32<\/sup>
\n= 25 – 9
\n\u2234 t = 16 = 4cm
\n\"Karnataka<\/p>\n

Question 4.
\nA solid piece of copper of dimension 24 \u00d7 49 \u00d7 33 cms is moulded and recast into a sphere. The radius of the
\nsphere formed is ________
\na) 49 cm
\nb) 24 cm
\nc) 21 cm
\nd)33 cm
\nAnswer:
\ne) 21 cm
\nl \u00d7 b \u00d7 h = \\(\\frac{4}{3} \\pi r^{3}\\)
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

Question 5.
\nThe degree of the polynomial in the graph given below is
\n\"Karnataka
\na) 4
\nb) 3
\nc) 1
\nd) 2
\nAnswer:
\na) 4
\nSolution:
\n4 since it is intersecting the x – axis at 4 points.<\/p>\n

Question 6.
\nThe sum of the n terms of an AP is 2n2 + 5n and its common difference is 6, then its first term is
\na) 0
\nb) 5
\nc) 2
\nd) 7
\nAnswer:
\nd) 7
\nSolution:
\nSn<\/sub> = 2n2<\/sup> + 5n
\nS1<\/sub> = 2(1)2<\/sup> + 5(1)
\n= 2 + 5 = 7<\/p>\n

\"Karnataka<\/p>\n

Question 7.
\nIn \u2206 PQR, PR = 12cm, QR = 6\u221a3 cm, PQ = 6cm. The angle Q is
\na) 45\u00b0
\nb) 90\u00b0
\nc) 30\u00b0
\nd) 120\u00b0
\nAnswer:
\nb) 90\u00b0
\nSolution:
\nPR2<\/sup> = PQ2<\/sup> + QR2<\/sup>
\n122<\/sup>= (6\u221a3 )2<\/sup> + 62<\/sup>
\n144 = 108 – 36
\n144=144
\n\u2234 \u2220Q = 90\u00b0
\n\"Karnataka<\/p>\n

Question 8.
\nA cube numbered 1 to 6 is thrown once, the probability of getting a number divisible by 3 is
\na) 2\/3
\nb) 0
\nc) 1\/3
\nd) 1
\nAnswer:
\nEvent A = {3,6}
\nn(s) = 6, n(A) =2
\nP(A) = \\(\\frac{n(A)}{n(S)}=\\frac{2}{6}=\\frac{1}{3}\\)<\/p>\n

II. Answer the following Questions : (1 x 8 = 8)<\/span><\/p>\n

Question 9.
\nGiven g(x) = 2x + 1, q(x) = (x3<\/sup> + 3x2<\/sup> – + 1), r(x) = 0, Find p(x)
\nAnswer:
\np(x) = g(x) q(x) + r(x) 1 650 1170
\n= (2x + 1) (x3<\/sup> + 3x2<\/sup> – x + 1) + 0
\np(x) = 2x (x3<\/sup> + 3x2<\/sup> -x + 1) + 1(x3<\/sup> + 3x2<\/sup> – x + 1)
\n= 2x4<\/sup> + 6x3<\/sup> – 2x2<\/sup>+ 2x + x3<\/sup> + 3x2<\/sup> – x + 1
\n= 2x4<\/sup> + 7x3<\/sup> + x2<\/sup> + x + 1<\/p>\n

Question 10.
\nIf the sum of first n odd natural number is 1225, find the value of n.
\nAnswer:
\nSum of \u201cn\u201d odd natural number = n2<\/sup> = 1225
\nn = \u221a1225
\nn = 35<\/p>\n

Question 11.
\nIn the fig. \u2220AOD is divided into 2 parts which are in A.P. the smallest angle \u2220AOB = 20\u00b0 . Find the common difference between each angle.
\nAnswer:
\n\u2220AOB +\u2220BOC + \u2220COD = 180\u00b0
\na + a + d + a + 2d= 180\u00b0
\n3a + 3d= 180\u00b0
\n3(20) + 3d = 180\u00b0
\n60 + 3d = 180\u00b0
\n3d = 180 – 60
\nd = 120\/3 = 40\u00b0
\n\"Karnataka<\/p>\n

Question 12.
\nIn \u2206 ABC, if DE || BC, then \\(\\frac{\\mathbf{A B}}{\\mathbf{A D}}=\\frac{\\mathbf{A C}}{\\mathbf{A} \\mathbf{E}}=\\frac{\\mathbf{B C}}{\\mathbf{D E}}\\) , state the theorem to justify this.
\nAnswer:
\nIn a triangle, if a line is parallel to one of the sides then the sides of given triangle are propotional to sides of the intercepted triangle.
\n\"Karnataka<\/p>\n

Question 13.
\nFind the largest number which divides 650 and 1170.
\nAnswer:
\nH.C.F of (650, 1170) = 130
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

Question 14.
\nIf sin \u03b8 = 7\/25, cos \u03b8 = 24\/25 find the value of sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8
\nAnswer:
\n\"Karnataka<\/p>\n

Question 15.
\nFind the value of cos60\u00b0 cos 30\u00b0 sin60\u00b0 sin 30\u00b0
\nAnswer:
\ncos 60\u00b0 cos30 – sin 60 sin 30
\n\"Karnataka
\n\"Karnataka<\/p>\n

OR<\/p>\n

This is in the form of
\ncos A cos B – sin A sin B = cos (A + B)
\ncos 60 cos 30 – sin 60 sin 30
\n= cos (60 + 30)
\n= cos (90)
\n– =0<\/p>\n

Question 16.
\nThe T.S.A of a solid hemisphere of radius 21 mm.
\nAnswer:
\nT.S.A of hemisphere = 3 \u03c0r\u00b2
\n= 3 \u00d7 \\(\\frac{22}{7}\\) \u00d7 21 \u00d7 21 = 4158 mm2<\/sup><\/p>\n

III. Answer the following : ( 2 x 8 = 16 )<\/span><\/p>\n

Question 17.
\nProve that if x and y are odd positive integers, then x2<\/sup> + y2<\/sup> is even but not divisible by 4.
\nAnswer:
\nLet x = 2m + 1 and y = 2n + 1 for some integers m and n.
\nx2<\/sup> + y2<\/sup> = (2m + 1)2<\/sup> + (2n + l)2
\nx2<\/sup> + y2<\/sup> = 4m2<\/sup> + 4m + 1 + 4n2<\/sup> + 1 + 4n
\n= 4m2<\/sup> + 4n2<\/sup> + 4m + 4n + 2.
\nx2<\/sup> + y2<\/sup> = 4(m2<\/sup> + n2<\/sup>) + 4 (m + n) + 2
\nx2<\/sup> + y2<\/sup> = 4 {(m2<\/sup> + n2<\/sup>) + (m + n)} +2
\nx2<\/sup> + y2<\/sup> = 4q + 2, where q = (m2<\/sup> + n2<\/sup> ) + (m+n)
\n\u21d2 x2<\/sup> + y2<\/sup> is even and leaves the remainder 2 when divided by 4.
\n\u21d2 x2<\/sup> + y2<\/sup> is even but not divisible by 4.<\/p>\n

Question 18.
\nSolve : 100x + 200y = 700
\n200x + 100y = 800
\nAnswer:
\nConsider 100x + 200y = 700 ….(1)
\n200x + 100y = 800 ….(2)
\nAdd (1) & (2) 300x + 300y= 1500
\nDivide by 300, x + y = 5 …… (3)
\nSubtract (1) and (2)
\n-100x + 100y= -100 Divide by 100.
\n-x + y = -1 …(4)
\nSolve (3) and (4)
\n\"Karnataka
\ny = 2
\nSubstitute the value of y in equation (3)
\nx + y = 5
\nx + 2 = 5
\nx = 5 – 2 = 3
\nx = 3<\/p>\n

\"Karnataka<\/p>\n

Question 19.
\nFind the roots of the quadratic
\nequation 3x2<\/sup> – 2\u221a6x + 2 = 0 by formula method.
\nAnswer:
\n\"Karnataka<\/p>\n

Question 20.
\nFind the value of x in which the points (1, -1) (x, 1) and (4, 5) are collinear.
\nAnswer:
\nA(1, -1) B(x, 1)C(4,5)
\nArea of the triangle = 0, when points are collinear.
\n0 = x1<\/sub>(y2<\/sub> – y3<\/sub>) + x2<\/sub>(y3<\/sub> – y1<\/sub>) + x3<\/sub> (y1<\/sub> – y2<\/sub>)
\n0 = (1) (1 – 5) + x(5 + 1) + 4(-1 – 1)
\n0 = -4 + 6x – 8
\n6x – 12 = 0
\nx = 12\/6
\nx = 2<\/p>\n

Question 21.
\nABC is a right angled triangle, having [B = 90\u00b0. If BD = DC, Show that AC2<\/sup> = 4AD2<\/sup> – 3AB2<\/sup>
\nAnswer:
\n\"Karnataka
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup> (Pythagoras)
\nAC2<\/sup> = AB2<\/sup> + 2(BD)2<\/sup>
\nAC2<\/sup> = AB2<\/sup> + 4BD2<\/sup>
\nIn right angled \u2206 ABD,
\nAD2<\/sup> = AB2<\/sup> + BD2<\/sup>
\nBD2<\/sup> = AD2<\/sup> – AB2<\/sup>
\nConsider AC2<\/sup> = AB2<\/sup> + 4BD2<\/sup>
\n= AB2<\/sup> + 4(AD2<\/sup> – AB2<\/sup>)
\n= AB2<\/sup> + 4AD2<\/sup> – 4AB2<\/sup>
\n= – 3AB2<\/sup> + 4AD2<\/sup>
\n= 4 AD2<\/sup> – 3AB2<\/sup>
\nAC2<\/sup> = 4 AD2<\/sup> – 3AB2<\/sup><\/p>\n

OR<\/p>\n

Prove that area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.
\nAnswer:
\n\"Karnataka
\nData : ABCD is a square.
\nEquilateral triangles \u2206 BCE and \u2206 ACF have been described on side BC and diagonal AC respectively.
\nT.P.T. : Area of ( \u2206 BCE) = 1\/2 Area of \u2206 ACF
\nProof : Since \u2206 BCE and \u2206 ACF are equilateral.
\n\u2234 They are equiangular.
\n\u2206 BCE ~ \u2206 ACF
\n\"Karnataka<\/p>\n

Question 22.
\nA box contains 90 dices which are numbered from 1 to 90. If one dise is drawn at random from the box, find the probability that it bears
\n(i) two digit number
\n(ii) a perfect square number.
\nAnswer:
\nThe numbers in the dise form the sample space
\nS = {1,2, 3,4,…….. 90}
\nOne dise can be drawn out of 90 in 90 ways. .
\nn(s) = 90<\/p>\n

i) There are 90 -9 = 81, two digit numbers, out of which one dise can be drawn in 81 ways.
\n\u2234 n(A) = 81
\nP(A) = \\(\\frac{n(A)}{n(S)}=\\frac{81}{90}=\\frac{9}{10}\\)<\/p>\n

ii) The perfect square numbers are B= {1,4,9, 16,25,36,49,64,81}
\n\u2234 n(B) = 9
\nP(B) = \\(\\frac{n(B)}{n(S)}=\\frac{9}{90}=\\frac{1}{10}\\)<\/p>\n

Question 23.
\nDraw a pair of tangents to a circle of radius 5cm which are inclined to each
\nother at an angle of 60\u00b0.
\nAnswer:
\n\"Karnataka<\/p>\n

Question 24.
\nProve that \\(\\frac{\\tan \\theta+\\sin \\theta}{\\tan \\theta-\\sin \\theta}=\\frac{\\sec \\theta+1}{\\sec \\theta-1}\\)
\nAnswer:
\n\"Karnataka
\nOR<\/p>\n

Prove that (cosec\u03b8 – cot\u03b8)2<\/sup> = \\(\\frac{1-\\cos \\theta}{1+\\cos \\theta}\\)
\nAnswer:
\nConsider (cosec\u03b8 – cot\u03b8)2<\/sup>
\n\"Karnataka
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

IV. Answer the following : ( 3 \u00d7 9 = 27 )<\/span><\/p>\n

Question 25.
\nThe sum of the ages of A and B is 85 years. 5 years ago, the age of A was twice that of B. Find the present ages.
\nAnswer:
\nLet the present age of A be x years.
\nLet the present age of B be y years.
\nSum of their ages = 85 x + y = 85 …….. (1)
\nFive years back, age of A was (x – 5) and that of B was (y – 5).
\nThe age A is twice that of B.
\n\u2234 x – 5 = 2(y – 5) ……(2)
\nx – 5 = 2y – 10
\nx – 2y = -10 + 5
\nx – 2y = -5 ……(2)
\nSolve (1) and (2)
\n\"Karnataka
\nSubstitute y in (1)
\nx + y=85
\nx + 30 = 85
\nx = 85 – 30 = 55
\nx = 55
\n\u2234 Present age of A is 55 years and B = 30 years.<\/p>\n

OR<\/p>\n

A piece of work can be done by 2 men and 7 boys in 4 days. The same piece of work can be done by 4 men and 4 boys in 3 days. How long it would take to do the same work by one man or one boy?
\nAnswer:
\nLet x and y be the number of days in which one man can complete the work.
\n\u2234 In 1 day a man can do 1\/x th work and
\nIn 1 day a man can do 1\/y th work.
\n2 men and 7 boys can complete the work in 4 days.
\n\u2234 They can complete in 1\/4 th of the work in one day.
\n\u2234 \\(\\frac{2}{x}+\\frac{7}{y}=\\frac{1}{4}\\) …… (1)
\n4 men and 4 boys together complete the work in 3days, and they can complete it in 1\/3 rd of work.
\n\"Karnataka
\n4(2y + 7x) = xy
\n8y + 28x = xy …….(3)
\n3(4y + 4x) = xy
\n12y + 12x = xy ……..(4)
\nSolve (3) and (4)
\n(8y + 2x = xy) x 12
\n(l2y+ 12x = xy) x 18
\n\"Karnataka
\n4 y = 240
\ny = 240\/4
\ny = 60<\/p>\n

3(4y + 4x) xy
\n12y+12x = xy ………..(4)
\n4y = 240
\ny = 240\/4 = 60
\nSubstitute y in equation (3)
\n8y + 28x = xy
\n8(60) + 28x = xy
\n480 + 28x = x(60)
\n480 + 28x = 60x
\n480 = 60x – 28x
\n480 = 32x
\nx = \\(\\frac{480}{32}=\\frac{120}{8}=\\frac{30}{2}\\) = 15
\nx = 15
\nThus, one man can complete the work in 15 days and one boy can do the work in 60 days.<\/p>\n

Question 26.
\nFind the zeroes of the polynomial p(y) = y3 – 5y2 – 2y + 24 if it is given that Sum of the two zeros is Answer:
\nLet \u03b1, \u03b2,\u03b3 be the zeros of the given polynomial.
\np(y) = y3<\/sup> – 5y2<\/sup> – 2y + 24
\n\u03b1 + \u03b2 = 7
\nSum of the zeros
\n\"Karnataka<\/p>\n

\u03b1 + \u03b2 + \u03b3 = 5
\n7 + r = 5
\nr = 5 – 7 = -2
\nr = -2
\nSum of product of zeroes taken two at a
\n\"Karnataka
\n\u03b1\u03b2 + \u03b2r + r\u03b1 = -2\/1 = – 2
\n\u03b1\u03b2 + r(\u03b2 + \u03b1) = -2
\n\u03b1\u03b2 + (7) (-2) = -2
\n\u03b1\u03b2-14 = -2
\n\u03b1\u03b2 = -2+ 14
\n\u03b1\u03b2 = 12
\n(\u03b1 – \u03b2 )2 <\/sup>= (\u03b1 + \u03b2)2<\/sup>– 4\u03b1\u03b2
\n= (7)2 <\/sup>– 4(12)
\n(\u03b1 – \u03b2)2<\/sup> = 49 – 48
\n\u03b1 – \u03b2 = \u00b11
\nSolve for \u03b1 and \u03b2. When \u03b1 – \u03b2 = +1
\n\u03b1 + \u03b2 = 7
\n\u03b1 – \u03b2 = 1
\n2\u03b1 =8
\n\u03b1 = 8\/2 = 4
\n\u03b1 = 4
\nSubstitute the value of a in \u03b1 + \u03b2 =7
\n4+p =7
\n\u03b2 =7-4 = 3
\n\u03b2 = 3
\nSolve for \u03b1 and \u03b2 when \u03b1 – \u03b2 = -1
\n\"Karnataka
\n\u03b1 = 6\/2 = 3
\n\u03b1 = 3
\n\u03b1 + \u03b2 = 7
\n3 + \u03b2=7
\n\u03b2 = 7 – 3
\na + \u03b2 = 7
\n\u03b2 = 4
\nWhen \u03b1 – \u03b2 = 1,
\nthe values are \u03b1 = 4, \u03b2 = 3, r = -2.
\nWhen \u03b1 – \u03b2 = -1,
\nThe values are \u03b1 = 3, \u03b2 = 4, r = -2.<\/p>\n

Question 27.
\nThe diagonal of a rectangular field is 60 metres more than the shorter side, if the longer side is 30 metres more than the shorter side, find the sides of the field.
\nAnswer:
\n\"Karnataka
\nLet the shorter D side be x mtr.
\nLonger side is 30m more than the shorter side. A Longer side = (x + 30)m.
\nDiagonal is 60m more than the shorter side.
\nDiagonal = (x + 60) m
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup> (Pythagoras)
\n(x + 60)2<\/sup> = (x + 30)2<\/sup> + x2<\/sup>
\nx2<\/sup>+(60)2<\/sup> + 2x (60) = x2<\/sup> + (30)2<\/sup> + 2(x)(30)+ x2<\/sup>
\nx2<\/sup> + 3600 + 120x = x2<\/sup> + 900 + 60x + x2<\/sup>
\n-x2<\/sup> + 60x + 2700 = 0
\nx2<\/sup> – 60x – 2700 = 0
\nx2<\/sup> – 90x + 30x – 2700 = 0
\nx(x – 90) + 30(x – 90) = 0
\n(x – 90) (x + 30) = 0
\nx – 90 = 0 or x + 30 = 0
\nx = 90 or x = – 30
\n\u2234 Shorter side = 90m = BC = x
\nLonger side = x + 30 = 90 + 30 = 120m
\nDiagonal = x + 60 = 90 + 60 = 150m<\/p>\n

OR<\/p>\n

Sum of the areas of two squares is 468m2. If the difference of their perimeters is 24m, find the sides of two squares.
\nAnswer:
\nLet the sides of the two squares be x&y. Sum of the areas of two squares = 468
\nx2<\/sup> + y2<\/sup> = 468 ……(1)
\nDifference of their perimeters = 24m.
\n4x – 4y = 24 x – y = 6
\nx = 6 + y …….(2)
\nSubstitute the value of x in (1)
\nx2<\/sup> + y2<\/sup> = 468
\n(6 + y)2<\/sup> + y2<\/sup> = 468
\n36 + y2<\/sup>+ 12y + y2<\/sup> = 468
\n2y2<\/sup> + 12y + 36 = 468
\nDivide by 2
\ny2<\/sup> +6y+ 18 = 234 y2 + 6y = 234 – 18
\ny2<\/sup> + 6y – 216 = 0
\ny2<\/sup>+18y – 12y -216 = 0
\ny(y + 18)-12(y + 18) = 0
\n(y + 18) (y – 12) = 0
\ny+18 = 0, y -12 = 0
\ny = -18,y = 12
\n\u2234 x = 6+y = 6 + 12 = 18m
\nThe sides of two squares are 18m and 12m.<\/p>\n

\"Karnataka<\/p>\n

Question 28.
\nShow that the points x(2, -2) y(-2, 1) and z(5, 2) are the vertices of a right angled triangle XYZ and also calculate its area.
\nAnswer:
\n\"Karnataka
\n\"Karnataka
\n\u21d2 XYZ is an isosceles right angled triangle.
\nArea of a right angled triangle = \\(\\frac{1}{2}\\) x base x height
\n\\(\\frac{1}{2}\\) \u00d7 5 \u00d7 5 = \\(\\frac{25}{2}\\) = 12.5 cm<\/p>\n

OR<\/p>\n

Find the values of k for which the points A(k + 1, 2k) B(3k, 2k+3) and C(5k – 1, 5k) are collinear.
\nAnswer:
\nA(k + 1, 2k) = (x1<\/sub>, y1<\/sub>)
\nB(3k, 2k + 3) = (x2,<\/sub> y2<\/sub>)
\nC(5k-1, 5k) = (x3<\/sub>, y3<\/sub>)
\nArea of the triangle = 0, for the points to be collinear.
\nArea =[x1<\/sub>(y2<\/sub> – y3<\/sub>) + x2<\/sub>(y3<\/sub>-y1<\/sub>) + x3<\/sub>(y1<\/sub>-y2<\/sub>)]
\n0 = \\(\\frac{1}{2}\\){(k + 1) (2k+3 – 5k)+3k(5k- 2k) + (5k-1) [2k-2k-3]}
\n0= \\(\\frac{1}{2}\\)]{(k+ 1) (-3k+ 3) + 3k (3k) + (5k- I)(-3)i
\n0 =\\(\\frac{1}{2}\\){-3k2<\/sup>-3k + 3k + 3 + 9k2 – 15k+3}
\n0= \\(\\frac{1}{2}\\){6k2<\/sup>– 15k + 6}
\n0 = 6k2<\/sup> – 15k + 6
\nDivide by 3
\n2k2<\/sup> – 5k + 2= 0
\n2k2<\/sup> – 4k – 1k + 2 = 0
\n2k(k – 2) -1(k – 2) = 0
\n(k – 2) (2k -1) = 0
\nk – 2 = 0, or 2k- 1=0
\nk = 2 or k = 1\/2<\/p>\n

Question 29.
\nProve that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
\nAnswer:
\n\"Karnataka
\nData: ‘O’ is the centre of the circle PA and PB are the two tangents drawn from an external point P. OA and OB are radii of the circle.
\nTo prove that: \u2220APB+ \u2220AOB = 180\u00b0
\nProof: In AAPO and ABPO
\n\u2220OAP = \u2220OBP = 90\u00b0
\n(\u2235 Angle between the radius and tangent at the point of contact is 90\u00b0)
\nOP = OP (\u2235 Common side)
\nOA = OB (\u2235 Radii of the same circle) According to RHS postulate AAPO ABPO
\n\u2220OAP = \u2220OBP = 90\u00b0
\n\u2220OAP + \u2220OBP = 90\u00b0 + 90\u00b0 = 180\u00b0
\n\u21d2 Opposite angles of OAPB quadrilateral are supplementary
\n\u2234 OAPB is a cylic quadrilateral
\n\u21d2 \u2220APB + \u2220AOB = 180\u00b0<\/p>\n

\"Karnataka<\/p>\n

Question 30.
\nFind the area of the shaded region where a circular arc of Radius 6 cm has been drawn with the vertex \u2018O\u2019 of an equilateral Triangle OAB of side 12cm as centre.
\nAnswer:
\nArea of the shaded Region = Area of circle area of sector OCDE + Area of equilateral
\n\u2206 le OAB
\n\"Karnataka
\n\"Karnataka<\/p>\n

OR<\/p>\n

From each corner of a square of side 4cm a quadrant of a circle of radius 1cm is cut and also a circle of diameter 2cm is cut as shown in Fig, Find the area of the remaining portion of the square.
\nAnswer:
\n\"Karnataka
\n\"Karnataka
\nArea of shaded region = Area of square – (Area of circle + Area of 4 quadrants)
\nThe following table gives production yield per hectare of wheat of 100 farms of a
\n\"Karnataka
\n\"Karnataka<\/p>\n

Question 31.
\nThe following table gives production yield per hectare of wheat of 100 farms of a
\nvillage.
\n\"Karnataka
\nChange the distribution to a more than type distribution and draw its ogive.
\nAnswer:
\n\"Karnataka<\/p>\n

Question 32.
\nIf the median of the distribution given below is 28.5. Find the valucs of x and y.
\n\"Karnataka
\nAnswer:
\nMeadian = 28.5, lies in the C.I , 20 – 30
\nThe median class = 20 – 30
\nl = 20, f = 20, c.f = 5 + h, h = 10, n = 60
\n\"Karnataka
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

Question 33.
\nConstruct a triangle ABC with side BC = 7cm, \u2220B = 45\u00b0, \u2220C = 105\u00b0. Then construct a triangle whose sides are 4\/3 times the corresponding sides of \u2206 ABC.
\nAnswer:
\n\"Karnataka<\/p>\n

V. Answer the following ( 4 \u00d7 4 = 16 )<\/span><\/p>\n

Question 34.
\nSolve the pair of equations graphically.
\nx+y=3 and 3x-2y=4
\nAnswer:
\nx + y = 3
\ny = 3- x
\n\"Karnataka
\n3x – 2y = 4
\n-2y = 4 – 3x .
\n\"Karnataka
\n\"Karnataka<\/p>\n

Question 35.
\nIf the sum of first 8 terms of an Arithmetic progression is 136 and that of first 15 terms is 465, then find the sum of first 25 terms.
\nAnswer:
\nGiven S8<\/sub> = 136, S15<\/sub> = 465, S25<\/sub> = ?
\nUse the formula Sn<\/sub> = \\(\\frac{n}{2}\\) [2a + (n -l)d]
\ns8<\/sub> = \\(\\frac{8}{2}\\)[2a + (s-l)d]
\n136 = 4 (2a + 7d)
\n\u2234 2a + 7d = 136\/4
\n2a + 7d= 34…….(1)
\nS15<\/sub> = \\(\\frac{15}{2}\\)[2a + (15-l)d]
\n465 = \\(\\frac{15}{2}\\)[2a + 14d]
\n465= \\(\\frac{15}{2}\\) x 2(a + 7d)
\n\u2234 a + 7d = 465\/15
\na + 7d = 31 …….(2)
\nFrom (1) and (2)
\n\"Karnataka
\nConsider a + 7d = 31
\n3 + 7d = 31
\n7d = 31 – 3
\nd = 28\/7
\nd = 4
\n\u2234S25<\/sub> = \\(\\frac{25}{2}\\)[2a + (25-1)d]
\nS25<\/sub> = \\(\\frac{25}{2}\\)[2a + (25-1)d]
\nS25<\/sub> = \\(\\frac{25}{2}\\)[2a + (24)d]
\nS25<\/sub> = \\(\\frac{25}{2}\\)[2(3) + 24(4)d]
\nS25<\/sub> = \\(\\frac{25}{2}\\) \u00d7 102
\nS25<\/sub> = 1275
\n\u20185 ‘ 9<\/p>\n

OR<\/p>\n

The sum of the 5th and 9th terms of an A.P is 40 and the sum of the 8th and 14th term is 64. Find the sum of the first 20 terms.
\nAnswer:
\nGiven T5<\/sub>+ T9<\/sub> = 40 and T8<\/sub> + T14<\/sub> = 64
\na + 4d + a + 8d=40
\n2a + 12d = 40….(1)
\na + 7d + a + 13d = 64
\n2a + 20d = 64…. (2)
\n\"Karnataka
\nConsider 2a + 12d = 40
\n2a + 12(3) = 40
\n2a + 36 = 40
\n2a = 40 – 36
\na = 4\/2
\na = 2
\nSn<\/sub> = \\(\\frac{n}{2}\\)[2a + (n-1)d]
\nS20<\/sub> = \\(\\frac{20}{2}\\)[2(2) + (20-1)3]
\n= 104(4 + 57)
\n= 10 \u00d7 61
\n= 610<\/p>\n

\"Karnataka<\/p>\n

Question 36.
\nTwo poles of equal heights are standing opposite to each other on either side of the road, which is 80m wide. From a point between them on to the road, the angles of elevation of the top of the poles re 60 and 30\u2019, respectively. Find the height of the poles and distances of the point from the poles.
\nAnswer:
\n\"Karnataka
\nIn \u2206 ABE.
\ntan60 = \\(\\frac{\\mathrm{AB}}{\\mathrm{BE}}\\)
\n\u221a3 = x\/BE
\nx = \u221a3BE ….. (1)
\ntan30 = DC\/EC
\n\\(\\frac{1}{\\sqrt{3}}=\\frac{x}{E C}\\)
\nEC = x\u221a3
\nEC = BE \u00d7 \u221a3 .\u221a3 [\u2235 x = \u221a3.BE]
\nFC = 3BE
\n\u2234 EC=3BF
\nWe know that
\nBE + EC = 80 rn
\nBE 3BE 80m
\n4BE = 80
\nBE = 80\/4 = 20
\n\u21d2 EC = 3BE 3(20)= 60m
\n\u2234 x = BE.
\nx = 20\u221a3
\n\u21d2 AB = CD = 20\u221a3 m
\n\u2234 The distance from the point to the pole arc 20m towards left and 60m towards the right
\npoles
\nHeight of the poles 20\u221a3m<\/p>\n

Question 37.
\nProve areas of similar triangles.
\nAnswer:
\nAreas of similar triangles are proportional
\nto the squares on the corresponding sides.
\n\"Karnataka
\nIn \u2206ABM and \u2206DEN
\n\u2220AMB = \u2220DNE = 90 [construction]
\n\u2220B =\u2220E (Data)
\n\u2220BAM = \u2220EDN (Remainingang1e)
\n\u2206ABM \u223c \u2206 DEN
\n\u2206ABM & \u2206DEN are equiangular
\n\"Karnataka
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

VI. Answer the following : ( 5 \u00d7 1 = 5 )<\/span><\/p>\n

Question 38.
\nA circus tent is made of canvas and is in the form of a right circular cylinder and a right circular cone above it. The diameter and height of the cylindrical part of the tent are 126m and 5m respectively. The total height of the tent is 21m. Find the total cost of the canvas used to make the tent when the cost per m2 of the canvas is ?15.
\nAnswer:
\n\"Karnataka
\nTotal canvas used=CS A of cylindrical part -1- CS A of conical part
\n=2\u03c0rh + \u03c0rl
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 63 \u00d7 5 + \\(\\frac{22}{7}\\) x 63 x 65
\n= 1980 m2<\/sup> + 12870 m2<\/sup>
\n= 14850 m2<\/sup>
\nTotal canvas used = 14850 m2<\/sup>
\n* Cost of canvas at the rate of \u20b916 per m2<\/sup>
\n= 14850 \u00d7 15= \u20b9 2,22.750.
\nCost of canvas = \u20b9 2,22,750<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Karnataka SSLC Maths Model Question Paper 3 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus SSLC Maths Model Question Paper 3 with Answers Time: 3 Hours Max Marks: 80 I. …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/6041"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=6041"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/6041\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=6041"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=6041"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=6041"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}