{"id":6035,"date":"2019-08-28T13:52:16","date_gmt":"2019-08-28T13:52:16","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=6035"},"modified":"2021-07-02T15:45:59","modified_gmt":"2021-07-02T10:15:59","slug":"karnataka-sslc-maths-model-question-paper-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/karnataka-sslc-maths-model-question-paper-2\/","title":{"rendered":"Karnataka SSLC Maths Model Question Paper 2 with Answers"},"content":{"rendered":"

Students can Download Karnataka SSLC Maths Model Question Paper 2 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 2 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.
\nFor some integer n every odd integer is of the form
\na) 2n + 1
\nb) n + 1
\nc) 2n
\nd) n
\nAnswer:
\n2n + 1<\/p>\n

Question 2.
\nThe value of sin215\u00b0 + sin225\u00b0 + sin265\u00b0 + sin275\u00b0 is
\na) 0
\nb) 1
\nc) 2
\nd) 3
\nAnswer:
\n2<\/p>\n

\"Karnataka<\/p>\n

Question 3.
\nIf chord AB subtends an angle 50\u00b0 at the centre of a circle then the angle between the tangents at A and B is
\na) 40\u00b0
\nb) 100\u00b0
\nc) 130\u00b0
\nd) 120\u00b0
\nAnswer:
\n130\u00b0<\/p>\n

Question 4.
\nThe formula used to find the volume of a sphere
\na) \\(\\frac{4}{3} \\pi r^{3}\\)
\nb) \\(\\frac{2}{3} \\pi r^{3}\\)
\nc) \\(\\frac{1}{3} \\pi r^{3}\\)
\nd) \u03c0r3<\/sup>
\nAnswer:
\na) \\(\\frac{4}{3} \\pi r^{3}\\)<\/p>\n

Question 5.
\n\u03b1 + \u03b2 are the zeroes of the polynomial x2<\/sup> – 6x + 4, then the value of \\(\\frac{(\\alpha+\\beta)^{2}-2 \\alpha \\beta}{\\alpha \\beta}\\) is
\na) 7
\nb) 8
\nc) -7
\nd) -8
\nAnswer:
\na) 7
\nSolution:
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

Question 6.
\nIf 29th term of an A.P is twice its 19th term, then the 9th term is
\na) -1
\nb) 0
\nc) 1
\nd) 2
\nAnswer:
\nb) 0
\nSolution:
\nT29<\/sub> = 2T19<\/sub>
\na + 28d= 2(a + 18d)
\na + 28d = 2a + 36d
\n0 = 2a + 36d-a-28d
\n0 = a + 8d
\nT9<\/sub> = 0<\/p>\n

Question 7.
\nIn \u2206 ABC, AB = 6\u221a3 cm, AC = 12 cm, BC = 6cm, The angle B is
\na) 45\u00b0
\nb) 90\u00b0
\nc) 60\u00b0
\nd) 30\u00b0
\nSolution:
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n122<\/sup>= (6\u221a3 )2<\/sup> + 62<\/sup>
\n144= 108 + 36 144=144
\n\u2234 \u2220B = 90\u00b0
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

Question 8.
\nIf the probability of an event is P(A) then the probability of its complimentary event will be
\na) 1+P(A)
\nb) 1 – P(A)
\nc) P(A) – 1
\nd) 1 \/ P(A)
\nSolution:
\nWe know that for any two complimentary events A and A\u0304
\nP(A) + P(A\u0304) = 1 \u2192 P(A\u0304) = 1 – P(A)<\/p>\n

II. Answer the following questions : ( 1 \u00d7 8 = 8 )<\/span><\/p>\n

Question 9.
\nIf \u03b1 and \u03b2 are the zeroes of the quadratic polynomial 2 – 3x – x2<\/sup> then what is the value of \u03b1 + \u03b2 + \u03b1\u03b2 ?
\nAnswer:
\n\u03b1 + \u03b2 = -b\/a , \u03b1\u03b2 = c\/a
\n= \\(\\frac{-(-3)}{-1}\\) \u03b1\u03b2 = \\(\\frac{2}{-1}\\) = – 2
\n= – 3
\n\u03b1 + \u03b2 + \u03b1\u03b2 = -3 -2 = -5<\/p>\n

Question 10.
\nWhat are the roots of the quadratic equation x2<\/sup> + (\u221a3 + 1)x + \u221a3 = 0?
\nAnswer:
\n2<\/sup> + (\u221a3 + 1)x + \u221a3 = 0 = 0
\nx2<\/sup> + x + \u221a3x + \u221a3 = 0
\nx(x + 1)+ \u221a3 (x + 1) = 0
\n(x + 1)(x + \u221a3) = 0
\nx + 1 = 0, x + \u221a3 = 0
\nx = -1, x = -\u221a3<\/p>\n

Question 11.
\nIf the nth terms of the two AP 9, 7, 5, ….. and 24, 21, 18, ……… are same. Find n.
\nAnswer:
\n9, 7, 5, …….
\na = 9, d = 7-9 = -2 T = a + (n – 1)d
\nTn<\/sub> = 9 + (n – 1) (-2)
\nTn<\/sub> = 9 + (-2n + 2) ‘
\nTn<\/sub> = 9 – 2n + 2
\nTn<\/sub> = 11 – 2n
\n24,21,18,
\na = 24, d = 21 – 24 = -3
\nTn<\/sub> = a + (n – 1) d T=24 + (n – 1)(-3)
\nTn<\/sub> = 24 – 3n + 3
\nTn<\/sub> = 27 – 3n
\nGiven : nth term of both A. P. is same.
\n\u2234 11 – 2n = 27 – 3n
\n-2n +3n = 27 -11
\nn = 16<\/p>\n

Question 12.
\nState A – A criterian theorem.
\nAnswer:
\n\u201cIf two triangles are equiangular, then the corresponding sides are proportional.\u201d<\/p>\n

Question 13.
\nFind the H.C.F of 455 and 42 with the help of Euclid\u2019s division algorithm.
\nAnswer:
\nStep (1): a = 455 b = 42
\na = bq + r
\n455 = 42 \u00d7 1 + 35<\/p>\n

Step (2): a = 42, b = 35
\na = bq + r
\n42 = 35 \u00d7 1 + 7<\/p>\n

Step (3): a = 35, b = 7
\na = bq + r
\n35 = 7 \u00d7 5 + 0
\n\u2234 H.C.F. (455, 42) = 7<\/p>\n

Question 14.
\nFind \u03b8 if sin (\u03b8 + 56) = cos\u03b8, where \u03b8 and ( \u03b8 + 56) are less than 90\u00b0.
\nAnswer:
\nsin (\u03b8 + 56) = cos \u03b8
\nsin (\u03b8 + 56) = sin (90 – \u03b8)
\n\u03b8 + 56 = 90 – \u03b8
\n\u03b8 + \u03b8 = 90 – 56
\n2\u03b8 = 34
\n\u03b8 = 34\/2
\n\u03b8 = 17<\/p>\n

Question 15.
\nIf x = a sin \u03b8 and y = b tan \u03b8, then find the value of \\(\\frac{a^{2}}{x^{2}}-\\frac{b^{2}}{y^{2}}\\)
\nAnswer:
\nConsider
\n\"Karnataka
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

Question 16.
\nCalculate the height of a right circular cone where C.S.A. and base radius are 12320 cm2 and 56 cms. respectively.
\nAnswer:
\n\"Karnataka
\nC.S.A of a cone = \u03c0 rl
\n12320 = \\(\\frac{22}{7}\\) x 56 x 1
\n1 = \\(\\frac{12320 \\times 7}{22 \\times 56}\\)
\n1 = 70 cm
\nBut 12<\/sup> = r2<\/sup> + h2<\/sup>
\n702<\/sup> = 562<\/sup> + h2<\/sup>
\nh22<\/sup> = 702<\/sup> – 562<\/sup>
\nh2<\/sup> = (70 + 56) (70 – 56)
\nh2<\/sup> =(126) (14)
\n\"Karnataka<\/p>\n

III. Answer the following ( 2 \u00d7 8 = 16 )<\/span><\/p>\n

Question 17.
\nShow that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5 where q is an integer.
\nAnswer:
\nLet nbe any positive, integer using division algorithm,
\na = bq + r
\nTaking a = n, b = 6
\nn = 6q + r
\nIf any number is divided by 6, then the possible remainders are 0, 1, 2, 3, 4 or 5.
\n\u2234If n is odd, then r = 1, 3, 5
\n\u21d2 6q + 1, 6q + 3, 6q + 5 are the positive odd integers.<\/p>\n

Question 18.
\nSolve : 2x + 3y = 9
\n3x + 4y = 5
\nAnswer:
\n\"Karnataka
\nSubstitute y = 17 in 2x + 3y = 9
\n2x + 3(17) = 9
\n2x + 51 =9
\n2x = 9 – 51
\nx = \\(\\frac{-42}{2}\\)
\n= -21<\/p>\n

Question 19.
\nSolve : (x – 2)2<\/sup> + 1 = 2x – 3
\nAnswer:
\nx2<\/sup> + 4 – 4x + 1 = 2x – 3
\nx2<\/sup> – 4x + 5 – 2x + 3 = 0
\nx2<\/sup> – 6x + 8 = 0 x2 – 4x – 2x + 8 = 0
\nx(x – 4) -2(x – 4) = 0
\n(x – 4) (x – 2) = 0
\nx – 4 = 0, x -2 = 0
\nx = 4 or x = 2<\/p>\n

Question 20.
\nShow that the points (-2, 1) (2, -2) and (5, 2) are the vertices of a right angled triangle.
\nAnswer:
\n\"Karnataka
\n\"Karnataka
\n\"Karnataka<\/p>\n

\u2234 \u2206 ABC is a right angled triangle at B.<\/p>\n

\"Karnataka<\/p>\n

Question 21.
\nThe equilateral triangles are drawn on the sides of a right triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of the triangles on the other two sides.
\nAnswer:
\n\"Karnataka
\n\"Karnataka
\n\u2234 Area of \u2206 XAB + Area of \u2206 YBC = Area of \u2206 ZAC .<\/p>\n

OR<\/p>\n

In the given figure, PA, QB and RC are each perpendicular to AC. Prove \\(\\frac{1}{x}+\\frac{1}{z}=\\frac{1}{y}\\)
\nAnswer:
\n\"Karnataka<\/p>\n

Question 22.
\nTwo dice are thrown simultaneously. Find the probability that the sum of the numbers on the faces is neither divisible by 4 nor by 5.
\nAnswer:
\nS = {(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)}
\nn(S) = 36.
\nAn event of getting sum of the numbers neither divisible by 4 nor by 5.
\nA={ 1,1) (1,2) (1,5) (1,6) (2,1) (2, 4)
\n(2, 5) (3, 3) (3, 4) (3, 6) (4, 2) (4,3) (4, 5) (5,1) (5, 2) (5, 4) (5, 6) (6,1) (6, 3) (6, 5)} n(A) = 20
\nn(A) = 20
\nP(A) = \\(\\frac{\\mathrm{n}(\\mathrm{A})}{\\mathrm{n}(\\mathrm{S})}=\\frac{20}{36}\\)<\/p>\n

Question 23.
\nDraw a circle of radius 3 cm. Take a point P outside the circle without using the centre of the circle, draw to tangents to the circle from an external point P.
\nAnswer:
\n\"Karnataka<\/p>\n

Question 24.
\nProve that (cosec\u03b8-cot\u03b8)2<\/sup> = \\(\\frac{1-\\cos \\theta}{1+\\cos \\theta}\\)
\nAnswer:
\nConsider LHS
\n(cosec\u03b8-cot\u03b8)2<\/sup> = cosec2<\/sup> \u03b8 +cot2<\/sup> \u03b8 – 2cosec\u03b8.cot\u03b8
\n\"Karnataka
\n\"Karnataka<\/p>\n

OR<\/p>\n

If sin\u03b8 + cos \u03b8 = \u221a2 sin (90 -\u03b8) determine cot \u03b8
\nAnswer:
\nsin \u03b8 + cos \u03b8 = \u221a2 sin (90 – \u03b8)
\nsin \u03b8 + cos \u03b8 = \u221a2 cos \u03b8
\nsin \u03b8 = ypi cos \u03b8 – cos \u03b8
\nsin \u03b8 = cos \u03b8 (\u221a2 -1)
\nsin\u03b8 \/cos\u03b8 = \u221a2 -1
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

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

Question 25.
\nAsha is 5 times as old as her daughter Usha, 5 years later Asha will be 3 times as old as her daughter Usha. Find the present ages of Asha and Usha.
\nAnswer:
\nLet the age of Asha be x years
\nLet the age of Usha be y years
\nAccording to question x=5y
\nx – 5y = 0 …….(1)
\nFive years later,
\n(x + 5) = 3(y + 5)
\nx + 5 = 3y + 15
\nx – 3y = 15 – 5
\nx – 3y = 15 – 5
\nx – 3y = 10 (2)
\nSubstitute x = 5y in equation (2)
\n5y – 3y = 10
\n2y = 10
\ny = 10\/2
\ny = 5
\n\u2234 x = 5y
\n= 5 \u00d7 5
\nx = 25
\n\u2234 The present age of Asha is 25 years.
\nThe present age of Usha is 5 years.<\/p>\n

OR<\/p>\n

The sum of 2 digits of a 2 digit number is 12 the number obtained by interchanging the digits exceeds by the given number by 18. Find the number.
\nAnswer:
\nLet the two digits be x and y
\nThe 2 digit number will be 10x + y
\nThe sum of digits = x + y – 12 ……..(i)
\nThe number obtained by interchanging the digits =10y + x
\n= 10x + y + 18
\n10y + x – 10x – y = 18
\n9y – 9x = 18
\nDivide by 9
\ny – x = 2
\ny = x + 2 ….(ii)
\nConsider equation (1)
\nx + y = 12
\nSubstitute y = x + 2
\nx + x + 2 = 12 .
\n2x =12 – 2
\n2x= 10
\nx= 5
\ny = x + 2
\ny = 5 +2
\ny = 7
\n\u2234 The number is : 10x +y
\n= 10(5)+7 = 50+ 7
\n= 57<\/p>\n

Question 26.
\nFind the other two zeroes of the polynomial y4<\/sup> + y3<\/sup> – 9y2<\/sup> – 3y + 18 if the
\nzeroes are \u221a3 and -\u221a3
\nAnswer:
\ny = \u221a3 and y = -\u221a3
\n(y + \u221a3) = o (y-V3) = o
\n(y + \u221a3 )(y – \u221a3 ) = o
\ny(y – \u221a3 ) + \u221a3 (y – \u221a3 ) = 0
\ny2<\/sup> – \u221a3y + \u221a3 y – 3 = o
\ny2<\/sup> – 3 = 0
\n\"Karnataka
\ny2<\/sup> + y – 6 = 0
\ny2<\/sup> + 3y – 2y – 6 = 0
\ny(y + 3)-2(y + 3) = 0
\n(y + 3) = 0 & (y – 2) = 0
\ny=-3 & y = 2
\nThe four zeros of polynomial are \u221a3 , -\u221a3 , -3 & 2<\/p>\n

Question 27.
\nSolve for x.
\n\\(\\frac{1}{a+b+x}=\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{x}\\) (Where a \u2260 0, b \u2260 0 x \u2260 0, x \u2260 -a,-b
\nAnswer:
\n\"Karnataka
\n(-a – b)ab = (a + b)x (a + b + x)
\n-a2<\/sup>b – ab2<\/sup> = (a + b) (ax + bx + x2<\/sup>)
\n-a2<\/sup>b – ab2<\/sup> = a2<\/sup>x + abx + ax2<\/sup> + abx + b2<\/sup>x + x2<\/sup>b
\n-a2<\/sup>b – ab2<\/sup>= a2<\/sup>x + 2abx + ax2<\/sup> + b2<\/sup>x + x2<\/sup>b
\n-ab(a+b) = ax(a+x) + xb(b+x)+2abx
\n=x[a(a+x)+b(b+x)+2ab]
\n=x[a2<\/sup> + ax+b2<\/sup> +bx + 2ab]
\n= x[(a+b)2<\/sup> + x(a+b)]
\n-ab(a + b) = x(a+b)[a+b+x]
\n-ab = x(a + b + x)
\n-ab = ax + bx + x2<\/sup>
\nx2 + x (a + b) + ab = 0
\nx2<\/sup> + ax + bx + ab = 0
\nx(x+a) + b(x + a) = 0
\n(x + a) (x +b) = 0
\n=>x + a = 0, x + b = 0
\n=> x = -a, x = -b<\/p>\n

OR<\/p>\n

The diagonal of a rectangular field is 60m more than the shorter side. If the larger side is 30m more than the shorter side, find the sides of the field.
\nAnswer:
\nLet the shorter side of the rectangular field be x m.
\nLonger side is x + 30 and the diagonal is x+60
\n\"Karnataka
\nIn \u2206 ABC,
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n(x + 60)2<\/sup> = ( x + 30)2<\/sup>+ x2<\/sup>
\nx2<\/sup> + 3600 + 120x = x2<\/sup> + 900 + 60x + x2<\/sup>
\nx2<\/sup> + 900 + 60x – 3600 – 120x = 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, x + 30 = 0
\nx = 90, x = – 30.
\n\u2234 The shorter side = x = 90m
\nThe longer side = x + 30 = 90+30 = 120m.<\/p>\n

\"Karnataka<\/p>\n

Question 28.
\nIf the points (7, -2) (5, 1) and (3, 5) are collinear. Find the value of k.
\nAnswer:
\nSince the give points are collinear, the area of the triangle formed by them must be O i.e.
\n= \\(\\frac { 1 }{ 2 }\\) [x1<\/sub>(y2<\/sub>-y3<\/sub>) + x2<\/sub>(y3<\/sub>-y2<\/sub>) + x3<\/sub>(y1<\/sub>-y2<\/sub>)]
\nx1<\/sub> = 7, x2<\/sub> = 5, x3<\/sub> = 3
\ny1<\/sub> = – 2, y2<\/sub> = 1, y3<\/sub> = k
\n= \\(\\frac { 1 }{ 2 }\\) [7(1 – k) + 5 (k + 2) + 3(-2-1)] = 0
\n= \\(\\frac { 1 }{ 2 }\\) [7 – 7k + 5k + 10 – 9]-0
\n= \\(\\frac { 1 }{ 2 }\\) [7 – 2k + 1] = 0
\n= \\(\\frac { 1 }{ 2 }\\) [8 – 2k] = 0
\n8-2k = 0 or 2k=8
\nk = 8\/2 = 4
\nk = 4<\/p>\n

OR<\/p>\n

Find the area of Rhombus if its vertices are (3, 0) (4, 5) (-1, 4) and (-2, -1) taken in order.
\nAnswer:
\n\"Karnataka<\/p>\n

Area of rhombus = \\(\\frac { 1 }{ 2 }\\) d1<\/sub> \u00d7 d2<\/sub>
\nLet A = (3, 0) B = (4, 5)
\nC = (-1, 4) D = (-2, -1)
\nDiagonal
\n\"Karnataka
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

Question 29.
\nProve the tangents drawn from an external point to a circle are equal.
\nAnswer:
\n\"Karnataka
\nData: A is the centre of the circle B is an external point.
\nBP and BQ are the tangents.
\nAP, AQ and AB are joined.
\nTo prove that: a) BP = BQ
\nb) \u2220PAB-\u2220QAB.
\nc) \u2220PBA = \u2220QBA
\nProof: In \u2206 APB and \u2206 AQB
\nAP = AQ [Radii of same circle]
\n\u2220 APB – \u2220AQB = 90\u00b0
\nAB = AB [common side]
\n\u2234\u00a0 \u2206 APB \u2245 \u2206 AQB [RHS]
\na) PB = QB
\nb) \u2220PAB = \u2220QAB
\nc) \u2220PBA = \u2220QBA<\/p>\n

Question 30.
\nFind the area of the shaded region in the figure, where ABCD is a square of side 14 cm
\nAnswer:
\n\"Karnataka
\nArea of square ABCD
\n= 14 \u00d7 14=196 cm2<\/sup>
\n\"Karnataka
\nHence, area of the shaded region
\n= Area of square – Area of four circles
\n= 196 – 154
\n= 42cm2<\/sup><\/p>\n

OR<\/p>\n

Find the area of the shaded regions. Given PQRS a square of sides 14cm. Soln: S R
\nAnswer:
\n\"Karnataka
\n\"Karnataka<\/p>\n

Question 31.
\nThe following table gives the weight of 120 articles :
\n\"Karnataka
\nChange the distribution to a \u201cmore than type\u201d distribution and draw its ogive.
\nAnswer:
\n\"Karnataka
\n\"Karnataka<\/p>\n

Question 32.
\nThe distribution below gives the weights of 30 students of a class. Find the median weight of the students.
\nWeight (in kg)
\n\"Karnataka
\nAnswer:
\n\"Karnataka
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

Question 33.
\nConstruct a right triangle whose hypotenuse and one side measures 10cm and 8cm respectively. Then construct another triangle whose sides are 4\/5 times the corresponding sides of the triangle .
\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 = 8 and x – y = -2
\nAnswe:
\nx + y = 8
\ny = 8 – x
\n\"Karnataka
\nx – y = -2
\nx = -2 + y
\n\"Karnataka
\nx = 6\/2 = 3
\nx + y = 8
\n3 + y = 8
\ny = 8 – 3
\ny = 5
\n\"Karnataka<\/p>\n

Question 35.
\nDevide 20 into four parts which are in arithmetic progression and such that the product of first and fourth is to the product of second and third in the ratio 2:3.
\nAnswer:
\nLet the four parts in A.P be :
\na – 3d, a – d, a +d, a + 3d
\nTheir sum = (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20
\n\u21d2 4a = 20 or a = 5
\nAlso, it is given that
\n\"Karnataka
\nor 75 -27d2<\/sup> = 50 – 2d2<\/sup>
\n25d2<\/sup> = 25 .
\n\u21d2 d2<\/sup> = 1 or d = +\u221a1 = \u00b11
\nWhen a = 5, d = 1, the four parts are 2, 4, 6,8
\nWhen a = 5, d = -1; the four parts are 8, 6, 4,2
\nHence the four parts are (2, 4, 6, 8) or (8, . 6,4,2)<\/p>\n

OR<\/p>\n

The angles of a quadrilateral are in AP such that the greatest is double the least calculate all the angles of the quadrilateral.
\nAnswer:
\nLet the quadrilateral be ABCD
\nThe angles of the quadrilateral are in AP.
\n\u2234 The four angles are A, B, C and D which are a -3d, a – d, a + d, a + 3d respectively.
\n\u21d2 The least angle is a – 3d and the greatest angle is a + 3d.
\nIt is given that
\na + 3d = 2(a – 3d)
\na + 3d = 2a – 6d
\n2a – a – 6d – 3d = 0
\na – 9d ……(1)
\n9d = a
\nBut \u2220A + \u2220B+\u2220C + \u2220D = 360\u00b0
\na – 3d + a – d + a + d + a +.3d = 360\u00b0
\n4a = 360 \/ 4 = 90\u00b0
\nBut 9d = a
\nd = \\(\\frac{a}{9}=\\frac{90}{9}\\) = 10\u00b0
\n\u2234 \u2220A=a- 3d = 90 – 3(10) = 90 -30 = 60\u00b0
\n\u2220B = a – d = 90 – 10 = 80\u00b0
\n\u2220C = a + d = 90 + 10= 100
\n\u2220D = a + 3d = 9 + 3(10) = 90 + 30 = 120<\/p>\n

\"Karnataka<\/p>\n

Question 36.
\nA person on the lighhouse of height 100m above the sea level observes that the angle of depression of a ship sailing towards the light house changes from 30\u00b0 to 45\u00b0. Calculate the distance travelled by the ship during the period of observation. (Take \u221a3 \u2248 1.73)
\n\"Karnataka
\nAnswer:
\nP is the top of the light house PQ. We are given that its height PQ = 100m PX is horizontal line through P
\n\u2220APX = 30\u00b0 and \u2220BPX =45\u00b0
\nThen, \u2220PAQ =30\u00b0 and \u2220PBQ = 45\u00b0
\nSuppose, AB = x metre and BQ = y metre
\nKarnataka SSLC Maths Model Question Paper 2 with Answers – 40
\n\u21d2 x + y =100\u221a3 = 100\u221a3 = 100\u00d7 1.732
\n\u21d2 x +y = 173.2
\n\u21d2 x + 100 = 173.2
\n\u21d2 x = 73.2
\nTherefore, the distance travelled = 73.2m<\/p>\n

Question 37.
\nProve that \u201cthe ratio of areas of two similar triangles is equal to the square of the ratio of their altitudes.
\nAnswer:
\n\"Karnataka
\nData : \u2206ABC \u223c \u2206DEF
\n\u2220A = \u2220D
\n\u2220B = \u2220E
\n\u2220C = \u2220F
\n\\(\\frac{A B}{D E}=\\frac{A C}{D F}=\\frac{B C}{E F}\\)
\nT.P.T : \\(\\frac{\\text { Area of } \\Delta \\mathrm{ABC}}{\\text { Area of } \\Delta \\mathrm{DEF}}=\\frac{\\mathrm{AB}^{2}}{\\mathrm{DE}^{2}}=\\frac{\\mathrm{AC}^{2}}{\\mathrm{DF}^{2}}=\\frac{\\mathrm{BC}^{2}}{\\mathrm{EF}^{2}}\\)
\nConstruction:Draw AM \u22a5 BC and AN \u22a5 EFim
\n\"Karnataka
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

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

Question 38.
\nThe radii of the circular ends of the frustrum of height – 6cm are 14 cm and 6cm respectively. Find the lateral surface area and total surface area of frustrum.
\nAnswer:
\nR = 14cm, r = 6cm and h =6cm
\nNow, let 1 be the slant height of the frustrum;
\n\"Karnataka
\nNow, lateral surface area of the frustrum
\n= \u03c0(R + r)l
\n= \u03c0 (14 + 6) 10
\n= 628.57 cm2<\/sup>
\nAnd total surface area of the frustrum = \u03c0[(R+r)l + R2<\/sup>+ r2<\/sup>]
\n= \u03c0 [(14 + 6)10 + (14)2<\/sup> + (6)2<\/sup>]
\n= \u03c0(200 + 196 + 36)
\n= \u03c0(432)
\n= 1357.71 cm2<\/sup>.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Karnataka SSLC Maths Model Question Paper 2 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 2 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\/6035"}],"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=6035"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/6035\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=6035"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=6035"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=6035"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}