3<\/sup> – 3abc = 0]<\/p>\nSolution 8. \nLet the first angles be x\u00b0. \nSecond angle = (x + 34)\u00b0 \n\u2234 x\u00b0 + (x + 34)\u00b0 = 180\u00b0 => 2x + 34\u00b0 = 180\u00b0 \n2x = 180\u00b0 – 34\u00b0 =>2x = 146\u00b0 \nx = 73\u00b0 \nThe first angle = 73\u00b0 \nSecond angle = 73\u00b0 + 34\u00b0 = 107\u00b0.<\/p>\n
Solution 9. \nIn \u2206 ABD, it is given that \nAB = AD ….(i) \nIn \u2206 ABC, AB + BC > AC \n=> AB + BC > AD + CD \n=> AB + BC > AB + CD [\u2235AD = AB] \n=> BC > CD \n <\/p>\n
Solution 10. \nIn \u2206ABC, AL is the corresponding height. \nBase = BC \nTo prove: ar(\u2206ABC) = \\(\\frac { 1 }{ 2 }\\) (BC x AL) \nConstruction: From point C and A draw CD || BA and AD || BC, which meet at D \n \n <\/p>\n
Solution 11. \nLet the third side of isosceles triangle be x. \nPerimeter = 30 cm => x + 12 + 12 = 30 \n=> x + 24 = 30 => x = 6 cm \n=> 2s = 30 cm => s = 15 cm \n <\/p>\n
Solution 12. \nThe first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 \n <\/p>\n
Solution 13. \nn = 3, n + 1 = 3 + 1 = 4 \n <\/p>\n
Solution 14. \nLet a\u00b2 – 2a = x \n(a\u00b2 – 2d)\u00b2 – 23 (a\u00b2 – 2d) + 120 = x\u00b2 – 23x + 120 \n= x\u00b2 – 15x – 8x + 120 \n= x(x – 15) – 8(x – 15) \n= (x – 15)(x – 8) \n= (a\u00b2 – 2a – 15)(a\u00b2 – 2a – 8) [Put x = a\u00b2 – 2a] \n= (a\u00b2 – 5a + 3a – 15) (a\u00b2 – 4a + 2a – 8) \n= [a(a – 5) + 3(a – 5)] x [a(a – 4) + 2(a – 4)] \n= (a – 5) (a + 3) x (a – 4) (a + 2) \n= (a + 2) (a + 3) (a – 4) (a – 5)<\/p>\n
Solution 15. \nAbscissa means x-axis (points on xx’ or x-axis) and ordinate means y-axis (points on yy’ or y-axis) \n(i) If ordinate is 5 and abscissa is – 3 => It represents the point (- 3, 5) \n=> Which is in II quadrant. \n(ii) If abscissa is -5 and ordinate is – 3 => It represents the point (-5, – 3) = III quadrant. \n(iii) If abscissa is – 5 and ordinate is 3 => It represents the point (- 5, 3) = II quadrant.<\/p>\n
Solution 16. \nBy the help of graph. \n <\/p>\n
Solution 17. \nPB = QC = DR => AP = PB, DR = RC, BQ = QC \nIn \u2206 PBQ and \u2206 QCR \n \n <\/p>\n
Solution 18. \nABCD is a rhombus. \nOA = OC [Diagonals of rhombus bisect each other] \nIn \u2206\u2019s AOB and COB \n \n \nSo, diagonals of rhombus are perpendicular to each other.<\/p>\n
Solution 19. \nGiven: AB = 6.3 cm, BC = 5.2 cm, CD = 5.6 cm, DA = 7.1 cm and \u2220 B = 60\u00b0. \nSteps of constructions: \nStep 1: Draw line segment AB = 6.3 cm. \nStep 2: Construct \u2220 ABC = 60\u00b0. \nStep 3: With centre B, draw an arc of radius 5.2 cm cutting the ray BX at the point C. \nStep 4: With centre C and radius 5.6 cm draw an arc. \nStep 5: With centre A and radius 7.1 cm draw an arc cutting the arc drawn in step no. 4 at D. \nStep 6: Join C and D, A and D, we get the quadrilateral ABCD. \nStep 7: Join B and D. \nStep 8: Draw a line CE || BD cutting ray AP at E. \nStep 9: Join D and E. We get the required triangle ADE. \n <\/p>\n
Solution 20. \nLet r be the radius of the hemisphere and S be its total surface area. Then S = 3\u03c0r\u00b2 \nThe diameter of the hemisphere is decreased by 30% \n \nThus if the diameter of a hemisphere is decreased by 30%, then its total surface area is decreased by 51%.<\/p>\n
Solution 21. \n(i) Surface area of sphere = 4\u03c0\u00b2 \n(ii) Height of cylinder = r + r = 2r \nRadius of cylinder = r \nC.S.A of the cylinder = 2\u03c0rh = 2\u03c0r(2r) = 4\u03c0r\u00b2 \n \n <\/p>\n
Solution 22. \nTotal number of students among whom the survey was conducted =100 \n40% of 50= \\(\\frac { 40 }{ 100 }\\) x 50 = 20 \nThe students scoring less than 40% marks are the students who scored less than 20 marks. From the given table it is observed that there are5 + 10 = 15 students who scored less than 40% marks. \n\u2234 Required probability = \\(\\frac { 15 }{ 100 }\\) = \\(\\frac { 3 }{ 20 }\\).<\/p>\n
Solution 23. \n \n <\/p>\n
Solution 24. \nIf x\u00b2 – 3x + 2 is a factor of x4<\/sup> – ax\u00b2 + b, then the remainder must be zero. \nx\u00b2 – 3x + 2 = 0 =>x\u00b2 – 2x – x + 2 = 0 \n=> (x – 2)(x – 1) = 0 =>x – 2 = 0 or x – 1 = 0 \n=> x – 1 = 0 => x = 1 \nor x – 2 = 0 => x = 2 \nP(x) = x4<\/sup> – ax2<\/sup> + b \nPutting x = 1, \nRemainder = P(1) = (1)4<\/sup> – a x (1)\u00b2 + b = 0 \n=> 1 – a + b = 0 =>a – b = 1 …(1) \nPutting x = 2, \nRemainder = P(2) = (2)4<\/sup> – a x (2)\u00b2 + b = 0 \n=> 16 – 4a + b = 0 =>4a – b = 16 …(2) \nSubtracting eq. (1) from eq. (2) we get \n \nPutting the value of a = 5 in equation (1) \na – b = 1 \n5 – b = 1 \n– b = – 4 \nb = 4 \na = 5, b = 4<\/p>\nSolution 25. \nStorage time of water = \u2018x\u2019 seconds \nAmount of stored water = y cubic cm \nWater stored per second = 30 cubic cm \nWater stored in \u2018x\u2019 second = 30x \n <\/p>\n
Solution 26. \n \nRadius of base of cone = \\(\\frac { r }{ 2 }\\) = R \nRadius of the base of cylinder = r \nHeight of conical portion = 12 cm \n=> Height of water in cylinder before cone taken out = 12 cm. \n\u2234 Volume of water left in the cylinder when cone is taken out \n \n= \u03c0r\u00b2 (12 – 1) \nHeight of the water left in cylinder = (12 – 1) = 11 cm.<\/p>\n
Solution 27. \nRay OS stands on the line POQ. \n\u2220POS = x \nTherefore \u2220POS + \u2220SOQ = 180\u00b0 \n\u2220POS = x \n <\/p>\n
Solution 28. \n \n(i) In \u2206 ABD and \u2206 ACD \nAB = AC (given) \nBD = CD (given) \nAD = DA (common) \n\u2206 ABD \u2245 \u2206 ACD (sss congruency) \n(ii) In \u2206 ABP and \u2206 ACP \nAB = AC \n\u2220BAP = \u2220CAP \nAP = AP \n=> \u2206 ABP \u2245 \u2206 ACP (sss congruency) \n(iii) From (i) \n\u2206ABD \u2245 \u2206ACD \n=> \u2220BAD = \u2220CAD (CPCT) \n=> \u2220BAP = \u2220CAP \n=> AP bisects \u2220A. \nIn \u2206BDP and \u2206CDP \nBD = CD (given) \nBP = CP [\u2235 \u2206ABP \u2245 \u2206ACP \u2235 BP = CP] \nDP = DP (Common) \n\u2206BDP \u2245 \u2206CDP (SSS congruency) \n=> \u2220BDP = \u2220CDP (CPCT) \n=> DP is bisector of \u2220 D. \n=> AP and DP are on same line segment AP. \nSo AP bisect \u2220A as well as \u2220D. \n(iv) \u2235 \u2206BDP \u2245 \u2206CDP \n=> BP = CP and \u2220BPD = \u2220CPD (CPCT) \n=> BP = CP and \u2220BPD = \u2220CPD = 90\u00b0 \n[ \u2235 \u2220BPD and \u2220CPD are linear pair] \n=> DP is the perpendicular bisector of BC. \n=> AP is the perpendicular bisector of BC.<\/p>\n
Solution 29. \n \nGiven: An arc PQ of a circle C(0, r) and two angles \u2220PRQ and \u2220PSQ are in the same segment of the circle. \nTo prove: \u2220PRQ = \u2220PSQ \nConstruction: Join OP and OQ \nProof: In figure 2 \n\u2220POQ = 2\u2220PRQ \n[ \u2235 The angle subtended by an arc at the centre is twice the angle subtended by it any point in the remaining part of the circle] \n\u2220POQ = 2\u2220PSQ \n=> \u2220 PRQ = \u2220PSQ \n \n <\/p>\n
Solution 30. \nHere the classes are of unequal width, so we use the adjusted frequencies instead of frequency. \n \n <\/p>\n
We hope the CBSE Sample Papers for Class 9 Maths Paper 5 help you. If you have any query regarding CBSE Sample Papers for Class 9 Maths Paper 5, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"
CBSE Sample Papers for Class 9 Maths Paper 5 is part of CBSE Sample Papers for Class 9 Maths\u00a0. Here we have given CBSE Sample Papers for Class 9 Maths Paper 5 CBSE Sample Papers for Class 9 Maths Paper 5 Board CBSE Class IX Subject Maths Sample Paper Set Paper 5 Category CBSE Sample […]<\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[6805],"tags":[],"yoast_head":"\n
CBSE Sample Papers for Class 9 Maths Paper 5 - A Plus Topper<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n