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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Find the total surface area of the cube whose volume is 343cm³ |
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Answer» PLEASE LIKE AND SHARE THIS APP |
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| 2. |
c) ABC is an equilateral triangle C is a mid point of DE. ADAC & A EBC are equal & supplementry. Prove ADACA EBC |
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| 3. |
ABCD is a quadrilateral in which AB = CD and AD =BC. Show that it is aparallelogram. [Hint: Draw one of the diagonals.] |
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| 4. |
Find the volume and total surface area of acube whose each edge is:(i) 8 cmfit) 2 m 40 cm |
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| 5. |
In the given figure, ABICD and BCI ED. Find the value of x.ure, AB | CD and BCI ED. Find the value of x.xo75% |
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Answer» Here angle EDC + angle DCB = 180° ( adjacent angles ) so, angle DCB = 180°-75° = 105° NOW angle DCB = Angle CBA (x).... ( alternate interior angles) so, x = 105° |
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| 6. |
12. Show that line joining (4, 1) and (6,0) is parallel to linéjoining (7-2) and (5,-3). |
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| 7. |
Divide 56 into 4 parts which are in AP such that the ratio of the products of its extremes to the products of its means is 5:6. |
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Answer» 4 no =a-3d,a-d,a+d,a+3dsum , 4a=56a=14(a-3d)(a+3d)/(a-d)(a+d)=5:66(a²-9d²)=5(a²-d²)a²=54d² - 5d²14² =49d²d²=14² /7²d=±14/7=±2for d=2, a-3d=14-6=8a-d=14-2=12a+d=14+2=16a+3d=14+6=20Numbers are 8,12,16,20 |
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| 8. |
1) How many rational numbers can you find between 3 and 4? |
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Answer» there are infinite numbers between to rational numbers |
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| 9. |
2) AB,I) AC18.The line joining (-5, 7) and (0, -2) is perpendicular to the line joining (1,-3) and (4,x),Find x. |
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| 10. |
16. In the adjoining figure, X and Y arerespectively two points on equal sides AB andAC of AABC such that AX AY. Prove that |
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| 11. |
16. In the adjoining figure, X and Y arerespectively two points on equal sides AB andAC of △ABC such that AX=AY. Prove thatCX = BY. |
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| 12. |
e adjoining figure, P and 9 are two potnts on equal sides ABof an isosceles triangle ABC such that AP AQ.prove that BO CP |
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| 13. |
27. In the figure, X and Y are the points on equal sides AB and AC ofa AABC such that AX-AY. Prove that XC-YB |
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| 14. |
16. In the adjoining figure, X and Y arerespectivel two points on equal sides AB andAC of ABC such that AX#AY. Prove thatCX = BY. |
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| 15. |
n Fig. 9.43, P and Q are two points on equal sides AB and AC of an isoseles triangleC such that AP AQ. Prove that BQ cP. |
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Answer» thanks |
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| 16. |
in the adjoining figure, P and Q are two points on equal sides AB and AC of an isosceles triangle ABCshat AP AQ. Prove that BQ- CP |
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| 17. |
tn the adjoining figure. P and 9 are two points on equal sides ABln AC of an isosceles triangle ABC such that APanAQ.Prove that BO CP |
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| 18. |
the identity for the addition of rational numbers is |
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Answer» thanks mayank kumar |
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| 19. |
Find the side of such, a cube whose totalsurface area is 600 cm^2 |
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| 20. |
2.AB such that AD = 4 cm. At B drawDraw a line segment AB = 6 cm. At A, draw ADBC 1 AB such that BC 4 cm. Join C and D. What type of figure is ABCD? |
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Answer» AB = CD = 6cm. BC = AD = 4cmThe angles are all right angles.Hence, the figure ABCD is a rectangle. PLEASE HIT THE LIKE BUTTON |
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| 21. |
AB = 6 सेमी०, AD = 4 सेमी०, BD = 5 सेमी०, AC = 6 सेमी तथा BC = 7 सेमी० |
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Answer» the answer of the question is 5 5cm the correct answer of the given question yes the answer is absolately 5 cm. |
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| 22. |
2. दिलेल्या आकृती 1.13 मध्ये 907 | 8,AD | AB,BC =4, AD = 8 A(A ABC)A_—_( A ADB) hlel. |
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| 23. |
22. In the figure, if AB II CD, <APO-50째 and ARD = 125째, find y-x.50째125째C QR D |
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Answer» As AB || CDy + 50° = 125 ( Alternate angle) so , y = 125° - 50° = 75° x = 50° (Alternate angle) So, y - x = 75° - 50° = 25° |
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| 24. |
9. If Q(0, 1 is equidistant from P(5, -3) and R(x, 6), find the values of x. Also find thedistances QR and PR. |
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| 25. |
Find $k$ so that $x^{2}+2 x+k$ is a factor of $2 x^{4}+x^{3}-14 x^{2}+5 x+6 .$ Also find all the zeros of the twopolynomial. |
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Answer» 1 2 |
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| 26. |
Theanglesofelevationofthetopof a TV tower from three points A, B and C in a straight li(in the horizontal plane) through the foot of tower are a, 2a and 3α respectively. IfAB-a, the heigh,of tower is:(A) a tan α(B) a sin α(C) a sin 2α(D) a sin 3α |
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| 27. |
Angle addition identity |
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Answer» Angle addition formulas express trigonometric functions of sums of angles alpha+/-beta in terms of functions of alpha and beta. The fundamental formulas of angle addition in trigonometry are given by sin(alpha+beta)=sinalphacosbeta+sinbetacosalpha(1)sin(alpha-beta)=sinalphacosbeta-sinbetacosalpha(2)cos(alpha+beta)=cosalphacosbeta-sinalphasinbeta(3)cos(alpha-beta)=cosalphacosbeta+sinalphasinbeta(4)tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)(5)tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta).(6)The first four of these are known as the prosthaphaeresis formulas, or sometimes as Simpson's formulas. The sine and cosine angle addition identities can be compactly summarized by the matrix equation [cosalpha sinalpha; -sinalpha cosalpha][cosbeta sinbeta; -sinbeta cosbeta]=[cos(alpha+beta) sin(alpha+beta); -sin(alpha+beta) cos(alpha+beta)]. (7)These formulas can be simply derived using complex exponentials and the Euler formula as follows. cos(alpha+beta)+isin(alpha+beta)=e^(i(alpha+beta))(8)=e^(ialpha)e^(ibeta)(9)=(cosalpha+isinalpha)(cosbeta+isinbeta)(10)=(cosalphacosbeta-sinalphasinbeta)+i(sinalphacosbeta+cosalphasinbeta).(11)Equating real and imaginary parts then gives (1) and (3), and (2) and (4) follow immediately by substituting -beta for beta. Taking the ratio of (1) and (3) gives the tangent angle addition formula tan(alpha+beta)=(sin(alpha+beta))/(cos(alpha+beta))(12)=(sinalphacosbeta+sinbetacosalpha)/(cosalphacosbeta-sinalphasinbeta)(13)=((sinalpha)/(cosalpha)+(sinbeta)/(cosbeta))/(1-(sinalphasinbeta)/(cosalphacosbeta))(14)=(tanalpha+tanbeta)/(1-tanalphatanbeta).(15)The double-angle formulas are sin(2alpha)=2sinalphacosalpha(16)cos(2alpha)=cos^2alpha-sin^2alpha(17)=2cos^2alpha-1(18)=1-2sin^2alpha(19)tan(2alpha)=(2tanalpha)/(1-tan^2alpha).(20)Multiple-angle formulas are given by sin(nx)=sum_(k=0)^(n)(n; k)cos^kxsin^(n-k)xsin[1/2(n-k)pi](21)cos(nx)=sum_(k=0)^(n)(n; k)cos^kxsin^(n-k)xcos[1/2(n-k)pi],(22)and can also be written using the recurrence relations sin(nx)=2sin[(n-1)x]cosx-sin[(n-2)x](23)cos(nx)=2cos[(n-1)x]cosx-cos[(n-2)x](24)tan(nx)=(tan[(n-1)x]+tanx)/(1-tan[(n-1)x]tanx).(25)TrigAnglesWeissteinThe angle addition formulas can also be derived purely algebraically without the use of complex numbers. Consider the small right triangle in the figure above, which gives a=(sinalpha)/(cos(alpha+beta))(26)b=sinalphatan(alpha+beta).(27)Now, the usual trigonometric definitions applied to the large right triangle give sin(alpha+beta)=(sinbeta+a)/(cosalpha+b)(28)=(sinbeta+(sinalpha)/(cos(alpha+beta)))/(cosalpha+sinalpha(sin(alpha+beta))/(cos(alpha+beta)))(29)cos(alpha+beta)=(cosbeta)/(cosalpha+b)(30)=(cosbeta)/(cosalpha+sinalpha(sin(alpha+beta))/(cos(alpha+beta))).(31)Solving these two equations simultaneously for the variables sin(alpha+beta) and cos(alpha+beta) then immediately gives sin(alpha+beta)=(cosalphasinalpha+cosbetasinbeta)/(cosalphacosbeta+sinalphasinbeta)(32)cos(alpha+beta)=(cos^2beta-sin^2alpha)/(cosalphacosbeta+sinalphasinbeta).(33)These can be put into the familiar forms with the aid of the trigonometric identities (cosalphacosbeta+sinalphasinbeta)(sinalphacosbeta+sinbetacosalpha)=cosbetasinbeta+cosalphasinalpha (34)and (cosalphacosbeta+sinalphasinbeta)(cosalphacosbeta-sinalphasinbeta)=cos^2alphacos^2beta-sin^2alphasin^2beta(35)=1-sin^2alpha-sin^2beta(36)=cos^2alpha-sin^2beta(37)=cos^2beta-sin^2alpha,(38)which can be verified by direct multiplication. Plugging (◇) into (◇) and (38) into (◇) then gives sin(alpha+beta)=sinalphacosbeta+sinbetacosalpha(39)cos(alpha+beta)=cosalphacosbeta-sinalphasinbeta,(40)as before. TrigAdditionSmileyA similar proof due to Smiley and Smiley uses the left figure above to obtain sinalpha=(sin(alpha+beta))/(cosbeta+(sinbetacosalpha)/(sinalpha)), (41)from which it follows that sin(alpha+beta)=sinalphacosbeta+sinbetacosalpha. (42)Similarly, from the right figure, (sinalpha)/(cosalpha)=(cosbeta)/(sinbeta+(cos(alpha+beta))/(sinalpha)), (43)so cos(alpha+beta)=cosalphacosbeta-sinalphasinbeta. (44)TrigSubtractionSmileySimilar diagrams can be used to prove the angle subtraction formulas (Smiley 1999, Smiley and Smiley). In the figure at left, h=(cosalpha)/(cosbeta)(45)x=hsin(alpha-beta)(46)=(sinalpha-hsinbeta)cosalpha,(47)giving sin(alpha-beta)=sinalphacosbeta-cosalphasinbeta. (48)Similarly, in the figure at right, h=(cosalpha)/(sinbeta)(49)x=hcos(alpha-beta)(50)=(sinalpha+hcosbeta)cosalpha,(51)giving cos(alpha-beta)=cosalphacosbeta+sinalphasinbeta. (52)TanSubtractionRenA more complex diagram can be used to obtain a proof from the tan(alpha-beta) identity (Ren 1999). In the above figure, let BF/BE=AD/DE. Then tan(alpha-beta)=(DE)/(BE)=(AD)/(BF)=(tanalpha-tanbeta)/(1+tanalphatanbeta). (53)An interesting identity relating the sum and difference tangent formulas is given by (tan(alpha-beta))/(tan(alpha+beta))=(sin(alpha-beta)cos(alpha+beta))/(cos(alpha-beta)sin(alpha+beta))(54)=((sinalphacosbeta-sinbetacosalpha)(cosalphacosbeta-sinalphasinbeta))/((cosalphacosbeta+sinalphasinbeta)(sinalphacosbeta+sinbetacosalpha))(55)=(sinalphacosalpha-sinbetacosbeta)/(sinalphacosalpha+sinbetacosbeta). |
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| 28. |
12.1. Write a quadratic polynomial, the sum and productof whose zeroes are 3 and -2 [Delhi 2008) |
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| 29. |
lillfA. B and C are the angles of a ÎABC, then dow hatB+C |
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| 30. |
Obain all zeroes of 3- 1513+25x-30, if two of is zeroes areand |
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Answer» Like if you find it useful |
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| 31. |
ohiObtain all zeroes of 3x-15+132+25-30, if two of its zeroes aread |
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| 32. |
13. If the zeroes of the polynomial 64x3 - 144x2 + 92x -15 are in A.P find the zeroes. |
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Answer» As three zeroes are in AP let three zeroes of 64x^3 - 144x^2 + 92x - 15 are (a - d), a, (a + d) (a-d) + a + (a+d) = - (-144)/643a = 9/4a = 3/4 Now, (a-d)*a*(a+d) = - (-15)/64(3/4 - d)*3/4*(3/4 + d) = 15/64(3/4 - d)(3/4 + d) = 5/169/16 - d^2 = 5/16d^2 = 4/16 = 1/4d = 1/2 When a = 3/4 and b = 1/2 roots are (a-d) = 3/4-1/2 = 1/4a = 3/4(a+d) = 3/4+1/2 = 5/4 |
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| 33. |
t. In the adjoining figure, P and 9 are two points on equal sidesand AC of an isosceles triangle ABC such that AP A.Prove that Bg CP |
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Answer» thanks |
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| 34. |
The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12, Findthe area of the triangle. Also, find the cost of cultivating the field at 24-60 per 100 ㎡ |
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| 35. |
by lfx in 12-x=x also find the value of 2x2 - 1. |
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Answer» Herex=12-xx=62(6 x 6)-171answer |
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| 36. |
Find the value of x for which the number x806 is divisible by 9. Also, find the number12.ind the vlue of zfo |
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| 37. |
Q,12, Sum of the first n terms of an APis52 3n Find the AV ard also find its 16" term |
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| 38. |
ate the process of addition. Or, for the non existing terms, we aero coefficients.mple 1.2ind out the sum of the polynomials 3x-y,2y-2x and x +yolutionow methodolumnmethod of addition3x -y)+ (2y |
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Answer» thank you 3x-y-2x+2y+x+y=2(x+y)=2x+2y. |
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| 39. |
what is addition? |
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Answer» adding numbers is called addition |
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| 40. |
10. Write quadratic polynomial whose zeroes are 2+1 |
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| 41. |
ABCD is a rectangle. M is the mid-point ofACand CPMN is a rectangle. Prove that:(i) P is the mid-point of CD(i) PN-AC2 |
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Answer» 1 2 |
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| 42. |
Ritika scored 320 marks in an examination. The minimum pass marks were 300. By what per cent00 are her marks over the minimum pass marks? |
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Answer» scored 320.which is 20 more than passing mark.so percentage is 6.67%so by 6.67% her marks more tha passing marks. |
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| 43. |
A shopkeeper marks his goods 20%above the cost price, but allows 30% discount for cash. What is hisnet loss per cent? |
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Answer» Let the cost price be x then marked Price = 1.2xDiscount = 30 % =. 36 xSelling price =. 84x.So net loss =. 16x = 16% |
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| 44. |
nhad is the per cent of these marks out of40-35, 36 25, 26,30 |
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Answer» 1)35/40*100=87.5%2)36/40*100=90%3)25/40*100=62.5%4)26/40*100=65%5)30/40*100=75% your overall percentage is 76% |
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| 45. |
find quadratic polynomial whose sum and product of zero are ½ and 4 |
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Answer» Given,Sum of zeroes = 1/2Product of zeroes = 4 Let a and b are zeroesa + b = 1/2ab = 4 Therefore,Quadratic equation is given byx^2 - (a + b)x + ab = 0x^2 - x/2 + 4 = 02x^2 - x + 8 = 0 Bhai Answer Hoga = 0 |
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| 46. |
Obtain a quadratic polynomial, whose addition ofzero is 2 and multiplication is -3 |
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| 47. |
EXAMPLE 3.55Find dy of V = s ezananz |
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| 48. |
obtain the quadratic polynomial whose addition of zero is 2 and multiplication is -3 |
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Answer» one more question |
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| 49. |
✓5v -VO0 ) V - V(i) In a cyclic quadrilateral ABCD, if ZABC = 72° then the value of ZCDA is |
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Answer» In cyclic quadrilateral the sum of each pair of opposite angle is 180, then 180-72 equal to 108 so angle CDA is of 108 degree 180°is the right answer |
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| 50. |
26. In a cyclic quadrilateral ABCD, cos A +cos B + cos C + cos DA) 90°B) 0°C) 80° |
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