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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. |
CD. If ZADC 2/ABC, ADa cm and CD- b cm, then the lenghABCD is a trapezium in which AB II(in cm) of AB is:22(A) +2b(B) a + b(C) a b(D) a+ 3b32 |
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| 2. |
AA BM and ΔΑΜC.)4. ABCD is a quadrilateral.Is AB + BC+ CD + DA AC + BD?ABCD is quadrilateral. IsAB + BC+ CD + DA <2 (AC+ BD)?5. |
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| 3. |
la b, f, d are in A.P., then true statements is :(A) ab> cd(B) ac > bd(C) ad> bc |
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Answer» hence ad>bc |
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| 4. |
Proove ihABC, if BD 1AC and BC 2AC. CD, then prove that AB -AC./ 4 + / D-900 Prove that AC, BD" |
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Answer» bc×bc = 2ac×cd not bc=2ac×cd |
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| 5. |
in fig, if AB | | CD, CD lIEF and x : y-3:2, find z. |
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| 6. |
3. In the given figure, CD LAB. IfAD = 4 cm. DB-6 cm and CD-3 cm, what is the value ofsec B?3cmA 4 cm D 6 cm B |
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| 7. |
IV.In Fig. 6.53, ABD is a triangle right angled at Aand AC丄BD. Show that(i) AB-BC. BD(ii) AC-BC, DC(iii) AD-BD, CDFig. 6.3 |
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| 8. |
13. In Î ABC, seg AD l seg BCDB = 3CD. Prove that :2AB2 = 2AC" + BC2 |
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| 9. |
6. In fig AC BD, then prove that AB CD |
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| 10. |
In Fig. 5.10, if AC BD, then prove that AB CD.Fig. 5.10 |
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| 11. |
6.In Fig. 5.10, if AC-BD, then prove that AB CDFig. 5.10 |
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Answer» Thank you |
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| 12. |
A saree 4m long at 1.25m wide has a border of width 25 cm all around it on inside find the cost of knitting the border at 20 paise per cmsquare ? |
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Answer» Area of saree = length *breadth= 4*1.25= 5 m^2 Area with border = length * breadth = (4 + .5)*(1.25 + .5)= 4.5*1.75= 7.875 m^2 Area of border = 7.875 - 5= 2.875 m^2 Cost of knitting border= 287.5*20= 5750 paise or Rs 57.50 |
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| 13. |
-6 In Fig. 5.10, if ACBD, then prove that AB-CDB.A-Fig. 5.10 |
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| 14. |
Q.21. Name one synthetic antioxidant.Ans. Butylated hydroxyanisole (BHA). |
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Answer» Butylated hydroxyanisole (E320) is asynthetic antioxidantwhich works by stabilising free radicals. Butylated hydroxytoluene (E321) or BHT is anothersynthetic antioxidant.Oneof the widely used is BHA. |
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| 15. |
. If the volume of cube is same as that of acuboid of dimensions 27 m x 9 m x 3 m. Find the edgeof the cube. |
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| 16. |
If α and β are the zeros of the quadratic polynomal j (x)-e quadraticpolynomial f(x)=x^2+x - 2, find theva is negative of th |
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| 17. |
3) In AABC seg AP is a median. If BC 18 and AB +AC 260, find AP. |
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Answer» In ∆ ABC , AP is a median , BC = 18 & AB² + AC² = 260In ∆ ABC , seg AP is a median. P is the midpoint of seg BCBP = PC = 1/2BCBP = PC = ½(18) = 9 AB² + AC² = 2AP² + 2PC² [ By APOLLONIUS THEOREM, this theorem tells us the relation among the sides and medians of a triangle.]AB² + AC² = 2(AP² + PC²)260 = 2(AP² + 9²)260/2 = AP² + 81130 = AP² +81130 - 81 = AP²49 = AP²AP = √ 49 = 7 AP = 7 Hence, the value of AP is 7 Like my answer if you find it useful! |
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| 18. |
3. A piece of cloth of length 8.5 m and width 2.5 mvhas a 25 cm wide border inside it. Find theVcost of embroidering the border at the rate of15.50 per sq.m. |
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| 19. |
x^3+4x^2-4x-16 Answer (x-2)(x+2)(x+4) |
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Answer» x^2(x+4)-4(x+4)=(x^2-4)(x+4)=(x-2)(x+2)(x+4) thanks |
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| 20. |
EXERCISE 7.3Δ ABC and Δ DBC are two isosceles triangles onthe same base BC and vertices A and D are on thesame side of BC (see Fig. 7.39). IfAD is extendedto intersect BC at P, show that1.LetunequalActivitC side BC(ii) Δ ΑΒΡ Δ ACP(iii) AP bisects 2 A as well as < D(iv) AP is the perpendicular bisector of BC.AD is an altitude of an isosceles triangle ABC in which ABFixat thehat pencilFig. 7.39-AC. Show l2.. Show |
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Answer» hit like if you find it useful |
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| 21. |
find the remainder when x cube + 4x square + 4x - 5 is divided by x |
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Answer» -5 will be the remainder |
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| 22. |
4x+4x=16 then x=? |
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Answer» x=2 is the right answer |
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| 23. |
2. In AABC, AP L BC, BQ L ACIf B-P-C, A-Q-C then show thatC.BC 12 then AC? |
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Answer» Like if you find it useful |
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| 24. |
कि का हर !5N~ | ही 00:09 0. Qum Da 2230 and -लि: o o eSO |
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Answer» 2x + 3y + (-5x - 2y) - ( 2x - 2y) = 2x + 3y - 5x - 2y - 2x + 2y = -5x + 3y |
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| 25. |
Exercise T(A)If for a particle in rectilinear motion, seso112, show that thevportional to time and acceleration is constant. |
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| 26. |
21. Name one synthetic antioxidant. |
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Answer» Butylated hydroxyanisole (E320) is asynthetic antioxidantwhich works by stabilising free radicals. Butylated hydroxytoluene (E321) or BHT is anothersynthetic antioxidant.Oneof the widely used is BHA. Like my answer if you find it useful! |
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| 27. |
2x²+9x+10 |
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Answer» 2x^2+9x+10=2x^2+4x+5x+10=2x(x+2)+5(x+2)=(2x+5)(x+2) |
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| 28. |
p(x)=4x+5/3x-9=f(4)/f(9) |
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| 29. |
0 b ंए92- [2v av S\_.. 190 Gहा W QY pLoeo1 Mw ap v 5 | 20 g&%19 Atyपाक 2 7‘ll |
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| 30. |
w AP L BC to show thats.ABC is an iis an isasceles triangle with AB AC. Draw AP |
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| 31. |
X+2x²+3x= |
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Answer» if equating to 0 sox +3x + 2x² =04x +2x² = 02x ( 2 + x) = 0so x = -2 |
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| 32. |
2x²+7x+3 |
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| 33. |
(2x+8)(2xâ8) |
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Answer» 4x²-16 is your Answer mate Simplify using identities 68 |
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| 34. |
x³-2x²-x+2 |
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| 35. |
(iv) 3x+2xâ [N (V) |
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| 36. |
T HINT 2xâ3 =y & |
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Answer» 2x - 3 = x + 2 2x - x = 2 + 3 x = 5 |
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| 37. |
Find the value of the equation 2x²-19x+35=0 by factorisation method. |
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Answer» 2x^2 -19x +35 = 0 2x^2 -14x -5x +35 = 0 (2x - 5)(x-7)=0X = 5/2;7 |
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| 38. |
The numerator and denominator of a rational number are in the ratidenominator is increased by 3, the ratio become:s3 :5. Find the ration |
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Answer» Let the numerator be x and denominator be yx/y=3/44x=3y4x - 3y=0 .....(1) x/y + 3= 3/55x=3(y+3)5x=3y+95x-3y=9 .....(2) Applying elimination in (1) and (2) 4x-3y=05x-3y=9- + -_______-x=-9x=9 Putting x=9 in (1)4(9)-3y=0-3y= -36y=12 Rational number is 9/12 |
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| 39. |
D. The Ryth ter of nAnd tk'WAP: 3, 15, 27, 39, .. . will be l 32 more than its 54th term?hich term of the / |
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| 40. |
VV hich term of the AP 21 ,,3.If P = ab, and Q = ab , a and b are the prime numbers what is the LCM of P and Q |
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Answer» not satisfied |
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| 41. |
째:4x-32x+2 + 25 =0 |
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| 42. |
Solve the following quadratic equations (real roots only) by the method ofcompleting the perfect square:(i) x2 +3x-10=0(iii). I2xn_ 17x + 6-0(ii) x2-6x+ 25 = 0(iv) 4x2-12x +1=0(vi) 62,2 tr-2=0 |
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| 43. |
2. If A = {0, 1} and B = { 1, 2, 31, show that A × BB x A |
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Answer» A X B = { (0,1) (0,2) (0,3) (1,1) (1,2) (1,3)}B X A = { (1,0) (1,1) (2,0) (2,1) (3,0) (3,1)} hence A X B is not equal to B X A |
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| 44. |
2x2-6x+3= 0 ସମୀକରଣର ମୂଳଦ୍ୟ ଓ ଓ } ହୁଏ ତେବେ (+1)(B+1) ର ମୂଲ୍ୟ2x2 – (p +1)x + p-1 = 0 ସମୀକରଣର ମୂଳଦ୍ବୟର ଅନ୍ତର ଓ ଗୁଣଫଳ ସମାନ ରେଣ କର ।x2 -3x -2 = 0 ସମୀକରଣର ମୂଳଦ୍ବୟ & ଓ B ହୁଏ, ତେବେ ପ୍ରମାଣ କର ଯେ ।1-2417x+କର= + 17x+6= 0 ସମୀକରଣର ମୂଳଦୟ ଓ ୫ B ହୁଏ ତେବେ ? ର ମୂଳ=16p = 0 ସମୀକରଣର ମୂଳୟ ଓ ଓ } ହେଲେ ପରିପ୍ରକାଶକୁ ୧ ମାଧବ |
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| 45. |
Consider the binary operation on Q defined byx^{*} y=1+12 x+x y, \forall x, y \in Q, then 2*3 equal's:a) 31(c) 43(b) 41(d) 51 |
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Answer» Here x=2 Andy=3By the given operation Answer is1+12*2+2*3 =1+24+6 =31 |
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| 46. |
Factorise 2x³-3x²-17x+30 |
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Answer» We can write the given equation as, =>(x-2)(2x²+x-15)=0=>(x-2)(2x²+6x-5x-15)=0=>(x-2)(x+3)(2x-5)=0Factorized |
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| 47. |
F() 3x^2 + 17x+24 () sqrt3x^2 +11x |
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| 48. |
реореж реорео ok g = )/l 2. e e kel Re A b |
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Answer» cosec²θ - cot²θ =1(cosec θ - cot θ)(cosecθ + cotθ) = 11/2 * (cosecθ + cotθ) =1(cosecθ + cotθ) = 2 |
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| 49. |
Example 4.5diA farmer has a field in the shape of a rhombus. The peris 400 m and one of its diagonal is 120 m. He wants to divide the fieldparts to grow two different types of vegetables. Find the area of the fieldeld |
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Answer» Like if you find it useful |
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| 50. |
Solve the following equations.1. x-2=7 |
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Answer» Sox-2= 7x= 2+7x= 9thanks how many edge in cube answer== x-2=7 so, x=7+2 x=9 the answer is equal to x=9 x-2=7×= 7+2 ×= 9it is the answer you can recheck |
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