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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. |
ABCD is a quadrilateral in which AD- BC andCBA Prove thatDAB(i)BD-AC |
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| 2. |
प्रश्नावली - 13.180 मी.बगल की आकृतियों में एक आयताकार।और एक वर्गाकार खेल के मैदान के माप ।दिए हुए हैं। यदि इनके परिमाप समान हैं।तो किस मैदान का क्षेत्रफल अधिक होगा?80 मी.80 मी.॥ 0980 मी.100 |
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Answer» area of square is greater than area of rectangle area of square is greater than area of rectangle |
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| 3. |
ŕ¤ŕ¤° 43 3-40 |
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Answer» 3x*x - 4√3x + 4 = 0 3x*x - 2√3x - 2√3x + 4 = 0 √3x ( √3x - 2 ) - 2 ( √3x - 2 ) = 0 (√3x - 2)(√3x - 2) = 0 x = 2/√3 If you find this answer helpful then like it. |
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| 4. |
TRहमर काone-1)=08—3d + X 2p2 T |
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| 5. |
जाँच कीजिए कि निम्न द्विवात समीकरण हैं या नहीं() x—2p+1=2x-3 ) xx+ D+8=(x+2) (x |
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Answer» hit like if you find it useful |
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| 6. |
19. In the adjoining figure, ABCD is a square of side 2 cm.Sides AB and BC are produced to P and Q respectivelysuch that BP = CQ = 1cm. AQ and DP are joined. IsAQ = DP ? Give reasons. |
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Answer» the cost of 12 pen is rupees 9 3/5 Yes.AQ= DP Traingle ABQ and Traingle DAP are congruent with AQ and DP being hypotenuse. cuxui8899cccfusxtzfiiiiibovo yes AQ = DP because tiangle ANQ is congruent to triangle DAP ab=bc( since it is a square)bp=cq (given) ab+bp=bc+cqap=bq=3 cm apply pythagorean theory for ADP triangle and ABQ trianglethen AQ=DP=√(2)sq+(3)sq=√13 the cost of pens its 9 3/5 RD Sharma ka question hai BC=AB,So BQ=AP,DA=AB,so by the cost of 12 pen is rupees 93/5 both are congruent by RHS 93/5 is the best answer As , ABQ is congruent to DAP So , AQ=DP Hypotenuses Triangle AQB Congruent Triangle ADP. (by SAS) AQ = DP (by CPCT) |
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| 7. |
12. For what value of k,しー //x + (2k + 1)y-4; kc + 6y_.k+6.the following pair of linear equations has infinite number of solutions: |
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| 8. |
8. In the figure, given below, ABCD is aparallelogram. P is a point on BC such thatBP: PC = 1 : 2. DP produced meets ABproduced at Q. Given the area of triangleCPQ = 20 cmCalculate :i) area of triangle CDP,(ii) area of parallelogram ABCD. |
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Answer» ii) is the correct answer |
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| 9. |
5. Two equal chords AB and CD of a circle withcentre O, intersect each other at point P insidethe circle. Prove that(ii) BP: DPd) |
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Answer» Kchxhkddlhdgshcccysyldjjxhstsyfkidudif Given:AB=CDto prove:PB=PDconst: draw OE and OQ perpendicular on AB and CD respectivelyproof:given AB and CD are two equal chords of the same circle.OE=OQ(equal chords of a circle are equidistant from the center.)now in triangle OEP and OQP,OE=OQOP=OP(common)angle OEP = OQP =90 degree,by constructiontherefore triangle OEP = OQP (RHS congruency)EP = QP (CPCT)also AE=EB=1/2 AB and CQ=QD=1/2CD (the line joining the center of the circle is perpendicular to the chord and bisects the chord.)Now AB = AC implies AE = EB= CQ=QD ....(1)therefore EP-BE =QP - BEEP - BE = QP - QD (FROM 1)BP = PDhence proved |
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| 10. |
Example 4 In Fig 10.33, ABCD is a cyclicquadrilateral in which AC and BD are its diagonalsIf Z DBC 55° and BAC 450, find Z BCD. |
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| 11. |
The sides of a triangle are 3 cm, 4 cm and 6 cmDetermine if it is a right-angled triangle. |
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| 12. |
9. Let Pex,y) be any point on the line jointing the points A(e,0) and B(0. b)Show that-1.a, 0) and |
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| 13. |
12. If LDBC = 70° and LBAC= 40°BCDfind70°A40° |
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Answer» angle BCA = angle BCD = 40 ... ( on the same segment BC) now in ∆ BCD , 70+40 + angle BCD = 180=> angle BCD = 180-70-40 = 70° |
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| 14. |
Find a Quadratic Polynominal whose zeroes are 3+ 75 and4-54. |
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Answer» 3+V5+4-V4=3+4-2+V5=5+V5; product of (3+V5)(4-V4)=12+4V4+4V5-V20=12+4(2)+4V5-2V5=12+8+4V5-2V5=20+4V5-2V5=20+2V5 |
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| 15. |
8. BCD is a rectangle in which diagonal AC bisects Z A as well asVi)ABCD is a square (ii) diagonal BD bisects Z B as well as ZD.C. Show |
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| 16. |
iqure, BnD 80 thendind <BCD |
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| 17. |
10. Find Root ja |
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Answer» √2= 1•414 is the right answer √2 =1.414 is right answer of this question |
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| 18. |
Calculate the area of the shaded region in each of the figures given below:1.5 m E3 m1.5 m1 m40 nm43 |
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| 19. |
for what value of p;x=2,y=3 is a solution ofc (p+1)x-(2p+3)y-1=0 and write the equation |
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| 20. |
Write a rational number between 2 and 3nts drawn at the ends of a diameter of a circle are parallel to et intn the shape of a ri |
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Answer» We know that√2=1.414 and√3=1.732so,rational number between √2 and √3=1.432=1.563=1.576=1.657=1.711 etc. |
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| 21. |
by 2 and 5.mber of natural numbersnts A (4, 3) and B (x.5) are on the circde with the centre 0(2.3), find thei notting the same numbervalue of x |
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| 22. |
A0 C anShew that LD0E İ,a avight angleJa 30 |
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| 23. |
ind the value of p and g for which the following system of linear equation has infinite numbersolutions:2x + 3y = 7 i (p + q)x + (2p-q)),21 |
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| 24. |
Find the area of shaded portion14 cm40 cm40 cm |
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Answer» Area of square As = 40*40= 1600 cm^2 Given, Radius of ciicle r = 14 cm Area of circle Ac = pi*r*r= 22/7*14*14= 22*2*14= 44*14 = 616 cm^2 Area of shaded portion= As - Ac= 1600 - 616= 984 cm^2 |
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| 25. |
\sqrt [ 3 ] { \frac { 1728 } { 2744 } } |
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Answer» m |
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| 26. |
( 12 ) ^ { 3 } = 1728 , \text { then } ( 1.2 ) ^ { 3 } = 1.72 |
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Answer» 1. T2. F , the smallest will be ³√0.008 = 0.2 |
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| 27. |
8.0Find a Quadratic Polynominal whose zeroes are 3+15 and4- JA |
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Answer» sum of the zeros = 3+√5+4-√4 = 7+√5-2 = 5+√5product of zeros= (3+√5)(4-√4) =( 3+√5)(4-2) = (3+√5)×2 = 6+2√5polynomial = x^2-x(sum of the zeros)+product of zerosx^2-x(5+√5)+6+2√5
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| 28. |
The cost of 12 toys is 1728. What is the cost of 1 toy? |
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Answer» PLEASE LIKE THE SOLUTION |
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| 29. |
40. In the given figure, APB and AQ0 are semicircles and AO OB. If theperimeter of the figure is 40 cm, find the area of the shaded region.[CBSE 2015) |
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| 30. |
. Calculate the area of the shaded region in each of the figures even below43 m40 m |
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| 31. |
12. The equation X-2x-n=0 where ne N and ne15.20). The total number ofdifferent values of n for which equation has integral roots is |
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Answer» your question is not clear Root will be an integer when the “square root of (1+n)” must be a perfect square. Between the numbers 5 to 100, perfect squares are 9,16,25,36,49,64,81,100 So, the values of n are 8,15,24,35,48,63,80,99 (so that n+1 is a perfect square) Therefore, there are 8 different values of n between 5 and 100 so that the roots are integers. Read more on Brainly.in - https://brainly.in/question/4907082#readmore |
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| 32. |
किसी समांतर श्रेणी के प्रथम n पदों के योग को S से निरूपितकरते हैं, यदि S, = rp और S = mp (m #n) ,तो(a) 5, =p (b) S, =2p(c) S, = p' (d) S, =2p |
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Answer» right answer is option c option c is a right answer |
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| 33. |
"9 100 - टण्ध न —_‘0100 |
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| 34. |
(1) (2p + q + 5)" |
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| 35. |
1) (2p+q+ 5)2 |
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Answer» (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) (2p + q + 5) ^2= (2p^2) + (q)^2 + (5)^2 + 2((2p)(q) + (q)(5) + (5)(2p))= 4p^2 + q^2 + 25 + 4pq + 5q + 20p |
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| 36. |
3p-2(2p-5)-2(p +3)-8 |
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Answer» 3p - 4p + 10 = 2p + 6 - 8 -3p = -12 so p = 4 |
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| 37. |
solve the equation 2p-6=8+5(p+9) |
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Answer» 2p-6=8+5p+452p-5p=53+6-3p=59p=-59/3 |
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| 38. |
JA Ify -ABe", show thatах |
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| 39. |
If p-25, then find the value of p-3p + 2p+5. |
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Answer» when p = -25p³ -3p² +2p +5(-25)³ - 3 × (-25)² + 2× (-25) +5 = -15,625 - 1875 -50 +5= −17,545 thnx |
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| 40. |
) सतह, + 3९९ Iy AQR - S करJa |
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| 41. |
Ja TCD3) volume of a cuboid sd |
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Answer» volume of cuboid =length *breadth*height lbh is the correct answer to the question |
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| 42. |
(ii) If log,ja 1728 = x, then find x. |
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| 43. |
ta rcached home, KUIthe equations below:(1) 2x +*+1=5the root of each equa |
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Answer» 11/5 is the correct answer of the given question n=11/5 ls the answer of qiestion 2x+(x-1)/2=54x+x-1=10x=11/5 (4x + x - 1) /2 = 5(5x - 1) /2 = 55x - 1 = 105x = 11x = 11/5 correct answer is x = 11/5 11/5 is the correct answer |
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| 44. |
8ft- A= SaWhat is the area of the shadedE 40 in1880 in160 in |
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Answer» 4800 is right answer. |
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| 45. |
In a polygon, there are 5 right angles and theremaining angles are equal to 195° each. Findthe number of sides in the polygon. |
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Answer» Let the number of sides of the polygon be n. therefore the sum of the interior angles of a polygon with sides n is (n-2)* 180. it is given that there are 5 right angles and remaining angles are equal to 195. therefore sum of interior angles = 5*90+(n-5)*195 hence,the above given picture is the solution thus the polygon has 11 sides. |
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| 46. |
1.If sin e = a. then the value of tan (6+) will be(a) + F(0) 1. Salich alphabet. |
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Answer» Answer:√a/√bExplanation option a is the right answer |
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| 47. |
Describe the following sets in Roster fori) ( is a letter of the word 'MARRIAGEi) {xhr is an integer, - < x <iii) fvIx = 2n, ne N}92 |
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Answer» In roster form , these sets are 1) x = {A,E,G,I,M,R}2) x = { 0,1,2,3,4}3) x = { 2,4 ,6... ∞} = { all even numbers} I want third question answer |
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| 48. |
5. In a right angled triangle, one of the angles otherthan the right angle is twice the other, write anequation to find the other two angles. |
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| 49. |
Prove that 5-V3 is an irrational number. |
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| 50. |
ng sets in the set-builder form:(ii){2, 4, 8, 16, 32)(v) (1, 4, 9,..,100) |
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