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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 parallelogram and linesegments AX, CY bisect LA and LCrespectively. Show that AX CY. |
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| 2. |
The length of a line segmentAB is 38 cm. A point X on it divides it in the ratio 9: 10. Findthe lengths ofthe line segments AX and XB. |
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| 3. |
Statewhichofthefollowingretriangles.a AB-7 cm, BC 8 cm, AC 7 cm |
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Answer» If you like the solution, Please give it a 👍 |
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| 4. |
| If the area of a circle is 154 cm", then its perimeter is(a) 11cm (b) 22cm (c) 44cm (d) 55cm |
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Answer» 2πr=2*22/7*7=44 cm is right answer area of circle=154let radius be r22/7r^2=154r^2=154*7/22= 7*7r=√7*7=7perimeter (circumference)=2*22/7*7=44cm πr²=154 22/7r²=154r=7cmperimeter=2πr=44 (c) is the right answer option C is a correct answer answer is c option correctly option (c) 44cm is the correct answer ....bcuz 2rootr =2×22/7×7=44 |
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| 5. |
5x+2y-10intersectboth axes.The sides of a triangle are 22cm, 20cm and 18 cm. Find its area. |
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Answer» S=22+20+18/2=60/2=30 Area=√(30(30-22)(30-18)(30-20) Area=√(30(8)(12)(10) Area=√3*5*2*2*2*2*3*2*2*5*2=5*3*8√2=120√2cm² |
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| 6. |
ABC is a triangle, right-angled at C. If AB 25cm and AC-7 cm, find BC. |
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| 7. |
4 Find the value i) sin(330°) i) tan(-945°) |
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Answer» sin(330) = sin(360-30) = -sin(30) = -1/2 tan(-945) = -tan(900+45) = -tan(45) = -1 thnxxxx......... |
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| 8. |
In Fig. 6.13, lines AB and CD intersect at O. IfOC + 2 ВОЕ 700 and < BOD-400, findI.BOE and reflex Z COEFig. 6.13 |
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| 9. |
11. The cost of 3 metres of cloth is 166 What is the cost of2. A cord of length 71 m has been cut into 26 pleth , has been cut into 26 peces of equal length. What is the length ofeach pteee?trehmetres, what is its length? |
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| 10. |
The length of a rectangular hall is 5 metres more than its breadth. Ii the phall is 74 metres, find its length and breadth.oh thot its length is |
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| 11. |
F. A rectangular surface has length 4601surface has length 4661 metres and breadth 3318 metres. On this areales are to be put. Find the maximum length of such tiles. |
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| 12. |
If two circles intersect at two points, prove that their centres lie on the perpendicularbisector of the common chord3. |
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| 13. |
From a rope of the length 40 metres, a man cuts some equal sizespieces can be cut if each piece is of metres length? |
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Answer» The right answer is 90. Total length=40m1 piece=4/9 mNo.of piece in 40 m=40/(4/9)=40*9/4=90 pieces |
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| 14. |
tan (-330°)=? |
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Answer» The value of it is 0.58 |
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| 15. |
3.Convert 330° into radians. |
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| 16. |
1. Which one shows the correct area of the book? Explain your aneua. 330 cmb. 330 cm?c. 74cmd. 74 cm22cm15 cm |
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Answer» The book is rectangular. Area of book = Area of rectangle = length × breadth = 22 × 15 = 330 cm^2 Unit of area will be in square centimeters. Therefore, the correct answer is(b) 330 cm^2 |
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| 17. |
Write the common factors of the following:(a) 35, 55(e) 126, 558(i) 546, 3738(b) 69, 237 (c) 54, 150(g) 350,91) 462, 1078(k) 330, 11(t) 90, 330 |
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Answer» 5 3 6 18 30 7 42 2 11 |
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| 18. |
9. In the given figure, two straight lines AB and CD intersect at a point O. cIf LAOC420, find the measure of each of the angles:(İ)LAOD(i) ZBOD(iii) <COB420zil |
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| 19. |
Thevalueofsin330 degreeis |
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| 20. |
9. In the given figure, two straight lines AB and CD intersect at a point O. CIf LAOC 42°, find the measure of each of the angles:ii) LCOB429(İ) LAOD(ii) /BOD |
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| 21. |
90^circ - sin(-330^circ)*cos(-300^circ) |
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| 22. |
EXERCISE 6.1I. In Fig. o.13, lines AB and CD intersect at O. If C2 BOE and reflex Z COEFig. 6.13 |
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Answer» Aoc + boe = 70 givenbod = 40 aoc = bod =40 (vertically opposite angle)so aoc + boe = 70boe = 70-40=30aob = 180 ( linear pair)aoc+ coe+ boe = 18040 + coe +30 =180coe=180-70=110 now reflex coe reflex coe= 360coe= 110reflex coe = 360-coe360-110=250 .. |
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| 23. |
49, Two circles whose centres are O and O' intersect at P. Through P, a lineI parallel to OO', intersecting the circle at C and D is drawn. Prove thatCD = 200".O' |
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Answer» Draw OL ⟂ CD and O'B ⟂ CDNow, OA ⟂ CD⇒ OA ⟂ CD⇒ CA = AP = 1/2 AP⇒ CP = 2AP ...i)Similarly, O'B ⟂ CD⇒ O'B ⟂ CD⇒ PB = BD = 1/2 PD⇒ PD = 2PB ...ii)Also, CD = CP + PD= 2AP + 2PB = 2(AP + PB) = 2ABCD = 2OO' [∵ OABO' is a rectangle] very very thanks |
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| 24. |
Two circles with center A and B of radii 3 and 4 respectively, intersect at twopoints C and D such that AC and BC are tangents to the two circles. Find thelength of the common chord CD |
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| 25. |
Theorem : The opposite angles formed by two intersecting lines are of equalare of equal mdeasGiven: Line AB and line CD intersect at point O such that A-0-B. C..To prove: (i) ZAOC- ZBOD(ii) ZBOC LAOD0 |
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Answer» Given two lines AB and CD intersect each other at the point O.To prove: ∠1 = ∠3 and ∠2 = ∠4Proof:From the figure, ∠1 + ∠2 = 180° [Linear pair] → (1)∠2 + ∠3 = 180° [Linear pair] → (2)From (1) and (2), we get∠1 + ∠2 = ∠2 + ∠3∴ ∠1 = ∠3Similarly, we can prove ∠2 = ∠4 also. |
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| 26. |
11. In the adjoining figure, D, E, F are the midpointsof the sides BC, CA and AB respectively, of△ABC. Show that EDF=<A, <DEF= B |
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| 27. |
11. In the adjdining figure, D, E, Fare the midpointsof the sides BC, CA and AB respectively, of△ABC. Show that LEDF =<A, <DEF=<Band DFE = C |
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| 28. |
22= 2 x x 1.4 (1.4 + 15)= 2 × 22 × 0.2 (16.4)= 144.32 m27Example 10: The diameter of a roller is 2.5 m and its length is 1.4 m. How much area vover in 7 revolutions?Radius of the roller (cylinder) =-mLength of the rollerCurved surface area of rollerolution:= 1.4m= 2trh22 2.5= 2×ー×ー×1,4 = 11m2We know that-Area covered in one revolution-Curved surface area of rollerTherefore |
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Answer» Radius of the roller (cylinder) = d/2 Length of the roller = 1.4m Curved surface area of the roller = 2πrh = 2 × 22/7 × 2.5/2 × 1.4 = 11 m² We know that, Area covered in one revolution = Curved surface area Therefore, Area covered in 7 revolution = 7 × Curved surface area = 7 × 11 = 77 m² Hence, the roller will cover 77m² in 7 revolution. |
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| 29. |
In each of the following figures ABCD. Find the value of x in each case.BHWCX7350>DPOEDugo<OSO5 ANGLE SUM PROPERTY OF A TRIANGLELet us now prove that the sum of the interior angles of a triangle is 180°.ACTIVITYaw and cut out a large triangle as shown in the figure.war and tear them off. |
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Answer» (i)45° ( ii) 60° (ii)90° is the correct answer.. |
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| 30. |
4x = 720) and 3rthe angles of the triangle are 720, 34Exercise 11.3Find the value of x in each of the following figures:110°50° |
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Answer» i) As opposite side are equal then opposite angle are also equal so x = 50° ii)Use exterior angle theorem x + x = 110° 2x = 110° x = 110°/2 x = 55° ans ans |
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| 31. |
Copy the triangle in each of the following figures on squared paper. In each casedraw the line(s) of symmetry, if any and identify the type of triangle. (Some ofyou may like to trace the figures and try paper-folding first!) |
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| 32. |
of the following figures, one side of a triangle has been produced. Find all the angles of thgle in each case.A50°(11)120° |
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Answer» (I) <x = 125° - 50° <B = 75°(ii). 2x + x = 120° x = 40°<A = 2 × 40 = 80°<C = 40° A.75°B.2x=80° x=40° is the correct answer. x=40digree is a correct answer i) 75° and (ii) ? par copy me bana ke dekna padega x=40° it is right answer I think 20 is a right answer x is 80 degree to Q 1 and x is 120 degree to Q2 1.sum of exterior angle property angle A+angle B=125 degree 50+x=125x=125-50x=752.sum of exterior angle property angle A+angle C =120 degree 2x+x=1203x=120x=120÷3x=40 x=75degree because sum of two interior angle =exterior angle x=40° is the correct answer h i) angle DCB = 125°So angle ACB = 180° - 125° = 55°Now,x = 180° - (50° + 55°)x = 180° - 105°x = 75°ans = the value of x angle is 75°. ii) angle DBC = 120° So angle ABC = 180° - 120° = 60°Now , 2x + x + 60° = 180° 3x + 60° = 180°3x = 180° - 60°3x = 120° x = 120°/ 3x = 40° x = 40°then, 2x = 2 × 40° = 80°ans = The angles of ∆ ABC are 80° , 40° and 60° respectively x + 50°= 125°x = 125°- 50°x = 75° |
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| 33. |
How many lines of symmetry do the following figures have?An isosceles triangle |
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| 34. |
toeĺąą |
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Answer» length of green line 9cm |
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| 35. |
2 tan 301+tan2 30ă) sin 60(B) cos 60°(C) tan 60(D) sin 30 |
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| 36. |
(2) If a tower 30 m high, casts a shadow 10/3 m long on the ground, then what isthe angle of elevation of the sun?(A) 60° (B) 30 (C) 45° (D) 90 |
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| 37. |
ActivityCalculate how many years and months ahead from nowdiary entry is.I. Answer these questions in a few words or a couple of senten1. How old are Margie and Tommy?What did Margie write in her diary?3. Had Margie ever seen a book before?4. What things about the book did she find strange?5. What do you think a telebook is?3) Where was Margie's school? Did she have any classma/ What subjects did Margie and Tommy learn?answer the following with reference to the story."I wouldn't throw it away."(i) Who says these words?Mot defoto |
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Answer» ok just wait for some days there are many questions which questions you want |
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| 38. |
23. The length of a rectangular hall is 5 metres more than its breadth. If the perimeter of thehall is 74 metres, find its length and breadth. |
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Answer» how are you set double x in this math |
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| 39. |
hall is 74 mA wire of lesthan its bread74 metres, find its length and breadth.of length 86 cm is bent in the form of a rectangle such that its lengits breadth. Find the length and the breadth of the rectangle so formeda rectangle such that its length is 7 cm more |
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| 40. |
36m, the given figure, ABCD is a rectangle in which AB 40 cm and BC 25 cm. If P9, R, S be midpoints of AB, BC, CD and DA respectively, find the area of the shaded region. |
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| 41. |
ABC is a triangle, right-angled at C. If AB 25cm and AC 7 cm, find BC. |
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| 42. |
ABC is a triangle, right-angled at C. If AB = 25cm and AC=7 cm, find BC. |
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Answer» This answers can be done by pythageorous theoram The formula is AC=BC+AB According to the questionBC²+AC²=AB BC=24hope this will help you like my answer |
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| 43. |
. Find the intervals in which the function Blvehde fid-sinxcos, 0sstrictlyfsinx-cosx, 0SXS2πntreasing or strictlydecreasing. |
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| 44. |
Very ShorT AHswer Type Questionsуре-α, β are the zeroes of the polynomial x2 + x + 1 then find the value of α + β |
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Answer» Given : x² + x + 1 so alpha + beta = -1/1 = -1 |
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| 45. |
Convert the following from sammaller unit to bigger unit :1. 135 cm in km |
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Answer» 135cm = 1.35m = 1.35/1000 km = 0.00135km thanks |
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| 46. |
Short Answer Type Questions (SLAB-I)1. A square OABC is inscribed in a quadrant OPBQof a circle as shown in the figure, if OA 14 cm.,(CBSE 2008)find area of shaded region. |
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| 47. |
The ratio between the length and breadth of arectangular field is 1 : 3. Its area is â hectares.What is the perimeter of the field?(a) 1000 m(b) 4000 m(c) 400 m(d) 20000 mLong Answer Type Questions8. Solve for x.2x+1417 - 3x.4x + 27x +14 17-38 - 63 - 4825E modeSolve the equations alven below and match the son |
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| 48. |
Example6: Show that the straight lines whosc direction costnes are given by the cquationsare perpandicular,(u+ ) 0 and paralled, fT W+-+--0.(Meerut 2001, 12; Kanpur 09, 11, 14,Kumaun 12, 14) |
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| 49. |
and A, theu find A and bA = 60b 30 |
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Answer» thankss |
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| 50. |
ofhours,spentbysul10UIDUThe numbergiven below:School Home Play Others TotalActivities Sleep42824Number of hours8Present the information in the form of a pie-chart. |
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