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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. |
for the linear equation 3x-5y-15=0 find the point where its graph intersect x axis and y axis find the area of figure |
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| 2. |
Q.3. Find the point where line 3x + 6y-40 intersectthe r-axis. |
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Answer» 3x+6y= 4dividing by 43x/4+6y/4= 1x/(4/3)+6/(4/6)=1intersection point will be 4/3 on x axis |
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| 3. |
QUADRILATERALS5. In a parallelogram ABCD, E and F are the Dmid-points of sides AB and CD respectively(see Fig. 8.31). Show that the line segments AFand EC trisect the diagonal BD.Fig. 8.31 |
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| 4. |
tan inverse (5x/ 1-6x^2) find the dy/ dx |
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Answer» thnx tan inverse (2x/1+15x^2) find dy/ dx |
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| 5. |
11. lia Δ ABC and Δ DEF,AB-, DE, ABIDE, BCaEFand BC II EE Vertices A, B and C are joined tovertices D, E and F respectively (see Fig 8.22).Show thato quadrilateral ABED is a parallelogram(a) quadrilateral BEFC is a parallelogram(ii) ADII CF and AD m CF(iv) quadrilateral ACFD is a parallelogramFig 8.22 |
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| 6. |
Rationalise the denominator ofRationalise the denominator of7+312 |
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| 7. |
Rationalise the denominator |
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| 8. |
4. Rationalise the |
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Answer» Thnks |
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| 9. |
5. In a parallelogram ABCD, E and F are themid-points of sides AB and CD respectively(see Fig. 8.31). Show that the line segments AFand EC trisect the diagonal BD.Fig. 8.31 |
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Answer» padhai karo agar padhai nahi karoge to pagal hein tu yea kay ha |
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| 10. |
8. In a parallelogram ABCD, AB = 27 cm, BC-18 cm, APL DC, and AQ LBC. If AP-9.6 cm.find the length of AQ.D P242 |
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Answer» area of ABCD = base×height =AB xAP = 27x 9.6 = 259.2this area is also equal = BCxAQ=259.218×AQ=259.2AQ = 259.2÷18AQ = 14.4 c.m |
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| 11. |
\sin 65^{\circ}+\cos 65^{\circ}=\sqrt{2} \cos 20^{\circ} \text { (ii) } \cos 18^{\circ}-\sin 18^{\circ}=\sqrt{2} \sin 27^{\circ} |
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| 12. |
E REAL1 Use Euclid's divisi(i) 135 and 225 |
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Answer» thnx |
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| 13. |
5. In a parallelogram ABCD, E and F are themid-points of sides AB and CD respectively(see Fig. 8.31). Show that the line segments AFand EC trisect the diagonal BD.Fig, 8.31 |
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| 14. |
\operatorname { cos } 6 x = 32 \operatorname { cos } ^ { 6 } x - 48 \operatorname { cos } ^ { 4 } x + 18 \operatorname { cos } ^ { 2 } x - 1 |
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Answer» Nice |
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| 15. |
6. In a parallelogram ABCD, E and F are themid-points of sides AB and CD respectively(see Fig. 8.31). Show that the line segments AFand EC trisect the diagonal BD.Fig. 8.31 |
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| 16. |
s In a parallelogram ABCD, E and F are themid-points of sides AB and CD respectively(see Fig. 8.31). Show that the line segments AFand EC trisect the diagonal BDFig 8.31 |
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| 17. |
( x %2B 5 ) ( x %2B 5 ) |
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Answer» ( x + 5)( x + 5) ( x + 5)² x² + 10x + 25 |
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| 18. |
23 - 4 - 3 - 34 - 3 |
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Answer» =23 - 4 - 3 -34 - 3=19 - 3 - 34 - 3=16 - 34 - 3=-21 |
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| 19. |
9.) In the Fig. APL BC. PC > PB and PQ PB. Show that AC> AB |
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| 20. |
PQ 11 BC and AP : PB = 1 : 2, Find area(AAPQ)area (Δ ABC)Fig. 7.238 |
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| 21. |
18. In the figure, biseciors of 4B and 4D of quadrilateral ABCD meet CD and Ab produced at aQ respectively. Prove ilat, .2243P D |
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Answer» InΔPBC we have,angle p +pbc+c=180p+1/2 B+C=180⇒eq 1 InΔqad we have Q+A+ADQ=180Q+A+1/2 D=180⇒eq 2 Adding 1 and 2P+Q+A+C+1/2 B+1/2 A=360 But,A+B+C+D=360P+Q+A+C+1/2(B +D)=A+B+C+D P+Q=1/2 (B+D)P+Q=1/2(ABC+ADC) thanks you |
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| 22. |
In te adjacent figure ΔΑ BC is isosceles as AB AC, BAand CA are produced to Q and P such that AO AP. Showthat PB QC(Hint: Compare ΔΑΡΒ and ΔΑΟΟ) |
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| 23. |
0 3'निकालें G ]343. तथा ? का मान हे पाला । |
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| 24. |
Find the height of the triangle whose area is 24? square units and base is Ba units? |
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Answer» 6a. thanks u r from where. |
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| 25. |
Prove that 2H2In the Fig. APL BC. PC > PB and PQ = PB. Show that AC > ABRC with AB - AC. Show that CD |
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| 26. |
n the given figure, a semicircle is drawn with O as centre and AB as diameter.Semi circles are drawn with AO and BO as diameter. If AB-28m, find theperimeter ofthe shaded region. (Use Ď = 22)24om ns m 6m 75 cm and |
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| 27. |
UITOl ltS Siues.ediagonals ofthe quadrilateral are perpendicular to each other. Is such a quadrilateraljways a rhombus? Draw a rough figure tojustify your answer. |
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| 28. |
uate the following:sin 7sB)g cos 18 cos 78 sin18cOS2. |
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Answer» sin A cos B - cos A sin B = sin ( A - B )so sin (78-18)= sin 60 =√3/2 thnx |
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| 29. |
2(23*34)-3(5+25) |
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Answer» 2(23×34)-3(5+25)=2(782)-3(30)=1564-90=1474 |
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| 30. |
TR 3+ 5) न 9 आणि 57 + 30 न 7, तर 3 + + ची |
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Answer» 3x + 5y = 9........ (1)5x + 3y = 7.........(2) Multiply eq(1) by 5 and eq(2) by 3 15x + 25y = 45.....(3)15x + 9y = 21.......(4) eq(3) - eq(4)16y = 24y = 24/16 = 3/2 Put value of y in eq(1)3x + 5*3/2 = 93x = 9 - 15/23x = 3/2x = 1/2 Therefore, x + y = 3/2 + 1/2 = 4/2 = 2 |
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| 31. |
art-A) Choose and write the correct optan (90° – 60°) CHI HA ETT :(34) 3 |
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Answer» tan (90° - 60°)= tan 30°= 1/√3 |
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| 32. |
LOODUDU, U IS ca.7. If the point P(3, 4) is equidistant from the points Aa + b, b - a) and Bla-b, a + b), then prove that3b - 4a = 0.COP |
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Answer» the intellectual and practical activity encompassing the systematic study of the structure and behaviour of the physical and natural world through observation and experiment. the amazon will have be so at the bed . The Amazon will have be sat the bed....athe answer is right Amazon will have be sat the bed....right answer answer jnvahjjhhhjhjihhh the amazon will have be so at the bed the amazon will have be so at the bed the amazon will have so at the bed volume of a cylinder good working learning is good for you 2/3 of a number is less than the original number by 20. 2 3 of number is less than original number by 20 the intellectual and practical activity encompassing the systematic study of the structure and behaviour of the physical and natural world through observation and experiment. anwe accoring to this the anwer is 1000000 2 3 number is less than original number by 20 the amazon will have be so at the bed... the amazon will be so at the bes. geoehepeheufciduridodhdjdod the amazon will have be so at the bed. I don't know bro please like I don't know please like The Amazon will have be so at the bed 😮😮😮 mark as best 👍💯👍💯👍💯👍💯 |
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| 33. |
Find HCF and LOM of 125 and 34 3. |
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Answer» 125=5×5×5=5^3343=7×7×7=7^3HCF=1LCM=125×343 HCF-1, LCM -42875 like HCF-1, LCM -42875 like it is easy.do prime factors HCF-1, LCM -42875 Answer HCF-1, LCM-42875 is the correct answer HCF 1. LCM 42875 hcf - 1,LCm - 42875 is the answer |
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| 34. |
In Fig. 6.70, bisector of ZB and D of a quadrilateral ABCD meetCD and AB produced atP and Q respectively. ProveFig. 6.70 |
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Answer» InΔPBC we have,angle p +pbc+c=180p+1/2 B+C=180⇒eq 1 InΔqad we have Q+A+ADQ=180Q+A+1/2 D=180⇒eq 2 Adding 1 and 2P+Q+A+C+1/2 B+1/2 A=360 But,A+B+C+D=360P+Q+A+C+1/2(B +D)=A+B+C+D P+Q=1/2 (B+D)P+Q=1/2(ABC+ADC) |
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| 35. |
The bisectors of angleB and angleC of quadrilateral ABCD meet at a point R inside the quadrilateral. If angleA=95 and angleD=111 , find angleBRC |
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Answer» Let the quadrilateral ABCD have the angles <A = 2a, <B = 2b, <C = 2c and <D = 2d. The bisectors of <B and <C meet at R. So <BRC = 180 -b-c. The sum of the four angles <A+<B+<C+<D + 360 deg = 2a+2b+2c+2d, or a+b+c+d = 180 deg …(1) we have seen that <BRC = 180-b-c = 180 -(b+c) …(2). Put the value of b+c from (1) in (2) to get <BRC = 180 -[180-a-d] = 180 -180+a+d, or <BRC = a+d = (1/2)[<A+<D]. =1/2(95+111)=1/2(206)=103° |
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| 36. |
5^(-x %2B 1) %2B 5^(x %2B 1)=26 |
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| 37. |
4. In the figure, Pol BC, if PQ = 2 , then AP = ?P.BC 5PB |
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| 38. |
2. The lidarectรกngular box ofsides 40 cm by 10 em is sealed all round widWhat is the length of the tape required? |
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| 39. |
If the interior and exterior angle of a regular polygonis equal. Find the number of sides of the pol |
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Answer» Interior angle of polygon= (n - 2)/n*180 Exterior angle of polygon= 180 - (n-2)/n*180 Where n = number of sides of polygon As per given condition 180 - (n-2)/n*180 = (n-2)/n*180 180 = 2(n - 2)/n *1802(n - 2) = n2n - 4 = n2n - n = 4n = 4 Therefore, Number of sides = 4 |
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| 40. |
8152/3) Find the value of t7 |
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| 41. |
13-lnatriangle, PQ, and R are the mid-points ofsides BC, CAandABrespectively. IfAC-21cm, BC-29and AB-30 cm, find the perimeter ofthe quadrilateral ARPQ. |
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| 42. |
triangle Whose siues 324. The interior angles of a polygon are in A.P. The smallest angle is 52 and the common differenceis 80 Find the number ofsides of the polygon2.dR, If AC> AB, then show that:AB2 AD2 |
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| 43. |
Example 7: AB is a line segment, P and Q arepoints on opposite sides of AB such that cach of themis cquidistant from the points A and B (see Fig. 7.37).Show that the line PO is the perpendicular bisectorof AB |
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| 44. |
4.What is the radius of a circle whose circumference is 44 cm?T7 m |
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Answer» Circumference of a circle = 2πrandGiven, 2πr = 44cm So,πr = 44/2 = 22 r = 22 × 7/22 r = 7cm is the answer. Thanks for the answer 😊 |
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| 45. |
3. AB and CD intersect at O. Find x,y and z.Amdm BOC |
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| 46. |
31 Kalve graphically the pain of linearequations and write the coordinates ofthe vertices of the triangle formed by thesetwo lines with K-aus.34 +-3=0 ,24-418 20 |
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Answer» 3x+y=3; 2x-y=-8; ; 2(3x+y=-3); 3(2x-y=-8); 6x+2y=-6; 6x-3y=24/5y=30; y=30/5=6; 3x+6=3; 3x=-3; x=0 |
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| 47. |
2. In the given figureare the bisectors of the exterior angles meeting eachother atO. If ZA-70 find BOc.70° |
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| 48. |
in Fig. 6.35,A ODC ~ Δ OBA, < BOC = 125。and < CDO = 70°. Find < DOC, < DCO and2OAB |
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Answer» Where is the figure? |
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| 49. |
The diagonals of a rectangle ABCD meet at O. IfBOC-50" then findODA.So |
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Answer» Given angle BOC = 50° and angle AOD = BOC ...( vertically opposite angles) now, in ∆AOD , since AO = OD ...( diagonal bisector) so, angle ODA = OAD , let say x . now adding all angles we get x+x + 50° = 180°=> 2x = 180°-50° = 130° =>x = 130°/2 = 65° which means angle ODA = 65° |
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| 50. |
4*(x*y^2) %2B y^2 %2B x^2 - 3*y^2 %2B 3*(x^2*y) %2B 5*(x^2*y) - 5*x^2 |
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