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. |
(vi)2 In the adjoining figure, show that ABCD isa parallelogram.Calculate the area of llgm ABCDD 5 cmA 5 cm B |
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| 2. |
. A 15 m tall boy is standing at some distance from a 30 m tall building. The angle ofclevation from his eyes to the top of the building increases from 30° to 60° as he walkstowards the building. Find the distance he walked towards the building. |
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| 3. |
EXERCISE 1.11. Use Euclid's division algorithm to find the HCF of:(ii) 196 and 38220135 and 225 |
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| 4. |
Euclid's division algorithm to find the HCF of:135 and 225(ii)196 and 38220 |
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| 5. |
Euclid's division algorithm to find the HC(1) 135 and 225(ii) 196 and 38220 |
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| 6. |
JUse Euclids division algorithm to find the HCF of135 an 22() 196 and 38220 |
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| 7. |
Use Euclid's division algorithm to find the HCF of:(i)135 and 225(ii)196 and 38220 |
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Answer» By euclid's division algorithm225=135×1+90135=90×1+4590=45×2+0hence the hcf is 45Please like the solution 👍 ✔️ As38220 = 196 x 195 + 0 H.C.F = 196 Please like the solution 👍 ✔️ |
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| 8. |
1. Use Euelid's division algorithm to find the HCF(1) 135 and 225(ii) 196 and 38220 |
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| 9. |
17. In a triangle ABC, the incircle (centre O)touches BC, CA and AB at points P, Q and Rrespectively. Calculate(i) ZQOR (ii) ZQPR;given that LA = 60°. |
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| 10. |
1.Use Euclid's division algorithm to find the HCF(i) 135 and 225(ii) 196 and 38220 |
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| 11. |
1. Use Euclid's division algorithm to find the HCF(i) 135 and 225(ii) 196 and 38220 |
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| 12. |
16 in the above right sided figure, a circle inscribed in triangle ABC touches its sides AB, BC andAC at points D, E and F respectively. If AB 12 cm, BC 8 cm and AC 10 cm, then find thelengths of AD, BE and CF |
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| 13. |
7.ABC is a triangle. A circle touches sides AB and AC produced and side BC at X, Y and Z respectively.Show that AXperimeter of AABC. |
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Answer» We get thatAX=AYBX=BZCY=CZnowper of triangle ABC =ab+bc+ac=ab+(bz+zc)+ac=ab+bx+ac+cy=ax+ay=2axper of triangle ABC =2axtherefore1/2perimeter of triangle =ax |
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| 14. |
6. The 4tevill7. ABC is a triangle. A circle touches sides AB and AC produced and side BC at X. Y and Z respectivelyShow that AXperimeter of AABC |
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Answer» AX=AYBX=BZCY=CZnowper of triangle ABC =ab+bc+ac=ab+(bz+zc)+ac=ab+bx+ac+cy=ax+ay=2axper of triangle ABC =2axtherefore1/2perimeter of triangle =ax . |
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| 15. |
2. The ratio of the corresponding sides oftwo similar triangles is 5:3. Then theratio of their areas is[ ]1) 5:3 2) 3:5 3) 6 : 10 49 25:9 |
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Answer» ans will be 25:9 formula- base×hight/2 |
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| 16. |
7.ABC is a triangle. A circle touches sides AB and AC produced and side BC at X, Y and Z respectively.Show that AXperimeter of AABC.perim-2 |
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Answer» AX=AYBX=BZCY=CZnowper of triangle ABC =ab+bc+ac=ab+(bz+zc)+ac=ab+bx+ac+cy=ax+ay=2axper of triangle ABC =2AXtherefore1/2perimeter of triangle =AX please put the fig |
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| 17. |
Obtain all other zeroes of $ 3 x^{4}+6 x^{3}-2 x^{2}-10 x-5, $ if two of its zeroes are $ \sqrt{\frac{5}{3}} $ and $ -\sqrt{\frac{5}{3}} $ |
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| 18. |
19. The number of chairs left over when1000 chairs are arranged in rows con-taining 48 chairs each |
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Answer» let x be the number of chairs left over 1 row contains 48 chairs 1000 chairs =48/1000 =20.833 rows so consider 20 rows which contain 48×20=960 chairstotally 1000 chair so 1000-960=40 chairs are kept left over |
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| 19. |
2.If two equations 5x + 3y + 7 = 0 andlx + my +9=0 are inconsistent, thenwhat is l: m?(A) 3:2 (B) 2:3(C) 5:3 (D) 3:5 |
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| 20. |
\begin{array}{l}{\text { Obtain all other zeroes of } 3 x^{4}+6 x^{2}-2 x^{2}-10 x-5, \text { if two of its zeroes }} \\ {\sqrt{\frac{5}{3}} \text { and }-\sqrt{\frac{5}{3}}}\end{array} |
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| 21. |
7. If two adjacent vertices of a parallelogram are (3, 2) and (1, 0) and the diagonals intersect at(2, -5), then find the coordinates of the other two vertices. |
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Answer» thank you |
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| 22. |
2. The value of 1- tan1S:1+tan 2 30° |
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Answer» 1 - tan² 30°-----------------1 + tan²30° 1 - (1/√3)²----------------1 + ( 1/√3)² 1 - 1/3-----------1 + 1/3 2/3-----4/3 1/2 |
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| 23. |
The value of1 1s- 10 |
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Answer» Please hit the like button if this helped you |
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| 24. |
In which quadrant the point (-2,-5) lies |
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| 25. |
what is the mirror image of the point 2, 5 in x axis |
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Answer» P (2,5)After reflection on x-axis, P' (2, -5). |
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| 26. |
If coordinates of two adjacent vertices of a parallelogram are (3, 2), (1, 0) anddiagonals bisect each other at (2,-5), find coordinates of the other two vertices.OR |
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Answer» Let the coordinates of C and D be (a, b) and (c, d) ∴ (3 + a)/2 = 2 ⇒ a = 1 and (2 + b)/2 = –5 ⇒ b = –12 Also (c + 1)/2 = –5 ⇒ c = 3 and (d + 0)/2 = –5 ⇒ d = –10 ∴ Coordinate of C and D are (1, –12) and (3, –10) |
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| 27. |
Example 13 The centroid of a triangle ABC is at the point (1, 1, 1). If the coordinatesof A and B are (3, -5, 7) and (-1, 7, - 6), respectively, find the coordinates of thepoint C. |
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| 28. |
A circle touches all the four sides of a quadrilateral ABCD. Prove thatAB + CD =B+ DA |
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| 29. |
DatePage+ 3.2L+9一皿一3.to-tifらWhat value ed Xand y |
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Answer» now substitute the value of x in any of the equation and find the value of y thanks its right way and right ans but equation 2-1,then value of x=200/7 |
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| 30. |
A circle touches all the four sides of a quadrilateral ABCD. Prove thatAB +CD=BC+DA |
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Answer» From the figure we observe that, DR = DS (Tangents on the circle from point D) … (i) AP = AS (Tangents on the circle from point A) … (ii) BP = BQ (Tangents on the circle from point B) … (iii) CR = CQ (Tangents on the circle from point C) … (iv) Adding all these equations, DR + AP + BP + CR = DS + AS + BQ + CQ ⇒ (BP + AP) + (DR + CR) = (DS + AS) + (CQ + BQ) ⇒ CD + AB = AD + BC |
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| 31. |
A circle touches all the four sides of a quadrilateral ABCD. Prove thatAB + CD BC+DA |
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| 32. |
A circle touches all the four sides of a quadrilateral ABCD. Prove thatAB +CD=BC + DA |
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| 33. |
A circle touches all the four sides of a quadrilateral ABCD. Prove thatAB + CDBC+ DA |
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| 34. |
1. 8 ball point pens cost & 64 What is the cost of 1pen 2 |
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Answer» cost of one pen is 64/8 = 8 rupees |
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| 35. |
1f2-2x-1 = 4, then the value of xisă |
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Answer» x=3, so 3 ki power 3 ka answer hai 27 |
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| 36. |
4x + 3)7(4x +3) |
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| 37. |
IVALLILLE8. The total cost of 24 chairs is 9255.60. Find the9255.60. Find the cost of each chair.oh shirts can be m: |
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| 38. |
7.Can altitude lie outside a triangle? |
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Answer» Analtitudeof atriangleis the perpendicular segment from a vertex of atriangleto the opposite side (or the line containing the opposite side). Analtitudeof atriangle can bea side or maylie outsidethetriangle. |
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| 39. |
15. Does the point (-2.5, 3.5) lie inside, outside or on the circle y25? |
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| 40. |
15. Does the point (-2.,5, 3.5) lie inside, outside or on the circle xy 252 |
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| 41. |
प्रश्न 4. सिद्ध कीजिए :९sinº - cost 8sin-a-cos-a-1. |
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| 42. |
8 pens cost rupees 144 how much the cost of 14 pens? |
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Answer» If cost of 8 pens = 144cost of 1 pen = 144/8Then,Cost of 14 pens = 144*14/8 = 18*14 = Rs 252 |
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| 43. |
One apple cost 1 rs then what is the cost of 8 apples |
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Answer» cost of 8 apples = 8 * cost of one apple = 8*1 = 8 rs. |
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| 44. |
The cost of 8 chairs is 2,120. Find the cost of 25 chairs. |
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Answer» Thanks |
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| 45. |
Find the equation of the line passing through the point of intersection of the lnes4xED,-3 = 0 and 2x-3y + l = 0 that has equal intercepts on the axes. |
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| 46. |
= cosec A + cot A, using the identity cosec A=丨cof A.coS A + sin A -1 |
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| 47. |
1+ cot Acosec A+ cot A using the identity cosec Acos A+ sin A 1 |
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Answer» Like if you find it useful |
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| 48. |
tan_TF cos A = m cos B!2 m- |
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Answer» Given Cos A = m Cos B ⇒ Cos A / Cos B = m / 1. ( apply componendo and dividendo) ⇒ (Cos A + Cos B) / (Cos A - Cos B) = ( m + 1) / (m - 1) ⇒ {(2 Cos ( A+B) / 2 × Cos (A-B) / 2 } / {(- 2 Sin ( A+B) / 2 × Sin (A-B) / 2 } = ( m + 1) / (m - 1) ⇒ {(2 Cos ( A+B) / 2 × Cos (B - A) / 2 } / {( 2 Sin ( A+B) / 2 × Sin (B - A) / 2 } = ( m + 1) / (m - 1) ⇒ {( Cot ( A+B) / 2 × Cot (B - A) / 2 } = ( m + 1) / (m - 1) ⇒ {( Cot ( A+B) / 2 = { ( m + 1) / (m - 1) } Tan (B - A) / 2 Here x = ( m + 1) / (m - 1). |
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| 49. |
| TFd-='HCF. (48,72). Eind-thevalue-of-01 |
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Answer» Factors of 48 List of positive integerfactors of 48 that divides 48 without a remainder. 1, 2, 3, 4, 6, 8, 12, 16,24 Factors of 72 List of positive integerfactors of 72that divides 72 without a remainder. 1, 2, 3, 4, 6, 8, 9, 12, 18,24, 36 So thegreatest common factor of 48 and 72is24. |
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| 50. |
cos A -sin A + 1cA Cosec A + cot A, using the identity cosec A-1+ cot A. |
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Answer» LHS:Cos A - Sin A + 1/ Cos A + Sin A - 1 dividing in numerator & denominator with Sin ACot A - 1 + Cosec A / Cot A - Cosec A + 1now putting 1= Cosec^2 - Cot ^2 = (Cot A + Cosec A) - (Cosec^2 A - Cot^2 A) / (Cot A - Cosec A + 1) = (Cot A + Cosec A) [1 - Cosec A + Cot A) / (Cot A - Cosec A + 1)] = (Cot A + Cosec A) = RHS Hence Proved |
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