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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. |
lu ave fdoar |
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Answer» This quadrilateral is a trapezium..and area is given by = 1/2*(sum of parallel sides)*(height)= 1/2*(28+40)*(9)= 306m² |
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| 2. |
(25) x78 x7 |
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| 3. |
From a point in the interior of an equilateral triangle, the perpendiculars are drawn on the three sides. Thelengths ofthe perpendiculars are 14 an, 10 crn and 6 on. Find the area of the triangle |
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| 4. |
19.A diagonal of a quadrilateravertices are 12.8 cm and 11.2 cm. Find the area of the quadrilateral.l is 26 cm and the perpendiculars drawn to it from the opposite |
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| 5. |
(2) x78' x7(i) |
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| 6. |
19. A diagonal of a quadrilateral is 26 cm and the perpendiculars drawn to it from the oppositelaonvertices are 12.8 cm and 11.2 cm. Find the area of the quadrilateral. 3 |
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| 7. |
5. Find the area of the quadrilateral in which the length of one of its diagonals is 36 cm andperpendiculars drawn to it from the opposite vertices are 10.5 cm and 5.5 cm respectively |
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| 8. |
Which of the following is equal to "x".a) 12√(x4)1 by 3b) x12/7-x7/12c) √(x3)2by3d) x12/7 × x7/12 |
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Answer» Option (D) |
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| 9. |
A2+ ÂŁâx7) (W) |
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| 10. |
x7+24 i=? |
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Answer» If you find this solution helpful, Please give it a 👍 |
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| 11. |
рем-рео -рд╢= AC + X7 |
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| 12. |
0(b)169 13(b) 1327(b) 372928, v 1.0816 =?(a) 0-1429、10009 =?(a) -0330. 10.4 =?(b) 1(b) .3 |
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Answer» 25. √ 25-9/81 = √16/81 = 4/9 26. √169+x/169 = 14/13 169+x = 196 x = 27 27. √784/729 = 27+x/27 27+x = √784 x = 28-27 = 1 28. √1.0816 = 1.04 29. √0.009 = 0.03 |
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| 13. |
ma AABC, BM and CN are perpendiculars from B and C respectively on any lineăpassing through A. If L is the mid-point of BC, prove that MAL-NL |
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| 14. |
1. Use Euclid's division algorithm to find the HCFp () f35 and 225 196 and 38220 |
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Answer» (i) 135 and 225Since 225 > 135, we apply the division lemma to 225 and 135 to obtain225 = 135 × 1 + 90Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain135 = 90 × 1 + 45We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain90 = 2 × 45 + 0Since the remainder is zero, the process stops.Since the divisor at this stage is 45,Therefore, the HCF of 135 and 225 is 45.(ii) 196 and 38220Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain38220 = 196 × 195 + 0Since the remainder is zero, the process stops.Since the divisor at this stage is 196,Therefore, HCF of 196 and 38220 is 196.please like the solution 👍 ✔️👍 |
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| 15. |
X is a point on the side BC of AABC XM and XN are drawn parald ACd atto AB anrespectively meeting AB in N and AC in M MIN produced meets CB produceProve that TX2 = TB × TC과떼 |
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Answer» XM || AB, XN || AC TX² = TB x TC BN || XM For ΔTXM we have BN || XM Now we will use BASIC PROPORTIONALITY THEOREM TB/TX = TN/TM --- (this is equation 1) XN || AC XN || CM In ΔTMC we have XN || CM Again we will use BASIC PROPORTIONALITY THEOREM TX/TC = TN/TM --- (this is equation 2) Now we will compare equation 1 and equation 2 TB/TX = TX/TC TX² = TB x TC (proved) |
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| 16. |
6. In a AABC, BM and CN are perpendicularsfrom B and C respectively on any linepassing through A. If L is the mid-point ofBC, prove that ML = NL. |
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| 17. |
BC, BM and CN are perpendiculars6. In aâłAfrom B and C respectively on any linepassing through A. If L is the mid-point ofBC, prove that ML NL. |
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| 18. |
13.In a AABC, BM and CN are perpendiculars from B and C respectively on any line passingthrough A. If L is the mid-point of BC, prove that ML = NL. |
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| 19. |
X is a point on the side BC of ΔABC. XM and XN are drawn parallel to AB and ACrespectively meeting AB in N and AC in M. MN produced meets CB produced at TProve that TX2 = TB × TC |
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Answer» ∆ΤXN ~ ∆TCM ⇒ TX / TC = XN / CM = TN / TM ⇒ TX × TM = TC × TN ....(i) Again, ∆TBN ~ ∆TXM ⇒ TB / TX = BN / XM = TN / TM ⇒ TM = (TN × TX) / TB ...(ii) Using (ii) in (i), we get TX2 × TN/TB = TC × TN ⇒ TX2 = TC × TB |
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| 20. |
6. In a AABC, BM and CN are perpendiculars from B and C respectively on any linepassing through A. If L is the mid-point of BC, prove that ML-NL |
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| 21. |
18. In a triangle ABC, BM and CN areperpendiculars from B and Crespectively on any line passingthrough A. If L is the mid-point of BC,prove that LM = LN |
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| 22. |
3. In Fig., BM and CN areAperpendiculars drawnto a line passingthrough the vertex Aof a triangle ABC. If L is Nacthe mid-point of BCprove that LM = LN. |
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| 23. |
Xis a point on the side BC of AABC. XM and XN are drawn parallel to AB and ACspecuvely meeting AB in N and AC in M. MN produced meets CB produced at T.REovethat TX2 = TB x TC |
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| 24. |
If both x – 2 and -- are factors of px2 + 5x + r show that p = r. |
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| 25. |
2 Find the value ofx for which the numbers (5x + 2), (4x-1)and (r+2in AP |
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Answer» thank you |
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| 26. |
X is a point on the side BC of AABC. XM andXN are drawn parallel to AB and ACrespectively meeting AB in N and AC in M. MN produced meets CB produced atProve that TX2 TB Ă TC |
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Answer» XM || AB, XN || AC TX² = TB x TC BN || XM For ΔTXM we have BN || XM Now we will use BASIC PROPORTIONALITY THEOREM TB/TX = TN/TM --- (this is equation 1) XN || AC XN || CM In ΔTMC we have XN || CM Again we will use BASIC PROPORTIONALITY THEOREM TX/TC = TN/TM --- (this is equation 2) Now we will compare equation 1 and equation 2 TB/TX = TX/TC TX² = TB x TC (proved) hit like if you find it useful |
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| 27. |
If both (x-2) and (x-1/2) are factor of px2 + 5x + r, show that p = r. |
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Answer» if x-2 and x-1/2 are the factors then x = 2 and x = 1/2 are the roots of equation px²+5x+r, whose product of roots = r/p = 2*1/2 = 1 => r/p = 1 => r = p |
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| 28. |
r+5x + 62 |
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Answer» Factorization x*x + 5x + 6 = x*x + 2x + 3x + 6 = x(x+2)+3(x+2) = (x+2)(x+3) If you find this answer helpful then like it. |
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| 29. |
8.X is a point on the side BC of AABC. XM and XN are drawn parallel to AB and ACrespectively meeting. AB in N and AC in M. MN produced meets CB produced at TProve that TX-TBxTC. |
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Answer» tnx |
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| 30. |
3. Find angle x if60°20Arh30 |
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| 31. |
4. If the radius of a circle is r units, then express the length of a diameterin terms of r. |
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Answer» diameter= 2 × radius =2r |
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| 32. |
R... sin 30° + tan 45° — cosec 60°करsec 30° + cos 60° + cot 45% |
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| 33. |
x is a point on the sideBC of Δrespectively meetiProve that TX TB x TCpoint on the side BC of AABC. XM and XN are drawn parailel to AB and ACparallel to AB and AÇAC in M. MN produccd meets CB produced at 1Tng AB in N and |
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Answer» XM || AB, XN || AC TX² = TB x TC BN || XM For ΔTXM we have BN || XM Now we will use BASIC PROPORTIONALITY THEOREM TB/TX = TN/TM --- (this is equation 1) XN || AC XN || CM In ΔTMC we have XN || CM Again we will use BASIC PROPORTIONALITY THEOREM TX/TC = TN/TM --- (this is equation 2) Now we will compare equation 1 and equation 2 TB/TX = TX/TC TX² = TB x TC (proved) |
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| 34. |
X is a point on the side BC of AABCrespectively meeting AB in N and AC in M. N produced meets CB produced al TProve that TXa_ TB TCXM and XN are drawn parallel to. AB and AC |
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Answer» Here is your answer: XM || AB, XN || AC TX² = TB x TC BN || XM For ΔTXM we have BN || XM Now we will use BASIC PROPORTIONALITY THEOREM TB/TX = TN/TM --- (this is equation 1) XN || AC XN || CM In ΔTMC we have XN || CM Again we will use BASIC PROPORTIONALITY THEOREM TX/TC = TN/TM --- (this is equation 2) Now we will compare equation 1 and equation 2 TB/TX = TX/TC TX² = TB x TC (proved) |
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| 35. |
X is a point on the side BC of AABC. XM and XN are drawn parallel to AB and ACrespectively meeting AB in N and AC in M. MN produced meets CB produced at TProve that TX2 TB x TC |
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Answer» XM || AB, XN || AC TX² = TB x TC BN || XM For ΔTXM we have BN || XM Now we will use BASIC PROPORTIONALITY THEOREM TB/TX = TN/TM --- (this is equation 1) XN || AC XN || CM In ΔTMC we have XN || CM Again we will use BASIC PROPORTIONALITY THEOREM TX/TC = TN/TM --- (this is equation 2) Now we will compare equation 1 and equation 2 TB/TX = TX/TC TX² = TB x TC (proved) Like my answer if you find it useful! |
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| 36. |
2. Let A be the set of all even whole numbers less than 10.(a) Write A in roster form.(b) Fill in the blanks with the approximate symbol e or :(i) ..... A (i) 10 ..... A (i) 3..... A (iv) 6... |
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Answer» Hi, Here is the answer to your question. |
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| 37. |
21. ABCD is a parallelogram. AB is produced to E so that BE=AB. EF meetsCB produced at F and is parallel to CA. Prove that AF is equal to EC |
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| 38. |
The side AB of a parallelogram ABCD is producedto any point P.A line through Aand parallel to CPmeets CB produced at Q and then parallelogramPBQR is completed (see Fig. 9.26). Show that(ABCD) ar (PBQR)Hint: Join AG and PO. Now compare ar (ACQ |
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| 39. |
Expand(3m+n)^2 |
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Answer» it will be 9m^2+n^2+6mnas it is just (a+b)^2 's formula |
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| 40. |
In the adjacent figure, the measure of ZA isa) 30°b) 45°c) 90°d) 60°5.120°120 |
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Answer» in the adjacent figure ,the measure of angle a is 60° |
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| 41. |
4 Find the value of x from the adjacent figure.32x5 |
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Answer» 2x +5x+3x=180(angle sum property) 10x=180 X=180/10 X=18 Hence,2x=2*18=36 3x=3*18=54 5x=5*18=90 2x Angle =60°3x angle =45°5x Angel =30° |
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| 42. |
७ केK4 se R R TNo =3 tono |
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| 43. |
5. Find the value of x from the adjacent figure.3.r25x |
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Answer» As we knowthat2x+3x+5x= 18010x= 180x= 18° 2x+3x+5x=180 (l.p)->10x=180->x=180/10-> x=18 Now, 2x=2*18=363x=3*18=545x=5*18=90 |
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| 44. |
ककeके ७ [TO oy QoWMo []8 Urssiun ol(1|k4> |
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| 45. |
5. In the adjacent figure, it is being given thatAO CD, OB CE and ZAOB-50°. Find themeasure of ZECD. |
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| 46. |
Find angle x in the adjacent figure |
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| 47. |
From the adjacent figure, express cosec R. |
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Answer» cosec R =q/r is the correct answer |
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| 48. |
EXERCISE 1.3I. A tailor stitches 2 shirts in half an hour. How many will he stitch in 2 hours? |
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| 49. |
: Ganga went to a tailor with 1 mt. cloth. She askedhim to make a blouse to her. The tailor used 0.75mts of cloth to make the blouse and returned theremaining cloth to Ganga. |
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Answer» ganga gave to tailor-1mttailor used-0.75 mts tailor give back to her-100cm_-75cm=25cm |
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| 50. |
10. Monu went to a tailor to get a shirtmade. The tailor showed him somedifferent kinds of buttons that wereround and yellow, purple and square,black and triangular, and simple whitein all shapes (round, square andtriangular)A. Write the sets formed for both the colours ofbuttons and the shapes of buttons in rosterformB.Write the cardinality of each set |
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