InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 41051. |
In an eqilateral triangle ABCD is a point on side BC such that BD=1/3BC prove that 9AD2=7AB2 |
| Answer» Given, {tex}\\triangle ABC{/tex}\xa0in which\xa0{tex}AB = BC = CA{/tex} and D is a point on BC such that BD ={tex}\\frac { 1 } { 3 }{/tex}BC.Prove: 9AD2 = 7AB2.Construction :Draw\xa0{tex}A L \\perp B C{/tex}\xa0Proof: In right triangles ALB and ALC, we have{tex}AB = AC {/tex}(given){tex}AL = AL {/tex}(common){tex}\\therefore \\triangle A L B \\cong \\triangle A L C{/tex}\xa0[by RHS axiom]So,\xa0{tex}BL = CL.{/tex}Thus, BD ={tex}\\frac { 1 } { 3 }{/tex}BC and BL ={tex}\\frac { 1 } { 2 }{/tex}BC.In\xa0{tex}\\triangle ALB{/tex},\xa0{tex}\\angle A L B = 90 ^ { \\circ }{/tex}By using pythagoras theorem, we get\xa0{tex}\\therefore{/tex}\xa0AB2 = AL2 + BL2 ...(i)In\xa0{tex}\\triangle ALD{/tex},\xa0{tex}\\angle A L D = 90 ^ { \\circ }{/tex}By using pythagoras theorem, we get\xa0{tex}\\therefore{/tex}\xa0AD2 = AL2 + DL2\xa0= AL2 + (BL - BD)2= AL2 + BL2 + BD2 -\xa0{tex}2 B L \\cdot B D{/tex}= (AL2 + BL2) + BD2 -\xa0{tex}2 B L \\cdot B D{/tex}= AB2 + BD2 -\xa0{tex}2 B L \\cdot B D{/tex}\xa0[using (i)]{tex}= B C ^ { 2 } + \\left( \\frac { 1 } { 3 } B C \\right) ^ { 2 } - 2 \\left( \\frac { 1 } { 2 } B C \\right) \\cdot \\frac { 1 } { 3 } B C{/tex}{tex}= B C ^ { 2 } + \\frac { 1 } { 9 } B C ^ { 2 } - \\frac { 1 } { 3 } B C ^ { 2 }{/tex}{tex}= \\frac { 7 } { 9 } B C ^ { 2 } = \\frac { 7 } { 9 } A B ^ { 2 }{/tex}\xa0[{tex}\\because{/tex}\xa0BC = AB]Therefore, 9AD2 = 7AB2. | |
| 41052. |
Sin theta is equal to cos theta then find the value of 2 tan theta plus Cos theta |
|
Answer» Thanks for answering my question Sin square theta=cos square theta (squaring on both sides) 1_ cos square theta= cos square theta (sin square theta =1_ cos square theta) 1= cos square theta +cos square theta 1=2cos square theta 1/2=cos square theta 1/underroot2=cos theta. 1/2=cos square theta 1/2=1_sin square theta. Sin square theta =1_1/2=1/2. Sin theta=1/underroot2. 2 tan theta+cos theta = 2sin theta /cos theta +cos theta =2*1/underroot2 /1/underroot2 +1/underroot2 =2*1+1/underroot2 =2+1/underroot2=2*underroot2 +1/underroot2 2+cos(theta) or 2+sin(theta) |
|
| 41053. |
solve x and y x+y=3 4x_3y=26 |
|
Answer» x + y = 3x = 3 - y........ (i)4x - 3y = 26...... (ii)Put (i) in (ii)4(3 - y) - 3y = 2612 - 4y - 3y = 26-7y = 26 - 12-7y = 14y = -2Y = -2 in (i)x = 3 - (-2)x = 3 + 2x = 5\xa0 X+y=3 .......... (1) 4x_3y=26..........(2). Multiply equation (1) by 3& equation (2) by1 to make coefficient of y equal 3x+3y=9...........(3). 4x_3y=26..........(4). From equation (3)&(4). X=5&y=_2 x=5, y=-2 |
|
| 41054. |
FIND A QUADRATIC POLYNOMIAL WITH ZEROS 2 root 7 minus 2 root 7 |
|
Answer» Zeroes are 2√7 and -2√7 . So sum of zeroes is 2√7 +(-2√7) --> 2√7-2√7 = 0.. product of zeroes=> 2√7(-2√7) = -28 .. formula for quadratic polynomial is x²- ( sum of zeroes ) + product of zeroes. So polynomial is x²-(0)x+(-28) => x²-28. Here, zeroes are 2√7 and -2√7 Sum of zeroes=2√7+(-2√7)= 2√7-2√7=0 Product of zeroes= 2√7(-2√7)=-28 New equation formed=xsquare -28 |
|
| 41055. |
The HCF of two number is 145 and their LCM is 2175.it one number is 725 then the other number is |
| Answer» Let the number be aL.C.M × HCF = product of number145 × 2175 = 725 × a315375 = 725 × aa = 315375/725a = 435\xa0 | |
| 41056. |
If sec theta + tan theta equals to p then find the value of cosec theta |
| Answer» 1/p(cos theta) - 1 | |
| 41057. |
If cotB»12/15 prove that tan²B-sin²B»sin²Bsec²B |
| Answer» | |
| 41058. |
Which sample paper book is best arihant ulike or etc.andwhy? |
|
Answer» Oswaal is the best NCERT+Ulike(if u are preparing for board exams) Arihant is the best . Because it has the updated sample papers with cbse examination paper and latest cbse sample paper Arihant or EAD Oswaal |
|
| 41059. |
Which sample paper are best arihant or ulike. |
|
Answer» ULIKE It is correct i have that latest edition and it best Oswal and Arihant Arihant |
|
| 41060. |
Find the coordinates of the point on y axis which is nearest to the point (-2;5) |
|
Answer» Any point on y axis is(0,y)d\u200b\u200b\u200b\u200b\u200b\u200b2=4+(5-y)2For d to be minimum y=5So point is (0,5) Is me distance formula lagega(-2,5) and (0,y) |
|
| 41061. |
Name the type of triangle PQR formed by the point p(√2,√2) ,q(-√2,-√2),R(-√6,√6) |
| Answer» | |
| 41062. |
If sin theta +tan theta =m and sin theta - tan theta =n. Prove that m^2 - n ^2 =4 under root "mn" |
|
Answer» Thank you for answers lHS=(sin+tan)^2-(sin^2-tan^2). A. (Sin^2+tan^2+2sin*tan)-(sin^2+tan^2-2sin*tan). B. Sin^2+tan^2+2sin*tan-sin^2-tan^2+2sin*tan. C. 4sin*tan D. RHS=4√(sin+tan)(sin-tan). A. =4√sin^2-tan^2. S. =4√sin^2-sin^2/cos^2. R. =4√sin^2*cos^2-sin^2. A. ------------------------------. F. Cos^2. A. =4√sin^2(cos^2-1). T. ---------------------. =4tan*sin. I. Cos^2 (TanA +sinA) ^2-2ho)(tanA-sinA)^2=√(tanA+sinA)(tanA-sinA) |
|
| 41063. |
Proof /5 is irrational. |
| Answer» Let us assume to the contary that √5 is a rational no. So , √5 =p/q where pand q are co primes and q not equal to zeroSquaring both the sides 5 = p^2 /q^2 5q^2 = p^2This means that 5q^2 is factors p and p^2 (5a)^2=5 q^25a^2= q^2 This means that 5a^2 is factor of q^2 This quantradicts that √5 is rational no. And p and q are co primes | |
| 41064. |
2x²-5x+6=0 |
|
Answer» It has no real roots It has no roots. u can solve this by quadratic formula .because discriminant is negitive. No real roots |
|
| 41065. |
Solve by cross multiplication (a+b)x+(a-b)y =2ab and( a-b)x+(a+b )y=a^2+b ^2 |
| Answer» x = a + b{tex}\\therefore{/tex} (a - b)(a + b) + (a + b)y = a2 - b2 - 2aba2 - b2 + (a + b)y = a2 - b2 - 2ab{tex}y = \\frac{{ - 2ab}}{{a + b}}{/tex} | |
| 41066. |
What is the formula of total surface area of semi circla |
|
Answer» Nhi πr^2/2 3 pie r sq. |
|
| 41067. |
solve for x and y bx+ ay=a+b and ax(1/ a-b - 1/a+b)+ by(1/b-a - 1/b+a) = 2a/a+b |
| Answer» | |
| 41068. |
Prove that; cotA-cosA/cotA+cosA=cosecA-1/cosecA+1 |
| Answer» We can write cot A minus Cos A upon cot A + cos A is equal to cosA upon sinA minus CosA.whole upon cos upon sinA + cosA now we will take cosA common from whole equation then we have 1upon sin minus 1 whole upon 1 upon sin + 1 then we know that one upon sin is equal to cosecA minus one whole upon 1 upon sin is equal to cosA plus 1 now we got our answer what we have to prove | |
| 41069. |
If alpha and beta are the zeros of the polynomial f(x)=x2+x-2,find the value of (1/aplha-1/beta) |
| Answer» Say A=Alfa,B=betasx\u200b\u200b\u200b\u200b\u200b\u200b2\u200b\u200b\u200b\u200b\u200b+x-2A+B= -1,AB=-21/A-1/B=(B-A)/AB{tex} = \\sqrt{((A+B)^2-4AB)} \\over AB{/tex}=√(1+8)/(-2)= -3/2 | |
| 41070. |
Find the sum of area of major sector of and their corresponding minor sector |
| Answer» | |
| 41071. |
Tell the important chapter of maths? |
|
Answer» sabhi hai ek do se kuch nhi hoga pura karna padega Quadratic equation and trigonometry Surface area volume Trigonometry , algebra , Tigonomerty both chapters and AP and circles or bhi hoge bt mere pass itna infrmtn nhi h |
|
| 41072. |
If secA+tanA=x than what is the value of secA |
|
Answer» No ans. Is xsquare _1/ 2 x It will be - xsquare +1/2 X-tanA may be?? |
|
| 41073. |
prove that: 1+ cosA +sinA/1+cosA -sinA= 1+sinA/cosA |
| Answer» Hfffyi | |
| 41074. |
Show that (a-b), a, (a+b) form consecutive terms of an A.P. |
|
Answer» D=a-(a-b)=a-a+b=b Another, D=a+b-a=b Hence ,common difference are same so it is in the form of an AP Subtract 2nd term from the 1st you will get b as the common diff. and again if you will subtract 2nd term from the 3rd term and again you will get b as the common diff.This shows that......... |
|
| 41075. |
If 7 times the 7th term is equal to the 11 times the 11th term. prove that its 18th term is zero |
|
Answer» Thanks garauv bhai.. whr frm u 7th term = (a+6d) 11th term = (a+10d)(a+6d)7=(A+10d)11 (given)7a + 42d =11a +110d (on solving)7a - 11a = 110d - 42d-4a = -68da= -17dwe know 18th term = (a+17d)........................... (i)now we substitute the value of a in (i)\xa0which emplies, -17d + 17d = 0hence proved |
|
| 41076. |
Triangle r similar how |
| Answer» By the equal ratio of their corresponding sides and equal angle | |
| 41077. |
If 2 power is x and added to 2whose power is 2x is equals3/4 |
| Answer» | |
| 41078. |
Hi gys who use sample paper in 2019 in osawal maths |
|
Answer» Plz go sample paper .1 and check 19 question plzzzzzzzzzzzzz solve Me use I use question bank of osawal. |
|
| 41079. |
A path of 4m runs\xa0 |
|
Answer» Kya hua???? My no.8076898823 puja ji Aap msg Karna mujhe ??? Hii puja ji ? Question toh Pura Likha Karo aap kamm se kamm ? |
|
| 41080. |
The zum of two numbers is 16 and sum of their reciprocals is 1/3. Find the numbers |
|
Answer» Let one no. Is xthe other is 16-xAnd 1/x + 1/(16-x)=1/33(16-x)+3x=x(16-x)48-3x+3x=16x-x2x\u200b\u200b\u200b\u200b\u200b\u200b2\xa0-16x+48=0(x-12)(x-4)=0X=12,4So no. are 12 and 4 12 and 4 LET THE NUMBER BE X. AND Y. THE QUESTION SAYS THAT X+Y=16. AND THE OF THEIR RECIPROCAL IS 1/X+1/Y=1/3SO THE SECOND EQUATIO N IS. 3X+3Y=1 |
|
| 41081. |
How i download oswal 2019 maths sample paper in free |
| Answer» Go to the site of oswal sample paper and then download | |
| 41082. |
Find the sum of 7,10 1/2,14 . .. 84 |
|
Answer» n=23Sn=1046 1/2 Here, a = 7, d = 3.5 and last term = 84Number of terms can be calculated as follows;{tex}a_n = a+(n−1)d{/tex}Or, {tex}84 =7+(n−1)3.5{/tex}Or, {tex}(n−1)3.5 = 84−7{/tex}Or, {tex}n−1 = 77 /3.5 = 22{/tex}Or, {tex}n = 23{/tex} |
|
| 41083. |
Bestest guide for maths....for board preparation |
|
Answer» Ncert exemplar Ok thnks....for this? Rd sharma R D Sharma RD Sharma & NCERT exemplar |
|
| 41084. |
TanA =nTanB and SinA=mSinB prove that cossquare A=msquareminus 1 upone nsquare -1 |
| Answer» | |
| 41085. |
यदि एक शंकु के आधार की त्रिज्या पी तथा उसकी ऊंचाई एस है तो शंकु की तिर्यक ऊंचाई क्या होगी। |
|
Answer» Here r=p h=sTriyak unchayee{tex}l=\\sqrt (p^2+s^2){/tex} Write in English please Ya, please write in English ? Bhai English me bata hindi me kuch samaj nahi aaya kya kehna chahta hai ???? |
|
| 41086. |
What is the probability that an ordinary year has 53 sundays..????? |
|
Answer» Thanks friends Please ans in detail. A year have have 52 weeks and 1day1 day can be1.sunday2.monday3.tuesday4.wednesday5.thrusday6.friday7.saturdaySo probability is 1/7 1/7 |
|
| 41087. |
Given a=8 ,an=62,Sn=210 find n and d |
|
Answer» Thanks to answer Sn=n/2(a+an)210=n/2(8+62)210=n/2*70210=n*35210÷35=nn=6 |
|
| 41088. |
(1+cotA- cosecA) (1+tanA +secA) =2 |
| Answer» (1+cosA/sinA-1/sinA)(1+sinA/cosA+2/cosA)=1/sinAcosA(sinA+cosA-1)(sinA+cosA+1)=1/sinAcosA{(sinA+cosA)2-1)}=1/sinAcosA(sin\u200b\u200b\u200b\u200b2\u200b\u200b\u200bA+cos\u200b\u200b\u200b\u200b2\xa0A+2sinAcosA-1)=2sinAcosA/sinAcosA=2 proved | |
| 41089. |
Maths k lia oswal paper sufficient h |
|
Answer» Yes it is better because it included all types of question. Firstly focus at ncert than extra book Yes but also NCERT in tips and rs I think so but complete yur ncert first...... |
|
| 41090. |
Cose 30 |
|
Answer» Cos or cosec Root 3/2 |
|
| 41091. |
I need hp board syllabus |
| Answer» There is no | |
| 41092. |
Prove that 2sec²A-sec⁴A-2cosec²A+cosec⁴A=cot⁴A-tan⁴A |
|
Answer» OK thanks puja and sorry for disturbness Actually I\'m unable to type the answer here, sorry First put the formula of sec square and cosec square and then solve it yuwill get yur answer. Nishant chauhan ? I think you got it ? Bakvas km kari chaiye Lol |
|
| 41093. |
is there anybody who studies in Sarvada modern secondary scschool? |
|
Answer» Noo.. Nope brother ??? |
|
| 41094. |
A sector is formed in a circle of diameter 21cm. If the angle of the sector 150°,find its area |
|
Answer» 144.375 144.375 144.375 |
|
| 41095. |
For what value of theta , sin theta =cos Theta where 0 |
|
Answer» Thnx to all 45° 45° 45 45 45° 45 degrees |
|
| 41096. |
Answers of cbsc sample paper math class10 |
| Answer» Teri mein na | |
| 41097. |
If point (x,y) is equidistant from point A (7,-2) and B(3,1).Express y in terms of x. |
| Answer» | |
| 41098. |
In a triangle ABCD is the point on side BC such that BD=1/3BC prove 9AD\'2=7AB\'2 |
| Answer» You can find on chrome its very long?? | |
| 41099. |
Find the sum of all natural numbers b/w 200 and 1502 which are exactly divisible by 3 |
| Answer» The AP is 201,204,207,......1500a=201 d= 3 l=1500Sn= a+(n-1)d1500= 201+(n-1)31500=201+3n-33n=1500-1983n=1302n=1302/3n=434Sn=N/2(a+l)a=201l=1500n=434Sn = 434/2 (201 + 15000Sn = 217 (1701)Sn = 369117Therefore Sn=369117 | |
| 41100. |
Volume and surface area extra questions |
| Answer» | |