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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.
| 41851. |
State and prove the Pythagoras theorm |
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Answer» Phythagoras th. - In a right ∆ the square of hypo. is equal to sum of squares of other 2 sides.Construct a ∆ABC in which D is perpendicular to AC.∆ADB~∆ABC (By AA similarity) ----(1)Similarly ∆BDC~ABC∆ ------(2)Now; AD/AB=AB/AC (from 1)So by cross multiplying AB2=AD×AC-----(3)Next; DC/BC=BC/AC (from 2)Again by cross multiplying BC2= AC×DC-----(4)Now from adding (3) and (4) we have;AB2+BC2=(AD×AC)+(AC×DC)AB2+BC2= AC(AD×DC)AB2+BC2=AC×ACAB2+ BC2= AC2 {Hence Proved} You can get this theorm in ncert book just open the book and see See from ncert |
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| 41852. |
if P is an prime number than p is a. Number |
| Answer» If p is a prime number, obviously it is also a whole number | |
| 41853. |
Sec square _ tan square is equal to |
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Answer» Sec square - Tan square = 1 It is one because sec2A- tan2A=1. Is it theta or x+1. |
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| 41854. |
prove that o/o=2 |
| Answer» Silly question that u will never get in class 10 | |
| 41855. |
Find the distance of a point (x,y) from the orgin |
| Answer» Root x²+y² | |
| 41856. |
Use euclid,s division algorithm to find the hcf of 726 and 275 |
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Answer» 726=275*2+176275=176*1+99176=99*1+7799=77*1+2277=22*3+1122=11*2+0Therefore, 11 is the H.C.F 11 |
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| 41857. |
Find the coordinates of the point on y_axis is nearest to the point (-2,5) |
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Answer» Q. Find the coordinates of the point on Y-axis which is nearest to the point (–2,5).Answer:It is clear from the figure that point (0, 5) is the nearest point. Coordinate of point is (0,5)..... |
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| 41858. |
Three consecutive vertices of a parallelogram are (-2,-1),(1,0),(4,3).Find fourth vertex? |
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Answer» Let A(-2, -l), B(1, 0). C(4, 3) and D(x, y) be the vertices of a parallelogram ABCD taken in order. Since, the diagonals of a parallelogram bisect each other. Therefore coordinates of the mid-point of AC = Coordinates of the mid-point of BD.(-2+4/2, -1+3/2) = (1+x/2, 0+y/2)(1,1) = (1+x/2 , y/2)x+1 = 2, y=2Therefore (1,2) is the fourth vertex. fourth vertex is (1,2) |
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| 41859. |
Ydx-xlogydy=0 |
| Answer» Ye 10 class me log kaha se a gaya ye 11 me hai differentiation me dy and dx aur log trigonometric function me hai | |
| 41860. |
Find distance of point p(x,y) |
| Answer» Write full question | |
| 41861. |
Find the distance between the point (x,y) from the origin |
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Answer» X, y √x2+y2 √x*x+y*y |
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| 41862. |
If 1+sin²theta=3sin theta ×cos theta Then find the valu of tan theta =? |
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| 41863. |
The mean of 5 number is 18. If one number is excluded then their mean is 16 find the excluded number |
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Answer» Is a 26 16 6 |
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| 41864. |
(7,-2) ,(5,1),(3,k) find the value of k. |
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Answer» K=5 No don\'t apply this method. First write full question Put formula of area of triangle |
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| 41865. |
X+1/2+y-1/3 =9;x-1/3+y+1/2=8 |
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| 41866. |
If the sum and the product of the roots of ax2 - 5x+c are both equal to the 10 find \'a\' and \'c\'. |
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Answer» So a= 1/2 and c= 5 Sum of roots = 5 \\a 5\\a= 10 (given that the sum and product =10)a = 1\\2. AnsNow,Product of zero:-C/a=10C=10×a C= 10× 1\\2C= 5 ans . Sum of zeros =- b/a then 10=-(-5)/a then a= 1/2 and product of zeros =c/a then 10=c/1/2 therefore c= 5 |
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| 41867. |
Find probability of 53 sunday in a leap year |
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Answer» 2/7 Totalbno of day in leap year=366366day=52weak and 2 dayRemaining day sundau-monday Monday -tuesday Total no of possible outcome 7 Fab no of out come 2 p(E)= 27 No .of day in aleap year =366 2/7... |
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| 41868. |
In a triangle RPQ, PS perpendicular QR and PS^2=QS.RS. Show that RPQ is a right triangle |
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| 41869. |
If cosA + sinA= root2 cosA, show that cosA - sinA=root2 sinA |
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Answer» In given take cosA to rhs take common, and then whatever you will get take, it to lhs again now rationalize the denominator and solve yu will get the proof............? CosA +sinA =\xa0√2cosAsquaring⇒(cosA + sin A)² = (√2cosA)²⇒cos²A + sin²A\xa0+ 2sinAcosA = 2cos²A⇒1\xa0-\xa0sin²A\xa0+ 1 -\xa0cos²A\xa0+ 2sinAcosA\xa0= 2cos²A⇒2 - 2cos²A =\xa0cos²A + sin²A -\xa02sinAcosA\xa0⇒2(1 - cos²A)= (cosA - sinA)²⇒\xa0cosA - sinA =\xa0√[2sin²A]⇒cosA-sinA =\xa0√2sinA |
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| 41870. |
Most important questions from all the subjects of class 10th board exams 2018-19 |
| Answer» Bpt , pythagoras , and compulsqry one theorem of triangles qnd circles is going to come, and one sum from ch . 1 that is of edl or root 2 to prove irrational | |
| 41871. |
Find hcf 726 |
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Answer» Complete yur question........? Write full question |
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| 41872. |
How can we know that which thita is to be used in trigonometry questions |
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| 41873. |
Tell me the important chapters plz |
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Answer» Real numbersProve that root 3,root 5,root 2 and termonating decimal expansion, eculid division algorithm show that 3q+1,3q+3 Is divisible by 6 Basically see the chapters like theorems Vol. Surf. Area-...circle...ap...coordinate ...triangle.... All chapters r important yrr |
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| 41874. |
Find the mean mode and median of folllowing data of Monthly consumption of electricity |
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Answer» ? Write the complete question.. The data is monthly consumption ( 65-85)(85-105)(105-125)(125-145)(145-165)(165-185)(185-205) & no.of consumer (4)(5)(13)(20)(14)(8)(4) Where is the data |
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| 41875. |
The lengths of tangents drwan from an external point...to a circle are equal prove it |
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Answer» Construct a circle with centre O. Draw 2 tangents namely TP and TQ which intersect each other at a point T on the circle. Also join OP and OQ which are radius of circle.Now in ∆OPT and ∆OPQ we haveOP=OQ (radius of the circle)Angle P=angleQ (angle between tangent and radius is 90)By AA similarity we have; ∆OPT~∆OPQBy cpct TP=TQ Hence proved. By construction join the two lines from the point of the tangents drawn to the centre of the circle . Then join line at the point of contact of tangents point and prove the two triangles congruent and prove the tangents by cpct or congruent part of congruence triangle. ..... |
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| 41876. |
Find the value of sin60° geometrically |
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Answer» Which formula Take equilateral triangle then take one of the sides name an angle 60 degree then find by formula |
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| 41877. |
Area of isoceles triangle |
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Answer» √3/4a ka hall square Root over sirf 4 b square - a square ke upar hoga 1/4a root 4 b square - a square ......, yad aa gya Ek to 1/2 × bh hai.... b= base, h= height.... iska particular formula yad nhi aa rha Nhi pata.search on Google |
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| 41878. |
ΔABC =ΔDEF having area of 64 and 81 and D= 9 find AB |
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Answer» DE=9,AB=?Now as area is different so triangle cannot be congruent\xa0so are similar{tex}\\begin{array}{l}Now\\;\\triangle ABC\\;\\sim\\triangle DEF\\\\\\frac{ar(\\triangle ABC)}{ar(\\triangle DEF)}=\\frac{AB^2}{DE^2}\\\\\\frac{64}{81}=\\frac{AB^2}{9^2}\\\\AB^2=64\\\\AB=8\\end{array}{/tex} Yes it mean congurent What yes....congurent ya similar??? Yes By " = " do you mean congurent or similar |
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| 41879. |
Find hcf of 3x*2y and 6xy3 |
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Answer» Sorry 3xy 2xy |
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| 41880. |
What is the work of modlus? |
| Answer» Modulus is meant by absolute value which only deals with the value irrespective of its sign. | |
| 41881. |
Check whether Pi minus 22/7 is a rational number. justify your answer |
| Answer» Its rational no. as pi is approximately equals to 22/7 and 22/7-22/7=0,which is a rational no. | |
| 41882. |
We have 6 subjects if we get 70+ in 5 only then did the result will include 6 or 5 |
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Answer» Otherwise only 5 No, the 6th one would only be displayed on in the marksheet...it won\'t be counted... If all 6 are there for board exam, then the marksheet will include all 6 |
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| 41883. |
If three numbers are in Ap.And the sum of their squares is 260.Find the numbers. |
| Answer» Tera question Sahi hai6 | |
| 41884. |
Ex13.2 quesn6 |
| Answer» Volume of CYL + volume of cyl ,,,,,,,,,,,πR^2h + πr^2h ,,,,,,,, 3.14×12×12×220 + 3.14×8×8×60 ===== 99475.2 + 12057.6 =====111532.8 ,,,,,,,,,,, 111532.8×8 ====892262.4 cm^3 ,,,,,,,,892.2624 km ^3 ?? | |
| 41885. |
Ex13.26 |
| Answer» | |
| 41886. |
find the zeros of the quadratic polynomial 8X² - 4 |
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Answer» 8x2 - 4 = 04(x2 - 2) = 0x2 - 2 = = 0x2 = 2x = √2 +/- under root 1/2 3/2 and 5/2. |
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| 41887. |
Are derivations of values of trigonometric ratios asked in exams? |
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Answer» No.... No...its never ask in exam at least in 2019 exam it doesnt be ask..sure... Please answer fast |
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| 41888. |
In a ⛛ ABC, angle C =3angleB= 2(angleA+angleB). Find 3 angles |
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Answer» Angle A=20,Angle B=40,angleC=120 A = 20°, B = 40°,C = 120°....... |
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| 41889. |
how we learn formulas |
| Answer» Right all the formulas on a chart paper and stick it at the place where you can see it anytime and before going to bed just read it and when you wake up you read it it will help you in learning the formulas without any efforts. | |
| 41890. |
Who are lmportant theorem in exam |
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Answer» Exam mein converse nhi aati h yaar I think acc to board exam all theoram of ch 6 and ch 10 some theoram like bpt and converse and Pythagoras theoram and converse and area realated to similar triangle and in one or two marks the fundamental theoram of arithmetic of ch 1 Mention questiom properly ☺ improve your english please. \'what\' should be used instead of \'who\' and theorems should be used instead of theorem |
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| 41891. |
If cos theta is equal to 7 by 25 find the values of all t ratios of theta |
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Answer» So b= 7 and h=25By Pythagoras p=24 Put cos theta=b/h and find p by using p.t |
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| 41892. |
tan 0-1+ sec 0 ÷tan 0+1-sec 0 =1÷sec 0- tan 0 |
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Answer» Mns theta Of nhii O |
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| 41893. |
2017 questions paper |
| Answer» Kounsa subject | |
| 41894. |
Let x & x+1 be two consecutive |
| Answer» Complete ur question | |
| 41895. |
Total surface area of a solid hemisphere is 462cm2 find its volume |
| Answer» TSA OF A HEMISPHERE = 3πr²3πr²=4623×22/7×r²=46266/7r²=462r²=462×7/66r²=49r=√49=7volume of hemisphere = 2/3πr³2/3×22/7×7×7×7=718.6cm³ | |
| 41896. |
Prove that A (2,-1) B (3,4) C (-2,3) D (-3,-2) are the vertices of rhombus ABCD a square? |
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Answer» Apply distance formula....if all the distances comes equal then it is a square or rhombus....and if the diagonals comes equal then it is square and if diagonals are not equal then it is rhombus By using diatance formula,√ [ (x2\xa0- x1)2\xa0= (y2\xa0- y1)2\xa0]AB = √ ( 3 - 2)2\xa0+ ( 4 + 1)2\xa0=√ 12\xa0+ 52\xa0=√ 1 + 25 =√26 ...1BC =√ ( -2 -3)2\xa0+ ( 3 - 4)2\xa0=√ 25 = 1 =√26 ...2CD =√ ( -3 + 2)2\xa0+ ( -2 -3)2\xa0=√ 25 + 1 =√26 ...3DA =√(2 + 3)2\xa0+ ( -1 + 2)2\xa0=√ 25 + 1 =√ 26 ...4from 1, 2, 3 and 4ABCD is a rhombus.For a rhombus to be a square, diagonals must be equalSo we need to prove that AB = CDAC =√( -2 - 2 )2\xa0+ (3 + 1)2\xa0=√ (-4)2\xa0+ (4)2\xa0=√ 16 + 16 =√32BD =√ (-3 - 3)2\xa0+ (-2 - 4)2\xa0=√ (-6)2\xa0+ (-6)2\xa0= √ 36 + 36 = √72Since AC is not equal to BD, it is not a square Use distance formula...... Dono, diagonal ka length equal hoga Kiske vertices prove karne hai. ? Rhombus or square? |
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| 41897. |
Prove that the (3,0) (6,4) (-1,3) are vertical of a right angle isosceles right tringle? |
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Answer» Use distance formula and prove two side equal to each other This question incorrect Find their lengths...2 sides should be equal and the sum of squares of two sides should be equal to the square of 3rd side....its proved. Find the distance between coordinates by using distance formula and prove Given A(3, 0), B(6, 4) and C(–1, 3)AB^2 = (3 – 6)^2 + (0 – 4)^2= 9 + 16 = 25\xa0BC^2 = (6 + 1)^2 + (4 – 3)^2= 49 + 1 = 50\xa0CA^2 = (–1 – 3)^2 + (3 – 0)^2\xa0= 16 + 9 = 25\xa0AB^2 = CA^2 ⇒ AB = CATriangle is isoscelesAlso,\xa025 + 25 = 50\xa0⇒ AB^2 + CA^2 = BC^2\xa0Since Pythagoras theorem is verified, thereforeTriangle is a right angled triangle |
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| 41898. |
Solve for x=====x+3÷x-2-1-x÷x=17÷x |
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Answer» Nahi x ke baad equal nahi h vo sirf alag karne k liye lagaya h Apke Que. M 2 equals h...please recheck it..☺️☺️ Plz check yur question again........?? My papet tomorrow plez |
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| 41899. |
If one of the zeroes of the quadratic equation(k-1)x^2×kx+1 is -3 find the value of k |
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Answer» Substitute -3 in the equation K =4/3 Put the value of x as -3and you will get k=4/3.. K=4/3 |
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| 41900. |
What is the difference between an and n in nth term of an ap |
| Answer» A.P. stands for Arithmetic progression.A sequence is in AP when the difference between it\'s termes are common.In AP number of terms is denoted by small n\'n\' - No. of terms. And\'an\' means nth term of that sequence. | |