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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.
| 39501. |
2+2=4 why .whats concept of it |
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Answer» Take 2 apple Then take another 2Count them (1+1)+(1+1)=2+2 |
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| 39502. |
Solve 1 by x-3 + 2 by x-2 = 8 by x |
| Answer» We have,{tex}\\frac{1}{{x - 3}} + \\frac{2}{{x - 2}} = \\frac{8}{x}{/tex}{tex}\\Rightarrow \\frac{{1(x - 2) + 2(x - 3)}}{{(x - 3)(x - 2)}} = \\frac{8}{x}{/tex}{tex}\\Rightarrow \\frac{{x - 2 + 2x - 6}}{{(x - 3)(x - 2)}} = \\frac{8}{x}{/tex}{tex}\\Rightarrow \\frac{{3x - 8}}{{{x^2} - 2x - 3x + 6}} = \\frac{8}{x}{/tex}{tex}\\Rightarrow \\frac{{3x - 8}}{{{x^2} - 5x + 6}} = \\frac{8}{x}{/tex}Cross multiply,{tex}\\Rightarrow{/tex}\xa0{tex}x(3x-8)=8(x^2-5x+6){/tex}{tex}\\Rightarrow{/tex}\xa0{tex}3x^2-8x=8x^2-40x+48{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}8x^2-40x+48-3x^2+8x=0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}5x^2-32x+48=0{/tex}Factorise the equation,{tex}\\Rightarrow{/tex} 5x2 - 20x - 12x + 48 = 0{tex}\\Rightarrow{/tex} 5x(x - 4) - 12(x - 4) = 0{tex}\\Rightarrow{/tex} (5x - 12) (x - 4) = 0{tex}\\Rightarrow{/tex} (5x - 12) = 0 or (x - 4) = 0{tex} \\Rightarrow x = \\frac{{12}}{5}{/tex} or x = 4 | |
| 39503. |
Find the sum of first 25 term of AP whose nth term is 1-4n |
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Answer» Take n is equal to 1 then 1 - 4n is equal to -3 the value of first term is -3. when we take n is equal to 2 then the value of second term is equal to -7.Now the common difference is 4 now using and sum formula -a25=93 Kese Take n =1,n=2 |
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| 39504. |
1/a+b+x=1/a+1+b+1+x |
| Answer» Given,{tex}\\frac { 1 } { ( a + b + x ) } = \\frac { 1 } { a } + \\frac { 1 } { b } + \\frac { 1 } { x }{/tex}{tex}\\Rightarrow \\quad \\frac { 1 } { ( a + b + x ) } - \\frac { 1 } { x } = \\frac { 1 } { a } + \\frac { 1 } { b } \\Rightarrow \\frac { x - ( a + b + x ) } { x ( a + b + x ) } = \\frac { b + a } { a b }{/tex}{tex}\\Rightarrow \\quad \\frac { - ( a + b ) } { x ( a + b + x ) } = \\frac { ( a + b ) } { a b }{/tex}On dividing both sides by (a+b){tex}\\Rightarrow \\quad \\frac { - 1 } { x ( a + b + x ) } = \\frac { 1 } { a b }{/tex}Now cross multiply{tex}\\Rightarrow{/tex}\xa0x(a + b + x) = -ab\xa0{tex}\\Rightarrow{/tex}\xa0x2 + ax + bx + ab = 0{tex}\\Rightarrow{/tex}\xa0x(x +a) + b(x +a) = 0{tex}\\Rightarrow{/tex}\xa0(x\xa0+ a) (x + b) = 0{tex}\\Rightarrow{/tex}\xa0x + a = 0 or x + b = 0{tex}\\Rightarrow{/tex}\xa0x = -a or x = -b.Therefore, -a and -b\xa0are the roots of the equation. | |
| 39505. |
2+14 |
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Answer» 16 16 |
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| 39506. |
If the zeroes of ax^3 +3bx^2 +3cx +d are in A.P prove that 2b^3 - 3abc+ a^2 d =0 |
| Answer» Given\xa0polynomial is f(x) = ax3 + 3bx2 + 3cx + d.Let p-q, p, and p+q\xa0be the zeros of the polynomial f(x). Then,Sum of the zeros =\xa0{tex}-\\frac{\\text { coefficient of } x^{2}}{\\text { coefficient of } x^{3}}{/tex}p - q\xa0+ p + p + q\xa0=\xa0{tex}\\frac{-3 b}{a}{/tex}{tex}3 p=\\frac{-3 b}{a}{/tex}p =\xa0{tex}\\frac{-b}{a}{/tex}Since p\xa0is a zero of the polynomial f(x).Therefore, f(p) = 0f(p)=ap3 + 3bp2 + 3cp\xa0+ d = 0{tex}\\Rightarrow{/tex}{tex}a\\left(\\frac{-b}{a}\\right)^{3}+3 b \\times\\left(\\frac{-b}{a}\\right)^{2}+3 c \\times\\left(\\frac{-b}{a}\\right)+d=0{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac{-b^{3}}{a^{2}}+\\frac{3 b^{3}}{a^{2}}-\\frac{3 c b}{a}+d=0{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac{-b^{3}+3 b^{3}-3 a b c+a^{2} d}{a^{2}}=0{/tex}{tex}\\Rightarrow{/tex}2b3 - 3abc +a2d = 0Therefore, 2b3 - 3abc + a2d = 0Hence proved. | |
| 39507. |
What is value of sin and cos |
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Answer» Sin is equal to ^1- cos2 and cos is equal to ^1- sin2 Sin equal to 1/cosecCos equal to 1/sec |
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| 39508. |
8x—22x—21=0 |
| Answer» We have,{tex}8 x ^ { 2 } - 22 x - 21 = 0{/tex}{tex}\\Rightarrow{/tex}\xa08x2 - 28x + 6x - 21 = 0{tex}\\Rightarrow{/tex}\xa04x(2x - 7) + 3 (2x - 7) = 0{tex}\\Rightarrow{/tex}\xa0(2x - 7) (4x + 3) = 0 {tex}\\Rightarrow{/tex}\xa02x - 7 = 0 or,\xa04x + 3 = 0\xa0{tex}\\Rightarrow{/tex}\xa0{tex}x = \\frac { 7 } { 2 } \\text { or } x = - \\frac { 3 } { 4 }{/tex}Thus,\xa0{tex}x = \\frac { 7 } { 2 } \\text { and } x = - \\frac { 3 } { 4 }{/tex} are two roots of the equation 8x2 - 22x - 21 = 0 | |
| 39509. |
Prove that tan10theta tan15theta tan75theta tan80theta=0 |
| Answer» Tan(90-80).tan(90-75).tan75.tan80Cot80. Cot75. Tan75.tan801/tan80. 1/tan75. tan75.tan80Cancel each. =0 | |
| 39510. |
3k square _k ,d=6 find first term |
| Answer» Let the sum of k terms of A.P. is Sn = 3k2\xa0- kNow kth term of A.P = Sn - Sn-1ak = (3k2 - k) - [3 (k - 1)2- ( k - 1)]= (3k2 - k) - [3 (k2- 2k + 1)- ( k - 1)]= 3k2 - k - [ 3k2\xa0-6k + 3 - k + 1]= 3k2 - k - [ 3k2\xa0-7k + 4 ]= 3k2\xa0- k - 3k2 + 7k- 4= 6k - 4first term = a =\xa0{tex}6 \\times 1 - 4 = 6 - 4 = 2{/tex} | |
| 39511. |
If the sum of the zeroes the polynomial 3 x square-kx+ 6 is 3 when find the value of k |
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Answer» Put x=3 in the eq.And value of k =11 9 |
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| 39512. |
How to solve the question of exercise 4.3 question number 2 ... 4xsquare+4root3x+3=0 |
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Answer» 4.3 exercise is not there. if it was there means u can solve by ur hand and buy ur mind Bro chek answer in last of book Give me the answer |
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| 39513. |
In CBSE board exam how much % of marks is asked from NCERT book? |
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Answer» 86% No , it\'s not 30% , last year more than 90% was from NCERT. 100 30 |
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| 39514. |
Arithmetic Progressions: How many three digit numbers are divisible by 7? |
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Answer» The three digits no. which are divisible by 7 are 128 128 There are 128 three digit no.divisible by 7 Uncountaible |
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| 39515. |
2,3 4,5 -6,7find a ratio |
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| 39516. |
X2+y=4 |
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| 39517. |
What is fundamental theorm of arithmatic |
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Answer» Every composite no can be written in product of primes irrespective of order and this factorisation is uniqe for ecery no Every composite no can be written in product of primes irrespective of order and this factorization is unique for every no |
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| 39518. |
Solve the equation graphically 2x+ 3y + 5=0 ; 3x-2y-12=0 |
| Answer» Value of x is 78by5 ; y is 71by5 | |
| 39519. |
How to set the timetable for exams of board in less time?? |
| Answer» Study daily.....Divide 8 hrs into 4 parts.....That means 2 hrs study at 1 hr interval.... | |
| 39520. |
Root 3 is irational prove |
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Answer» Bhai itna Bhadrak mat PTA nhi Hoga bechaare ko Let , √3 be rational number.so it can be written in the form of a/b whereb≠0 (a and b are co prime numbers).√3=a/b, squaring on both sides gives us 3=a²/b² .. 3b²=a² it means a² is divisible by 3 ,so a is also divisible by 3.let a= 3c.. putting value of a ..3b²=9c²..b²=3c².. it means b² is divisible by 3 , so b is also divisible by 3..... So our assumption is wrong as they have a common factor 3 so √3 is irrational number तुझे नहीं पता |
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| 39521. |
If a=b=60verify that cos(a-b)=cosA ×cosB +sinA× sinB |
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Answer» 1=1 What is the answer of the question |
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| 39522. |
For what value of k do the equation 3x-y+8=0 and 6x-ky=-16 represent coincident lines |
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Answer» Your Quostion is Not Valid Becouse There is value of X is not Given a1/a2=b1/b2=c1/c2= 3/6=-1/-k=8/16=3/6=1/k= 1/2= camparing = 3k=6=K =2 |
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| 39523. |
Which book is best for board exam except ncert textbook |
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Answer» Oswall practise or que.bank is best for exam. Maths---RD Sharma 1.golden books2.R.D sharma3.fully marks4.all in one Do practice \'exemplar\' guide questions for maths and science. \'R.D. Sharma\' is also useful for maths. For social science refer any good guides available like \'full mark\' or \'golden\'. But, being thorough with the ncert textbook is the first thing that you have to do to score good marks for board exam! |
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| 39524. |
Why cot ^2 A= cos^2A/sin^2A |
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| 39525. |
Sin =what |
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Answer» A word P/b Beta tum phale question thick se likhooo..... Ok Aur sin ka matlb x ke kaisa hai jaiye mathematics me hum log koi angle ya phir valu ko x mante hai usi thar se ye bhi wohii hai P/h Sin=perpendicular/hypotenuse |
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| 39526. |
With the help of venn diagram prove that A union B -A intersaction B = A-B union B-A |
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| 39527. |
How can we check divisiblity of 7 |
| Answer» By divide the number by7 | |
| 39528. |
Find the value of 5 cosec square x minus 5 cot square x |
| Answer» 5 cosec^2 x - 5 cot^2 x5 ( cosec^2 x - cot^2 x )5 ( 1 ) [since: cosec^2 theta - cot^2 theta=1]Therefore, Answer = 5 | |
| 39529. |
Prove Angel Bisector Theorem |
| Answer» Given:AD bisects angle ATo prove:BD/DC=AB/AC Construction:Extend AB to E such that CE parallel to AD.Proof:angle BAD=angle DAC (given)AD paralel CEangle BAD=angle AEC (corresponding angle)angle DAC=angle AECangleDAC=angleACE(alt.int.angle)angle ACE=angle AECSo,AC=AEIn ∆BECAD parallel CEBD/DC=BA/AE(by bpt)BD/DC=AB/AC(proved) | |
| 39530. |
Exercise 4.1 |
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Answer» Yes 1 que |
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| 39531. |
Find value of x & y5x+4/y=97x-2/y=5 |
| Answer» x=1. Y=1 | |
| 39532. |
Stap divisen method some questiom |
| Answer» RQuestion-1The following table shows marks secured by 140 students in an examination: Marks number of student 0-10 2010-20 2420-30 4030-40 3640-50 20Calculate mean marks by using all the three methods, i.e., direct method, assumed mean method and step-deviation method. | |
| 39533. |
Step deviation method some questions |
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| 39534. |
Determine if the points (1,5),(2,3) and (-2,-11) are collinear. |
| Answer» If these points are collinear then the area of triangle formed by the points is 0 so by using the formula of area of triangle we can detect | |
| 39535. |
Sin30+ tan45-cosec60/sec30+ cos60+ cot45 |
| Answer» 1 | |
| 39536. |
Cos45/sec 30+cosec30 |
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Answer» Wrong answer √3+4√2/2√3 |
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| 39537. |
Show that square of any odd positive integer is in form 4m+1 for some integer m |
| Answer» Let a be any odd positive integer, then on dividing a by b, we have\xa0a = bq+r,{tex}\u200b\u200b\u200b\u200b0 \\leqslant r < b{/tex}.....(i) [by Euclid\'s division lemma]On putting b = 2 in Eq.(i), we get\xa0a = 2q+r,\xa0{tex}0\\leqslant r < 2{/tex}\xa0⇒ r = 0 or 1If r = 0, then a = 2q, which is divisible by 2. So, 2q is even.If {tex}r = 1{/tex}, then {tex}a = 2q + 1{/tex}, which is not divisible by 2. So {tex}(2q+1){/tex} is odd.Now, as a is odd, so it cannot be of the form 2q.Thus, any odd positive integer a is of the form (2q+1).Now, consider a2\xa0= (2q+1)2\xa0\xa0= 4q2\xa0+ 1 + 4q {tex}\\lbrack\\therefore(x+y)^2=x^2+y^2+2xy\\rbrack{/tex}= 4( q2\xa0+ q ) + 1\xa0= 4m + 1, where m = q2\xa0+ qHence, for some integer m, the square of any odd integer is of the form 4m+1.Hence Proved. | |
| 39538. |
Qhat is a integer |
| Answer» An integer is a number that can be written without a fractional component. | |
| 39539. |
Prove that root 6 is an irrational number |
| Answer» We have to prove that {tex} \\sqrt6{/tex}\xa0is an irrational number.Let\xa0{tex}\\sqrt6{/tex}\xa0be a rational number.{tex}\\therefore \\quad \\sqrt { 6 } = \\frac { p } { q }{/tex}where p and q are co-prime integers and\xa0{tex}q \\neq 0{/tex}On squaring both the sides, we get,or,\xa0{tex}6 = \\frac { p ^ { 2 } } { q ^ { 2 } }{/tex}or, p2 = 6q2{tex}\\therefore{/tex}\xa0p2 is divisible\xa0by 6.p is divisible by 6........(i)Let p = 6r for some integer ror, p2\xa0= 36r26q2= 362 [∵ p2 = 6q2]or, q2 = 6r2or, q2\xa0is divisible by 6.{tex}\\therefore{/tex}q is divisible by 6..........(ii)From (i) and (ii)p and q are divisible by 6, which contradicts the fact that p and q are co-primes.Hence, our assumption is wrong.{tex}\\therefore{/tex}\xa0{tex}\\sqrt6{/tex}\xa0is irrational number. | |
| 39540. |
How to find roots of a quadratic equation by completing square?? |
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| 39541. |
3x*3x+11x+10 by method of completing square |
| Answer» The given equation is{tex}\\Rightarrow{/tex}\xa03x2 +11x + 10 = 0Now, we have to find two numbers such that their sum is 11 and product is 3{tex}\\times{/tex}10 = 30. Clearly, 5 +6 =11 and 5{tex}\\times{/tex}6 = 30{tex}\\Rightarrow{/tex}\xa03x2 + 6x + 5x + 10 = 0{tex}\\Rightarrow{/tex}\xa03x(x + 2) +5(x + 2) = 0{tex}\\Rightarrow{/tex}\xa0(3x+5)(x+2) = 0Either, 3x+5 =0\xa0{tex}\\Rightarrow{/tex}\xa0{tex}{/tex}\xa0{tex}x= - \\frac{5}{3}{/tex}Or, x+2 =0 {tex}\\Rightarrow{/tex}\xa0x = -2Hence, the required roots of given equation are -2 and\xa0{tex}-\\frac{{5}}{3}{/tex} | |
| 39542. |
2019 blue print of maths and science |
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Answer» For maths Number system - weightage 06Algebra - weitage 20Co-ordinate geometry - weightage 06Geometry - weightage 15Trignometry - weightage 12Mensuration - weightage 10Statistics and probability - weightage 11Total - 80 marks1 marks - 6 ques.2 marks - 6 ques.3 marks - 10 ques.4 marks - 8 ques. Total - 30 ques.These blue print is not officially declared but it is 95% correct 2019 Mujha mathematics ka cbsc paper bata do bhai |
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| 39543. |
21rt6rey |
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| 39544. |
If two vertics of equilateral ail triangle(3,0) and(6,0)find cordinate of third vertices |
| Answer» Jo angle nahi pta usse x and y let kar lo or usse A maan lo A (x,y )or fir (3,0 ) or (6,0 )ko a sath alag alag distance formula se solve kar lo fir jo bhi aaye use ek dosre ke equal rahk lo because it is equilateral triangle so it\'s all sodes are equal toh jab tumm un dono ko ek dusre ke equal ŕakho ge then you get the answer ☺ | |
| 39545. |
If A,B,C are interior angles of a triangles ABC then show that Sin(B+C)by 2 =cos A by 2 |
| Answer» \xa0A, B, C, are interior angles of a\xa0{tex}\\Delta {/tex}{tex}\\because A + B + C = 180 ^ { 0 }{/tex}{tex}\\Rightarrow B + C = 180 ^ { 0 } - A \\Rightarrow \\frac { B + C } { 2 } = 90 ^ { 0 } - \\frac { A } { 2 }{/tex}{tex}\\Rightarrow \\sin \\frac { \\mathrm { B } + \\mathrm { C } } { 2 } = \\sin \\left( 90 ^ { \\circ } - \\frac { \\mathrm { A } } { 2 } \\right) \\left[ \\because \\sin \\left( 90 ^ { 0 } - \\theta \\right) = \\cos \\theta \\right]{/tex}{tex}\\Rightarrow \\sin \\frac { \\mathrm { B } + \\mathrm { C } } { 2 } = \\cos \\frac { \\mathrm { A } } { 2 } \\text { proved }{/tex}LHS = RHS | |
| 39546. |
Class 10 exercise 6.4 |
| Answer» Check NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 39547. |
Given tan A=4/3 Find the other trigonometric ratios of the angel A |
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Answer» Abhishek ir answer is wrong but √25 is 5 no ? AB=4,BC=3,AC=25(by pythagoras theorem) SinA=3/25,cosA=4/25,cotA=3/4,secA=25/4,cosecA=25/3. |
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| 39548. |
Some applications of trigonometry |
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Answer» Comedy Calculator?? |
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| 39549. |
Prove √ 2 is irrational |
| Answer» Let , us assume that √2 is rational number then it can be written in the form of a/b where b≠0 and (a,b are co prime numbers). So, √2=a/b, now squaring on both sides gives 2=a²/b²... 2b²=a² so we can say thay a² is divisible by 2 , therefore a is also divisible by 2. Then take a=2c, now putting value ofa and squaring on both sides giv gives us ...... 2b²=4c² => b²=2c² so b² is divisible by 2 therefore b is also divisible by 2. Hence our assumption is wrong , they have other common factor as 2. There fore √2 is irrational.hence, proved | |
| 39550. |
Prove that ratio of area of two similar triangle is equal to ratio of square of angle bisector |
| Answer» Given: {tex}\\triangle {/tex}ABC {tex} \\sim {/tex}{tex}\\triangle {/tex}DEFAM is the bisector of{tex}\\angle{/tex} BAC and is the corresponding bisector of {tex}\\angle{/tex} EDFTo prove:{tex}\\frac{{area\\vartriangle ABC}}{{area\\vartriangle DEF}} = \\frac{{A{M^2}}}{{D{N^2}}}{/tex}Proof: {tex}\\frac{{area\\vartriangle ABC}}{{area\\vartriangle DEF}} = \\frac{{A{B^2}}}{{D{E^2}}}{/tex}.......(i) [Area theorem]Now {tex}\\triangle {/tex}ABC {tex} \\sim {/tex}{tex}\\triangle {/tex}DEF{tex}\\Rightarrow {/tex}{tex}\\angle{/tex} A = {tex}\\angle{/tex} D {tex}\\Rightarrow {/tex}{tex}\\frac{1}{2}\\angle A = \\frac{1}{2}\\angle D{/tex}{tex}\\Rightarrow {/tex}{tex}\\angle{/tex}BAM = {tex}\\angle{/tex}EDNIn {tex}\\triangle {/tex}ABM and DEN {tex}\\angle{/tex} B = {tex}\\angle{/tex} E and {tex}\\angle{/tex} BAM= {tex}\\angle{/tex}EDN{tex}\\Rightarrow {/tex}{tex}\\triangle {/tex}ABM {tex} \\sim {/tex}{tex}\\triangle {/tex}DEN{tex}\\Rightarrow {/tex}{tex}\\frac{{AB}}{{DE}} = \\frac{{AM}}{{DN}} \\Rightarrow \\frac{{A{B^2}}}{{D{E^2}}} = \\frac{{A{M^2}}}{{D{N^2}}}{/tex}...................(ii){tex}\\Rightarrow {/tex}{tex}\\frac{{area\\vartriangle ABC}}{{area\\vartriangle DEF}} = \\frac{{A{M^2}}}{{D{N^2}}}{/tex} [From (i) and (ii)] Proved. | |