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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.
| 43351. |
If x square minus 5x plus 1 the value of x plus one upon x is |
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| 43352. |
१+१+१+१+१+१+२+६+७+८+? |
| Answer» 29 | |
| 43353. |
Find the value of under root 6+ root 6 + root 6 + ... |
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Answer» Given : x = √6 + √6 + √6 +... ---(i)Then,x = √6 + x --- From Eqn. (i)On squaring both sides,⇒ x² = 6 + x⇒ x² - x - 6 = 0⇒ x² - 3x + 2x - 6 = 0⇒ x(x-3) + 2(x-3) = 0⇒ x = 3 Or x = -2But x is a natural number.∴ x = 3\xa0\xa0 Can you write it step-by-step!By the way sis not bro. |
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| 43354. |
How do you factoriee 2x√2-x+1/8 |
| Answer» Question :\xa0Factorise 2x^2 - x + 1/8 =0 using splitting the middle term.Answer :Given Quadratic equation:2x²-x+1/8\xa0=\xa00Multiply each term by\xa08\xa0we get=>\xa016x²\xa0-\xa08x\xa0+\xa01\xa0=\xa00Splitting the middle term,we get=> 16x²\xa0-\xa04x\xa0-\xa04x\xa0+ 1 = 0=> 4x(4x-1)-1(4x-1)=0=>\xa0(4x-1)(4x-1)=0=>\xa04x-1=0\xa0or\xa04x-1 = 0=>\xa04x\xa0=\xa01\xa0or 4x = 1=>\xa0x\xa0=\xa01/4\xa0or x = 1/4Therefore,Roots of the quadratic equation are\xa01/4\xa0,\xa01/4 | |
| 43355. |
About trigonometry |
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| 43356. |
Hlw,frnds anyone of u like to talk with me |
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Answer» Aap kha se ho brijesh Osm r u shivani Yes |
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| 43357. |
Factorization x2-9x+18 |
| Answer» => x² + 9x + 18=> x² + 6x + 3x +18=> x ( x+6 ) + 3 ( x+6 )=> ( x+6 ) ( x+3 )Therefore, ( x+6 ) ( x+3 ) are the factors . | |
| 43358. |
Are some topics removed from the Circles topic? |
| Answer» No | |
| 43359. |
The quadratic polynomial, the sum of whose zeroes is -5 and their product is 6 , is(1 Point) |
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Answer» Sum of zeroes = -5Product of zeroes = 6 So, we know that when we have to find the quadratic equation then we use this formulas : x2-(sum of zeroes) x+product Now, x2-(-5)x+6. Required quadratic equation is --> x2+5x+6. Let the quad. polynomial be ax^2+bx+c and whose zeroes are alpha and beeta. According to question- sum of zeroes=-b/a=-5/1 products of zeroes =c/a=6/1 from here, a=1, b=5 and c=6 thus, required quadratic equation is x^2+5x+6 |
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| 43360. |
Can anyone have the idea that do we guys have project in class 11 |
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| 43361. |
What five letter word becomes shorter when you add 2 letters in it |
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Answer» ?? The five letter word is SHORT. You add two letters E and R and it becomes SHORTER. The word “short” becomes “shorter” when two (2) letters are added to it. C.H. |
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| 43362. |
What are the uses and applications of frequency polygon |
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Answer» Frequency polygons are a graphical device for understanding the shapes of distributions. They serve the same purpose as histograms, but are especially helpful for comparing sets of data. Frequency polygons\xa0are a graphical device for understanding the shapes of distributions. They serve the same purpose as histograms, but are especially helpful for comparing sets of data.\xa0Frequency polygons\xa0are also a good choice for displaying cumulative\xa0frequency\xa0distributions |
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| 43363. |
Find the root of the quadratic equations by using the quadratic formula :- 2x^-3x-5=0 |
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Answer» here ,, discriminant = (b^2 - 4ac ) = (-3 )^2 + (4×2×5) = 49Quadratic formula:x = (-b +or- root of B^3 - 4ac )÷ by 2a1. X = (3 + root of 49)÷ 4 =2.52. X= ( 3- root of 49) ÷ 4 = -4÷4 -1 a=2 , b=-3 , c=-5D=b^-4acD=2^-4×2×-5D=4+40D=44,D<0,two distinct rootQuadratic formula = -b±√D/2a= -(-3)±√44/2×2X=3+2√11/4 , X=3-2√11/4 |
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| 43364. |
Worried about 10 result |
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Answer» I got 93 percent Kaisa raha bhaiya aapka result ? Kitni percentage bani? Don\'t worry Nd don\'t worry u will be definitely pass nd with a wonderful %age Dear, it will be coming today |
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| 43365. |
Hi, everyone |
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Answer» Hllo Hlw Hello Hi Hi dear? |
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| 43366. |
9x-10y=-122x+3y=13Solve by substituting method |
| Answer» 9x - 10y = - 12X = ( 10y - 12)÷92 ( 10y - 12)/9 + 3y = 1347y - 24 = 11747y = 141Y = 3X = ( 10 × 3 - 12)÷9 = 2 | |
| 43367. |
LCM of 289 |
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Answer» 78897 Least Common Multiple of 273 and 289. Least common multiple (LCM) of 273 and 289 is 78897. |
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| 43368. |
Sin(b+c)/2*cosA/2+cos(b+c)/2*Sina/2 |
| Answer» We knoe that A+B+C=180 B+C=180-A (B+C)/2=(180-A)/2 (B+C)/2=90-A/2 Cos(B+C)/2=Cos(90-A/2) Cos(B+C)/2=Sin A/2Or, Sin(B+C)/2=Sin(90-A/2) Sin(B+C)/2=Cos A/2Now, According to questionSin(b+c)/2 * cos A/2 +cos(b+c)/2 *sin A/2Cos A/2 *cos A/2 + sinA/2 *sin A/2 Cos^2 A/2+sin^ A/2 1 is ur answer | |
| 43369. |
X^2-3√5x+10 = 0 |
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Answer» x²-3√5x+10=0Splitting the middle term, we\xa0get=> x²-2√5x-√5x+10=0=> x²-2√5x-√5x+2×5=0=> x²-2√5x+√5x +2×√5×√5=0=> x(x-2√5)-√5(x-2√5)=0=> (x-2√5)(x-√5)=0=> x-2√5 =0 Or x-√5 = 0=> x = 2√5 Or x = √5Therefore,Roots of given Quadratic equation are:2√5 Or √5\xa0 x²-3√5x+10=0.........Splitting the middle term, weget......=> x²-2√5x-√5x+10=0........=> x²-2√5x-√5x+2×5=0.........=> x²-2√5x+√5x +2×√5×√5=0........=> x(x-2√5)-√5(x-2√5)=0........=> (x-2√5)(x-√5)=0........=> x-2√5 =0 Or x-√5 = 0.......=> x = 2√5 Or x = √5........Therefore,........Roots of given Quadratic equation are.......:2√5 Or √5......••••• |
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| 43370. |
Express the trigonometric ratios sinA, secA and tanA in terms of cotA |
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Answer» Right answer [ i ] sinA in terms of cotA :sinA = 1 / cosecAWe know that: cosec²A = 1 + cot²AsinA = 1 / √1 + cot²A[ ii ] secA in terms of cotA :We know that: 1 + tan²A = sec²AsecA = 1 + tan²AsecA = 1 + 1 / cot²AsecA = √(cot²A + 1 ) /cotA[ iii ] tanA in terms of cotA :tanA = 1 / cotA |
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| 43371. |
Why 9÷3×3=0 |
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Answer» Yes,but why 9 is answer ??? Your answer is incorrect mahi Your answer is incorrect You can see in calculator Answer is 9 3×3=9 then 9÷9=1 not 0 (is my answer is correct or not) |
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| 43372. |
prove that quadratic equation ax^2+bx+c =-b+-√b^2 -4ac |
| Answer» Quadratic Formula DerivationConsider the equation\xa0ax2+bx+c\xa0= 0, a ≠ 0.Dividing the equation by a gives,\xa0x2+ b/a x+c/a\xa0= 0By using method of completing the square, we get(x+b/2a)2\xa0– (b/2a)2\xa0+ c/a = 0(x+b/2a)2\xa0– [(b2-4ac)/4a2]= 0(x+b/2a)2\xa0= (b2-4ac)/4a2Roots of the equation are found by taking the square root of RHS. For that b2-4ac\xa0should be greater than or equal to zero.When b2-4ac ≥ 0,(x\xa0+\xa0b2a)\xa0=\xa0±\xa0b2\xa0−\xa04ac√2ax\xa0=\xa0−b\xa0±\xa0b2\xa0−\xa04ac√2a\xa0—-(1)Therefore roots of the equation are,\xa0−b\xa0+\xa0b2\xa0−\xa04ac√2a\xa0and\xa0−b\xa0−\xa0b2\xa0−\xa04ac√2aThe equation will not have real roots if b2-4ac < 0, because square root is not defined for negative numbers in real number system.Equation (1) is a formula to find roots of the quadratic equation\xa0ax2+bx+c\xa0= 0, which is known as quadratic formula. | |
| 43373. |
Find the value of the following up to three places of decimal /5 |
| Answer» We are supposed to find the square root of 5 up to 3 decimal placesNumber = 5Group the digits into pairsFor digits to the left of the decimal point, pair them from right to left.For digits after decimal point, pair them from left to rightSo, we have, 05Refer the attached figure for the steps to be followedSo, √5 = 2.236Hence the value of root 5 up to 3 decimal places is 2.236 | |
| 43374. |
The cost of 2kg apples and 1kg grapes on a day was rs 160 |
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| 43375. |
135 +567 |
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Answer» 702 702 702 702 |
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| 43376. |
Show that x=-bc/ad is a solution of the quadratic equation ad²(ax/b+2c/d)x+bc²=0. |
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| 43377. |
Cbsce last year 10 exam papers |
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Answer» Question paper Year 10 Last A |
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| 43378. |
Fundamental theoram of arithmetics |
| Answer» The fundamental theorem of arithmetic says that every composite number can be factorized as a product of primes actually it says more. it says that given any composite number it can be factorized as a product of prime numbers in a unique way , except for the order in which the primes occur . | |
| 43379. |
If 2 X + 1 upon 3 y is equals to 2 and 1 upon 3 |
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| 43380. |
1/4x_ 1/9y=763/x+99/y=67Solve this equation by reducing them to a pair of linear equation. |
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| 43381. |
What is alzebra |
| Answer» Algebra spelling is wrong | |
| 43382. |
3x-5y=76x-10y=3 |
| Answer» 3x-5y=7--->16x-10y=3--->2Multiple eq 1 with 2Such that,\xa06x-10y=14Now, substrat 1 from 26x-10y=14-(6x-10y=3)--------------12x=3X=3/12X=⅓Sub X value in any equation,6(⅓)-10y=32-10y=3-10y=3/2-y=3/2/10-y=3/20y=-3/20\xa0 | |
| 43383. |
The ratio length of a rod and it\'s shadow is 1:√3 .the angle of elevation of the sum is |
| Answer» Let AB be the pole and BC be its shadow.Given that AB/BC = 1/√3Let ACB be the angle of elevation.In triangle ABC,X is equal to thetatan x = AB/BCtan x = 1/√3tan x = tan 30X = 30\xa0Hence angle of elevation\xa0=30∘ | |
| 43384. |
Form a quadratic equation whose roots are 3+√5 and 3-√5 |
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| 43385. |
2x+5y-13=03x+5y=12 |
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Answer» X=-1,y=3 Get the variable with the smaller coefficient by itself:2x+5y−13=02x+5y−13=0⟹2x=−5y+13⟹2x=−5y+13⟹2x=−4y+12+1−y⟹2x=−4y+12+1−y⟹2x2=−4y\xa02+122+1−y2⟹2x2=−4y\xa02+122+1−y2⟹x=−2y+6+1−y2.⟹x=−2y+6+1−y2.Next lettingt=1−y2t=1−y2⟹2t=1−y⟹2t=1−y⟹y=1−2t.⟹y=1−2t.Let\xa0tt\xa0be an integer, which means that\xa0yy\xa0is an integer. Thenx=−2y+6+1−y2x=−2y+6+1−y2=−2(1−2t)+6+t=−2(1−2t)+6+t=−2+4t+6+t=−2+4t+6+t=4+5t.=4+5t.Check these two answers:2x+5y−13=02x+5y−13=0⟹2(4+5t)+5(1−2t)−13⟹2(4+5t)+5(1−2t)−13=8+10t+5−10t−13=8+10t+5−10t−13=13+10t−10t−13=0✓=13+10t−10t−13=0✓∴∴\xa0integer solutions to the given equation have the formx=4+5t,\xa0y=1−2tx=4+5t,\xa0y=1−2twhere\xa0tt\xa0is an integer.So there is a countable infinitude of integer solutions to the given equation and more than\xa04.4.Choose integer values of\xa0tt\xa0to obtain specific answers. For example,\xa0t=10t=10\xa0yields the answerx=4+5(10)=4+50=54,\xa0y=1−2(10)=1−20=−19.x=4+5(10)=4+50=54,\xa0y=1−2(10)=1−20=−19.Select four different values of\xa0tt\xa0to get four distinct solutions.To obtain positive answers for\xa0xx\xa0and\xa0y,y,\xa0must havex=4+5t>0\xa0and\xa0y=1−2t>0x=4+5t>0\xa0and\xa0y=1−2t>0⟹5t>−4\xa0and\xa01>2t⟹5t>−4\xa0and\xa01>2t⟹t>−45\xa0and\xa0t<12⟹t>−45\xa0and\xa0t<12⟹−45 |
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| 43386. |
Find the value of k if a(2, 3), b(5, k), c(6, 7) are collinear points. |
| Answer» Given 3 points A(2,3), B(5,k) and C(6,7) which are collinear we have to find the value of k.Points are collinear if the slopes of any two pairs are equal.Slope of AB = y2 -y2/x2 - x1= k-3/5-2Slop of BC=\xa0y2 -y2/x2 - x1= 7-k/6-5As lines are collinear slopes are equal⇒k-3/5-2= 7-k/6-5⇒ k- 3= 2(7-k)⇒ k= 6The value of k is 6.\xa0 | |
| 43387. |
Prove that sin2 theta + cos2 theta =1 |
| Answer» Step-by-step explanation:cos = b/h and sin = b/hLHSSin2 +cos2={p/h}2 +[b/h]2=p2 +b2 /h2=h2/h2=1=RHS.......{...HENCE PROVED} | |
| 43388. |
Topper students please tell how to make my mathematics better |
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Answer» Baby, if u r very week in maths then give near about an hour for solving one exercise of ncert nd do extra questions too of that ch...... HOPE YOU WILL SCORE 100 IN MATHS IN THIS CLASS BY FOLLOWING MY THESE INSTRUCTIONS...... ?? Practice daily for 30 minutes and practice solved example from NCERT textbook. Really how many hours Pado or padao In practice |
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| 43389. |
Please tell how to solve system of equations graphically, please tell some tricks to solve it |
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Answer» Ya thanks ??? Baby, check on YouTube rashmi maths teacher channel u will get a great help from there..... Nd u will understand the ch Or questions in which u r getting confused..... ??HOPE MY ADVICE WILL HELP U.... ? |
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| 43390. |
By elimination method x_y=2 |
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| 43391. |
Mathematics ka syllabus kya hai 2020 ka please bata dijiye |
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Answer» Maths topics are deleted I-NUMBER SYSTEMS REAL NUMBERS \uf0b7 Euclid’s division lemma UNIT II-ALGEBRA POLYNOMIALS \uf0b7 Statement and simple problems on division algorithm for polynomials with real coefficients. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES \uf0b7 cross multiplication method QUADRATIC EQUATIONS \uf0b7Situational problems based on equations reducible toquadratic equationsARITHMETIC PROGRESSIONS \uf0b7Application in solving daily life problems based on sumto n termsUNIT III-COORDINATE GEOMETRY COORDINATE GEOMETRY \uf0b7 Area of a triangle. UNIT IV-GEOMETRY TRIANGLES Proof of the following theorems are deleted\uf0b7The ratio of the areas of two similar triangles is equal tothe ratio of the squares of their corresponding sides.\uf0b7 In a triangle, if the square on one side is equal to sumof the squares on the other two sides, the angleopposite to the first side is a right angle.CIRCLES No deletion CONSTRUCTIONS \uf0b7 Construction of a triangle similar to a given triangle. UNIT V- TRIGONOMETRY INTRODUCTION TO TRIGONOMETRY \uf0b7 motivate the ratios whichever are defined at 0o and 90oTRIGONOMETRIC IDENTITIES \uf0b7Trigonometric ratios of complementary angles.HEIGHTS AND DISTANCES No deletion UNIT VI-MENSURATION AREAS RELATED TO CIRCLES \uf0b7 Problems on central angle of 120° SURFACE AREAS AND VOLUMES \uf0b7 Frustum of a cone. UNIT VI-STATISTICS & PROBABILITY STATISTICS \uf0b7 Step deviation Method for finding the mean \uf0b7 Cumulative Frequency graph PROBABILITY No deletion http://cbseacademic.nic.in/Revisedcurriculum_2021.htmlGo on this link and select Math. Sir math ka syllabus kya board exam ka 2020 2021 Search on youtube 30% topic Kam hua hai |
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| 43392. |
Euclid\'s division lemmaProve it.... |
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Answer» Hum..... Topic reomved for this year |
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| 43393. |
Solve by substitution method3x-y=39x-3y=9Plz solve this ?????? |
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Answer» 3x-y=3........19x-3y=9..(divided by 3)3x-y=3......23x=3-yX=3-y/3sub X in 13x-y=33(3-y/3)-y=33-y-y=3-2y=3-3y=0sub y in 23x-y=33x-0=33x=3X=1 3x-y=3........(i)9x-3y=9..(divided by 3)3x-y=3......(i)3x=3-yX=3-y/3sub X in (i)3x-y=33(3-y/3)-y=33-y-y=3-2y=3-3y=0sub y in (ii)3x-y=33x-0=33x=3X=1 Put the value,Y=3x-3 in eq (2) 9x-3(3x-3)=9 => 9x-9x+9=9.. hence, proved |
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| 43394. |
Given 15 cot A =8, find sin A and sec A. |
| Answer» CotA=8/15O=15k. A=8kBy Pythagoras the theorem,h=√225+64 =√289 =17kSo, sinA=15/17 SecA=17/8 | |
| 43395. |
If the difference of cube of two no. is 331 and sum of thier square is 221. Then find the no.s |
| Answer» Question glat hai two consective number hoga | |
| 43396. |
4root3xsquare plus 5x minus 2root3=0, find \'x\' |
| Answer» Please question in a right way????? | |
| 43397. |
The distance of the point (-5,-6) from Y-axis is(a)”√61 |
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| 43398. |
Solve the following for x and y by using method of elimination : |
| Answer» What is the equation here ??? | |
| 43399. |
Kx+2y=53x-4y=10 |
| Answer» The given system of equations:\xa0kx + 2y = 5\xa0⇒ kx + 2y - 5 = 0 ….(i)\xa03x - 4y = 10\xa0⇒3x - 4y - 10 = 0 …(ii)\xa0These equations are of the forms:Thus for all real values of k other than −3/2 , the given system of equations will have a unique solution.\xa0(ii) For the given system of equations to have no solutions, we must have: | |
| 43400. |
If angle A and angle B are acute angles such that cosA =cosB, show that angle A= angle B. |
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Answer» Hey anshika An In right triangle ABCCosA= AC/AB AND CosB= BC/ABBUT CosA= CosB (given) AC/AB=BC/ABAC=BC THEREFORE Angle A = Angle B ( angle opposite to equal sides are equal Hii Anshika Let us consider two right angle triangles right angled at Q and S respectively.Now,But cos A = cos B\xa0(given)∴ Let In ΔRAS,Using Pythagoras theorem, we haveIn APBQ,Using Pythagoras theorem, we haveSo,\xa0Comparing (i) and (ii), we getSo, by using SSS similar condition∴ |
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