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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.
| 151. |
In fig. 6.38 altitudes AD and CE intersect each other at the point P. Show that |
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Answer» AD and CE are altitudes, which intersect each other at P.(i) In ∆AEP and ∆CDP∠AEP = ∠CDP =\xa090°\xa0[given]and ∠APE = ∠CPD[vertically opposite angles]Therefore, by using AA similar condition∆AEP ~ ∆CDP.(ii) In ∆ABD and ∆CBE∠ADB = ∠CEB = 90° [given]and ∠B = ∠B [common]Therefore, by using AA similar condition∆ABD ~ ∆CBE.(iii) In ∆AEP and ∆ADB∠AEP = ∠ADB = 90° [given]and ∠PAE = ∠DAB [common]Therefore, by using AA similar condition∆AEP ~ ∆ADB(iv) In ∆PDC and ∆BEC∠PDC = ∠CEB = 90° [given]∠PCD = ∠ECB [common]Therefore, by using AA similar condition∆PDC ~ ∆BEC. In the given Fig,\xa0altitudes AD and CE of\xa0\xa0intersects each other at the point P. Show that:(i) ∆AEP ~ ∆CDP(ii) ∆ABD ~ ∆CBE(iii) ∆AEP ~ ∆ADB(iv) ∆PDC ~ ∆BEC. \xa0AD and CE are altitudes, which intersect each other at P.(i) In ∆AEP and ∆CDP∠AEP = ∠CDP =\xa090°\xa0[given]and ∠APE = ∠CPD[vertically opposite angles]Therefore, by using AA similar condition∆AEP ~ ∆CDP.(ii) In ∆ABD and ∆CBE∠ADB = ∠CEB = 90° [given]and ∠B = ∠B [common]Therefore, by using AA similar condition∆ABD ~ ∆CBE.(iii) In ∆AEP and ∆ADB∠AEP = ∠ADB = 90° [given]and ∠PAE = ∠DAB [common]Therefore, by using AA similar condition∆AEP ~ ∆ADB(iv) In ∆PDC and ∆BEC∠PDC = ∠CEB = 90° [given]∠PCD = ∠ECB [common]Therefore, by using AA similar condition∆PDC ~ ∆BEC. |
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| 152. |
Prove that root 7 is a irrational number |
| Answer» let us assume that √7 be rational.then it must in the form of p / q.As definition of rational number says.. P is whole number q is non zero whole number.. And p and q is simplest ratio which is expressed.. That means there exists no prime factor common in p and q.√7 = p / q√7 x q = psquaring on both sides7q² = p² ------1.p is divisible by 7p = 7c [c is a positive integer] [squaring on both sides ]p²= 49c²subsitute p² in eqn(1) we get7q² = 49 c²q² = 7c²q is divisble by 7thus q and p have a common factor 7.there is a contradiction to our assumptionas our assumsion p & q are co prime but it has a common factor.so that √7 is an irrational. | |
| 153. |
In taingle ABC,DE parallel too AC and DC parallel AP. Prove thatBE/EC=BC/CP |
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Answer» Kuch bhi maat likha kro plzzzzz?? O bhai ye sagar dhadke kya h ye I love you In\xa0△BPA, we have DC∣∣AP [Given]Therefore, by basic proportionality theorem, we haveBC/\xa0CP \u200b= BD\u200b/DA ........ (I)In\xa0△BCA, we have\xa0DE∣∣ACTherefore, by basic proportionality theorem, we haveBE/EC \u200b = BD\u200b/DA........ (iI)From (i) and (ii), we get\xa0BC/CP \u200b= BE/ EC \u200b\xa0or\xa0BE/\u200bEC = BC/CP\u200b [Hence\xa0proved] |
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| 154. |
Half of |
| Answer» Please write the full questions | |
| 155. |
What are the different trignometric ratiis? |
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Answer» sine:\xa0Sine of an angle is defined as the ratio of the side opposite(perpendicular side) to that angle to the hypotenuse.cosine:\xa0Cosine of an angle is defined as the ratio of the side adjacent to that angle to the hypotenuse.tangent:\xa0Tangent of an angle is defined as the ratio of the side opposite to that angle to the side adjacent to that angle.cosecant:\xa0Cosecant is a multiplicative inverse of sine.secant:\xa0Secant is a multiplicative inverse of cosine.cotangent:\xa0Cotangent is the multiplicative inverse of the tangent. \tsine:\xa0Sine of an angle is defined as the ratio of the side opposite(perpendicular side) to that angle to the hypotenuse.\tcosine:\xa0Cosine of an angle is defined as the ratio of the side adjacent to that angle to the hypotenuse.\ttangent:\xa0Tangent of an angle is defined as the ratio of the side opposite to that angle to the side adjacent to that angle.\tcosecant:\xa0Cosecant is a multiplicative inverse of sine.\tsecant:\xa0Secant is a multiplicative inverse of cosine.\tcotangent:\xa0Cotangent is the multiplicative inverse of the tangent. |
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| 156. |
Determine the AP. whose 3rd term is 5 and 7th term is 9 |
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Answer» 3rd term of the AP: a+2d = 5 …(1)7th term: a+6d = 9 …(2)Subtract (1) from (2)4d = 4, ord = 1 and so a = 3.The AP is 3,4,5,6,…. 3rd term of the AP: a+2d = 5 …(1)7th term: a+6d = 9 …(2)Subtract (1) from (2)4d = 4, ord = 1 and so a = 3.The AP is 3,4,5,6,…. Here a3= a+2d =5___(1) a7= a+6d =9___(2) Subtracting 2 from 1a+2d-a-6d = 5-9=> -4d = -4 d= 1Now a+2d = 5 a = 5-2dPutting the value of (d) a= 5- 2(1) a = 3The required AP is 3,4,5,6,7...... |
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| 157. |
Give example of two irrationals whose sum is rational |
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Answer» (3+√2) and (3-√2) Root3+4 and root4-3 Root3 +4 and root 4-3 |
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| 158. |
Similar triangle theorem |
| Answer» Theorem: If two\xa0triangles\xa0are\xa0similar, then the ratio of the area of both\xa0triangles\xa0is proportional to the square of the ratio of their corresponding sides. This proves that the ratio of the area of two\xa0similar triangles\xa0is proportional to the squares of the corresponding sides of both the\xa0triangles. | |
| 159. |
Solve the equation: 4x ‒ 3 = 52?+3 , x ≠ 0, 32 |
| Answer» Given :\xa0Expression\xa0Cross multiply,Apply middle term split,Therefore, The solution of the expression is x=-2 and x=1. | |
| 160. |
Find each of the products 3x(-1) |
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Answer» -3x _3x -3x -3x -3x |
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| 161. |
5+5= |
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Answer» 5+5=10I hope it will help you!?? 10 10 10 10 |
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| 162. |
2x+3y=94x+6y=18 solve in substantial method |
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Answer» It\'s ncert just do it by yourself Unique siruliont 2x=9-3yx=9-3y/2Now put value of x in 2 equation4 (9-3y/2)+6y=18(18-6y)+6y=180This equation is unique solution |
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| 163. |
9sec2A-tan2A=? |
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Answer» 9 I think 9 is the answer lol ?ok I will try it again Ans is wrong 47.255817 A2 |
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| 164. |
2minus8 |
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Answer» Hnji? Free fire khelte ho You play Ffy -6 -6 |
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| 165. |
The larger of two supplementary angles exceed the smaller by 18 find the angles |
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Answer» Let the angle be xother angle (it\'s supplement) =( 180-x)x+18= (180-x)2x = 162x = 162/2 = 81other angle(its supplement) = 180- x =180-81 = 99 lol ??????? See in your Book............,,,,,,,.,.,.,.,.,., |
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| 166. |
Find the zeros of polynomial x2-9x-8 |
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Answer» Yes their is an mistake in this Q The zeros of the polynomial is (9+√113)/2 , (9- √113)/2 Their is some problem in this question. Check the signs of this question |
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| 167. |
Chapter 4 exercise 4.1 |
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Answer» NCERT Solutions for Class 10 Maths Exercise 4.1 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 10 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.Click on the linkhttps://mycbseguide.com/blog/ncert-solutions-class-10-maths-exercise-4-1/ Go to mathematics ch.4 ex. 4.1 and u will find ur answer Number kiya hai |
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| 168. |
Ex.2.4 4 |
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Answer» In chap2 only ex 2.1 to 2.3 alone... L ,,,,,d |
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| 169. |
What is sin∅ |
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Answer» Sin theta = Perpendicular/ Hypotenuse Perpendicular / hypotenuse = sin ∅ . Sin thetha=opposite side/ hypotenuse Sin theta= opposite side / hypotenuse sin ∅ = opposite\xa0side/hypotenuse\xa0 |
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| 170. |
Sir good morning |
| Answer» Good evenimg | |
| 171. |
Give me some guidance to solve problem in maths . |
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Answer» Problems will definitely there because maths is a subject where you will face difficulties ; if you will work on it with keen interest than gradually you will not face difficulty.As per my advice you should 1st solve all doubts of ncert with the optional ones than you should solve sums from practice books i.e. RD Sharma,Navneet etc.... than don\'t forget to solve last 5 years paper. You can also practice with Gala. I hope it will help you! ?? (Prothom Boss) RS Aggarwal is better than ncert because it gives all questions of ncert Maths is not a subject which can be mastered in a day or two.It requires lots of practise.If you realy want to score good in maths than firstly go through all questions in ncert also examples given book(excluding the optional one).Once you are done with ncert and have enough time than solve refresher.But most importantly dont forget to solve the previous years question paper and also go through the latest pattern of exam paper which will be provided by CBSE on ots sites. |
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| 172. |
(2,3) , (4,1) |
| Answer» But write your question properly ( with whole question ) . | |
| 173. |
Value of X |
| Answer» It depends on number which number are there in question | |
| 174. |
Factories √3x-4x-4√3 |
| Answer» √3x^2 -4x -4√3=√3x^2 -6x +2x -4√3 =√3x(x-2√3) +2(x-2√3)=(√3x + 2) (x - 2√3) | |
| 175. |
Solve by substitution method : 2x + 3y =9 ; 4x+ 6y =18 |
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Answer» Step 1: 2x+3y=9 ..............(eqn 1) 4x-6y=18 ............(eqn 2)divide equation 2 by 2 on both sides, we get: 2x - 3y = 9 2x = 9 + 3y x = (9+ 3y)/2 ..........(eqn 3) step 2:now put the value of x (from equation 3) in equation 1, we get: 2((9+ 3y)/2) + 3y = 9 (9 + 3y) +3y = 9 9 + 6y =9 6y = 9-9 6y = 0 y = 0 .........................(eqn 4) step 3:Now put the value of y = 0 (from equation 4) in equation 3, we get x = (9+ 3(0))/2 x = 9/2 x = 4.5 .......................(eqn 5)so the answer is x = 4.5 and y = 0. step 4:Verification:to verify the answer we can put these values of x and y in any one of the equation 1 or equation 2.put x = 4.5 and y = 0 in equation 1, we get 2x+3y=9 2(4.5) + 3(0) = 9 9 + 0 = 9 9 = 9 L.H.S. = R.H.S. Step 1: 2x+3y=9 ..............(eqn 1) 4x-6y=18 ............(eqn 2)divide equation 2 by 2 on both sides, we get: 2x - 3y = 9 2x = 9 + 3y x = (9+ 3y)/2 ..........(eqn 3)\xa0step 2:now put the value of x (from equation 3) in equation 1, we get: 2((9+ 3y)/2) + 3y = 9 (9 + 3y) +3y = 9 9 + 6y =9 6y = 9-9 6y = 0 y = 0 .........................(eqn 4)\xa0step 3:Now put the value of y = 0 (from equation 4) in equation 3, we get x = (9+ 3(0))/2 x = 9/2 x = 4.5 .......................(eqn 5)so the answer is x = 4.5 and y = 0.\xa0step 4:Verification:to verify the answer we can put these values of x and y in any one of the equation 1 or equation 2.put x = 4.5 and y = 0 in equation 1, we get 2x+3y=9 2(4.5) + 3(0) = 9 9 + 0 = 9 9 = 9 L.H.S. = R.H.S.It means our answer is right. |
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| 176. |
whtejt |
| Answer» See in your Book....... | |
| 177. |
Prove that 3root is a irrational number |
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Answer» Let us assume to the contrary that √3 is a rational number.It can be expressed in the form of p/qwhere p and q are co-primes and q≠ 0.⇒ √3 = p/q⇒ 3 = p2/q2\xa0(Squaring on both the sides)⇒ 3q2\xa0= p2………………………………..(1)It means that 3 divides p2\xa0and also 3 divides p because each factor should appear two times for the square to exist.So we have p = 3rwhere r is some integer.⇒ p2\xa0= 9r2………………………………..(2)from equation (1) and (2)⇒ 3q2\xa0= 9r2⇒ q2\xa0= 3r2Where q2\xa0is multiply of 3 and also q is multiple of 3.Then p, q have a common factor of 3. This runs contrary to their being co-primes. Consequently, p / q is not a rational number. This demonstrates that √3 is an irrational number. Hajjsv |
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| 178. |
Use Euclid\'s division lema find hcf 11. And 121 |
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Answer» 11 square is 121 By CBSE according to new syllabus This portion is deleted |
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| 179. |
anyone hlp me plZ see my previous question its urgent |
| Answer» What was it.!? | |
| 180. |
Distance between (2, 4) (-1, 5) |
| Answer» 3 units | |
| 181. |
If 1 is a zero of the polynomial ax²-3(a-1)x-1 then find the value of a. |
| Answer» x=1Now put the value of x in given polynomial we got,a×1×1-3(a-1)×1-1=0a-3a+3-1=0-2a+2=0-2(a-1)=0(a-1)=0a=1So we got the value of a is 1 | |
| 182. |
Write a quadratic polynomial whose zeroes are 2-2 |
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Answer» The zeroes 2 and -2.\xa0∴ Sum of the zeroes = 2 +(-2) = 0∴ Product of the zeroes = 2\xa0× (-2) = -4\xa0The required polynomial is x2\xa0- (sum of the zeroes) x + Product of the zeroes\xa0⇒ x2\xa0- (0) x + (-4)x2\xa0- 4\xa0 Here given zeroes of polynomial (2,-2)That means, alpha is 2 and Beta is -2Now alpha+beta=2+(-2)=0 and alpha×beta=2×-2=-4Then put these value in x2-(alpha+beta)x+alpha×betax2-0x+(-4)X2+4 it is a polynomial |
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| 183. |
The smallest number is divisible by 15,24and30 |
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Answer» 120 The smallest number is 3 3 The smallest no. Is 3 120 |
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| 184. |
If n is a natural no. then 9²n -4²n is divisible by__________.(2n is the exponent) |
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| 185. |
If α and β are two zeroes of the polynomial x2-8x +15, then find 1?+1? |
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| 186. |
Show that 5-√3 is irrational |
| Answer» Let us assume that 5 -√3 is rational. Then it can be written in the form5 - √3 = p/qor 5 - p/q = √ 3It implies √3 is a rational number [Since 5 - p/q are rationalsBut this contradicts the fact that √3 is irrational. Hence our supposition was wrong. Therefore 5 -√ 3 is irrational. | |
| 187. |
2+2 =x+y +1 |
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Answer» x=1Y=2 2+2=X+y+1 4=X+y+1 4-1= X+y 3=X+y x= y-3. ----eq(1) x+y=3y-3+y=3 2y=3+3 2y=6 y=3So, putting the value of y in eq(1) x =0 |
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| 188. |
If sec 5A equal to cosec (A+30) where 5A is an acute angle . Find value of A is |
| Answer» Follow me on Instagram @yahiya_khan786 I will explain properly on that platform | |
| 189. |
Find distance between (√5-1,√2+2) and (√5-1,√3+1) |
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| 190. |
If cos A = 2 / 5 , find the value of 4 + 4tan2A.\xa0?? |
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Answer» How 25 ???? Can you explain this . 25 |
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| 191. |
The angles of triangle are in A,P, the least being half the greatest. Find the angles |
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Answer» Here Sum of all angles of a triangle =180 degree -(1) It is given that the three angles are in APSo we consider the three angles as a,a+d,a+2dFrom (1) a+a+d+a+2d=180a+a+a+3d=1803a+3d=1803(a+d)=180a+d=60 -(1)And a=30 -(2)Substitute (2) in (1)30+d=60 d=30 Soa+d=30+30=60a+2d=30+2(30)=30+60=90 Therefore The three angles of a triangle are in AP are 30,60,90. Hi Akansha here is your answer. But I don\'t know why this OBJ boxes have come, so neglect it.THANK YOU It is given that the angles are in AP.Let the angles be a-d, a and a+d.The sum of angles is 180 degree.The value of a is 60.The least being half the greatest find the angle.The value of d is 20.Therefore, the measure of angles are 20, 60, 100. |
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| 192. |
How many formula are there |
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| 193. |
Check whether the following are quadratic equations ??1. (x + 1)² = 2(x-3)2. (x-3) (2x+1)=x(x+5) |
| Answer» 1. Yes 2. Yes | |
| 194. |
Find the 20th term from the last of the AP 3,8,13........253 |
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Answer» 158 is the right answer \xa03,8,13.......,253\xa0To find 20th term form the last term we consider the sequence as followsa = 253 and d = 3 - 8 = -5an\xa0= a + (n - 1)da20\xa0= a + (20 - 1)d = 253 + 19\xa0×\xa0(-5) = 253 - 95 = 158Hence, the 20th term from the last term is 158. 158 368 |
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| 195. |
Check whether following are quadratic equations (X+1)^2=2(x-3) |
| Answer» Yes it is quadratic equation | |
| 196. |
Solve the following pair of linear pair by submission method :(1)=x plus y=14 |
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| 197. |
Most repeatedly given questions in boa |
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| 198. |
Ye ch4 me kon se que remove huehai |
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| 199. |
Y no one answering my question?? |
| Answer» What is your question? I will answer. | |
| 200. |
Kx-y=26x-2y=3Ke kitne hal honge |
| Answer» For a unique solution, we have\xa0a1/a2 ≠ b1/b2\xa0∴k/6 ≠ −1/2⇒ k≠3\xa0 | |