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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.
| 11101. |
To show experimentaly first twin of cone experimentaly |
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| 11102. |
If A and B are (1,4) and (5,2) respectively, find the coordinates of P when AP/BP=3/4. |
| Answer» Let A(1, 4) and B(5, 2) be the given points.We know that\xa0{tex}\\frac { A P } { B P } = \\frac { 3 } { 4 }{/tex}or, AP : BP\xa0= 3:4Coordinates of P are{tex}\\left( \\frac { 3 \\times 5 + 4 \\times 1 } { 3 + 4 } , \\frac { 3 \\times 2 + 4 \\times 4 } { 3 + 4 } \\right){/tex}{tex}= \\left( \\frac { 19 } { 7 } , \\frac { 22 } { 7 } \\right){/tex} | |
| 11103. |
What is the last prime no |
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| 11104. |
Formula of mean |
| Answer» mean or average valuemean =\xa0sum of all the elements of a set / number of elementse.gmean of 4 , 6 , 8 = 4 + 6\xa0+ 8 / 3 = 18/3 = 6\xa0 | |
| 11105. |
Solve 6x2-x-2=o by factorisation method ( x this is exe not multiplication sign ) |
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| 11106. |
What is the portion for half yearly exams? |
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| 11107. |
5,?,?,91/2 |
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| 11108. |
(A-P)(B-P)(C-P).............(Z-P)equals |
| Answer» It is zero because (P-P) will be zero and multiplication of zero with any number yields zero. | |
| 11109. |
Exercise 7.2 all question and answer |
| Answer» Check NCERT solutions here\xa0:\xa0https://mycbseguide.com/ncert-solutions.html | |
| 11110. |
Prove that 9AD=7AD |
| 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. | |
| 11111. |
If a tree of height 90m breaks due to storm and forms 30°. Find distance of base.\xa0 |
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Answer» 30root under 3 90root under3 |
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| 11112. |
Sin theta+2cos theta=1 prove that 2sin theta-cos theta=2 |
| Answer» Given, sin θ + 2 cos θ = 1\xa0On squaring both sides, we get{tex}(sin\\theta+2cos\\theta)^2=1{/tex}{tex}\\Rightarrow sin^2\\theta+4cos^2\\theta+4sin\\theta cos\\theta=1{/tex}⇒ 1 – cos2\xa0θ + 4 (1 – sin2\xa0θ) + 4 sin θ cos θ = 1\xa0{tex}[\\because sin^2\\theta+cos^2\\theta=1]{/tex}⇒ 1 – cos2\xa0θ + 4 – 4 sin2\xa0θ + 4 sin θ cos θ = 1⇒ –cos2\xa0θ – 4 sin2\xa0θ + 4 sin θ cos θ = –4{tex}\\Rightarrow -(cos^2\\theta+4sin^2\\theta-4sin\\theta cos\\theta)=-4{/tex}⇒ cos2\xa0θ + 4 sin2\xa0θ – 4 sin θ cos θ = 4⇒ (cos θ)2\xa0+ (2 sin θ)2\xa0– 2(cos θ) (2 sin θ) = 4⇒ (2 sin θ – cos θ)2\xa0= 22{tex}\\Rightarrow{/tex}2 sin θ – cos θ = 2{tex}{/tex}Hence\xa0proved. | |
| 11113. |
Tangent to the circle with Centre O V biceps AP prove that triangle AEO is similar to triangle ABC |
| Answer» According to question we are given that BC is tangent with centre O and OE bisects APIn {tex}\\Delta{/tex}AOP,OA = OP (radii) {tex}\\Delta{/tex}AOP is an isosceles triangle. OE is a median.Since a perpendicular to a circle from a center to a chord bisects it.{tex}\\therefore \\angle OEA = 90^\\circ {/tex}In {tex}\\Delta{/tex}AOE and {tex}\\Delta{/tex}ABC,{tex}\\angle ABC = \\angle OEA = 90^\\circ{/tex}{tex}\\angle A{/tex} is common.{tex}\\Delta{/tex}AEO ~ {tex}\\Delta{/tex}ABC ..…(AA test) | |
| 11114. |
sec210 --cosec280 --sin15cos75 +sin75cos15÷costhetasin (90_theta) --tan10tan20tan30tan70tan80 |
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| 11115. |
What is chord |
| Answer» The line segment which cut the circle at two different points. | |
| 11116. |
How to practice maths in board exam |
| Answer» Check last year papers :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 11117. |
Cot(a -b) ,cot(a+b)=45 find sina.tana+cota.cosec |
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| 11118. |
how we find the quardratic equation if digit is in root |
| Answer» Let α & β are the roots then quadratic epuation bex^2-[α+β]x + [αβ] | |
| 11119. |
Find the value of p for which the difference between the roots of the equation x^2 +px+8=ois2 |
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| 11120. |
Find value of p for which the difference between the roots of the equation X^2 +px+8=o is2 |
| Answer» P= -2Use α+β=-b/a | |
| 11121. |
2+5 |
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| 11122. |
X-1upon =3 |
| Answer» 3x-1÷3 | |
| 11123. |
√2x+√9+x=13 |
| Answer» √2x+√9=13-x Squaring both sides 2x+9=(13-x) square and then solve it | |
| 11124. |
What is composite numbers |
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| 11125. |
Prove that 2x+3=0 |
| Answer» it can be zero if the value of x is minus 3 upon 2 | |
| 11126. |
(X-a)(x-b)+(x-b) x-c) +(x-c)(x-a) |
| Answer» Its hard question | |
| 11127. |
Represent the following pair of linear equation graphically x - 5y = 6 2x - 10y = 12 |
| Answer» Given, x - 5y = 6 or x = 6 + 5y\tx61-4y0-1-2\tThus when x = 6, y = 0when x = 1, y = -1when x = -4, y = -2and 2x - 10y = 12 or x = 5y + 6\tx61-4y0-1-2\twhen x = 6, y =0when x = 1, y = -1when x = -4,y = -2Since the lines are coincident, so the system of linear equations is consistent with infinite many solutions | |
| 11128. |
If X+Y+Z =98; and the X:Y =2:3 ,Y:Z= 5:8. Find the value of Y |
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Answer» Y=30 Given : X+Y+Z =98 .......(1)and the X:Y =2:3 ,Y:Z= 5:8.Now, {tex}{X\\over Y}={2\\over 3}\\\\X = {2Y\\over 3}{/tex}Also,\xa0{tex}{Y\\over Z}={ 5\\over 8}\\\\Z= {8Y\\over 5}{/tex}Put values of X and Z in (1), we get{tex}{2Y\\over 3}+Y+{8Y\\over 5}= 98\\\\=> {10Y+15Y+24Y\\over 15}=98\\\\=> {49Y\\over 15}=98{/tex}=> Y = 30 |
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| 11129. |
Find the 31st term of an ap whose 11th term is 38 &16th term is 73 |
| Answer» {tex} a_n=a+(n-1)d{/tex}{tex}a_{11}=a+(11-1)d{/tex}{tex}38=a+10d-------eq(i){/tex}{tex}a_{16}=a+(16-1)d{/tex}{tex}73=a+15d -------eq(ii){/tex}Subtract eq(i) from eq(ii){tex}35=5d{/tex}{tex}d=\\frac{35}{5}=7{/tex}Putt value of d in eq(I){tex}38=a+10×5{/tex}{tex}a=38-50=-12{/tex}Now\xa0{tex}31^{st}{/tex}\xa0term{tex}a_{31}=-12+(31-1)×7=198{/tex} | |
| 11130. |
25786547865/78 |
| Answer» 330596767.5 | |
| 11131. |
Proof of Pythagoras theorem |
| Answer» Pythagoras theoremIn a right triangle the square of the hypotenuse is equal to the square of other two sides Given -ABC is a right triangle Right angled at BTo prove -AC2=AB2+BC2Construction -BD prepndicular ACProof- In triangle ABC and triangle ABD Angle A is common Angle ADB=angle B (both common) Therefore triangle ABM similar to triangle ABC AM/BC=BC/AC BC2=AM×AC ......eq1 Similarly, in triangle BMC & triangle ABC CM/AB =AB/AC AB2=CM×AC........eq 2 From 1 and 2AB2+BC2=AM×AC+CM×ACAB2+BC2=AC(AM +MC)AB2+BC2=AC×ACAB2+BC2=AC2(Proved) | |
| 11132. |
Where implies that use? |
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| 11133. |
If SinA+tana =m and sina-tana=n show that m2-n2=4√mn |
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| 11134. |
What is meant by Dharacharya rule? |
| Answer» shri Dharacharya formula (rule) also known as quadratic formula is used to solve quadratic equationse.glet us consider quadratic equation ax2\xa0+ bx + c=0then shri Dharacharya rule is {tex}x = {-b \\pm \\sqrt{b^2-4ac} \\over 2a}{/tex} | |
| 11135. |
Prove that diagonals of square are equal |
| Answer» Given that ABCD is a square.To prove : AC = BDProof:\xa0(i) In a Δ ABC and Δ BAD,AB = AB ( common line)BC = AD ( opppsite sides of a square)∠ABC = ∠BAD ( = 90° )Δ ABC ≅ Δ BAD ( By SAS property)AC = BD ( by CPCT) | |
| 11136. |
What\'s ionic bound |
| Answer» Ionic bonding\xa0is the complete transfer of valence electron(s) between atoms. It is a type of chemicalbond\xa0that generates two oppositely charged\xa0ions. Inionic bonds, the metal loses electrons to become a positively charged cation, whereas the nonmetal accepts those electrons to become a negatively charged anion | |
| 11137. |
How we solve big roots only in 2 mints |
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| 11138. |
Tan60= |
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Answer» Answer is √3 √3 |
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| 11139. |
Trgnometry |
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| 11140. |
Ch 7 n c e r t ( example - 1) |
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| 11141. |
56 |
| Answer» Do the prime factorisation. Or apply the method for taking out root which we have studied in class 8. | |
| 11142. |
Why math is make |
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| 11143. |
Which term of the AP 3,8,13,18,....,is 78? |
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Answer» Let a=3\xa0d = 5, (8-3)=5an = 7878 = a+(n-1)d78 = 3+ (n-1)578 = 3+5n-578 = -2 +5n78+2 = 5n80 = 5nn = 80/5n =\xa016 an=a+(n-1)d d= a2-a1 = 8-3 = 5if an =78 an=a+(n-1)d 78=3+(n-1)5 78=3+5n-5 78=-2+5n 80=5n therefore n=80÷5 16 |
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| 11144. |
What is the probability of getting 53 Sunday in a year or leap year |
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Answer» In a year - 1 Sunday because 52 complete weeks and 1 day is extra In a non leap year - 2 Sundays because 52 complete weeks and 2 days are extra A year =27.5 /182.5 |
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| 11145. |
Proving Pythagoras thearem |
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| 11146. |
sinA+2cosA=1 prove that2sinA+cosA=2 |
| Answer» sinA+2cosA=1or,sinA=(1-2cosA)2sinA+4cosA=2(1-2cosA)+4cosA=2-4cosA+4cosA=2 | |
| 11147. |
Name the good learning apps which u have installed in your phone except BYJU\'S |
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| 11148. |
Sample papers for 2017 18 |
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| 11149. |
Cbse sample paper for sep 2017 for class 10 |
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| 11150. |
4+4 |
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