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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.
| 28301. |
PQ is a tangent at a point C toa circle with center O. AB is a diameter and CAB=30° find PCA |
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Answer» PCA=60° I think 60° answer hai? PCA= 60° |
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| 28302. |
Root 3 is irranional |
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Answer» Pooja r u mad NCERT book ka examples me dekho yrr In THE maths book answer is their Book bhi khola kro...plzzz, answer bahut bada hai, kaise type krun.... hope yu can understand ? |
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| 28303. |
Find the area of triangle ABC with A1,-4 and the mid point of side through A being 2,-1 and 0,-1 |
| Answer» 12 sq. units | |
| 28304. |
(7,2),(5,1),(3,k) points are Collinear. Find the value of k |
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Answer» A(7,-2), B(5,1), C(3,2K)Δ = 1/2[(x₁y₂+x₂y₃+x₃y₁)-(x₂y₁+x₃y₂+x₁y₃)Δ = 1/2[(7+10K-6)-(-10+3+14K)] = 1/2[1+10K+7-14K] = 1/2[8-4K] = 4-2KGIVEN THE POINTS ARE COLLINEARΔ = 04-2K = 02K = 4K = 2 Area of ∆ABC=01/2{x1(y2-y3)+x2(y3-y2)+x3(y1-y2)=01/2{7(1-k)+5(k-2)+3(2-1)=07-7k+5k-10+3=02k=0k=0 Put area of triangle as 0...then write the formula for area of triangle =0 put all the values of x1, x2, x3, y1, y2, y3.....then yu will get d answer |
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| 28305. |
Derive frustum formula |
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Answer» Let V be the volume of the frustum of cone. Then,V = Volume of cone V AB– Volume of cone VA ‘B’Thus, the volume of the frustum of the cone is given by\xa0 Cbse never ask derivation only question of it What? |
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| 28306. |
Abcd is parallelogram find x OA = 2x-y OB= 16 OC= 6 OD= 3x-5y |
| Answer» x= 13/14 | |
| 28307. |
If secA+tanA=pFind the value of cosecA |
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Answer» Please saw me how you solved this question Ohh...sry by mistake... square lagana rh gaya..sachin ji ap Eeya ji k answer ko refer kr lena mene to galat type kar dia...?? 1+p/1-p cosecA= p^2+1/p^2-1,-1 |
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| 28308. |
Pooja ji tum online ho abhi |
| Answer» Nhi hun..... | |
| 28309. |
in ∆abc, ab=3, bc=4, ac=5, then the radius of the circle touching all the three sides is ? |
| Answer» | |
| 28310. |
Tan 48 tan 23 tan 42 tan 67 |
| Answer» Tan 48 tan 42 tan23 tan67. = Tan(90-42) tan42 tan23 tan(90-23) = Cot42 tan 42 tan23 cot 23 = 1 | |
| 28311. |
Total surface area of hemisphere |
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Answer» Hi sneha jii. 3πr² 3 pi r2 |
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| 28312. |
Hcf of 28 |
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Answer» 28 itself 7 |
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| 28313. |
If sin tetha =x and sec theta=y tehn find the value of cot thetha |
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Answer» So Cot Q= sinQ/cosQSo cotQ=1/xy SinQ=x and secQ=y |
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| 28314. |
If d=4 and a7=4 find first term of an A.P |
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Answer» A= -20 A=-20 a= - 20... a = -20 |
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| 28315. |
If a tangent drawn from external point are equal |
| Answer» Yes | |
| 28316. |
Define lemna |
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Answer» A lemma is a proven statement used for proving another statement. Lemma is a proven statement which is used to prove another statement Not lemna,it\'s lemma |
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| 28317. |
Theoram triangle 10 class cbse |
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Answer» Page number 124 of our NCERT textbook.? NCERT dekho |
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| 28318. |
In ΔABC, right angle triangle at A, tanC=√3Find the value of sin B. cosC + cosB. sinC |
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Answer» sinB.cosC+cosB.sinCAC/BC×AC/BC+AB/BC×AB/BC=1/2×1/2+√3/2×√3/2=1/4+3/4=4/4=1 1 1 1+ root3 |
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| 28319. |
Use Euclid\'s algorithm to find HCF of 1651 n 2032 express the HCF in the form 1651m+2030n |
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Answer» It\'s difficult to type here sorry..... U can refer Google Pl show method H. C. F=127....m=5 and n=-4 |
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| 28320. |
difference between tan p and tan r |
| Answer» Figure bhi hoga | |
| 28321. |
What is the formula for area of sector and segment |
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Answer» Formula of segment.....?? Pie r^2theta/360-[1/2r^2sinTheta] I think it will help you a lot Area of Segment = θ − sin(θ)2 × r2 (when θ is in radians)Area of Segment = ( θ × π360 − sin(θ)2 ) × r2 Area of Sector = θ × π360 × r2 (when θ is in degrees) |
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| 28322. |
Maths NCERT of class 7 of chapter perimeter and area |
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Answer» Jao Class 7 mai jau Class 7.... nhi ye class 10 hai??? |
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| 28323. |
If x=y^z , y=z^x , z=x^y then find xyz |
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Answer» {tex}\\begin{array}{l}\\mathrm x=\\mathrm y^{\\mathrm z},\\mathrm y=\\mathrm z^{\\mathrm x},\\mathrm z=\\mathrm x^{\\mathrm y}\\\\\\mathrm{xyz}=\\mathrm x^{\\mathrm y}\\mathrm y^{\\mathrm z}\\mathrm z^{\\mathrm x}\\\\\\mathrm{xyz}=\\mathrm{xyz}(\\mathrm x^{\\mathrm y-1}\\mathrm y^{\\mathrm z-1}\\mathrm z^{\\mathrm x-1})\\\\\\mathrm{or}\\;\\mathrm x^{\\mathrm y-1}\\mathrm y^{\\mathrm z-1}\\mathrm z^{\\mathrm x-1}=1\\\\\\frac{\\mathrm x^{\\mathrm y}\\mathrm y^{\\mathrm z}\\mathrm z^{\\mathrm x}}{\\mathrm{xyz}}=1\\\\\\mathrm{So}\\;\\;\\mathrm{xyz}=\\mathrm x^{\\mathrm y}\\mathrm y^{\\mathrm z}\\mathrm z^{\\mathrm x}\\end{array}{/tex} Kya Tujya i chi bhook |
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| 28324. |
ptove that (sin4+cos4)/1-2sin2×cos2 = 1 |
| Answer» Prove to ho gya, bt apko bataun kaise | |
| 28325. |
In an AP,if the common difference=2 and sixth term a6 is 20,then find the first term |
| Answer» a=10 | |
| 28326. |
If the point P(x,y) is equidistant from the pointsA(5,1) and B(-1,5) .then prove that 3x=2y |
| Answer» Apply distance formula ,(AP)^2=(BP)^2ND get ur ans..... | |
| 28327. |
Solve x2+4x-5=0 by completing the square method. |
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Answer» Are you Anushka Sharma from dichoan x2+4x−5=0,x2+4x+4−9=0,(x+2)²=9,(X+2)=(3)²X+2=3X=3-2X=1 |
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| 28328. |
Hy iss bar ka maths ka paper mostly NCERT m si hoga kya...-? |
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Answer» Up board walo ka ncert se kam question aya the Yes 70% chance Hha sayad .!! |
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| 28329. |
Prove that length of tangents from an external point are equal |
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Answer» Pg no. 211 Ncert m h yai U can read from 10.2 |
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| 28330. |
(SinA-secA)^2+(cosA-cosecA)^2=(1-secA.cosecA)^2 |
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Answer» I know yr........Isliye pucha It is complicated |
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| 28331. |
For what value of k, x=a is a solution of equation x²-(a+b)x+k=0 ? |
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Answer» The value of K is AB ... How please tell me K = ab K=ab |
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| 28332. |
How can we save iron from ruting write 5ways? |
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Answer» (¡) Oiling...(¡¡) Painting...(¡¡¡) Galvanisation...(¡v) Zinc coating...(v) Greasing... Sry, apne 5 pucha tha.. maine glti se 6 bta diya, ap isme se 5 choose kr lijiyega....??? * oiling, *electroplating, *painting, *galvanisation, *greasing, *chrome plating..... 1. By painting.......2. by preventing it from moisture......3. by galvanisation.....4. Anodising......5. oil coating... Give me ans pls... |
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| 28333. |
If the first two terms of an ap is -3 and 4 find its 21 terns |
| Answer» an = a + ( n - 1)d = -3 +( 21 -1) 7 = -3 +140 = 137 | |
| 28334. |
Sec =13/5,show that 2sin-3cos/4sin+9cos |
| Answer» | |
| 28335. |
10.1 theorm full |
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Answer» 10.2 exam m hamesha aata cbse board m The tangent from any point of a circle is perpendicular to the radius through the point of contact. |
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| 28336. |
alfa |
| Answer» ã | |
| 28337. |
495÷7Plz urgent Explain it in detail plzzz |
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Answer» 71 Sorry ;;;. ....the answer is 70.7142857 71 |
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| 28338. |
Prove that the points (3,0), (6,4), (-1,3) are the vertices of a right angled isosceles triangle. |
| Answer» Apply distance formula and find the measure of all sides and then use pyth. theorem for proofing it. | |
| 28339. |
Solve value of x :√(2x+9) +x-13=0 |
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Answer» By quadratic formula solve it Shift x-13 to rhs and square the rhs |
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| 28340. |
ncert exercise 4.3 ka 6 question |
| Answer» | |
| 28341. |
How we will devide 220/3.4 |
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Answer» 2200÷34 it is very easy question ? 220/3.4=2200/34=64.70588 2200/34 2200/34 64.7 |
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| 28342. |
Prove that cos u\\1+sin u =1-sin u\\cos u |
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Answer» Hope it will help u??? By cross multiplication.....-------> cosU × cosU = (1-sinU)(1+sinU)....-------> cos^2U = 1-sin^2U....-----> cos^2U = cos^2U.......proved |
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| 28343. |
If 1, 2, 4, 1, x, 6 and 3, 5 are the vertices of a parallelogram taken in order find X and Y |
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Answer» Same here question clearly likho bhn Apply the formula (x1+x3)-x2=x4 Yur question is nt clear Isme Y Kahan hai aur vertices kuch samajh me nhi aa rha?????? |
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| 28344. |
Prove that parallelogram circumscribing a circle is a rhombus |
| Answer» Given ABCD is a ||gm\xa0such that its sides touch a circle with centre O.∴ AB = CD and AB || CD,AD = BC and AD || BCNow, P, Q, R and S are the touching point of both the circle and the ||gmWe know that, tangents to a circle from an exterior point are equal in length.∴ AP = AS [Tangents from point A] ... (1)\xa0BP = BQ [Tangents from point B]\xa0... (2)\xa0CR = CQ [Tangents from point C]\xa0... (3)\xa0DR = DS [Tangents from point D]\xa0... (4)On adding (1), (2), (3) and (4), we getAP + BP + CR + DR = AS + BQ + CQ + DS⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)⇒ AB + CD = AD + BC⇒ AB + AB = BC + BC [∵ ABCD is a ||gm .\xa0∴ AB = CD and AD = BC]⇒ 2AB = 2BC⇒ AB = BCTherefore, AB = BC impliesAB = BC = CD = ADHence, ABCD is a rhombus.\xa0In rhombus, it is not necessary that diagonals are equal. If they are equal, then rhombus is considered as a square whose diagonals are always equal. So, there isn\'t any use of proving that the diagonals of a rhombus are equal. | |
| 28345. |
Find the value of k if points A(2,3),B(4,k),C(6,-3) are collinear |
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Answer» When points are collinear area of triangle is 0 Then put values inX1(y2-y3) +x2(y3-y1) +x3(y1-y2),Aur solve kar lo Bhai isme triangle ka formula = 0 rakh de ar(ABC) = 0So, k = 0 |
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| 28346. |
what formulas can help me to clear algebra part?? |
| Answer» (a+b)^2=a^2+2ab+b^2 | |
| 28347. |
Theorm from circle will definate come |
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Answer» Thm 8 Circle or from triangles definitely ?% Ya bt only 1 and 2 |
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| 28348. |
In an equilateral triangle ABC ,AD perpendicular to BC .prove that 3BC2=4AD2 |
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Answer» Thank you both AB=BC=AC , BD=DC=BC/2,. AB ^2=BD^2+AD^2 bus issi mein AB ke jagah BC aur BD ke jagah BC/2 put kar do aur value nikal lo Solution kaisr bhejun........ Please answer fast |
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| 28349. |
Given that √2 is irrational, prove that (5+3√2) is an irrational |
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Answer» Let 5+3√2 is rational number say p/q where q is not equal to zero and p and q is coprime.5+3√2=p/q3√2=p/q-53√2=p-5q/q√2=p-5q/3q√2is irrational and p-5q/3q is rational.Therefore our supposition is wrong. 5+3√2 is irrational Let assume √2 as a rational no. It can be wriien as a/b where a,b both r integers so a/b=5+3√2 .now a/b -5=3√2,now a-5b/b= 3√2and a-5b/3b =√2 bt it can\'t happen bcz a,b both r integers so r.h.s not equals to l.h.s so assumption is wrong and hence it is proved that 5+3√2 is an irrational no. |
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| 28350. |
If Cos (q+r)=0, then find sin(q-r) |
| Answer» | |