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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.
| 61401. |
in quadrilateral ABCD,AB=CD,BC=AD show that ∆ABC=~∆CDA consider ∆ABC and ∆CDA |
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Answer» tqq |
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| 61402. |
show that in a quadrilateral ABCD, AB +BC+CD+AD is greater than AC + BD |
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Answer» kutte bar bar galat answer bhej raha hai |
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| 61403. |
4. In the quadrilateral ABCD, AD = CD and<A = 90g= <C. Prove that AB = BC. |
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| 61404. |
show that exactly one of the numbers n,n+2 or n+4 is divisible by 3 |
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Answer» Answer.We applied Euclid Division algorithm on n and 3.a = bq +r on putting a = n and b = 3n = 3q +r , 0<r<3i.e n = 3q -------- (1),n = 3q +1 --------- (2), n = 3q +2 -----------(3)n = 3q is divisible by 3or n +2 = 3q +1+2 = 3q +3 also divisible by 3or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3Hence n, n+2 , n+4 are divisible by 3. Hit like if you find it useful |
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| 61405. |
13.Show that exactly one of the numbers n, n + 2 or n + 4 is divisible by 3. |
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Answer» Ans :- Since n is a positive integer taking b =3 We can write n = 3q + r , where q is some integer n = 3q , n = 3q + 1 , n = 3q + 2 Case 1 = When n = 3q, n + 2 = 3q + 2 and n + 4 = 3q + 4 clearly only 3q is divisible by 3 Case 2 = When n = 3q + 1, n + 2 = 3q + 3 and n + 4 = 3q + 5 . Here also only n + 2 = 3q + 3 = 3(q + 1) is divisible by 3. Other two namely n and n + 4 are not divisible by 3. Case 3 = When n = 3q + 2, n + 2 = 3q + 4 and n + 4 = 3q + 6 and in this case, only n + 4 = 3(q + 2) is divisible by 3 Hence only one out of n, n + 2 and n + 4 is divisible by 3 for any positive integer n . |
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| 61406. |
\frac{\sqrt{288}}{\sqrt{128}}= |
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Answer» this question has answer is one because Rock 288 by root what 128 is equal to 1 |
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| 61407. |
128/(-8 %2B 16) |
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Answer» according to bodmas128÷16-88-8= 0 16 is the correct answer of thish question 16 is the correct answer of the given question 16 is right answer of this question. please like my answer correct answer is 16 16 is the correct answer following 16 is the correct answer =128÷(16-8)=128÷16=16 16 is the correct answer to your question 128÷(16-8)=128÷8=16 is the answer Your correct answer is 16 128÷(16-8)128÷8=16 |
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| 61408. |
QI26 1281 |
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Answer» the ans is 21×26 =546 26×29= 754 21*26= 54626*29= 754 21*26= 54626*29= 754 |
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| 61409. |
100Exercise 1.39 0 -1)2. Verify the following3. Find out by which number (21) homild bemultiplied to get the following product0b11 15 154. Using the properties of multiplicate d thefollowing products**(-512) 25d. 2*(-379) 50(10)1401 -82) • (-1401) 18-19925-1)-9 113. Starting fromthe |
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Answer» plz capture photo clearly |
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| 61410. |
128(8) 160,--(x+3) (x4 |
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| 61411. |
82. The total number of 5-digit numbers that canbe composed of distinct digits from 0 to 9 is |
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| 61412. |
\begin{array}{l}{x^{2}+\frac{y}{x^{2}}=\frac{82}{9}} \\ {\text { value } x^{3}-\frac{y}{x^{3}}=?}\end{array} |
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| 61413. |
9. यदि समीकरण निकाय kx + 2y = 5 और 3x + y=1 का एक अद्वितीयहल है, तो[पॉलिटेक्निक 2013](a) k = 6(b) k+6(c) k = 3(d)k #3 |
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Answer» hx + 2y = 5; (3x +y = 1)2 = 6x + 2y = 2; hx + 2y = 5; 6x + 2y = 2/ 6h = 3; h = 3/6=2 option (b) is the right answer of question option b.is the right answer options b is the right answer Option b is right answer |
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| 61414. |
4)Find k if the following equations areconsistenti) x + 3y + 2 = 0, 2x + 4y - k = 0,x-2y – 3k=0 |
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| 61415. |
बे 2 R e R R -Y. (k=-2y=%) |
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Answer» Please don't post the same question again and again.Thankyou. |
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| 61416. |
The value of k, such that the equation2x2+ 2y?-6x + 8y +k 0 represents a point circle, is(a) 0(b) 252525 |
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Answer» 2x²+2y²-6x+8y+k = 0 => x²+y²-3x+4y +k/2= 0 comparing with => x²-3x+(3/2)²-(3/2)² + y²+4y+4-4 = 0=> (x-3/2)²+(y+2)² - 4-9/4 = 0=> (x-3/2)² +(y+2)² -25/4 = 0 so k/2 = -25/4=> k = -25/2 option d |
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| 61417. |
एक चतुर्भुज & ४07) खींचा गया है। =AD+BCrilateral ABCD is drawn to circumscribeProve that AB+CD=AD+BC |
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| 61418. |
In A ABC (Figure 3), AD BC. Prove thatB DFigure 3 |
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Answer» In the right angle triangle ABD, we have AD² = AB² - BD²In the right angle triangle ACD we have AD² = AC² - CD² So AC² - CD² = AB² - BD² => AC² = AB² + CD² - BD² => = AB² + (CD + BD)* (CD - BD ) = AB² + BC * [(BC - BD) - BD] = AB² + BC * [ BC - 2 BD ] = AB² +BC² - 2 BC * BD |
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| 61419. |
2thegiven figure of AD BC, Prove that AB+CD-BD+ACe. |
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| 61420. |
AB-CD-AD BCquadrilateral ABCD is drawn to circumscribe a circle. Prove thAB-CD AD BC |
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| 61421. |
11. In the given figure, ABCD is a quadrilateral in which AB || DC andAD || BCProve that ZADC = LABC. |
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Answer» In a ||gm opposite angles are equal so,angle adc=angle abc |
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| 61422. |
Q.20) The angles of a quadrilateral are in AP, whose commondifference is 10. Find the angles.21 ) It the numbers n--2, An-1 and 5n + 2 are in AP, theniefind the value of rn |
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| 61423. |
4 सका s o PO हा उज (1. है। कं e e sl e stde 7de 11 Ad ik Ad A A | skt |
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| 61424. |
11. In an equilateral triangle ABC, AD is drawn perpendicular to BC meeting BC in D. Prove thatAD 3BD2A R |
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| 61425. |
then prove that AC2 = AD2 + BC Ă DM +d BC247. If in ABC, AD is median and AMä¸BC, |
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Answer» Solution:-D is the midpointof BC and AM⊥ BC.In right angled triangleABM,AB² = AM² + BM² ....(1) - Pythagoras TheoremIn right angled triangle ADM,AD² = AM² + MD² ....(2) - Pythagoras TheoremFrom (1)and(2), we getAB² = AD² - MD² + BM²⇒ AB² = AD² - DM² + (BD - DM)²⇒ AB² = AD² - DM² + BD² + DM² - 2BD× DM⇒ AB²= AD² - 2BD× DM + BD²⇒ AB² = AD² - 2(BC/2)× DM + (BC/2)² {∵ BD = DC = BC/2}⇒ AB² = AD² - BC× DM + BC²/4Hence proved. |
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| 61426. |
ORIn ÎABC, ifAD is the median, then show that AB2+AC22(AD2-BD) |
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| 61427. |
4. Find the roots of the equation-82,, 90. |
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Answer» ß |
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| 61428. |
(9)2048, 512, 128....(A) 32,8(C) 36, 18(B) 82, 28(D) 12,8 |
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Answer» a. 32 and 8 is the correct answer of the given question. |
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| 61429. |
५ ७9५»८ 0-9 0०28 - 40 दि 82 सोनी को पट -(७86 १ 5५५71(5 |
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Answer» Like if you find it useful |
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| 61430. |
Two lines are given by (x-2y), k (x-2y) = 0. The value of k, so that the distancebetween them is 3, is(A) k#025.(B) k 315(D) k 3 |
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| 61431. |
If the point P(k-1, 2) is equidistant from the points A(3, k) andB(k,5), find the values of k.CBSE 2014] |
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| 61432. |
० Sda +और 82 4 हि, कौ जोड़िए।दाहरण 3 3x+8yB it N Nl 9 |
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Answer» (3x+8y)+(8x+5y)=3x+8x+8y+5y=11x+13y |
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| 61433. |
यदि ८०५4 + cos? A =1 dd sin? A + sin* A का मान होगा : |
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Answer» Given: CosA + Cos^2A = 1CosA = 1 - Cos^2ATherefore, Sin^2A = CosA Thus, Sin^2A + Sin^4A= Sin^2A + ( Sin^2A)^2= CosA + Cos^2A . (AS Sin^2A = CosA )= 1
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| 61434. |
ETE NSRSy बी T Ywr4-7. +46 पार है3 S 1 |
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Answer» |x² -1|+|-2x 3| |-1 2||2 -3| |4 5|=|6 2| x²-2x=-1x²-2x+1=0(x-1)²=0x=1 |
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| 61435. |
In the given figure, find the length AC12 |
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Answer» 10² = ad²+8² ad² = 100-64 ad² = 36 ad = 6 17² = dc²+8² dc² = 289-64 dc² = 225 dc = 15 ac= ad+dc = 6+15 = 21 cm |
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| 61436. |
MATHEMATnicsNow consider the follois. In the adjoining figure, it is given that AB CD and AD BC Prove that AADC |
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Answer» AC is common because both triangles have a side AC. |
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| 61437. |
For a sequence, if S 3(2" -1), find rnterm, hence show that it is a G.P. Also findthe common ratio. |
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| 61438. |
in the figure given, ABCD is a parallelogram and E is the midpoint of side BC. DE and AB whenproduced meet at F, prove that AF = 2AB |
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| 61439. |
BC2 BA BD. BDAC2 AB. AD ADUT 11: |
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| 61440. |
2. In ΔABC, APBC, BQAC.If B-P-C, A-Q-C then show thatACPA ACQB. If AP 7, BQ 8,BC 12 then AC ?C. |
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| 61441. |
In the adjoining figure, AB AC and AP-AQ. Provethat( AAPC AAQB(ii) CP- BQ(iii) ZAPC-ZAQB. |
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| 61442. |
In AABC, mZA-90, AD LBC. Prove thatAD2 AB AC2 |
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| 61443. |
Example 9 : Find the roots of 4r +3x +5 0 by the method of completing thesquare. |
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| 61444. |
x + \operatorname { tan } x = k , \text { then show that } \operatorname { sin } x = \frac { k ^ { 2 } - 1 } { k ^ { 2 } + 1 } |
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Answer» secx+tanx=k [1/cosx+sinx/cosx]=k (1+sinx)/cosx=k (1+sinx)=kcosx (1+sinx)²=k²cos²k 1+2sinx+sin²x=k²(1-sin²x) 1+2sinx+sin²x=k²-k²sin²x (k²+1)sin²x+2sinx+1-k²=0 use the quadratic formula to solve: sinx=t (k²+1)t²+2t+1-k²=0 -2±√(2)²-4(k²+1)(1-k²) / 2k²+2 -2±√4k⁴/ 2k²+2 =>-2±2k² / 2k²+2 =>-1±k² / k²+1 => -1-k² / k²+1 or -1+k² / k²+1 In conclusion sinx=t =>sinx= (k²-1)/(k²+1) |
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| 61445. |
\operatorname { sec } x + \operatorname { tan } x = k , \text { then show that } \operatorname { sin } x = \frac { k ^ { 2 } - 1 } { k ^ { 2 } + 1 } |
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| 61446. |
\left. \begin{array} { l } { 2 x + 3 y = 7 } \\ { ( k - 1 ) x + ( k + 2 ) y = 3 k } \end{array} \right. |
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Answer» comparing the equations- k -1 = 2=> k = 3 k + 2= 3 k = 1 3k = 7=> k = 7/3 |
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| 61447. |
5. In adjoining figure PQ丄BC,ADL BC then find following ratiosA(A( Δ PBC)POD)(ii) A(AABOA(Δ ABC)A( AADC)A(Δ ADCAPOC)Fig. 1.16(iii)(iv) |
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| 61448. |
1. Prove that sqrt5 is an irrational DD |
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| 61449. |
ven figure. ABC is a triangle with BD and CE perpendicularand AB respectively, such that BD = CE. Prove thatIn the given figure10 AC and ABBCD = ACBEBin the mid points of sides AB, BC |
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Answer» it is given that BD is perpendicular to AC and CE is perpendicular to a b now in angle abc and ABC be equal to CD given യുവർ ടെസ്റ്റ് 1+1=2 അതുപോലെ നൂക്കുംനെ Given: In ∆ABC, BD and CE perpendicular to AC and AB , respectively. And, BD = CE To Prove: ∆BCD is congruence to ∆CBE. Proof: In ∆BCD and ∆CBE, <BDC = <CEB = 90° BD = CE [Given] BC = BC [Common side]Therefore, by RHS congruency rule∆BCD is congruent to ∆CBE. |
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| 61450. |
4. In the adjoining figure, it is given that AB CD and AD - BC Prove that AADCetemp |
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