Maths last 10 year papers

5 2020 FO ELBAT CONTENTS Answers Begin at page 15 40 2019 Answers Begin at page 47 79 2018 Answers Begin at page 86 122 2017 Answers Begin at page 128 154 2016 Answers Begin at page 161 187 2015 Answers Begin at page 193 212 2014 Answers Begin at page 218 245 2013 Answers Begin at page 252 278 2012 Answers Begin at page 284 300 2011 Answers Begin at page 307

CBSE Ma hema ic 2020 Ge e a I c: . ()A e aec ( )T e aec f 30 e d ded f ec A, B, C a d D. ( ) Sec A c a 6 e f 1 a eac . Sec Bc a 6 e f 2 a eac , Sec C c a 10 e f3 a eac . Sec D c a 8 e f 4 a eac . ( )T ee ea c ce. H e e , a e a c ce a bee ded f e f 3 a eac a d 3 e f4 a ae eac . Y a e e f ea e a e ac e. ( ) U e f ca c a e ed. Downloaded from www.padhle.in

Qe in Sec i n-A (1 Ma k Each) 1. ​If one of the eroes of the quadratic pol nomial x2 + 3x + k is 2, then the value of k is: (a) 10 (b) 10 (c) 7 (d) 2 2. T​ he total number of factors of a prime number is: (a) 1 (b) 0 (c) 2 (d) 3 3. T​ he quadratic pol nomial, the sum of whose eroes is 5 and their product is 6, is (a) x2 + 5x + 6 (b) x2 5x + 6 (c) x2 5x 6 (d) x2 + 5x + 6 4. T​ he value of k for which the s stem of equations x + 4 = 0 and 2x + k =3 has no solution, is (a) -2 (b) 2 (c) 3 (d) 2 5. ​The HCF and the LCM of 12, 21, 15 respectivel are Downloaded from www.padhle.in

(a) 3140 (b) 12,420 (c) 3,420 (d) 4,203 6. The value of x for which 2x,(x + 10) and ( 3x + 2 ) are the three consecutive terms of an AP, is (a) 6 (b) 6 (c) 18 (d) 18 7. ​The first term of an AP is p and the common difference is q, then its 10th term is (a) q + 9p (b) p 9q (c) p + 9q (d) 2p + 9q 8. T​ he distance between the points (a cos + b sin , 0) and (0, a sin b cos ), is (a) a2 + b2 (b) a2 b2 (c) a2 +b2 (d) a2 - b2 9. ​Find the ratio in which the segment joining the points (1, -3) and (4, 5) is divided b x-axis? Also, find the coordinates of this point on the x-axis. (a) 1 (b) 2 (c) 2 (d) 1 10. ​The value of p, for which the points A(3, 1), B(5, p) and C(7, 5) are collinear, is (a) 2 (b) 2 (c) 1 (d) 1 11. In Fig. 1, ABC is circumscribing a circle, the length of BC is _______cm. Downloaded from www.padhle.in

12. ​Given ABC PQR, if AB/PQ = ⅓ , then a( ABC ) = a( P QR ) 13. ​ABC is an equilateral triangle of side 2a, then length of one of its altitude is __________ 14. c 80° +​ ​ ​c 59 c ec 31 = _______. 10° 15. ​The al e f ( in2​ ​ + 1/ 1 + an2​ ​ ) = _______. OR The al e f (1 + an2 ) (1 in ) (1 + in ) =_______. 16. T​ he ratio of the length of a vertical rod and the length of its shadow is 1: 3 . Find the angle of elevation of the sun at that moment? Downloaded from www.padhle.in

17. ​Two cones have their heights in the ratio 1:3 and radii in the ratio 3:1. What is the ratio of their volumes? 18. ​A letter of English alphabet is chosen at random. What is the probabilit that the chosen letter is a consonant. 19.​ A die is thrown once. What is the probabilit of getting a number less than 3? OR If the probabilit of winning a game is 0.07, what is the probabilit of losing it? 20. ​If the mean of first n natural number is 15, then find n. Sec i n-B (2 Ma k Each) 21.Sh ha (a b)​2​ , (a​2​ + b​2)​ and (a + b)2​ ​ a e in AP. 22. ​In the given Fig. DE AC and DC AP. Prove that BE = BC EC CP Downloaded from www.padhle.in

OR In he gi en Fig. , angen TP and TQ a e d a n a ci cle i h cen e O f m an e e nal in T. P e ha ∠PTQ = 2∠OPQ. ​23. ​The rod AC of a TV disc antenna is fixed at right angle to the wall AB and a rod CD is supporting the disc as shown in Fig. 4. If AC = 1.5m long and CD = 3m, find (i) tan (ii) sec + cosec . Downloaded from www.padhle.in

24. A iece f i e 22 cm l ng i ben in he f m f an a c f ci cle b ending an angle f 60 a i cen e. Find he adi f he ci cle. ( U e = 22/7 ) 25. If a n mbe i ch en a and m f m he n mbe 3, 2, 1, 0, 1, 2, 3. Wha i he babili ha ​2​ 4? 26. ​Find the mean of the following distribution: Sec i n- C (3 Ma k Each) 27. F​ ind the quadratic pol nomial whose eroes are reciprocal of the eroes of the pol nomial f (x) = ax2 + bx + c, a 0, c 0. OR Downloaded from www.padhle.in

Divide the pol nomial f(x) = 3x2 x3 3x + 5 b the pol nomial g(x) = x 1 x2 and verif the division algorithm. 28. D​ etermine graphicall the coordinates of the vertices of a triangle, the equations of whose sides are given b 2 x = 8, 5 x = 14 and 2x = 1. OR If 4 is the ero of the cubic pol nomial x2 3x2 10x + 24, find its other two eroes. 29. ​In a flight of 600 km, an aircraft was slowed due to bad weather. Its average speed for the trip was reduced to 200 km/hr and time of flight increased b 30 minutes. Find the original duration of flight. 30. Find the area of triangle PQR formed b the points P( 5, 7), Q( 4, 5) and R(4, 5). OR If the point C( 1, 2) divides internall the line segment joining A(2, 5) and B(x, ) in the ratio 3 : 4, find the coordinates of B. 31. In Fig.5, ∠D = ∠E and AD/DB = AE/EC , e ha BAC i an i cele iangle. Downloaded from www.padhle.in

32. In a iangle, if a e f ne ide i e al he m f he a e i e he fi f he he ide , hen e ha he angle ide i a igh angle. 33. If in + c = 3 , hen e ha an + c = 1 34. A c ne f ba e adi 4 cm i di ided in a b d a ing a i ba e. lane h gh he mid in f i heigh and a allel C m a e he l me f he a. Sec i n- D (4 Ma k Each) 35. Sh ha he a e f an i i e in ege cann be f f m (5 + 2) (5 +3) f an in ege . OR P e ha ne f e e h ee c n ec i e i i e in ege i di i ible b 3. Downloaded from www.padhle.in

36. The m f f c n ec i e n mbe in AP i 32 and he a i f d c f he fi and la e m he d c f middle e m i 7:15. Find he n mbe . 37. D a a line egmen AB f leng h 7 cm. Taking A a cen e, d a a ci cle f adi 3 cm and aking B a cen e, d a an he ci cle f adi 2 cm. C n c angen each ci cle f m he cen e f he he ci cle. 38. A e ical e and n a h i n al lane and i m n ed b a e ical flag- aff f heigh 6 m. A a in n he lane, he angle f ele a i n f he b m and f he flag- aff a e 30 and 45 e ec i el . Find he heigh f he e . (Take 3 = 1.73) 39. A b cke in he f m f a f m f a c ne f heigh 30 cm i h adii f i l e and e end a 10 cm and 20 cm e ec i el . Find he ca aci f he b cke . Al find he al c f milk ha can c m le el fill he b cke a he a e f R . 40 e li e. ( U e = 22/7 ) 40. +The f ll ing able gi e d c i n ield e hec a e (in in al ) f hea f 100 fa m f a illage: Downloaded from www.padhle.in

An 1). (b) 10 An 2). (c) 2 An 3). (a) ​2​ + 5 + 6 An 4). (d) 2 An 5). (c) 3,420 An 6). (a) 6 An 7). (c) + 9 An 8). (c) a​2 ​+ b2​ An 9). (d) 1 An 10). (a) 2 An 11). 10cm i he leng h f BC. An 12). We kn ha a ea f imila iangle a e e al he i n f he a e f i nal Side . S​ , a ABC/a PQR= = 1/9 Downloaded from www.padhle.in

An 13). ​Gi en : ABC i an e ila e al iangle f ide 2a. In e ila e al iangle, △ADB = △ADC & ∠ADB = ∠ADC (B h 90 deg ee a AD ⊥ BC) Th , △ ADB ≅ △ ADC (B R. H. S. C ng enc ) ​A e CPCT i.e. he a e c e nding a f c ng en iangle, B​ D = DC U​ ing P hag a he em, Hence, Downloaded from www.padhle.in

Th , he al i de f he gi en e ila e al iangle i e al An ​ H​ ence, leng h f ne f he al i de i a 3 An 14). c (90 10) + c 59 10 c ec 31 c (90 10) + c 59 10 c ec 31 ​ 10 + c 59 10 c ec 31 ​1+c 59 / in(90-31) ​1+c 59/c 59 ​1+1 = 2 Hence ed. An 15). Downloaded from www.padhle.in

= ( in2​ ​ + c ​2​ ) =1 An 16). f he leng h f a e ical d and he leng h f i G​ I en: The a i had i 1: 3 . ​S l i n: an = 1/ 3 = 30 An 17). ​ ​Gi en: T c ne ha e hei heigh in he a i 1:3 and adii in he a i 3:1. ​ ​S l i n: 1 = 3 = 1 .2 .1 .3 Ra i f l me : 1​ ​2​h1​ ​ / ​22​ h​ ​2 ​= 3:1 An 18). The e a e 26 le e . S , he babili ha he ch en le e i a c n nan i : Downloaded from www.padhle.in

​ P(c n nan ) = 21 .26 An 19). We kn in a die, he al n . f c me d e n e ceed 6. S, P (f n mbe le han 3) = Fa ab e ce Ta .f ce P (f n mbe le han 3) = 2 = 6 OR An 20). P (l ing) = Fa ab e ce Ta .f ce P (l ing) 1 0.07 = 0.93 An 20). Gi en : Mean f fi n na al n mbe i 15 We kn , S m f fi n na al n mbe = [ ( +1) ] 2 Downloaded from www.padhle.in

Mean f fi n na al n mbe = S m f b e a i n / T al n mbe f be ai n = [ ( +1) ] 2 = (n + 1)/2 = 15 n + 1 = 30 n = 29 An 21). Gi en: (a-b)2​ ,​ (a​2​ + b​2​) and (a+b)2​ ​ a e in A.P. S , (a2​ ​ + b2​ ​) - (a-b)​2​ = 2ab ame. (a+b)​2 ​- (a​2​ + b​2)​ = 2ab A e can ee, C mm n diffe ence i An Hence, he gi en e m a e in AP An 22). Gi en: In ABC, DE AC T P e: BE/EC = BC/CP S l i n: In ABC, DE AC, BD = BE DA EC In ABP, DC AP, BD = BC DA CP Downloaded from www.padhle.in

N ,f m (i) & (ii), BE = BC EC CP An Hence, ed. OR Le ∠OPQ = ∠TPQ = ∠TQP = 90 In TPQ, 2(90 ) + ∠PTQ = 180 ∠PTQ = 2 = 2∠OPQ An Hence, P ed. An 23). Gi en: AC = 1.5m l ng and CD = 3m I ​ ACD , C​ D2​ ​ = AC2​ ​ +AD​2 ​Tan T​ an (ii) sec + cosec Downloaded from www.padhle.in

An ​ Hence, he al e f ​Tan = 1/ 3 and he al e f ec + c ec i 2​​( 1 + 3 ) 3 An 24). Gi en: Leng h f he i e = 22 cm, b end an angle f 60 a i cen e Sl i n: 2 ​( 22 ) ​ 60° = 22 7 360° 2 ​( 22 ) ​ 1 = 22 7 6 = 21cm f he ci cle i 21cm. An The adi An 25). T al n mbe f c me = 7 Downloaded from www.padhle.in

Fa able c me a e 2, 1, 0, 1, 2, i.e., 5 P( 2​ ​ 4) = 5/7 An 26). Cla F e enc (f) X X*f 20 3-5 54 60 80 5-7 10 6 70 ​ 96 7-9 10 8 ​ X*f = 326 9 - 11 7 10 11 - 13 ​8 ​ 12 ​ f = 40 ​ Mean f he da a, = ∑f = 326 = 8.15 ∑f 40 OR Gi en acc ding he able: M dal cla : 60 80 F m la f m de: ​ ​l + f1 f0 h 2f 1 f0 f2 On ing he al e : ​60 +​ 12 10 6 20 24 10 Downloaded from www.padhle.in

= 60 + 5 = 65 An The m de f he f ll ing da a i 65. An 27). f he l n mial f( ) = a ​2​ + b + c. ​Le and a e he e e ​ Acc ding he e i n, 1/ and 1/ a e he e e f he e i ed ad a ic l n mial Downloaded from www.padhle.in

E a i n f he e i ed ad a ic l n mial O (c ​2 +​ b +a ) An 28). Le 2 - =8 .. e ​n​ 1 5 - =14 .. e ​n ​2 2 - = -1 .. e n​ ​3 On l ing 1 and 2, e ge = -4 and = 2 C dina e f B = (-4,2) Downloaded from www.padhle.in

On l ing 2 and 3: We b ain =1 and =3 C dina e f C = (1,3) On l ing 1 and 3: =2 and =5 Hence C dina e f A=(2, 5) An S , C dina e f he e ice f he iangle a e A( 4, 2), B(1, 3) and C(2, 5) Downloaded from www.padhle.in

OR Gi en: 1). Gi en ha he c bic l n mial i 2). 4 i a e f he gi en l n mial T find : I he e e . S l i n: Since, 4 i a e f he gi en l n mial, ( - 4) i i fac . On f he l ing, ​2​ + + 6 = ( + 3)( - 2) An The he e e f he l n mial a e 3 and 2. An 29). Le he eed f ai c af be km/h Downloaded from www.padhle.in

600 _​ 600 = 30/60 200 ​2​ - 200 - 240000 = 0 ( 600) ( + 400) = 0 = 600, Since, 400 cann be acce ed 1/2 S eed f ai c af = 600 km/h ∴ D a i n f fligh = 1 h An 30). ​S l i n: . ni P = (-5 , 7) Q = ( -4 , - 5) R = ( 4 , 5) A ea f T iangle PQR = (1/2) -5( -5 - 5 ) -4( 5 - 7 ) + 4(7 + 5) = (1/2) 50 + 8 + 48 = (1/2) 106 = 106/2 = 53 A​ ea f T iangle PQR = 53 ni An 31). Downloaded from www.padhle.in

C​ dina e fCa e ( 3 +8 , 3 + 20 ) = ( 1, -2) 7 7 ​ = 5, = 2 S , ​ ​C dina e f B a e ( 5, 2) An 31). ​Gi en: ∠D = ∠E , hich mean AE = ED AD = AE DB = EC DB EC AD + DB = AE + EC AB = AC A , ​AB = AC , BAC i an i cele iangle. An BAC i an i cele iangle. Downloaded from www.padhle.in

An 32). Gi en:- ABC i a iangle AC​2 =​ AB​2 ​+BC2​ T e:- ∠B=90 C n c i n:- C n c a iangle PQR igh angled a Q ch ha , PQ=AB and QR=BC P f:- In △PQR PR2​ =​ PQ​2 ​ + QR2​ (B hag a he em) PR​2​ =AB​2​ +BC2​ ​ .....(1) a AB=PQ and QR=BC AC2​ =​ AB2​ ​+BC​2.​ ...(2)(Gi en) F m e a i n (1)&(2), e ha e AC​2 =​ PR2​ AC=PR.....(3) N , in △ABC and △PQR AB=PQ BC=QR AC=PR(F m (3)) △ABC≅△PQR(B SSS c ng enc ) The ef e, b C.P.C.T., ∠B=∠Q ∠Q=90 ∠B=90 Hence ed. Downloaded from www.padhle.in

An 32). Gi en in + c = 3 ​T e: an + c = 1. S l i n: ​ in + c = 3 ​ ( in + c )​2​ = ( 3)2​ ​ in2​ ​ + c ​2​ + 2 in c =3 ​ in c = 1 = 1​ RHS L.H.S = an + c = c +c = 1 c H​ ence, P ed. An 33). ​Le heigh f he c ne i h N B i he mid in f AC. Downloaded from www.padhle.in

L​ e V1 and V2 be he l me f e and l e a f he c ne e ec i el In ABE and ACD ​ ∠B=900​ =​ ∠C ∠​ A=∠A (C mm n angle) ​ ​ ABE∼ ACD ​AB/BE = AC/CD h/2 = h/4 BE B​ E = 2 cm Ra i f l me f a: = 4/28 = 1/7 a i 1:7. Hence, he a i f l me f An 35). i i e in ege . Le a be an he di i . Take b = 5 a Downloaded from www.padhle.in

​ ​ a = 5m + , = 0,1,2,3,4 1 C​ a e-1 : a = 5m ⇒ a2 = 25m2 = 5(5m2) = 5 C​ a e-2 : a = 5m+1 ⇒ a2 = 5(5m2 + 2m) + 1 = 5 + 1 ​ Ca e-3 : a = 5m+2 ⇒ a2 = 5(5m2 + 4m) + 4 = 5 + 4 C​ a e-4 : a = 5m+3 ⇒ a2 = 5(5m2 + 6m + 1) + 4 = 5 + 4 ​ Ca e-5 : a = 5m+4 ⇒ a2 = 5(5m2 + 8m + 3) + 1 = 5 + 1 An ​ ​Hence ed, be f he f m (5 + 2) (5 + 3) a e f an i i e in ege cann . f an in ege OR ​Le n be an i i e in ege . Di ide i b 3. ​n=3 + ​ = 0, 1, 2 1 C​ a e-1 : n = 3 (di i ible b 3) n+1=3 +1 n+2=3 +2 Ca e-2 : n = 3 + 1 n+1=3 +2 n + 2 = 3 + 3 (di i ible b 3) Ca e-3 : n = 3 + 2 n + 1 = 3 + 3 (di i ible b 3) n+2=3 +4 Downloaded from www.padhle.in

An 36). GI en: S m f f c n ec i e n mbe in AP i 32 Le he f c n ec i e n mbe in AP be (a 3d),(a d),(a+d) and (a+3d) a 3d+a d+a+d+a+3d=32 4a=32 a=32/4 a=8 ----> e ​n (​ 1) N , (a 3d)(a+3d)/(a d)(a+d)=7/15 15(a 9d )=7(a d ) 15a 135d =7a 7d 15a 7a =135d 7d 8a =128d P ing he al e f a=8 in ab e e ge . 8(8) =128d ae 128d =512 d =512/128 d =4 d=2 S , he f c n ec i e n mbe 8 (3 2) 8 6=2 8 2=6 8+2=10 8+(3 2) Downloaded from www.padhle.in

8+6=14 F c n ec i e n mbe a e 2,6,10and14 OR Gi en : a = 1, d = 3 We kn ha , S(n) = n/2 [2a + (n - 1)d] P ing all he al e , e ge 287 = n/2[2 1 + (n - 1) (3) ] 287 = n/2[2 + (n - 1) 3] 574 = 3n - n 3n - n - 574 = 0 3n - 42n + 41n - 574 = 0 3n(n - 14) + 41(n - 14) = 0 n = 14, - 41/3 ( Since, n can' be nega i e) n = 14 We kn ha , a + (n - 1)d = 1 + (14 - 1) (3) = 3 1 + 13 (3) = 3 = 40. Downloaded from www.padhle.in

An The al e f i 40. An 37). F ll ing a e he e f c n c i n: 1. Take AB = 7 cm. 2. Wi h A a cen e and 3 cm a adi , d a a ci cle. 3. Simila l , i h B a cen e and 2 cm a adi , d a a ci cle. 4. N , d a he e endic la bi ec f AB and ma k he in f in e ec i n O. 5. Wi h O a cen e and OA a adi , d a a ci cle. Ma k he 2 in he e he ci cle i h cen e O and A mee a Q and R. Simila l , ma k he in he e he ci cle i h cen e O and B mee a S and T e ec i el . 6. J in BR and BQ a ell a AS and AT. N , BR, BQ, AS and AT a e he e i ed angen . Downloaded from www.padhle.in

An 38). Le BC be he heigh f he e and DC be he heigh f he flag - aff. In .△ABC, AB = BC c 30 AB = BC 3 -----> e ​n ​(i) In .△ABD, AB = BD c 45 AB = (BC + CD) c 45 AB = (BC + 6) -----> e n​ (​ ii) E a ing (i) and (ii) (BC + 6) = BC 3 BC( 3-1) = 6 Downloaded from www.padhle.in

BC = 6/0.73 = 9.58 m e i 9.58 m. The ef e he heigh f he An 39). f he e end f he f m f c ne = R = 20 cm Gi en: Radi e end f he f m f c ne = = 10 cm adi f he l H = 30 cm Sl i n: V l me = 1 h[R + + R* ] 3 = 1 22 30 [20 + 10 + 20*10] 3 7 = 660/21 [400 + 100 + 200] = (660 700)/21 = 22000 cm​3 Hence he ca aci = 22L S , he C f milk = ​₹​ ​40 ₹​ 2​ 2 = ​₹8​ 80 An 40). P d c i n ield / hec a e N . ff m Downloaded from www.padhle.in

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CBSE Ma hema ics 2019 Ge e a I c i : . (i) A e i a e c (ii) Thi e i aec i f 30 e i di ided i f ec i A, B, C a d D. (iii) Sec i A c ai 6 e i f 1 a each. Sec i B c ai 6 e i f 2 a each, Sec i C c ai 10 e i f3 a each. Sec i D c ai 8 e i f 4 a each. (i ) The e i e a ch ice. H e e , a i e a ch ice ha bee ided i f e i f 3 a each a d 3 e i f 4 a a e e f he a e a i e i a ch each. Y ha e ei . ( ) U e f ca c a i e i ed. Downloaded from www.padhle.in

Q es ions Sec ion-A (1 Mark Each) 1. ​Find he coo dina e of a oin A, he e AB i diame e of a ci cle ho e cen e i (2, -3) and B i he oin (1, 4). 2. ​Fo ha al e of k, he oo of he e a ion 2 + 4 + k = 0 a e eal? [1] OR Find he al e of k fo hich he oo of he e a ion 3 2 10 + k = 0 a e eci ocal of each o he . 3. F​ ind A if an 2A = co (A 24 ) OR Find he al e of ( in2 33 + in2 57 ) 4. H​ o man o digi n mbe a e di i ible b 3? 5. ​In Fig., DE BC, AD = 1 cm and BD = 2 cm. ha i he a io of he a ( ABC) o he a ( ADE) Downloaded from www.padhle.in

6. F​ ind a a ional n mbe be een 2 and 3. Sec ion-B (2 Marks Each) 7. F​ ind he HCF of 1260 and 7344 ing E clid algo i hm. he e OR Sho ha e e o i i e odd in ege i of he fo m (4 + 1) o (4 + 3), i ome in ege . 8. W​ hich e m of he A.P. 3, 15, 27, 39, ill be 120 mo e han i 21 e m? [2] OR If Sn, he m of fi e m of an A.P. i gi en b Sn = 3n2 4n, find he n h e m. 9. ​Find he a io in hich he egmen joining he oin (1, -3) and (4, 5) i di ided b -a i ? Al o, find he coo dina e of hi oin on he -a i . 10. A​ game con i of o ing a coin 3 ime and no ing he o come each ime. If ge ing he ame e l in all he o e i a cce , find he obabili of lo ing he game. 11. A​ die i h o n once. Find he obabili of ge ing a n mbe hich (i) i a ime n mbe (ii) lie be een 2 and 6. 12. F​ ind c if he em of e a ion c + 3 + (3 c) = 0, 12 + c c = 0 ha infini el man ol ion ? Downloaded from www.padhle.in

Sec ion- C (3 Marks Each) 13.P​ o e ha 2 i an i a ional n mbe . 14. F​ ind he al e of k ch ha he ol nomial 2 (k + 6) + 2(2k 1) ha m of i e o e al o half o hei od c 15. ​A fa he age i h ee ime he m of he age of hi o child en. Af e 5 ea hi age ill be o ime he m of hei age . Find he e en age of he fa he . A f ac ion become OR become hen 1 i hen 2 i b ac ed f om he n me a o and i b ac ed f om he denomina o . Find he f ac ion. 16. F​ ind he oin on -a i hich i e idi an f om he oin (5, -2) and (-3, 2). OR The line egmen joining he oin A(2, 1) and B(5, -8) i i ec ed a he oin P and Q ch ha P i nea e o A. If P al o lie on he line gi en b 2 + k = 0, find he al e of k. 17. ​P o e ha ( in + co ec )2 + (co + ec )2 = 7 + an2 + co 2 . [3] OR P o e ha (1 + co A co ec A) (1 + an A + ec A) = 2. Downloaded from www.padhle.in

18. I​ n Fig. PQ i a cho d of leng h 8 cm of a ci cle of adi 5 cm and cen e O. The angen a P and Q in e ec a oin T. Find he leng h of TP. 19.I​ n Fig. ∠ACB = 90 and CD AB, o e ha CD2 = BD AD. OR If P and Q a e he oin on ide CA and CB e ec i el of ABC, igh -angled a C, o e ha (AQ2 + BP2 ) = (AB2 + PQ2). 20. ​Find he a ea of he haded egion in Fig. if ABCD i a ec angle i h ide 8 cm and 6 cm and D i he cen e of he ci cle. 21. ​Wa e in a canal, 6 m ide and 1.5 m dee , i flo ing i h a eed of 10 km/ho . Ho m ch a ea ill i i iga e in 30 min e , if 8 cm anding a e i needed? 22. ​Find he mode of he follo ing f e enc di ib ion. Downloaded from www.padhle.in

Sec ion- D (4 Marks Each) 2​ 3. T​ o a e a oge he can fill a ank in 1 7/8 ho . The a i h longe diame e ake 2 ho le han he a i h a malle one o fill he ank e a a el . Find he ime in hich each a can fill he ank e a a el . 24. ​If he m of fi fo e m of an A.P. i 40 and ha of fi 14 e m i 280. Find he m of i fi n e m . 25. 26. A​ man in a boa o ing a a f om a ligh ho e 100 m high ake 2 min e o change he angle of ele a ion of he o of he ligh ho e f om 60 o 30 . Find he eed of he boa in me e e min e. [U e 3 = 1.732] 27. ​Con c a ABC in hich CA = 6 cm, AB = 5 cm and ∠BAC = 45 . Then con c a iangle ho e ide a e ⅗ of he co e onding ide of ABC. Downloaded from www.padhle.in

28. A​ b cke o en a he o i in he fo m of a f m of a cone i h a ca aci of 12308.8 cm3. The adii of he o and bo om of ci c la end of he b cke a e 20 cm and 12 cm e ec i el . Find he heigh of he b cke and al o he a ea of he me al hee ed in making i . 29. P​ o e ha in a igh -angle iangle, he a e of he h o en e i e al he m of a e of he o he o ide . 30. enc di ib ion i 32.5. Find he If he median of he follo ing f e al e of f1 and f2. OR den of a cla in an e amina ion a e The ma k ob ained b 100 gi en belo . Downloaded from www.padhle.in

D a a le han e c m la i e f e enc c e (ogi e). Hence find he median. Ans ers An 1). Le he coo dina e of oin A be ( , ) and oin O (2, -3) be oin he cen e, hen B mid oin fo m la, Downloaded from www.padhle.in

The coo dina e of oin A a e (3, -10) Ans 2). ​ ​The gi en e a ion i 2 + 4 + k = 0 e ge On com a ing he gi en e a ion i h a 2 + b + c = 0, a = 1, b = 4 and c = k F​ o eal oo , D 0 o b2 4ac 0 o 16 4k 0 ok 4 Fo k 4, e a ion 2 + 4 + k ill ha e eal oo . OR ​The gi en e a ion i 3 2 10 + k = 0 On com a ing i i h a 2 + b + c = 0, e ge a = 3, b = -10, c = k Le he oo of he e a ion a e and 1/ P od c of he oo = c/ . 1/ = k/3 Downloaded from www.padhle.in

o k=3 An 3). G​ i en, an 2A = co (A 24 ) o co (90 2A) = co (A 24 ) [ ​ an = co (90 )] o 90 2A = A 24 o 3A = 90 + 24 o 3A = 114 A = 38 ​ in233 + in257 57 ) + co 2 OR = in2 33 + co 2(90 [ ​ in2 = 1] = in2 33 + co 2 33 =1 An 4). The o-digi n mbe di i ible b 3 a e 12, 15, 18, 99 ​Thi i an A.P. in hich a = 12, d = 3, an = 99 ​an = a + (n 1) d 99 = 12 + (n 1) 3 87 = (n 1) 3 o n 1 = 29 o n = 30 So, he e a e 30 o-digi n mbe di i ible b 3. An 5). G​ i en, AD = 1 cm, BD = 2 cm AB = 1 + 2 = 3 cm Downloaded from www.padhle.in

​Al o, DE BC (Gi en) ∠ADE = ∠ABC (i) (co e onding angle ) I​ n ABC and ADE ∠A = ∠A (common) ∠ABC = ∠ADE [b e a ion (i)] ABC ADE (b AA le) No , ar(AB C ) = (AB/AD)​2 ar(ADE) ​ ABC ADE ar(AB C ) = (3/1)2​ ​ = 9:1 ar(ADE) An 6). We kno , 2=1.414 and 3=1.732 so,ra ional n mber be een 2 and 3 =1.432 =1.563 =1.576 =1.657 =1.711 e c. An 7). Downloaded from www.padhle.in