# CBSE MATHEMATICS (SET-2)_2019

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1 CBSE 09 MATHEMATICS (SET-) (Solutions). OC AB (AB is tangent to the smaller circle) In OBC a b CB CB a b CB a b AB CB (Perpendicular from the centre bisects the chord) AB a b. In PQS PQ 4 (By Pythagoras theorem) PQ 5 PQ 5 cm In PQR tan 5 tan 9 PQ RQ OR 5 tan let AB 5k and BC k 5 AC k k AC 69k AC k AC sec BC CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

2 . x 5 5x x x x x x 0x 0 a, b 0, c 4. Discriminant b 4ac Number will terminate after 4 decimal places OR CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

3 5. The list of first 0 multiples of 6 is 6,, 8,.., 60 which forms an AP S S 0 n AB 0 m 5 0 (by distance formula) 8 m 9 6m 0 Squaring both sides m m 6m m 7 0 m m m m m 9 0 m 9 Or m positive value of m is. 7. Possible outcomes,,, 4, 5, 6 Probability of an event Number of outcomes favourable to event Total number of equally likely outcomes (i) Favourable outcomes = {4, 6} P (a composite number) = 6 (ii) Favourable outcomes = {,, 5} P (a prime number) = 6 CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

4 8. Possible outcomes = {7, 8, 9, 0,.., 40} Total outcomes Probability of an event Favourable outcomes = {7, 4,, 8, 5} P (multiple of 7) = Mid point of AC Mid point of BD Number of outcomes favourable to event Total number of equally libely outcomes (Diagonals of a parallelogram bisect each other) a b 5 4,, a 9 b and a 6 and b (By mid point formula) OR P divides AB in the ratio : Coordinates of 4 8 Coordinates of P, Q divides AB in the ratio : Coordinates of 6 Coordinates of Q, P, Q, (By section formula) CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

5 0. x5y 4..(i) 9xy 7..(ii) Multiply (i) by x5y 4..(iii) Solving (iii) and (ii) 9x5y 9x y7 y 5 5 y Substituting this value of y in (i) we get 9 x. Using Euclid s division algorithm to find HCF of 65 and (i) (ii) (iii) HCF = from (ii) 65 5 (iii) from (i) put in (iii) (iv) on comparing (iv) with 65n 7 we get n CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

6 OR Minimum distance each should walk so that each can cover the same distance in complete steps is LCM of 0, 6 and LCM 60 0, 6, 40 5 Minimum distance = 60 cm. kx 6x 0 Roots are not real Discriminant < 0 b ac 4 0 k k 0 4k 6 k 9 This equation will have no real roots for all real values of k less than -9. Given: (i) A B C 80 To prove: sin B C cos A B C Proof: LHS: sin 80 A sin A sin90 A B C 80 CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

7 sin90 cos A RHS Hence Proved cos (ii) A 90 B C80 90 BC 90 B C 90 tan tan tan 45 tan AB A B (i) tan AB A B 0 (ii) Adding (i) & (ii) A 75 OR 75 A Putting this value in (i) we get B 5 CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

8 4. Join OT. Let it intersect PQ at the point R. Then TPQ is isosceles and TO is the angle bisector of PTQ. So, OT PQ and therefore, OT bisects PQ PR RQ 4cm. Also, OR OP PR 5 4 cm cm Now, TPR RPO 90 TPR PTR So, RPO PTR. TRP PRO (by AA similarity) TP RP PO RO i.e TP TP cm. OR Given: ABCD is circumscribing a circle AB, BC, CD, AD are tangents To prove: AOB COD 80 and BOC AOD 80 Proof: Join OH, OG, OF, OE In AOH and AOG AGO AHO (tangent is perpendicular to the radius at the point of contact) OG AO OH (radii) AO (Common) AOH AOG ( RHS ) AOH AOG x(cpct) BOH BOE Similarly, BOH BOE y COE COF p COE COF FOD GOD q FOD GOD x x y y p p q q 60 (angle at O) CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

9 x y p q 60 x y p q 80 AOB x y AOB COD 80 COD p q Similarly, it can be proved that BOC AOD Hence Proved Number of days Number of students( Mean fx i i fi f i ) x i fx i i Mean number of days is Total area cleaned Area cleaned by wiper r cm CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

10 7. Given : AD BC To prove : BD CD AB AC BC Proof : let CD x, BD x BC 4x In ABD AB AD BD AB AD BD (Multiply both sides by ) AD AD 8x AC CD x AC AD DC AC x x 8 8 AC 6x AC 4 x x AC BC Hence proved OR Given : ABC PQR To prove : AD and PM are medians AB PQ AD PM Proof : B Q ABC PQR (i) AB BC PQ QR AB PQ BC QR AB BD.(ii) PQ QM CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

11 from (i) & (ii) ABD PQM (SAS criteria) AB PQ AD PM 8. x x 5 x x x x 4 x x x x 5 x x x x As remainder is, g x is not the factor of px 9. Let us assume that p q is a rational number pq, are coprime Squaring both sides and q 0 p q q p..(i) divides p divides p p m Putting this value in (i) q q m m divides q CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

12 divides q Therefore p and q have as a common factor. But this contradicts our assumption that p and q are co prime. This contradiction has arisen due to our wrong assumption that is rational Therefore is an irrational number. OR Let the largest number be a according to Euclid s division lemma 5 aq aq 50..(i) 977 aq' aq' 975..(ii) 568 aq'' aq'' 565.(iii) from (i), (ii) and (iii) a HCF (50, 975, 565) HCF 50, 975, a Let coordinates of Bx, y x y4 E, E, x y 4 and x and y B, [as E is mid point of AB] Similarly, since F is the mid point of C, coordinates of C will be, CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

13 ar 4 4 ABC Numerical value of Numerical value of square units. 4. Let the numbers are 5, 6 According to question 5x 7 4 6x 7 5 x x x x 7 x x 5x 5; 6x 4 Numbers are 5, 4 x. Let radius of pipe is ' r ' m Volume of water flowing through the pipe in half an hour Volume of water in the cylindrical tank of height.5m Volume flowing through pipe in hour r Volume flowing through pipe in half an hour r 50 Volume of water in cylindrical tank of height.5 m.5 Now, r r CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

14 4 r 0000 r m 00 4 d m or 00 4cm. Total surface area of decorative block = Total surface area of cube + curved surface (6) 6 (.) cm area of hemisphere area of circle Volume of decorative block = volume of cube + Volume of hemisphere 4. = cm Let r, r 0 Or h r r r r Volume of frustum =.4h h h 5cm CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

15 l 5 (0 ) 7 cm Area of metal sheet = curved surface area of frustum + area of base l( r r ) r.4 7(0 ) 60.cm 4. Given : DE BC To prove : AD AE DB EC Join BE and CD and then draw DM Now, area of So AC and EN AB ADE base height AD EN ar( ADE) AD EN Similarly, ar( BDE) DB EN ar(a DE) AE DM and ar( DEC) EC DM ar( ADE) AD EN AD ar( BDE) DB EN DB ar( ADE) AE DM AE ar( DEC) EC DM EC BDE and DEC are on the same base DE and between the same parallels BC and DE. So ar( BDE) ar( DEC) ( iii ) (i) (ii) From (i), (ii) and (iii), we have CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

16 AD AE DB EC OR Given :- ABC is a right angled triangle. To prove :- AC AB BC Construction:- Draw BD AC Proof:- In ADB and ABC A A (Common) ADB ABC (each 90 ) ADB ABC (AA similarity) So, AD AB (Sides are proportional) AB AC AD. AC In BDC (i) AB and ABC C C (common) BDC ABC (each 90 ) BDC ABC (By AA) So, CD BC BC AC CD. AC (ii) BC Adding (i) and (ii) AD. AC CD. AC AB BC AC( AD CD) AB BC AC. AC AB BC AC AB BC Hence proved. CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

17 5. Observations More than or equal to 0 More than or equal to 0 More than or equal to 40 More than or equal to 50 More than or equal to 60 More than or equal to 70 More than or equal to 80 Cumulative frequency (cf) Let h be the height of the tower CD 40m In ABC AB tan60 BC h BC h BC.(i) CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

18 In AB ABD tan0 BD h BC 40 (from (i) ) h h 40 h h 40 h h 40 h 40 h h h 4.64 m Height of the tower is 4.64 m 7. Let first term be a and common difference be d a a m d m a a n d n According to question ma m na n ( ) m a m d n a n d ma m md na n nd 0 a( m n) d ( m ) m ( n ) n 0 a m n d m m n n ( ) 0 a m n d m n m n ( ) ( ) 0 CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

19 a( m n) d ( m n) ( m n) (m n) 0 a( m n) d m n m n 0 ( m n) a ( m n ) d 0 mn 0 or a ( m n ) d 0 m n a ( m n ) d 0 But m n (given) am n 0 Or Let three numbers in AP be a d, a, a d ( a d) a( a d) 8 a 8 a 6 Numbers are 6 d,6, 6 d (6 d)(6 d) 5d 6 d 5 d d d d d d ( d 9)( d 4) 0 d 9 or d 4 Numbers are Numbers are 6 ( 9),6,6 ( 9) 6-4, 6, 6+4 5, 6, -, 6, 0 8. Let number of books he buys is ' x ' Cost of each book ` 80 x According to question x 4 x CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

20 80 80 x 4 x 0 x x 4 x 4x 0 0 x x x x x x 0 or (Rejected) Number of books is 6 x 6 9. TP and TQ are required tangents. CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

21 sin cos sec cosec 0. LHS sin cos cos sin cos sin sin cos cos sin sin cos sin sin cos cos sin cos sin cos R. H. S Hence Proved CBSE MATHEMATICS (SET-)_09 SCO 6, Sec 8 C&D, Chd. Ph

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