MT - GEOMETRY - SEMI PRELIM - I : PAPER - 1

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Roger Stevenson
5 years ago
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1 07 00 MT A.. Attempt ANY FIVE of the following : (i) Slope of the line (m) 5 intercept of the line (c) B slope intercept form, The equation of the line is m + c 5 () + ( ) 5 MT - GEOMETRY - SEMI PRELIM - I : Time : Hours Model Answer Paper Ma. Marks : 0 The equation of the given line is 5 (ii) cot sin cot cosec [ + cot cosec cot cosec ] (iii) Equation of a line parallel to X-ais and passing through the point (5, 7) is 7 (iv) sin cos 0 sin cos sin cos tan (v) 5 ( 7) Comparing with the equation of a line in slope point form, m ( ) m Slope of the line 5 ( 7) is

2 / MT (vi) + 90º [Given] tan [Given] cot tan [ cot tan (90 )] cot A.. Solve ANY FOUR of the following : (i) (Analtical figure) P. cm R P. cm R (ii) The terminal arm passes through P (, ) and r ()() mark for constructing the circle mark for constructing the tangent at R r 5 units

3 / MT Let the angle be sin r 5 cosec r 5 cos r 5 sec r 5 tan cot (iii) Let O (0, 0) (, ) A (, 5) (, ) The equation of line OA b two point form is, 0 0 ( ) The equation of the line passing through the origin and the point (, 5) is (iv) (Analtical figure) L N.6 cm M A

4 / MT L.6 cm O N M A mark for drawing circle mark for drawing tangent (v) sin + sin [Given] sin sin sin cos...(i) sin cos cos cos cos + cos cos + cos (vi) P (, ), Q (, 6), R (8, ), S (0, k) Line PQ is parallel to line RS Slope of line PQ Slope of line RS 6 k 0 8 k

5 5 / MT k k + k 5 Value of k is 5. A.. Solve ANY THREE of the following : (i) Analsis : In PQR, P + Q + R Q Q 80 0 Q 50º R P (Analtical figure) R 0º 50º Q 6 cm P 0º 6 cm 50º Q mark for drawing PQR mark for drawing perpendicular bisectors mark for drawing circumcircle (ii) cot 7 tan tan 7 cot + tan sec + 7 sec

6 6 / MT sec sec sec sec cos cos 5 sec sin cos tan sin tan cos sin sin 5 (iii) V ( 7, 8) (, ), W ( 5, ) (, ), U (, 6) (, ) Slope of line VW 8 5 ( 7) Slope of line WU 6 ( 5) 5 8 Slope of line VW and slope of line WU are not equal. Points V, W and U are not collinear.

7 7 / MT (iv) 8 sin cos 8 sin cos... (i) sin + cos sin + (8 sin ) [From (i)] sin + 6 sin 6 sin + 6 sin + 6 sin 6 sin sin 6 sin sin + sin 6 sin sin 6 sin sin 9 sin 5 sin sin (5 sin ) 5 (5 sin ) 0 (5 sin ) ( sin 5) 0 5 sin 0 or sin sin or sin 5 sin 5 or sin 5 (v) Let, A (, ), B ( 9, 6), C (, ), D (6, 9) Slope of a line Slope of line AB 6 9 ( ) Slope of line AB 5 8 Slope of line CD 9 6 ( ) 5 6 Slope of line CD 5 8 Slope of line AB and slope of line CD are equal. line AB line CD The line joining (, ) and ( 9, 6) is parallel to the line joining (, ) and (6, 9).

8 8 / MT A.. Solve ANY TWO of the following : (i) cosec + cot...(i) cosec cot...(ii) Multipling (i) b, cosec + 6 cot...(iii) Multipling (ii) b, cosec 9 cot...(iv) Subtracting (iv) from (iii), cosec + 6 cot ( cosec 9 cot ) cosec + 6 cot cosec + 9 cot 5 cot cot 5 Substituting cot cosec in equation (i) cosec cosec cosec ( ) 5 cosec cosec cosec 5 We know, cosec cot 5 5 ( )( ) Multipling throughout b 65, ( + ) ( ) 65

9 9 / MT (ii) R (Analtical figure) R 6 cm I 6.5 cm 6 cm 6.5 cm I S M 7 cm T S 7 cm T mark for drawing triangle mark for drawing angle bisectors mark for drawing perpendicular mark for incircle (iii) L.H.S. tan cot cot tan sin cos cos sin cos sin sin cos sin cos cos sin sin cos cos sin sin cos sin cos cos(sin cos) sin(cos sin) sin cos cos(sin cos) sin(sin cos) sin cos sin cos cos sin sin cos sin cos cos sin sin cos sin sin cos cos sin cos cos sin

10 0 / MT sin sin. cos cos cos. sin sin. cos cos. sin [ sin + cos ] sin. cos cos. sin cos. sin cos sin sec. cosec + sec, cosec cos sin R.H.S. tan cot + cot tan + sec. cosec. A.5. Solve ANY TWO of the following : (i) T T E (Analtical figure) E.9 cm A 0º A A H 6. cm M.9 cm A A A 5 A6 A 0º 6. cm H M A 7 A A A A A 5 A6 mark for drawing analtical figure mark for AMT mark for constructing 7 congruent parts mark for constructing HA 5 A MA 7 A mark for constructing EHA TMA A 7

11 / MT (ii) Let seg AB represent the tree seg BC represent width of river Let BC m C and D represent the initial and final positions of the observer DC 0 m ACB and ADB are the angles of elevation m ACB 60º and m ADB 0º In right angled ACB, tan 60º AB BC AB [B definition] AB m...(i) In right angled ADB, tan 0º AB DB AB 0 D 0º 60º 0 m C [B definition] A B 0 AB m...(ii) From (i) and (ii) we get, BC 0 m AB 0 m [From (i)] AB 0.7 AB.6 m Height of tree is.6 m and width of river is 0 m.

12 / MT (iii) Let, P (, ), Q (0, ), R (, ), S (0, ) Slope of a line Slope of line PQ 0 ( ) P (, ) S (0, ) 0 Q (0, ) R (, ) Slope of line PQ Slope of line RS 0 Slope of line RS Slope of line PQ Slope of line RS line PQ line RS...(i) Slope of line QR 0 Slope of line QR Slope of line PS 0 ( ) 0 Slope of line PS Slope of line QR Slope of line PS line QR line PS...(ii) In PQRS, side PQ side RS side QR side PS [From (i)] [From (ii)] PQRS is a parallelogram [B definition] The points (, ), (0, ), (, ) and (0, ) are the vertices of parallelogram.

MT - GEOMETRY - SEMI PRELIM - I : PAPER - 2

07 00 MT MT - GEOMETRY - SEMI PRELIM - I : PAPER - Time : Hours Model Answer Paper Ma. Marks : 40 A.. Attempt ANY FIVE of the following : (i) Slope of the line (m) 4 intercept of the line (c) 3 B slope