Time : 2 Hours Preliminary Model Answer Paper Max. Marks : 40. [Given] [Taking square roots]

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Valentine Phelps
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1 .P. SET CODE MT - w MT - w - MTHEMTICS (7) GEOMETRY - (E) Time : Hours Preliminary Model nswer Paper Max. Marks : 40.. ttempt NY FIVE of the following : (i) BC ~ PQ [Given] ( BC) ( PQ) BC PQ [reas of similar triangles] 4 BC PQ [Given] BC PQ [Taking square roots] (ii) M In MQNP, m MPN 40º [Given] N m PMQ 90º [Radius is perpendicular to the tangent] m PNQ 90º m MQN 40º [Remaining angle] P 40º Q (iii) Since the initial arm rotates in Y clockwise direction and the angle is more than 80º but less than 70º, the terminal arm lies in II quadrant. O X 0º X Y

2 / MT - w (iv) Inclination of the line 45º Slope of the line tan tan 45º Slope of the line is (v) rea of the sector 360 r rea of the sector is.83 cm. (vi) rea of a triangle base height (PQR) PR QS Q 4 PR 8 4 PR 4 P S 8 cm R PR 4 4 PR 6 cm.. Solve NY FOUR of the following : (i) In BC and ED, BC DE [Common angle] BC ED [ Each is 90º] BC ~ ED [By test of similarity] B E C D E 8 6 [c.s.s.t.] [Given] E 6 8 B C E 4 cm D E

3 3 / MT - w (ii) D M B CB BDC...(i) [ngles inscribed in the same arc are congruent] In CM and BDM, CM BDM [From (i), - M - B and D - M - C] MC DMB [Vertically opposite angles] CM ~ BDM [By test of similarity] CM BM C BD [c.s.s.t.] CM BD BM C (iii) (, ) (x, y ) B (0, 3) (x, y ) Slope of line B x x y y C 3 0 ( ) Slope of line B is (iv) tan + tan tan tan 4 [Squaring both sides] tan + tan. tan tan 4 tan + + tan 4 tan + tan 4 tan + tan

4 4 / MT - w (v) Let, (, 3) (x, y ) B (4, 7) (x, y ) The line passes through points and B The equation of the line by two point form is x x x x x 4 x y y y y y y (x ) (y 3) 4x 8 y 6 y 4x y 4x y x [Dividing throughout by ] y x is the equation of the line passing through (, 3) and (4,7) (vi) (Rough Figure) l l O.9 cm M O.9 cm M mark for rough figure mark for circle mark for drawing perpendicular

5 5 / MT - w.3. Solve NY THREE of the following : (i) Given : In BC, ray D is the bisector of BC such that B - D - C. x x E To Prove : BD DC B C B D C Construction : Draw a line passing through C, parallel to line D and intersecting line B at point E, B - - E. Proof : In BEC, line D side CE [Construction] BD DC B E line CE line D On transversal BE,...(i) [By B.P.T.] [Construction] BD EC...(ii) [Converse of corresponding lso, On transversal C, angles test] DC CE...(iii) [Converse of alternate angles test] But, BD DC...(iv) [ ray D bisects BC] In EC, EC CE [From (ii), (iii) and (iv)] seg C seg E [Converse of Isosceles C E...(v) triangle theorem] BD DC B C [From (i) and (v)]

6 6 / MT - w (ii) D C M B - M - B [If two circles are touching circles then the common point lies on the line joining their centres] M D 6 cm...(i) BM BC 9 cm...(ii) [Radii of the same circle] B M + MB [ - M - B] B [From (i) and (ii)] B 5 cm...(iii) In BC, m CB 90º [Radius is perpendicular to the tangent] B² C² + BC² [By Pythagoras theorem] 5² C² + 9² [From (ii) and (iii)] 5 C² + 8 C² 5 8 C² 44 C cm [Taking square roots] In DB, m DB 90º [Radius is perpendicular to the tangent] B D + BD [By Pythagoras theorem] BD [From (i) and (iii)] BD BD 5 36 BD 89 BD 9 BD 3 cm. [Taking square roots] The lengths of seg C and seg BD are cm and 3 cm respectively.

7 7 / MT - w (iii) (Rough Figure) L L K 60º 55º 7 cm M m O K 60º 55º 7 cm M l mark for rough figure mark for drawing KLM mark for drawing perpendicular bisectors mark for drawing circumcircle (iv) sec cos 3 sec 3 3 cos sin + cos sin + 3

8 8 / MT - w sin sin 4 sin sin 4 sin [Taking square roots] cosec sin cosec is in IV quadrant cosec cos ec cos ec ( ) ( ) cos ec cos ec cos ec cos ec 3 (v) (, 6), B (3, 4) Point P divides seg B internally in the ratio : 3 Let, P (x, y) By section formula for internal division, x mx nx my + ny m + n y m + n (3) + 3( ) ( 4) x 0 y P (0, ) The line having slope 3 passes through the point P (0, )

9 9 / MT - w The equation of the line by slope point form is, (y y ) m (x x ) (y ) 3 (x 0) (y ) 3x y 4 3x 3x y The equation of the required line is 3x y Solve NY TWO of the following : (i) Let seg B represents the tree seg BC represents width of river Let BC x m C and D represents the initial and final positions of the observer DC 40 m CB and DB are the angles of elevation m CB 60º and m DB 30º In right angled CB, tan 60º B BC D 30º 60º B 40 m C ( mark for figure) [By definition] 3 B x B 3 x m...(i) 3 In right angled DB, tan 30º B DB B 40 x B 40 x 3 From (i) and (ii) we get, 3 x 40 x 3 3x 40 + x 3x x 40 [By definition] m...(ii)

10 0 / MT - w x 40 x 0 BC 0 m B 0 3 m [From (i)] B 0.73 B 34.6 m Height of tree is 34.6 m and width of river is 0 m. (ii) ( mark for figure) Given : BCD is a cyclic B D To Prove : m BC + m DC 80º m BD + m BCD 80º C Proof : m BC m (arc DC)...(i) [Inscribed angle m DC theorem] m (arc BC)...(ii) dding (i) and (ii), we get m BC + m DC m (arc DC) + m (arc BC) m BC + m DC [m (arc DC) + m (arc BC)] m BC + m DC 360º [ Measure of a circle is 360º] m BC + m DC 80º...(iii) In BCD, m BD + m BCD + m BC + m DC 360º [ Sum of measure of angles of a quadrilateral is 360º] m BD + m BCD + 80º 360º [From (iii)] m BD + m BCD 80º

11 / MT - w (iii) side of cubical wooden block Volume of cubical wooden block l 3 m 00 cm (00) cm 3 cylindrical hole is bored through the cubical wooden block Height of cylindrical hole (h) m 00 cm Diameter of cylindrical hole 30 cm Its radius (r) 30 5 cm Volume of cylindrical hole r h cm 3 Volume of the object so formed Volume of cubical wooden block Volume of cylindrical hole cm 3 Volume of the object so formed is cm Solve NY TWO of the following : (i) (a) rea of a triangle ( BC) ( BC) lso, ( BC) base height B CD p c p...(i) BC C ( BC) a b...(ii) From (i) and (ii) we get, c p a b cp ab b C c D a B

12 / MT - w (b) cp ab cp ab c p a b [By Invertendo] [Squaring both sides] p c...(iii) a b In CB, m CB 90º [Given] B C + BC [By Pythagoras theorem] c b + a...(iv) p b a [From (iii) and (iv)] a b p b a a b a b p + a b (ii) T (Rough Figure) T 4.9 cm E E 4.9 cm 0º H 6.3 cm M 0º 6.3 cm H M mark for rough figure 4 mark for HE mark for constructing 7 congruent parts mark for constructing H 5 M 7 mark for constructing HE MT mark for required MT

13 3 / MT - w (iii) Diameter PR 6 units Its radius (r ) 3 units Diameter PQ 8 units Its radius (r ) 4 units In PQR, m RPQ 90º...(i) [ngle subtended by a semicircle] QR PR + PQ [By Pythagoras theorem] QR QR QR 00 QR 0 units [Taking square roots] Diameter QR 0 units Its radius (r 3 ) 5 units PQR is a right angled triangle [From (i)] R P Q ( PQR) product of perpendicular sides PR PQ sq. units. rea of shaded portion rea of semicircle with diameter PR + rea of semicircle with diameter PQ + rea of PQR rea of semicircle with diameter QR r + r + 4 r 3 r r r 4 3 (r + r r 3 ) + 4

14 4 / MT - w 3.4 ( ) ( ) (0) sq. units rea of shaded portion is 4 sq.units

Time : 2 Hours (Pages 3) Max. Marks : 40. Q.1. Solve the following : (Any 5) 5 In PQR, m Q = 90º, m P = 30º, m R = 60º. If PR = 8 cm, find QR.

Time : 2 Hours (Pages 3) Max. Marks : 40. Q.1. Solve the following : (Any 5) 5 In PQR, m Q = 90º, m P = 30º, m R = 60º. If PR = 8 cm, find QR. Q.P. SET CODE Q.1. Solve the following : (ny 5) 5 (i) (ii) In PQR, m Q 90º, m P 0º, m R 60º. If PR 8 cm, find QR. O is the centre of the circle. If m C 80º, the find m (arc C) and m (arc C). Seat No. 01