MT - w A.P. SET CODE MT - w - MATHEMATICS (71) GEOMETRY- SET - A (E) Time : 2 Hours Preliminary Model Answer Paper Max.

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1 .P. SET CODE.. Solve NY FIVE of the following : (i) ( BE) ( BD) ( BE) ( BD) BE D 6 9 MT - w MT - w - MTHEMTICS (7) GEOMETRY- (E) Time : Hours Preliminary Model nswer Paper Max. Marks : 40 [Triangles with common base] B E ( BE) ( BD) D (ii) Let r 5 cm and r cm The circles are touching internally. Distance between their centres r r 5 cm O T (iii) Given : 45º tan ( ) tan tan ( 45º) tan 45º tan ( 45º) (iv) m 5, c By slope point form, the equation of line is y mx + c y 5x 5x y 0

2 / MT - w (v) radius 7 cm Circumference r cm Circumference is 44 cm. (vi) Height of an equilateral triangle side 6 Height of an equilateral triangle cm.. Solve NY FOUR of the following : (i) PE EQ PE 4 0 EQ PE EQ PE EQ Q E...(i) P 4 8 F 9 R PF FR 8 9 In PQR,...(ii) PE EQ PF FR [From (i) and (ii)] line EF side QR [By converse of B.P.T.] (ii) Chords B and CD intersect each other internally at point P C P PB PC PD PD PD 4 8 PD units D P B

3 / MT - w (iii) Curved surface area of a cone 640 cm its radius (r) 40 cm. Curved surface area of a cone rl l 640 l 40 l 4 cm Now, r + h l 40 + h 4 h 4 40 h h 8 h 9 cm [Taking square roots] Height of a cone is 9 cm. (iv) Since the initial arm rotates in clockwise direction and the angle ( mark for figure) Y is more than 80º but less than 70º, the terminal arm lies in II quadrant. X 0º O X (v) Radius of a right circular cylinder cm Y its height (h) 7cm (a) Curved surface area of a cylinder rh 7 7 Curved surface area of a cylinder cm (b) Total Surface area of a cylinder r (r + h) 7 ( + 7) 7 0 Total surface area of the cylinder is cm cm

4 4 / MT - w (vi) 8. cm B mark for drawing seg B mark for drawing perpendicular bisector of seg B.. Solve NY THREE of the following : (i) In BC, seg P is the median [Given] B + C P + BP [By pollonius theorem] 60 (7) + BP [Given] 60 (49) + BP BP BP BP 6 BP 6 B C P BP 8 BP 9 units [Taking square roots] BP BC [ P is the midpoint of seg BC] 9 BC BC 8 units

5 5 / MT - w (ii) P PB 5 P 5 B Q BQ CQ [The lengths of the two Let, tangent segments to a circle C drawn from an external S SD x point are equal] CR DR y PQRS is a parallelogram [Given] S R D PQ SR [ Opposite sides of a parallelogram are congruent] PB + BQ SD + DR [P - B - Q and S - D - R] 5 + x + y x + y 8...(i) PS QR [ Opposite sides of a parallelogram are congruent] P +S QC + CR [ P - - S and Q - C - R] 5 + x + y x y 5 x y...(ii) dding (i) and (ii) x + y + x y 8 + ( ) x 8 x 6 x PS P + S [ P - - S] PS 5 + x PS 5 + PS 8 units (iii) (nalytical Figure) Z X 4.6 cm Y P

6 6 / MT - w Z X 4.6 cm Y P mark for drawing chord XY mark for drawing tangent at X mark for drawing tangnet at Y (iv) tan sin cos sin cos...(i) + tan sec + () sec + sec sec sec [Taking square roots] cos sec cos sin [From (i)]

7 7 / MT - w cosec sin cosec sin cos sec cos ec 4 sin cos sec cos ec (v) Let,, 5 (x, y ) B, k C 4, 0 5 (x, y ) (x, y ) Points, B and C are collinear Slope of line B Slope of line BC y y y y x x x x k 0 k k 0 k 0

8 8 / MT - w k 0 0 k k k k + k 4k k 4 The value of k is 4.4. Solve NY TWO of the following : (i) B represents the height of the tree The tree breaks at point D D is the broken part of tree which then takes the position of DC D DC mdcb 0º BC 0 m In right angled DBC, tan 0º DB BC DB 0 [By definition] ( mark for figure) D B 0º 0 m C DB 0 DB 0 DB 0 m cos 0º BC DC 0 DC DC 0 DC 0 DC 0 m [By definition]

9 9 / MT - w D DC 0 m B D + DB [ - D - B] B B 0 m B 0.7 B 5.9 m The height of tree is 5.9 m. (ii) Given : BCD is a cyclic To Prove : m BC + m DC 80º Proof : m BD + m BCD 80º ( mark for figure) m BC B m (arc DC)...(i) D [Inscribed angle m DC }theorem] m (arc BC)...(ii) C 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 60º [ Measure of a circle is 60º] m BC + m DC 80º...(iii) In BCD, m BD + m BCD + m BC + m DC 60º [ Sum of measure of angles of a quadrilateral is 60º] m BD + m BCD + 80º 60º [From (iii)] m BD + m BCD 80º

10 0 / MT - w (iii) (4, 7), B (, ), C (0, ) Let, seg D, seg BE and seg CF be the medians on sides BC, C and B respectively. D, E and F are the midpoints of sides BC, C and B respectively. (4, 7) By midpoint formula, D E F x + x y + y, + 0 +, 4, (, ) x + x y + y, 4 0 7, 4 8, (, 4) x + x y + y, 4( ) 7, 4 0,, 5 (, 5) By two point form, The equation of median D, x x y y x x y y x 4 4 ( ) y 7 7 x 4 4 y 7 5 x 4 y F B (, ) C (0, ) D E

11 / MT - w x 4 y 7 x y x y + 0 The equation of the median BE x x y y x x y y x ( ) y 4 x 4 y x + 4 (y ) x + 4y x 4y x 4y The equation of the median CF x x y y x x y y x 0 0 y 5 x y 4 4x y 4x y + 0 The equation of the medians of BC are x y + 0, x 4y and 4x y Solve NY TWO of the following : (i) Given : In BC, (i) Line l side BC (ii) Line intersects sides B and C at points D and E respectively. - D - B, - E - C D To Prove : DB E EC B Construction : Draw seg BE and seg CD. Proof : DE and BDE have a common vertex E and their bases D and BD lie on the same line B. Their heights are equal ( DE) ( BDE) D...(i) [Triangles having equal heights] DB DE and CDE have a common vertex D and their bases E and EC lie on the same line C. Their heights are equal. ( mark for figure) D E C

12 / MT - w ( DE) ( CDE) E CE...(ii) [Triangles having equal heights] line DE side BC [Given] BDE and CDE are between the same two parallel lines DE and BC. Their heights are equal. lso, they have same base DE. (BDE) (CDE)...(iii) [reas of two triangles having equal bases and equal heights are equal] ( DE) ( DE) ( BDE) ( CDE)...(iv) [From (i), (ii) and (iii) ] D E DB EC [From (i), (ii) and (iv)] (ii) nalysis : BC ~DEF [Given] B DE BC EF C DF B E 45º B DE 5. DE...(i) [c.s.s.t.] [From (i)] BC EF [c.a.s.t.] 4.6 EF 5.6 DE.8 EF DE 7.8 cm EF 6.9 cm [From (i)] Information for constructing DEF is complete. (Given triangle) 5. cm B 45º 4.6 cm C

13 / MT - w (Required triangle) D 7.8 cm E 45º 6.9 cm F (iii) 0 cm 0 cm 0 cm 60 toy is a combination of cylinder, hemisphere and cone, each with radius 0 cm r 0 cm Height of the conical part (h) 0 cm Height of the hemispherical part its radius 0cm Total height of the toy 60cm Height of the cylindrical part (h ) cm l r + h l l l 00 l 00 [Taking square roots] l 0 cm Slant height of the conical part ( l) cm

14 4 / MT - w Total surface area of the toy Curved surface area of the conical part + Curved surface area of the cylindrical part + Curved surface area of the hemispherical part rl + rh + r r (l + h + r).4 0 ( ).4 ( ) cm Total surface area of the toy is cm.

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

Time : 2 Hours Preliminary Model Answer Paper Max. Marks : 40. [Given] [Taking square roots] .P. SET CODE MT - w 05 00 - 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