MT - MATHEMATICS (71) GEOMETRY - PRELIM II - PAPER - 3 (E)

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1 Seat No. MT - MTHEMTICS (71) GEOMETY - PELIM II - (E) Time : Hours (Pages 3) Max. Marks : 40 Note : (i) Q.1. Solve NY FIVE of the following : 5 (i) ll questions are compulsory. Use of calculator is not allowed. Lines PM and PN are tangents to the circle with centre O. If PM 7 cm, find PN. P M N O The radius of the base of a cone is 7 cm and its height is 4 cm. What is its slant height? (iii) State the value of tan ( 60). (iv) (v) The slope of line B is. What is the slope of line DE which is parallel to 3 line B? adius of a circle is 10 cm. The length of an arc of this circle is 5 cm. What is the area of the sector? (vi) Find tan, for the angle, whose terminal arm passes through (3, 4). Q.. Solve NY FOU of the following : 8 (i) The ratio of the areas of two triangles with the common base is 6 : 5. Height of the larger triangle is 9 cm. Then find the corresponding height of the smaller triangle.

2 / MT In the adjoining figure, E is a point on side CB produced of an isosceles BC with B C. If D BC and EF C, prove that BD ~ ECF. F E B D C (iii) In the adjoining figure, point is the centre of the circle. N 10 cm. Line NM is tangent at M. Determine the radius of the circle if MN 5 cm. M N (iv) (v) (vi) Draw a circle of radius 3.6 cm, take a point M on it. Draw a tangent to the circle at M without using centre of the circle. If the terminal arm passes through the point (1, 1) making an angle find the value of sec. Eliminate, if x a sec, y b tan Q.3. Solve NY THEE of the following : 9 (i) Find the length of the altitude of an equilateral triangle, each side measuring a units. In the adjoining figure, BC and B are tangents to circle. Prove that OD is perpendicular bisector of C, where O is the centre of the circle. C O D B (iii) Construct the incircle of ST in which S 6 cm, ST 7 cm and T 6.5 cm. (iv) Find x if the slope of line joining (x, ) and (8, 11) is 3 4.

3 3 / MT (v) In the adjoining figure, P is the centre of the circle with radius 18 cm. If the area of the PQ is 100 cm and area of the segment QX is cm. Find the central angle. ( 3.14) Q 18 cm P X Q.4. Solve NY TWO of the following : 8 (i) In the adjoining figure, point is a common point of contact l of two externally touching circles and line l is a common tangent to both circles touching at B and C. Line m is another common tangent at and it intersects BC at D. Prove that (i) BC 90º Point D is the midpoint of seg BC. B m D C (iii) If P (, 4), Q (4, 8), (10, 5) and S (4, 1) are the vertices of a quadrilateral show that it is a parallelogram. Eliminate, if x 3 cosec + 4 cot, y 4 cosec 3 cot Q.5. Solve NY TWO of the following : 10 (i) Prove : In a triangle, the angle bisector divides the side opposite to the angle in the ratio of the remaining sides. SH ~ SVU, In SH, SH 4.5 cm, H 5. cm, S 5.8 cm and SH SV 3 ; construct SVU. 5 (iii) In the adjoining figure, seg Q is a tangent to the circle with centre O. Point Q is the point of contact. adius of the circle is 10 cm. O 0 cm. Find the area of the shaded region. ( 3.14, ) Best Of Luck O 10 cm Q T

4 Seat No. MT - MTHEMTICS (71) GEOMETY - PELIM II - (E) Time : Hours Prelim - II Model nswer Paper Max. Marks : ttempt NY FIVE of the following : (i) PM PN [Length of the two tangent segments from an external point to a circle are equal] But, PM 7 cm [Given] PN 7 cm adius of base of cone (r) 7 cm its height (h) 4 cm l r + h l l l 65 l 5 [Taking square roots] Slant height of cone is 5 cm (iii) tan ( 60) tan 60 3 tan ( 60) 3 (iv) line DE line B [Given] Slope of line DE slope of line B But, slope of line B 3 [Given] Slope of line DE 3 (v) adius of circle (r) 10 cm Length of arc (l) 5 cm r rea of sector l

5 / MT cm The area of the sector is 15 cm. (vi) The terminal arm passes through (3, 4) x 3 y 4 tan y x tan Solve NY FOU of the following : (i) Let the areas of the larger and the smaller triangle be 1 and respectively. Let their heights be h 1 and h respectively and h 9 cm [Given] 1 The two triangles have a common base [Given] 1 h 1 h h h [Triangles with common base] h 15 h 7.5 The corresponding height of the smaller triangle is 7.5 cm. In BC, seg B seg C [Given] BC CB...(i) [Isosceles triangle theorem] 1 F In BD and ECF, BD FCE [From (i) and B - D - C, - F - C, C - B - E] DB EFC [ each is 90º] E B D C BD ~ ECF [By test of similarity] 1

6 3 / MT (iii) In MN, m MN 90º [adius is perpendicular to the tangent] N² M² + MN² [By Pythagoras theorem] 10² M² + 5² [Given] 100 M² + 5 M N M² M² 75 M 75 M 5 3 M 5 3 cm. adius of the circle is 5 3 cm. (iv) L (ough Figure) L N M N M mark for rough figure 1 mark for drawing NM NLM mark for tangent at M.

7 4 / MT (v) The terminal arm passes through point (1, 1) x 1 and y 1 r x y r (1) ( 1) r 1 1 r units Let the angle be sec sec r x 1 sec (vi) x a sec sec x a...(i) y b tan tan y b tan sec 1 + y b x a [From (i) and ] 1 + y x b a x a y b 1.3. Solve NY THEE of the following : (i) Given : BC is an equilateral triangle. B BC C a seg D side BC To find : D Sol. BC is an equilateral triangle a a B BC C a...(i) [Given] In DB, m DB 90º [Given] B C D a

8 5 / MT m BD 60º [ngle of an equilateral triangle] m BD 30º [emaining angle] DB is a 30º º triangle By 30º - 60º - 90º triangle theorem, D D D 3 3 B [Side opposite to 60º] a [From (i)] 3 a units. O OC...(i) [adii of same circle] BC B... C [The lengths of the two tangent segments to a circle drawn from O an external point are equal] D B Points O and B are equidistant from the end points and C of seg C. [From (i) and ] Points O and B lie on the perpendicular bisector of seg C. [By perpendicular bisector theorem] seg OB is the perpendicular bisector of seg C. seg OD is the perpendicular bisector of seg C. [ O - D - B] (iii) (ough Figure) 6 cm 6.5 cm 6 cm 6.5 cm O S 7 cm T S 7 cm T mark for rough figure mark for drawing ST 1 mark for drawing the angle bisectors 1 mark for drawing the incircle

9 6 / MT (iv) Let, (x, ) (x 1, y 1 ) B (8, 11) (x, y ) Slope of line B 3 4 [Given] Slope of line B y y1 x x ( ) 4 8 x x x 3 (8 x) x 36 3x x 1 x 1 3 x 4 The value of x is 4. (v) adius of a circle (r) 18 cm rea of PQ 100 cm rea of the segment QX cm rea of sector P-QX rea of PQ + rea of rea of sector P-QX cm rea of sector segment QX Q r Central angle is 40º. 18 cm P X

10 7 / MT.4. Solve NY TWO of the following : (i) In BD, m DB D...(i) l B [The lengths of the two tangent D C segments from an external point to a circle are equal] DB DB [Isosceles triangle theorem] Let, m DB m DB xº... In DC, D DC...(iii) [The lengths of the two tangent segments from an external point to a circle are equal] DC DC [Isosceles triangle theorem] Let, m DC m DC yº...(iv) m BC m DB + m DC[ngle ddition Property] m BC (x + y)º...(v) [From and (iv)] In BC, m BC + m CB + m BC 180º [ Sum of the measures of the angles of a triangle is 180º] x + y + x + y 180 [From, (iv), (v) and B - D - C] x + y 180 (x + y) x + y x + y 90 m BC 90º [From (v)] From (i) and (iii) we get, DB DC D is the midpoint of seg BC. P (, 4), Q (4, 8), (10, 5), S (4, 1) Slope of a line y y1 x x1 P (, 4) S (4, 1) 8 4 Slope of line PQ 4 ( ) 4 4 Q (4, 8) (10, 5) 4 6

11 8 / MT Slope of line PQ 3 Slope of line S Slope of line S 3 Slope of line PQ Slope of line S line PQ line S...(i) 5 8 Slope of line Q Slope of line Q Slope of line PS 4 ( ) Slope of line PS Slope of line Q Slope of line PS line Q line PS... In PQS, side PQ side S [From (i)] side Q side PS [From ] PQS is a parallelogram [By definition] 1 (iii) x 3 cosec + 4 cot...(i) y 4 cosec 3 cot... Multiplying (i) by 4, 4x 1 cosec + 16 cot...(iii) Multiplying by 3, 3y 1 cosec 9 cot...(iv) Subtracting (iv) from (iii), 4x 3y 1 cosec + 16 cot (1 cosec 9 cot ) 4x 3y 1 cosec + 16 cot 1 cosec + 9 cot 4x 3y 5 cot

12 9 / MT cot 4x 3y 5 Substituting cot 4x 3y 5 in equation (i) x 4x 3y 3cosec x 1y x 3cosec x 1y x 3cosec 5 5x 16x 1y 5 3cosec 9x 1y 5 3cosec 3 (3x 4y) 3 5 cosec 3x 4y cosec 5 We know, cosec cot 1 3x 4y 4x 3y (3x 4y) (4x 3y) Multiplying throughout by 65, (3x + 4y) (4x 3y) Solve NY TWO of the following : (i) Given : In BC, ray D is the bisector E of BC such that B - D - C. To Prove : BD DC B x x C Construction : Draw a line passing through C, parallel to line D and B D C intersecting line B at point E, B - - E. ( marks for figure) Proof : In BEC, line D side CE [Construction] BD DC B...(i) [By B.P.T.] E line CE line D [Construction] On transversal BE,

13 10 / MT BD EC... [Converse of corresponding angles test] lso, On transversal C, DC CE...(iii) [Converse of alternate angles test] But, BD DC...(iv) [ ray D bisects BC] In EC, EC CE [From, (iii) and (iv)] seg C seg E [Converse of Isosceles triangle theorem] C E...(v) 1 BD DC B C (ough Figure) U [From (i) and (v)] U 5.8 cm 5. cm S 4.5 cm H V 5.8 cm 5. cm S 4.5 cm H V S 1 S 1 mark for SH S 3 1 mark for constructing 5 congruent parts 1 mark for constructing VS 5 S HS 3 S 1 mark for constructing UVS HS 1 mark for required SVU S 4 S 5

14 11 / MT (iii) In OQ, m OQ 90º [adius is perpendicular to the tangent] OQ + Q O [By Pythagoras theorem] 10 + Q 0 Q Q 300 O Q 300 T Q Q Q 10 3 Q 10 (1.73) Q 17.3 cm 10 cm rea of OQ 1 1 Product of Perpendicular sides OQ Q cm In OQ, m OQ 90º OQ 10 cm O 0 cm OQ 1 O By converse of 30º - 60º - 90º triangle theorem. m OQ 30º m QO 60º [emaining angle] Now, For sector O-QXT Measure of arc () 60º adius (r) 10 cm rea of Sector O-QXT 360 r rea of sector O-QXT 5.33 cm rea of shaded region rea of OQ rea of sector O-QXT cm rea of the shaded region is cm

MT - MATHEMATICS (71) GEOMETRY - PRELIM II - PAPER - 4 (E)

MT - MATHEMATICS (71) GEOMETRY - PRELIM II - PAPER - 4 (E) 014 1100 Seat No. MT - MTHEMTIS (71) GEMETY - ELIM II - (E) Time : Hours (ages 3) Max. Marks : 40 Note : (i) ll questions are compulsory. (ii) Use of calculator is not allowed..1. Solve NY FIVE of the