MT - MATHEMATICS (71) GEOMETRY - PRELIM II - PAPER - 4 (E)
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1 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 following : 5 (i) (ii) Two circles with centres and having diameter 5 cm and 15 cm respectively touch each other externally at. Find the distance between and. The corresponding central angle of an arc is 90º. What is the length of this arc, if the radius of the circle is 14 cm? For the angle in standard position, if the initial arm rotates 340º in the anticlockwise direction, state the quadrant in which the terminal arm lies. (iv) Find the slope of a line having inclination 60º. (v) What is the volume of a cube with side 5 cm? (vi) If 3 sin 4 cos 0, what is the value of tan?.. Solve NY FU of the following : 8 (i) If E 4 cm, E 4.5 cm, F 8 cm, and F 9 cm, then state whether EF. 4 8 E F 4.5 9
2 / MT (ii) ay T is the angle bisector of. Find the value of x and the perimeter of. 5.6 cm x 4 cm 5 cm T (iv) If two circles touch externally then show that the distance between their centers is equal to the sum of their radii. 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. (v) If sin + sin 1, prove that cos + cos 4 1. (vi) Eliminate, if x a sec, y b tan.3. Solve NY THEE of the following : 9 (i) djacent sides of a parallelogram are 11 cm and 17 cm. If the length of one of its diagonals is 6 cm. Find the length of the other. 11 cm 17 cm D (ii) hords and D of a circle intersect in point in the interior of a circle as shown in figure, if m (arc D) 5º and m (arc ) 31º, then find. D onstruct incircle of SGN such that SG 6.7 cm, S 70º,G 50º and draw incircle of SGN. (iv) Using slope concept, heck whether the points (7, 8), ( 5, ) and (3, 6) are collinear. (v) The length, breadth and height of a cuboid are in the ratio 5:4:. If the total surface area is 116 cm, find the dimensions of the solid.
3 3 / MT.4. Solve NY TW of the following : 8 (i) In the adjoining figure, points, and are points of contact of the respective tangents. line is parallel to line. If 7. cm, 5 cm, find the radius of the circle. (ii) If the points (1, ), (4, 6), (3, 5) are the vertices of a triangle. Find the equation of the line passing through the mid points of and. Two buildings are in front of each other on either side of a road of width 10 metres. From the top of the first building, which is 30 metres high, the angle of elevation of the top of the second is 45º. What is the height of the second building?.5. Solve NY TW of the following : 10 (i) (ii) rove : In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the remaining two sides. Draw a triangle, right angled at such that, 3 cm and 4 cm. Now construct a triangle similar to, each of whose sides is 7 5 times the corresponding side of. In the adjoining figure, and S are two diameters of the circle. If 8 cm and S 14 3 cm, find (i) rea of triangle S (ii) The total area of two shaded segments. ( ) 10º S est f Luck
4 Seat No. MT - MTHEMTIS (71) GEMETY - ELIM II - (E) Time : Hours relim - II Model nswer aper Max. Marks : ttempt NY FIVE of the following : (i) Diameter of bigger circle 5 cm Its radius (r 1 ) 1.5 cm Diameter of smaller circle 15 cm Its radius (r ) 7.5 cm The two circles with centres and touch each other externally at. Distance between and r 1 + r cm The distance between and is 0 cm. (ii) Measure of central angle () 90º adius (r) 14 cm Length of the arc (l) r 360 l l Length of the arc is cm. For the standard angle, if the initial arm rotates 340º in the anticlockwise direction then the terminal arm lies in the IV quadrant. 1 (iv) Inclination of the line () 60º Slope of the line tan tan 60 3 Slope of the line is 3.
5 / MT (v) Side of a cube (l) 5 cm Volume of cube l 3 (5) 3 15 cm 3 Volume of the cube is 15 cm 3. (vi) 3 sin 4 cos 0 3 sin 4 cos sin 4 cos 3 4 tan 3.. Solve NY FU of the following : (i) E E E E E E E 8 E F...(i) E F 8...(ii) F 9 In, E E F F [From (i) and (ii)] line EF side [y converse of..t.] 4 8 (ii) In, ray T bisects T T [roperty of angle bisector of a triangle] 5.6 x 4 5 x [Given] 5.6 cm x 4 cm 5 cm T
6 3 / MT x 7 7 cm T + T [ - T - ] cm erimeter of erimeter of 1.6 cm ( mark for figure) Given : Two circles with centres and touch each other externally at point T. T To rove : T + T roof : - T - [If two circles are touching circles then the common point lies on the line joining their centres] T + T [ - T - ] (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) sin + sin² 1 [Given] sin 1 sin² sin + cos 1 sin cos 1 sin cos sin cos 4 [Squaring both sides] 1 cos cos 4 cos² + cos 4 1 (vi) x a sec sec x a y b tan tan y b 1 + tan sec y b x a y b y b x a x a...(i)...(ii) sin + cos 1 1 cos sin [From (i) and (ii)] 1.3. Solve NY THEE of the following : (i) D is a parallelogram [Given] D 1 D...(i) [ Diagonals of parallelog ram bisec t each other] D 1 6 [Given] D 13 cm In D, seg is the median [From (i) and by definition] + D + [y ppollonius theorem] (11) + (17) + (13) (169) cm 17 cm D
8 5 / MT 6 cm [Taking square roots] 1 [ Diagonals of parallelog ram bisec t each other] cm Length of other diagonal is 1 cm. (ii) onstruction : Draw seg m (arc D) 5º [Given] D m D 1 m(arc D) [Inscribed angle theorem] m D 1 5º m D 1.5º m 1.5º [ D - - ] m (arc ) 31º [Given] m 1 m(arc ) [Inscribed angle theorem] m 1 31º m 15.5º m 15.5º [ - - ] is an exterior angle of, m m + m [emote interior angle theorem] m 15.5º + 1.5º m 8º (ough Figure) N S 70º 50º 6.7 cm G
9 6 / MT N S 70º 6.7 cm 50º G mark for rough figure mark for drawing SGN 1 mark for angle bisectors 1 mark for the incircle (iv) (7, 8) (x 1, y 1 ) ( 5, ) (x, y ) (3, 6) (x 3, y 3 ) Slope of line y y 1 x x Slope of line y 3 y x 3 x 6 3 ( 5) 4 3 5
10 7 / MT Slope of line and slope of line are equal and point is a common point for both the lines oints, and are collinear. (v) atio of the length, breadth and height of a cuboid is 5 : 4 : Let the common multiple be x Length of the cuboid 5x cm its breadth 4x cm and its height x cm Total surface area of a cuboid 116 cm Total surface area of a cuboid (l b + bh + l h) 116 [(5x) (4x) + (4x) (x) + (5x) (x)] 116 0x + 8x + 10x x x x 16 x 4 [Taking square roots] Length of the cuboid 5x 5 (4) 0 cm its readth 4x 4 (4) 16cm and its height x (4) 8 cm Dimensions of the cuboid are 0 cm, 16 cm and 8 cm..4. Solve NY TW of the following : (i) onstruction : Draw seg D line, - - D D 7. cm...(i) [The lengths of the two tangent 5 cm...(ii) segments to a circle drawn from an external point are equal]
11 8 / MT + [ - - ] [From (i) and (ii)] 1. cm... In D, m 90º m D 90º [adius is perpendicular to the tangent] m D 90º [onstruction] md 90 0 [emaining angle] D is a rectangle [y definition] D 7. cm...(iv) [pposite sides of a rectangle are D...(v) congruent] D + D [ - - D] D [From (ii) and (iv)] D 7. 5 D. cm...(vi) In D, m D 90º [onstruction] D + D [y ythagoras theorem] (1.) D + (.) [From and (vi)] D (1.) (.) D (1. +.) (1..) D D 144 D 1 cm [Taking square roots] 1 cm [From (v)] is a diameter of the circle adius of the circle is 6 cm (ii) (1, ), (4, 6), (3, 5) Let D and E be the midpoints of sides and respectively oint D is the midpoint of side D x1 x y1 y, 1 4 6, 5 8, 5, 4
12 9 / MT oint E is the midpoint of side E x1 x y1 y, , 4 7, 7, The required line passes through points D and E The equation of the line by two point form, x x1 y y1 x 1 x y 1 y x 5 y x 5 y x 5 (y 4) x 5 (y 4) x 5 y 8 x y x y The equation of the required line is x y ( mark for figure) seg and seg D represents the two buildings 30 m seg D represents the width of the road 45º E D 10 m 1 represents the position of observer. 30 m E is the angle of elevation m E 45º D DE is a rectangle 10 m DE 30 m [pposite sides of rectangle] D E 10 m In right angled E, E tan 45º [y definition] E
13 10 / MT 1 E 10 E 10 m D E + ED [ - E - D] D D 40 m The height of second building is 40 m..5. Solve NY TW of the following : (i) Given : In, m 90º To rove : ² ² + ² D onstruction : Draw seg D side such that - D -. roof : In, m 90º [Given] ( mark for figure) seg D hypotenuse [onstruction] ~ D ~ D...(i) [Similarity in right angled triangles] 1 ~ D [From (i)] [orresponding sides of similar triangles] D ² D...(ii) ~ D [From (i)] D [orresponding sides of similar triangles] ² D... dding (ii) and we get, ² + ² D + D ² + ² (D + D) ² + ² [ - D - ] ² + ² ² ² ² + ² (ii) (ough Figure) 3 cm 4 cm
14 11 / MT 3 cm 4 cm mark for 1 mark for constructing 7 congruent parts 1 mark for constructing mark for constructing 1 mark for required (i) Draw seg M side S 1 M S [adius is half of diameter] 10º cm seg M chord S [y construction] M 1 S [The perpendicular drawn from the centre of a circle to a chord bisec ts M the chord] 7
15 1 / MT M 7 3 cm In M, M 90º [y construction] M + M [y ythagoras theorem] M M M 49 M 7 cm [Taking square roots] rea of S 1 base height rea of S 1 S M (1.73) rea of S cm rea of sector (-S) r cm (ii) rea of segment S (rea of sector -S) (rea of S) cm Similarly we can prove, rea of segment cm Total area of two shaded segments cm rea of S is cm and total area of two shaded segments is 41.1 cm.
MT - MATHEMATICS (71) GEOMETRY - PRELIM II - PAPER - 3 (E)
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