MT - GEOMETRY - SEMI PRELIM - II : PAPER - 4

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Milo Blake
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1 MT.1. ttempt NY FIVE of the following : (i) In STR, line l side TR S SQ T = RQ x 4.5 = x = MT - GEOMETRY - SEMI RELIM - II : ER - 4 Time : Hours Model nswer aper Max. Marks : [y..t.] x = x = 1.5 T l 4.5 x S 1.3 Q 3.9 R (ii) hords and D intersect each other at point outside the circle. = D 6 3 = 4 D = = 9 = 4.5 units (iii) F + V = E + F + 6 = 1 + F + 6 = 14 F = 14 6 F = 8

2 / MT ER - 4 (iv) (v) Given : Two circles with centres O and touch each other externally at point T. T O To rove : O = OT + T roof : O - T - [If two circles are touching circles then the common point lies on the line joining their centres] O = OT + T [ O - T - ] X In XYZ, ray YM bisects XYZ XY YZ = XM MZ Y Z [roperty of angle bisector of a triangle] 1 = XM [ XY = YZ] MZ XM = MZ (vi) Volume of cuboid = Volume of cube 3 9 x = (6) x = x = x = 8.. Solve NY FOUR of the following : (i) In O, m O = 90º [Radius is perpendicular O m O = 90º to the tangent] m = 90º m O = 90º [Remaining angle] O is a rectangle [y definition] seg O seg O [Radii of the same circle] O is a square [ rectangle in which adjacent sides M are congruent is a square]

3 3 / MT ER - 4 (ii) Mark point X as shown in the figure D is a square X side = 8 cm Radius (r) = side of a square r = 8 cm D Measure of arc () = 90º [ngle of a square] 8 cm sin rea of the segment X = r sin 90 = = = = = cm rea of shaded region = rea of segment X = = cm rea of shaded region is cm. (iii) (61) = (i) (11) + (60) = = (ii) (61) = (11) + (60) [From (i) and (ii)] The given sides form a right angled triangle. [y onverse of 1 ythagoras theorem] (iv) 60º 7 cm For a sector, Measure of arc () = 60º Radius (r) = 6 cm

4 4 / MT ER - 4 (a) urved surface area of the cheese = Length of arc height = r h = = 44 cm The curved surface area of the cheese is 44 cm. (v) (vi) Line is a secant intersecting the circle at points and and line is a tangent to the circle at point. = ² [Tangent secant property] 10 = (15)² 10 = 5 = 5 10 =.5 units = + [ - - ].5 = + 10 =.5 10 = 1.5 units 1 In, seg is the median + = + [y ppollonius theorem] 60 = (7) + 60 = (49) + 60 = = = 16 = 16 = 81 = 9 units [Taking square roots] = 1 9 = 1 [ is the midpoint of seg ] = 18 units

5 5 / MT ER Solve NY THREE of the following : (i) D is a trapezium side side D On transversal, O D [onverse D of alternate angles test] O DO...(i) [ - O - ] In O and OD, O DO [From (i)] O OD [Vertically opposite angles] 1 O ~ OD [y test of similarity] O O = O DO [c.s.s.t.] O O = O DO [y lternendo] 1 (ii) Diameters of circular ends of frustum are 18 cm and 8 cm r 1 = 18 = 9 cm and r = 8 = 4 cm 1 Slant height (l) = 13 cm urved surface area of frustum of frustum = (r 1 + r ) l = (9 + 4) 13 = = 169 cm Radius of a cylinder (r ) = 4 cm Its height (h) = 10 cm urved surface area of a cylinder = rh = 4 10 = 80 cm Surface area of tin required to make the funnel = urved surface area of frustum + curved surface area of cylinder = = 49 cm The surface area of the tin required to make the funnel is 49 cm. (iii) onstruction : Draw seg. roof : Take points R and S on the tangent at as shown in the figure line DE line RS On transversal D, ED DR [onverse of D E alternate angles test] R S

6 6 / MT ER - 4 ED R...(i) [ - D - ] R...(ii) [ngles in alternate segment] ED...(iii) [From (i) and (ii)] Similarly, we can prove that DE...(iv) In, seg seg...(v) [Isosceles triangle theorem] In DE, ED DE [From (iii), (iv) and (v)] seg D seg E [onverse of isosceles 1 triangle theorem] D = E (iv) In, = 5 units = 6 units = 7 units erimeter of QR = 360 units Q + QR + R = (i) ~ QR Q = QR = R 5 Q = 6 QR = 7 R 5 Q = 6 QR = 7 R = Q QR R 5 Q = 6 QR = 7 R = Q = 6 QR = 7 R = Q = 1 0 Q = 100 units 6 QR = 1 0 QR = 10 units 7 R = 1 0 R = 140 units [c.s.s.t.] [y theorem on equal ratios] [From (i)]...(ii) [From (ii)] [From (ii)] [From (ii)]

7 7 / MT ER - 4 (v) Q Let, = Q = x...(i) R [The lengths of the two tangent = R = y...(ii) segments to a circle drawn from R = Q = z...(iii) an external point are equal] = + [ - - ] = x + y...(iv) [From (i) and (ii)] Similarly, = y + z = 1...(v) = x + z...(vi) erimeter of = 44 cm + + = 44 x + y + y + z + x + z = 44 [From (iv), (v), and (vi)] x + y + z = 44 (x + y + z) = 44 x + y + z = x + 1 = [From (v)] x = 1 x = 10 = Q = 10 cm [From (i)] Length of a tangent segment from to the circle is 10 cm..4. Solve NY TWO of the following : (i) D Given : ~ QR. ( ) To rove : ( QR) = = = Q QR R onstruction : (i)draw seg D side, - D - (ii)draw seg S side QR, Q - S - R roof : ( ) ( QR) = D QR S Q [ The ratio of the areas of two triangles is equal to ratio of the products of a base and its corresponding height ] S R

8 8 / MT ER - 4 (ii) ( ) = D...(i) ( QR) QR S ~ QR Q QR...(ii) [c.s.s.t.] lso, Q...(iii) [c.a.s.t.] In D and SQ, D SQ [Each is a right angle] Q [From (ii)] D ~ SQ [y - test of similarity] D S Q...(iv) [c.s.s.t.] ( ) ( QR) = Q Q [From (i), (ii) and (iv)] ( ) ( QR) = Q Similarly we can prove...(vi) ( ) ( QR) = = QR R...(vii) ( ) ( QR) = Q = QR = R [From (vi) and (viii)] Given : D is a cyclic To rove : m + m D = 180º m D + m D = 180º roof : D m = 1 [Inscribed m (arc D)...(i) } angle m D = 1 m (arc )...(ii) theorem] 1 dding (i) and (ii), we get ( mark for figure) m + m D = 1 m (arc D) + 1 m (arc ) m + m D = 1 [m (arc D) + m (arc )] m + m D = 1 360º [ Measure of a circle is 360º] m + m D = 180º...(iii) In D, m D + m D + m + m D = 360º [ Sum of measure of angles of a quadrilateral is 360º] m D + m D + 180º = 360º [From (iii)] m D + m D = 180º

9 (iii) 10 cm 60 cm 9 / MT 10 cm 10 cm ER - 4 toy is a combination of cylinder, hemisphere and cone, each with radius 10 cm r = 10 cm Height of the conical part (h) = 10 cm Height of the hemispherical part = its radius = 10cm Total height of the toy = 60cm Height of the cylindrical part (h 1 ) = = 60 0 = 40 cm 1 l = r + h l = l = l = 00 l = 00 [Taking square roots] l = 10 cm Slant height of the conical part ( l) = 10 = = 14.1 cm Total surface area of the toy = urved surface area of the conical part + urved surface area of the cylindrical part + urved surface area of the hemispherical part = rl + rh 1 + r = r (l + h 1 + r) = ( ) = 31.4 ( ) = = cm Total surface area of the toy is cm..5. Solve NY TWO of the following : (i) onstruction :Draw seg E side, such that - D - E -. roof : is an equilateral triangle. = =...(i) [Sides of an equilateral triangle] In ED, m ED = 90º [onstruction] D E D² = E² + DE²...(ii) [y ythagoras Theorem]

10 10 / MT ER - 4 In E, m E = 90º [onstruction] m E = 60º [ngle of an equilateral triangle] m E = 30º [Remaining angle] E is a 30º - 60º - 90º triangle. y 30º - 60º - 90º triangle theorem, E = 3...(iii) [Side opposite to 60º] E = 1...(iv) [Side opposite to 30º] DE = E D [ - D - E] DE = 1 1 [From (iv) and Given] 3 DE = [From (i)] DE = 3 6 DE = 1...(v) 1 6 D² = [From (ii), (iii) and (v)] D² = 3 4 ² ² 7 ² + ² D² = 36 D² = 8 ² 36 D² = 7 9 ² 9D² = 7² (ii) Radius of the cylinder (r) = 1 cm 6.75 cm spherical iron ball is dropped into the cylinder and the water 0 cm level rises by 6.75 cm Volume of water displaced = volume of the iron ball Height of the raised water level (h) = 6.75 m Volume of water displaced = r h = cm 3

11 11 / MT ER - 4 Volume of iron ball = cm 3 ut, Volume of iron ball = 4 r = 4 3 r = r 4 r 3 = r 3 = r 3 = r = [Taking cube roots] r = 3 3 r = 9 Radius of the iron ball is 9 cm. 3 (iii) In, and Q are midpoint of seg and seg seg Q seg [y midpoint theorem] seg Q seg R [ - R - ] Similarly, seg QR seg RQ is a parallelogram [y definition] Q In S, m S = 90º seg S is median to hypotenuse S R S = 1...(i) [In a right angled triangle the median drawn to the hypotenuse is half of it] ut, = 1...(ii) [ is the midpoint of side ] In S, S = [From (i) and (ii)] m S = m S [Isosceles triangle theorem] RQ is a parallelogram m R = m QR [Opposite angles of a parallelogram are congruent] m S = m QR...(iv) [ - S - R] m S = m QR...(v) [From (iii) and (iv)] ut, m S + m SR = 180º [Linear pair axiom] 1 m QR + m SR = 180º [From (v)] QRS is cyclic [If opposite angles of a 1 quadrilateral are supplementary then it is a cyclic quadrilateral]

MT - GEOMETRY - SEMI PRELIM - II : PAPER - 5

017 1100 MT MT - GEOMETRY - SEMI PRELIM - II : PPER - 5 Time : Hours Model nswer Paper Max. Marks : 40.1. ttempt NY FIVE of the following : (i) X In XYZ, ray YM bisects XYZ XY YZ XM MZ Y Z [Property of