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

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1 MT MT - GEOMETRY - SEMI PRELIM - II : PPER - 5 Time : Hours Model nswer Paper Max. Marks : ttempt NY FIVE of the following : (i) X In XYZ, ray YM bisects XYZ XY YZ XM MZ Y Z [Property of angle bisector of a triangle] 1 XM [ XY YZ] MZ XM MZ (ii) hords and intersect each other at point E inside the circle E E E E E E E E 8 units (iii) F + V E + F F F 14 6 F 8 M (iv) Given : Two circles with centres O and T touch each other externally at point T. O To Prove : O OT + T Proof : O - T - [If two circles are touching circles then the common point lies on the line joining their centres] O OT + T [ O - T - ]

2 / MT PPER - 5 (v) (vi) ( E) ( ) E [Triangles with common base] ( E) ( ) 6 E 9 ( E) ( ) 3 cylinder and cone have equal height and equal radii Volume of cone 1 volume of cylinder cm 3 Volume of the cone is 100 cm 3... Solve NY FOUR of the following : (i) m 1 m 1 80 m(arc P) [Tangent secant theorem] m 40º P m 1 m (arc Q) [Inscribed angle theorem] Q 30 1 m (arc Q) m (arc Q) 30 m (arc Q) 60º (ii) Mark point X as shown in the figure is a square X side 8 cm Radius (r) side of a square r 8 cm Measure of arc () 90º [ngle of a square] 8 cm sin rea of the segment X r sin

3 3 / MT PPER cm rea of shaded region rea of segment X cm rea of shaded region is cm. (iii) (37) (i) (1) + (35) (ii) (37) (1) + (35) [From (i) and (ii)] The given sides form a right angled triangle. [y onverse of 1 Pythagoras theorem] (iv) 6 cm 60º 7 cm For a sector, Measure of arc () 60º Radius (r) 6 cm (a) urved surface area of the cheese Length of arc height r h cm The curved surface area of the cheese is 44 cm.

4 (v) 4 / MT PPER - 5 In 9 m 90º [ngle of a rectangle] + [y pythagoras theorem] X cm [Taking square roots] m 90º [ngle of a rectangle] line is a tangent to the circle at point [ line perpendicular to the radius at its outer end is a tangent to the circle] Line is a tangent and line X is a secant intersecting at points X and X. [Tangent secant property] 1 X X. 15 X X 9.6 cm (vi) In, seg Q is the median Q Q 1 Q Q 1 10 Q Q Q 5 units...(i) + Q + Q [y ppollonius theorem] 1 Q + (5) [From (i) and given] 1 Q + (5) 1 Q + 50 Q 1 50 Q 7 Q 36 [Taking square roots] Q 6 units

5 5 / MT PPER Solve NY THREE of the following : (i) In, seg PQ side P P Q Q...(i) [y.p.t.] Q R 1 In, seg PR side P P R R...(ii) [y.p.t.] 1 In, Q Q R R [From (i) and (ii)] seg QR side [y converse of.p.t.] 1 (ii) urved surface area of the frustum of a cone 180 cm Perimeters of circular bases are 18 cm and 6 cm r (i) r 6...(ii) dding (i) and (ii), we get r 1 + r (r 1 + r ) 4 (r 1 + r ) 4 (r 1 + r ) 1...(iii) 1 urved surface area of the frustum of a cone (r 1 + r ) l 180 (r 1 + r ) l l [From (iii)] 1 l 15 cm Slant height of the frustum of a cone is 15 cm. P (iii) (1 mark for figure) E F E...(i) [ngles in alternate segment] 1 F...(ii) ut,...(iii) [ Ray bisects ] E F [From (i), (ii) and (iii)] 1

6 6 / MT PPER - 5 (iv) P (v) In PQR, seg PT is the median Q T R PQ + PR PT + QT...(i) [y ppollonius theorem] In PQT, m PQT 90º PT PQ + QT [y Pythagoras theorem] QT PT PQ...(ii) 1 PQ + PR PT + (PT PQ ) [From (i) and (ii)] PQ + PR PT + PT PQ PR 4PT PQ PQ PR 4PT 3PQ onstruction : raw seg. M N Proof : M is cyclic [y definition] m M + m M 180º...(i) [Opposite angles of a cyclic quadrilateral are supplementary] 1 N is cyclic [y definition] M N...(ii) [The exterior angle of a cyclic quadrilateral is equal to its interior opposite angle] 1 m M + m N 180º [From (i) and (ii)] m M + m N 180º [ - - ] seg M seg N [y Interior angles test] 1.4. Solve NY TWO of the following : (i) P Given : ~ PQR. ( ) To Prove : ( PQR) PQ QR PR Q S R

7 7 / MT onstruction : (i)raw seg side, - - (ii)raw seg PS side QR, Q - S - R Proof : ( ) ( PQR) ( ) ( PQR) QR PS QR PS...(i) [ The ratio of the areas of two triangles is equal to ratio of the products of a base and its corresponding height ] PPER - 5 (ii) ~ PQR PQ QR...(ii) [c.s.s.t.] lso, Q...(iii) [c.a.s.t.] In and PSQ, PSQ [Each is a right angle] Q [From (ii)] ~ PSQ [y - test of similarity] PS PQ...(iv) [c.s.s.t.] ( ) ( PQR) PQ PQ [From (i), (ii) and (iv)] ( ) ( PQR)...(vi) PQ Similarly we can prove ( ) ( PQR) QR...(vii) PR ( ) ( PQR) PQ QR PR [From (vi) and (viii)] Given : is a cyclic To Prove : m + m 180º m + m 180º Proof : m 1 m (arc )...(i) [Inscribed } 1 m 1 angle m (arc )...(ii) theorem] dding (i) and (ii), we get ( mark for figure) m + m 1 m (arc ) + 1 m (arc ) m + m 1 [m (arc ) + m (arc )]

8 8 / MT PPER - 5 m + m 1 360º [ Measure of a circle is 360º] m + m 180º...(iii) In, m + m + m + m 360º [ Sum of measure of angles of a quadrilateral is 360º] m + m + 180º 360º [From (iii)] m + m 180º (iii) iameter PR 6 units Its radius (r 1 ) 3 units iameter PQ 8 units Its radius (r ) 4 units In PQR, m RPQ 90º...(i) [ngle subtended by a semicircle] QR PR + PQ [y Pythagoras theorem] QR QR QR 100 QR 10 units [Taking square roots] iameter QR 10 units Its radius (r 3 ) 5 units PQR is a right angled triangle [From (i)] R P Q ( PQR) 1 1 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 1 r r r 3 1 r 1 r 1 1 r3 4

9 9 / MT PPER (r 1 + r r 3 ) ( ) ( ) (0) sq. units rea of shaded portion is 4 sq.units.5. Solve NY TWO of the following : (i) E L M In EL and L, EL L [From (i) and E - L - ] LE L [Vertically opposite angles] EL ~ L [y test of similarity] 1 EL L E is a parallelogram...(i) [c.s.s.t.] seg seg [y definition] seg E seg [ - - E] On transversal E, E E...(ii) [onverse of alternate angles test] In ME and M, side M side M ME M [Vertically opposite angles] EM M [From (ii) and - - E, - M - E] ME M [y S test of congruence] 1 E...(iii) [c.s.c.t.] ut,...(iv) [Opposite sides of a parallelogram] E...(v) [From (iii) and (iv)] EL L E [From (i) and E - - ] 1

10 10 / MT PPER - 5 EL L [From (v)] EL L EL L 1 EL L (ii) Height of the cylindrical container (h) 14cm Its radius (r) 6 cm Volume of cylindrical container r h cm 3 ut, volume of ink filled in the cylindrical container 91% of cm Length of ball pen refill (h 1 ) 1m its inner diameter mm Its radius (r 1 ) 1 mm 1 10 cm Volume of the refill r 1 h cm ut, volume of ink filled 84% of cm Number of refills that can be filled with ink Volume of ink filled in the cylindrical container Volume of ink filled in each refill Number of refills that can be filled with this ink is 4550.

11 11 / MT PPER - 5 (iii) Given : (i) is cyclic. Q R (ii) Ray P, ray Q, ray R and P 1 S ray S are the bisectors of,, and respectively. To Prove : PQRS is cyclic. Proof : P P [ray P bisects ] Let, m P m P aº...(i) Similarly, m P m P bº...(ii) m R m R cº...(iii) m R m R dº...(iv) In Q, m Q + m Q + m Q 180º [Sum of the measures of angles of a triangle is 180º] m Q + b + c 180 [From (ii) and (iii)] m Q (180 b c)º m PQR (180 b c)º...(v) [ - P - Q and - R - Q] Similarly, we can prove m PSR (180 - a - d)º...(vi) dding (v) and (vi), m PQR + m PSR 180 b c a d m PQR + m PSR 360 a b c d m PQR + m PSR 360 (a + b + c + d)...(vii) In, m + m + m + m 360º [ Sum of the measures of angles of a quadrilateral is 360º] m P + m P + m P + m P m Q + m Q + m R + m R [ngle addition property] a + a + b + b + c + c + d + d 360 [From (i), (ii), (iii) and (iv)] a + b + c + d 360 (a + b + c + d) 360 a + b + c + d (viii) m PQR + m PSR [From (vii) and (viii)] m PQR + m PSR 180º PQRS is cyclic. [If opposite angles of a quadrilateral are supplementary, then quadrilateral is cyclic]

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

MT - GEOMETRY - SEMI PRELIM - II : PAPER - 4 017 1100 MT.1. ttempt NY FIVE of the following : (i) In STR, line l side TR S SQ T = RQ x 4.5 = 1.3 3.9 x = MT - GEOMETRY - SEMI RELIM - II : ER - 4 Time : Hours Model nswer aper Max. Marks : 40 4.5 1.3