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 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 3 4 6 E 4 6 3 E 8 units (iii) F + V E + F + 6 1 + F + 6 14 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 - ]
/ 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 3 1 300 3 100 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 360 3.14 90 sin 90 8 360
3 / MT PPER - 5 64 1.57 1 64 1.57 1 64 0.57 36.48 cm rea of shaded region rea of segment X 36.48 36.48 cm rea of shaded region is 36.48 cm. (iii) (37) 1369...(i) (1) + (35) 144 + 15 1369...(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 360 60 6 7 360 7 44 cm The curved surface area of the cheese is 44 cm.
(v) 4 / MT PPER - 5 In 9 m 90º [ngle of a rectangle] + [y pythagoras theorem] 1 + 9 1 X 144 + 81 5 15 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. 15 144 X. 15 X 144 15 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 / MT PPER - 5.3. 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 1 18...(i) r 6...(ii) dding (i) and (ii), we get r 1 + r 18 + 6 (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 180 1 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 / 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 / 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 / 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 6 + 8 QR 36 + 64 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 1 6 8 4 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 + 1 1 r + 4 1 r 3 1 r 1 r 1 1 r3 4
9 / MT PPER - 5 1 (r 1 + r r 3 ) + 4 1 3.14 (3 + 4 5 ) + 4 1 3.14 (9 + 16 5) + 4 1 3.14 (0) + 4 0 + 4 4 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 / 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 6 6 14 504 cm 3 ut, volume of ink filled in the cylindrical container 91% of 504 91 504 cm 3 100 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 1 1 10 1 10 1 1 cm 3 100 ut, volume of ink filled 84% of 1 100 84 1 cm 3 100 100 Number of refills that can be filled with ink Volume of ink filled in the cylindrical container Volume of ink filled in each refill 91 504 100 84 1 100 100 91 504 100 100 100 84 1 4550 Number of refills that can be filled with this ink is 4550.
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 + 180 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 + 360 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 180...(viii) m PQR + m PSR 360 180 [From (vii) and (viii)] m PQR + m PSR 180º PQRS is cyclic. [If opposite angles of a quadrilateral are supplementary, then quadrilateral is cyclic]