Time : 2 Hours (Pages 3) Max. Marks : 40. Q.1. Solve the following : (Any 5) 5 In PQR, m Q = 90º, m P = 30º, m R = 60º. If PR = 8 cm, find QR.

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1 Q.P. SET CODE Q.1. Solve the following : (ny 5) 5 (i) (ii) In PQR, m Q 90º, m P 0º, m R 60º. If PR 8 cm, find QR. O is the centre of the circle. If m C 80º, the find m (arc C) and m (arc C). Seat No MT - z MTHEMTICS (71) GEOMETRY - PPER D (E) Time : Hours (Pages ) Max. Marks : 40 (iii) The terminal arm is in II quadrant, what are the possible angles? O C (iv) line has the equation y x. State its y-intercept. (v) C ~ PQ, if ( C) ( PQ) 1 4, find C PQ. (vi) Find the area of a circle with radius 7 cm. Q.. Solve the following : (ny 4) 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. (ii) In the adjoining figure, if m (arc PC) 60º and m C 80º Find (a) C (b) m (arc QC). Q 80º C P

2 / MT - z (iii) Prove : sec + cosec sec. cosec (iv) (v) (vi) Check whether points (4,-5), (7,8) and (-,-) are collinear. sin 4 cos 0, then find the values of tan and sec where is an acute angle. Write the equation x+y-7 0 in double intercept form and write x-intercept and y-intercept. Q.. Solve the following : (ny ) 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. (ii) In the adjoining figure, line is tangent to both the circles touching at and. O 9, P 18, OP 61 then find. 9 O P (iii) (iv) (v) Draw tangents to the circle with centre P and radius.9 cm. From a point Q which is at a distance 8.8 cm from the centre. Show that the line joining ( 1, 1) and ( 9, 6) is parallel to the line joining (, 14) and (6, 9). Construct a circumcircle of C such that 5 cm, C 1 cm, C 90º. Q.4. Solve the following : (ny ) 8 (i) Prove : The lengths of the two tangent segments to a circle drawn from an external point are equal. (ii) (iii) LMN ~ XYZ, In LMN, LM 6 cm, MN 6.8 cm, LN 7.6 cm and LM XY 4 ; construct XYZ. roller of diameter 0.9 m and length 1.8 m is used to press the ground. Find the area of ground pressed by it in 500 revolutions. (Given.14)

3 / MT - z Q.5. Solve the following : (ny ) 10 (i) Prove : In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the remaining two sides. (ii) person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60º. When he moves 40 m away from the bank, he finds the angle of elevation to be 0º. Find the height of the tree and the width of the river. ( 1.7) (iii) In the adjoining figure, PR 6 units and PQ 8 units. Semicircles are draw taking sides PR, RQ and PQ as diameters as shown in the figure. Find out the area of the shaded portion. (.14) R P Q est Of Luck

4 .P. SET CODE MT - z MTHEMTICS (71) GEOMETRY - PPER D (E) Time : Hours Prelim - I Model nswer Paper Max. Marks : 40 Q.1. Solve the following : (ny 5) (i) In PQR, m P 0º m R 60º [Given] m Q 90º PQR is a 0º - 60º - 90º triangle y 0º - 60º - 90º triangle theorem, QR 1 PR [Side opposite to 0º] QR 1 8 QR 4 cm (ii) m C 1 m (arc C) [y Inscribed angle theorem] 80º 1 m (arc C) m (arc C) 160º m (arc C) 60º m (arc C) m (arc C) 00º Y (iii) The terminal arm is in II quadrant, the angle is in between 90º and 180º if the initial arm rotates anticlockwise direction or the angle is between 70º and 180º if the initial arm rotates clockwise. X O X 1 Y (iv) Equation of the line is y x Comparing the given equation with slope-intercept form y mx + c, c y intercept of the line is.

5 / MT - z (v) C ~ PQ [Given] ( C) ( PQ) C PQ [reas of similar triangles] 1 4 C PQ [Given] C PQ 1 [Taking square roots] (vi) Radius of circle (r) 7 cm rea of the circle r The area of a circle is 154 cm cm Q.. Solve the following : (ny 4) (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 [Triangles with common base] h h h 15 h 7.5 The corresponding height of the smaller triangle is 7.5 cm. (ii) (a) m C 1 m(arc PC) [Inscribed angle theorem] m C 1 60 m C 0º (b) m C 1 m(arc QC) [Inscribed angle theorem]

6 / MT - z 80 1 m(arc QC) m(arc QC) 80 m(arc QC) 160º (iii) L.H.S. sec + cosec 1 1 cos sin 1 1 sec, cos ec cos sin sin + cos cos. sin 1 cos. sin [ sin + cos 1] sec. cosec R.H.S. sec + cosec sec. cosec (iv) Let, (4, 5) (x 1, y 1 ) (7, 8) (x, y ) C (, ) (x, y ) y y Slope of line x x Slope of line C ( 5) y y x x Slope of line and slope of line C are not equal. The points (4, 5), (7, 8) and (, ) are not collinear.

7 (v) sin 4 cos 0 sin 4 cos sin cos 4 tan 4 4 / MT - z 1 + tan sec sec sec sec sec sec 5 [Taking square roots] (vi) x + y 7 0 x + y 7 Dividing throughout by 7, x y x y x intercept of line x + y 7 0 is 7 1 y intercept of line x + y 7 0 is 7

8 5 / MT - z Q.. Solve the following : (ny ) (i) 17 cm D 11 cm O CD is a parallelogram [Given] O OD 1 D...(i) [ Diagonals of parallelog ram bisec t each other] O OD 1 6 [Given] O OD 1 cm In D, seg O is the median [From (i) and by definition] + D O + O [y ppollonius theorem] (11) + (17) O + (1) O + (169) 410 O O 7 O O 6 O 6 cm [Taking square roots] O 1 C [ Diagonals of parallelog ram bisec t each other] 6 1 C C 1 cm Length of other diagonal is 1 cm. (ii) Construction : Draw seg PM seg O, - M - O. Proof : In PM, m P 90º m M 90º [Radius is perpendicular to the tangent] m PM 90º [Construction] m MP 90º [Remaining angle] 9 M O P C

9 (iii) 6 / MT - z PM is a rectangle [y definition] P M 18 units [Opposite sides of a rectangle] O OM + M [ - M - O] 9 OM + 18 OM 9 18 OM 11 units In PMO, m PMO 90º [Construction] OP OM + PM [y Pythagoras theorem] (61) (11) + PM PM PM PM 600 PM 60 units ut, PM [Opposite sides of a rectangle] 60 units (Rough Figure).9 cm P.9 cm 8.8 cm Q.9 cm P.9 cm M 8.8 cm Q mark for rough figure mark for drawing the circle of radius.9 cm mark for drawing the perpendicular bisector of seg PQ mark for drawing the circle with centre M 1 mark for drawing both the tangents from point Q

10 7 / MT - z (iv) Let, ( 1, 1), ( 9, 6), C (, 14), D (6, 9) 1 Slope of a line y y x x 1 Slope of side ( 1) Slope of line 5 8 Slope of line CD ( ) 5 6 Slope of line CD 5 8 Slope of line and slope of line CD are equal. line line CD The line joining ( 1, 1) and ( 9, 6) is parallel to the line joining (, 14) and (6, 9). (v) (Rough Figure) 5 cm 1 cm C

11 8 / MT - z 5 cm O 1 cm C mark for rough figure mark for drawing C 1 mark for perpendicular bisectors 1 mark for the circumcircle Q.4. Solve the following : (ny ) (i) Given : (i) circle with centre O. (ii) P is a point in the exterior of the circle.

12 9 / MT - z (iii) Points and are the points of contact of the two tangents from P to the circle. [ mark for figure] To Prove : P P O P Construction : Draw seg O, seg O and seg OP. Proof : In PO and PO, m PO m PO 90º[Radius is perpendicular to the tangent] Hypotenuse OP Hypotenuse OP [Common side] seg O seg O [Radii of same circle] PO PO [y hypotenuse - side theorem] seg P seg P [c.s.c.t] P P (ii) nalysis : LMN ~ XYZ [Given] LM XY MN YZ LN XZ 4... (i) [c.s.s.t.] LM XY 4 [From (i)] MN YZ 4 [From (i)] LN XZ 4 [From (i)] 6 XY YZ XZ XY YZ.8 4 XZ XY 4.5 cm YZ 5.1 cm XZ 5.7 cm Information for constructing XYZ is complete.

13 10 / MT - z (Given triangle) L marks for nalysis mark for drawing LMN 1 mark for XYZ 6 cm 7.6 cm M 6.8 cm N (Required triangle) X 4.5 cm 5.7 cm Y 5.1 cm Z (iii) Diameter of the roller 0.9 m its radius (r) m its length (h) 1.8 m Curved surface area of the roller rh m rea of the ground pressed by the roller in 1 revolution curved surface area of roller rea of the ground pressed in one revolution m rea of the ground pressed in 500 revolution m rea of the ground pressed by the roller is 54.4 m.

14 11 / MT - z Q.5. Solve the following : (ny ) [ mark for figure] (i) Given : In C, m C 90º To Prove : C² ² + C² D Construction : Draw seg D side C such that - D - C. Proof : In C, C m C 90º [Given] seg D hypotenuse C [Construction] C ~ D ~ DC...(i) [Similarity in right angled triangles] C ~ D [From (i)] D C [Corresponding sides of similar triangles] ² C D...(ii) C ~ DC [From (i)] C DC C C [Corresponding sides of similar triangles] C² C DC...(iii) dding (ii) and (iii) we get, ² + C² C D + C DC ² + C² C (D + DC) ² + C² C C [ - D - C] ² + C² C² C² ² + C² (ii) Let seg represents the tree seg C represents width of river Let C x m C and D represents the initial and final positions of the observer DC 40 m C and D are the angles of elevation m C 60º and m D 0º In right angled C, tan 60º C D 0º 60º 40 m C [ mark for figure] [y definition] 1 x x m...(i) In right angled D, tan 0º D [y definition]

15 1 40 x 40 x m From (i) and (ii) we get, x 40 x x 40 + x x x 40 x 40 x 0 C 0 m...(ii) 0 m [From (i)] m 1 / MT - z Height of tree is 4.6 m and width of river is 0 m. (iii) Diameter PR 6 units Its radius (r 1 ) units Diameter PQ 8 units Its radius (r ) 4 units P In PQR, R Q 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] Diameter QR 10 units Its radius (r ) 5 units PQR is a right angled triangle [From (i)] (PQR) 1 1 product of perpendicular sides PR PQ sq. units.

16 1 / MT - z 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 1 1 r 1 r 1 1 r 4 1 (r 1 + r r ) ( ) ( ) (5 5) (0) sq. units rea of the shaded portion is 4 sq. units.

Time : 2 Hours Preliminary Model Answer Paper Max. Marks : 40. [Given] [Taking square roots]

Time : 2 Hours Preliminary Model Answer Paper Max. Marks : 40. [Given] [Taking square roots] .P. SET CODE MT - w 05 00 - MT - w - MTHEMTICS (7) GEOMETRY - (E) Time : Hours Preliminary Model nswer Paper Max. Marks : 40.. ttempt NY FIVE of the following : (i) BC ~ PQ [Given] ( BC) ( PQ) BC PQ [reas