04 00 eat No. MT - MTHEMTI (7) GEOMETY - PELIM II - PPE - (E) Time : Hours (Pages 3) Max. Marks : 40 Note : (i) ll questions are compulsory. Use of calculator is not allowed. Q.. olve NY FIVE of the following : 5 M (i) Lines PM and PN are tangents to the circle with centre O. P O If PM 7 cm, find PN. Using Euler s formula, find F, if V 6 and E. N (iii) (iv) (v) If x-coordinate of point is negative and y-coordinate is positive, then in which quadrant point lies? If m 5 and c 3, then write the equation of the line. The area of a circle is 34 cm and the area of its minor sector is 3.4 cm. Find the area of its major sector. (vi) Find the value of 3sin + 3cos. Q.. olve NY FOU of the following : p q 8 (i) In the adjoining figure, line l line m line n. Lines p and q are transversals. l 8 From given information m find T. 0 n T
/ MT PPE - is a right angled at. D is any point on. DE. If D 6 cm, cm, 8 cm, find E. D E (iii) (iv) (v) In the adjoining figure, seg and seg D are chords of the circle. be a point on tangent to the circle at point. If m (arc P) 80º and D 30º, then find (i) m (arc QD) Draw a tangent at any point M on the circle of radius.9 cm and centre O. Eliminate, if, x a sec, y b tan D P Q (vi) If sin + sin, prove that cos + cos 4. Q.3. olve NY THEE of the following : 9 (i) In the adjoining figure, LMN 90º and LKN 90º, seg MK seg LN. Prove that is the midpoint of seg MK. L M N In the adjoining figure, two circles intersect each other in two points and. eg is the chord of both circles. Point is the exterior point of both the circles on the line. From the point tangents are drawn to the circles touching at M and N. Prove that M N. K M N (iii) onstruct the incircle of N, such that N 5.9 cm, 4.9 cm, 95º.
3 / MT PPE - (iv) (v) Write down the equation of a the line whose slope is 3 and which passes through P where P divides the line segment joining (, 6) and (3, 4) in the ratio : 3. The radius of a circle is 3.5 cm and area of the sector is 3.85 cm. Find the length of the corresponding arc and the measure of arc. Q.4. olve NY TWO of the following : 8 (i) (iii) uppose and are equal chords of a circle and a line parallel to the tangent at intersects the chords at D and E. Prove that D E. Find the equation of the straight line passing through the origin and the point of intersection of the lines x + y 7 and x y 4. Eliminate, if x cos 3 sin, y cos + sin Q.5. olve NY TWO of the following : 0 (i) Prove : The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. H ~ VU, In H, H 4.5 cm, H 5. cm, 5.8 cm and H V 3 ; construct VU. 5 (iii) In the adjoining figure, P and Q are two diameters of the circle. If P 8 cm and P 4 3 cm, find (i) rea of triangle OP The total area of two shaded segments. ( 3.73) P Q 0º O est Of Luck
04 00 eat No. MT - MTHEMTI (7) GEOMETY - PELIM II - PPE - (E) Time : Hours Prelim - II Model nswer Paper Max. Marks : 40.. ttempt NY FIVE of the following : (i) M PM PN [Length of the two tangent segments from an external point to a circle are equal] ut, PM 7 cm [Given] PN 7 cm F + V E + F + 6 + F + 6 4 F 4 6 F 8 (iii) 30º [Given] sin sin ( 30) sin 30 P N O sin ( 30) (iv) m 5, c 3 y slope point form, the equation of line is y mx + c y 5x 3 5x y 3 0
/ MT PPE - (v) rea of major sector rea of circle rea of minor sector 34 3.4 8.6 cm The area of the major sector is 8.6 cm. (vi) 3sin + 3cos 3 (sin + cos ) 3 () [ sin + cos ] 3.. olve NY FOU of the following : (i) p q line l line m line n On transversals p and q, T 8 0 T T 0 8 [Given] [ y Property of Intercepts made by three parallel lines] [Given] T 5 units l m n 8 0 T In and ED, DE [ommon angle] ED [ Each is 90º] ~ ED [y test of similarity] E E E 8 6 D [c.s.s.t.] [Given] E 6 8 E 4 units D
3 / MT PPE - (iii) m m(arc P) D [Tangent secant theorem] Q m 80 P m 40º m D m (arc QD) [Inscribed angle theorem] 30 m (arc QD) m (arc QD) 30 m (arc QD) 60º (iv) (ough Figure) O.9 cm M O.9 cm M mark for rough figure mark for circle mark for drawing perpendicular (v) x a sec sec x a y b tan tan y b...(i)...
4 / MT PPE - + + tan sec x a y b + y b y b x a x a [From (i) and ] (vi) sin + sin² [Given] sin sin² sin cos sin + cos sin cos sin cos 4 [quaring both sides] cos cos 4 cos² + cos 4 sin + cos cos sin.3. olve NY THEE of the following : (i) In LMN, m LMN 90º [Given] M seg M hypotenuse LN [Given] M L N...(i) [y property of geometric mean] L N In LKN, m LKN 90º [Given] seg K hypotenuse LN [Given] K K L N... [y property of geometric mean] M K [From (i) and ] M K [Taking square roots] is the midpoint of seg MK Line is a secant intersecting the circle at points and and line M is a tangent to the circle at point M. M²...(i) [Tangent secant property] Line is a secant intersecting the M N circle at points and and line N is a tangent to the circle at point N.
5 / MT PPE - N²... [Tangent secant property] M² N² [From (i) and ] M N [Taking square roots] (iii) (ough Figure) 4.9 cm O 4.9 cm O 95º 5.9 cm N 95º N 5.9 cm mark for rough figure mark for drawing N mark for drawing the angle bisectors mark for drawing the incircle (iv) (, 6), (3, 4) Point P divides seg internally in the ratio : 3 Let, P (x, y) y section formula for internal division, x mx nx my + ny m + n y m + n (3) + 3 ( ) ( 4) 3 6 + 3 3 6 6 8 + 8 5 5 0 0 5 5 0 P (0, ) The line having slope 3 passes through the point P (0, ) The equation of the line by slope point form is,
6 / MT PPE - (y y ) m (x x ) (y ) 3 (x 0) (y ) 3x y 4 3x 3x y + 4 0 The equation of the required line is 3x y + 4 0 (v) adius of a circle (r) 3.5 cm rea of the sector 3.85 cm rea of sector r l 3.85 3.5 l 3.85 3.5 l 385 0 00 35 l l 0 l. cm rea of sector r 360 3.85 3.5 3.5 360 7 3.85 35 35 360 7 0 0 385 35 00 360 0 385 360 0 00 35 36º Length of arc is. cm and measure of an arc is 36º..4. olve NY TWO of the following : (i) onstruction : Draw seg. Proof : Take points and on the tangent at as shown in the figure line DE line [Given] D E On transversal D, ED D [onverse of alternate angles test] ( marks for figure)
7 / MT PPE - ED...(i) [ - D - ]... [ngles in alternate segment] ED...(iii) [From (i) and ] imilarly, we can prove that DE...(iv) In, seg seg [Given]...(v) [Isosceles triangle theorem] In DE, ED DE [From (iii), (iv) and (v)] seg D seg E [onverse of isosceles triangle theorem] D E Let line x + y 7 and x y 4 intersect at point x + y 7...(i) x y 4... ubtracting from (i), x + y 7 x y 4 ( ) (+) ( ) 3y 3 y ubstituting y in equation, x 4 x 4 + x 5 (5, ) The straight line passes through (5, ) and O (0, 0) The equation of the line by two point form, x x y y x x y y x 5 y 5 0 0 x 5 y 5 x 5 5 (y ) x 5 5y 5 x 5y 5 + 5 0 x 5y 0 The equation of the line passing through the origin and the point of intersection of the lines x + y 7 and x y 4 is x 5y 0.
8 / MT PPE - (iii) x cos 3 sin...(i) y cos + sin... Multiplying by, y cos + 4 sin...(iii) ubtracting (iii) from (i), x y cos 3 sin ( cos + 4 sin ) x y cos 3 sin cos 4 sin x y 7 sin sin (x y) 7...(iv) ubstituting sin y in equation 7 x y y cos + 7 x y y cos 7 x y y + cos 7 7y x y cos 7 7y x 4y cos cos We know, 7 x 7 3y sin + cos (x y) x 3y 7 7 (x y) (x 3y) 49 49 Multiplying throughout by 49, (x y) + (x + 3y) 49.5. olve NY TWO of the following : (i) Given : ~ PQ. To Prove : ( ) ( PQ) Q PQ P
9 / MT PPE - onstruction : (i) Draw seg D side, - D - Draw seg P side Q, Q - - ( mark for figure) P Proof : ( ) ( PQ) D Q P...(i) D Q ~ PQ [ The ratio of the areas of two triangles is equal to ratio of the products of a base and its corresponding height ] [Given]... [c.s.s.t.] PQ Q P lso, Q...(iii) [c.a.s.t.] In D and PQ, D PQ [Each is a right angle] Q [From ] D ~ PQ [y - test of similarity] D D...(iv) [c.s.s.t.] P Q PQ D...(v) [From and (iv)] P Q ( ) ( PQ) ( ) ( PQ) D Q P [From (i)] [From (v)] Q Q ( ) ( PQ)...(vi) Q (Δ) (ΔPQ) ² Q² ² PQ² ² [From and (vi)] P²
0 / MT PPE - (ough Figure) U U H V 5.8 cm 5. cm 4.5 cm H V 3 mark for H mark for constructing 5 congruent parts mark for constructing V 5 H 3 mark for constructing UV H mark for required VU 4 5 (iii) Draw seg OM side P OP P M P [adius is half of diameter] 0º OP 8 Q OP 4 cm seg OM chord P [y construction] PM P [The perpendicular drawn from the centre of a circle to a chord bisec ts the chord] O
/ MT PPE - PM 4 3 PM 7 3 cm In OMP, OMP 90º [y construction] OM + PM OP [y Pythagoras theorem] OM + 7 3 4 OM OM 96 47 49 OM 7 cm [Taking square roots] rea of OP base height rea of OP P OM 4 3 7 49 3 49 (.73) rea of OP 84.77 cm rea of sector OP r 360 0 4 4 360 7 66 3 05.33 cm rea of segment P rea of sector OP rea of OP 05.33 84.77 0.56 cm imilarly we can prove, rea of segment Q 0.56 cm Total area of two shaded segments 0.56 + 0.56 4. cm rea of OP is 84.77 cm and total area of two shaded segments is 4. cm.