MAHESH TUTORIALS. Time : 1 hr. 15 min. Q.1. Solve the following : 3


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1 S.S.. MHESH TUTRILS Test  II atch : S Marks : 30 Date : GEMETRY hapter : 1,, 3 Time : 1 hr. 15 min..1. Solve the following : 3 The areas of two similar triangles are 18 cm and 3 cm respectively. What is the ratio of their corresponding sides? Two circles with centres and having diameter 5 cm and 15 cm respectively touch each other externally at, then what is the distance between and? If the circumcircle lies in the interior of the triangle, then it is... angled triangle... Solve the following : 6 From the information given in the adjacent figure, state whether the triangles are similar with reasons. In the adjoining figure, if m (arc ) = 60º and m = 80º Find (a) (b) m (arc ). 80º R onstruct an arc M such that seg M of length 6. cm subtends an angle of 40º on it..3. Solve the following : 9 onstruct the incircle of SRN, such that RN = 5.9 cm, RS = 4.9 cm, R = 95º.
2 D is a trapezium in which D and its diagonals intersect each other at the point. Show that = D. In the adjoining figure, and are centers of two circles touching each other at M. Line and line D are tangents. If D = 6 cm and = 9 cm then find the length of seg and seg D. D M.4. Solve the following : 1 In the adjoining figure, Y and Y divides the triangular region into two equal areas. Determine :. Y is inscribed in a circle with centre, seg is a diameter of the circle with radius r. seg D seg. rove that (a) ~ D, (b) () = abc 4r. (a is side opposite to,...) D onstruct YZ such that Y = 9.5 cm, ZY = 115º, Z is median. Z = 3.3 cm. est f Luck
3 S.S.. MHESH TUTRILS Test  II atch : S Marks : 30 Date : GEMETRY hapter : 1,, 3 MDEL NSWER ER Time : 1 hr. 15 min..1. Solve the following : Let ~ R ( ) ( R) = [reas of similar triangles] 18 3 = [Given] 9 16 = = 3 4 [Taking square roots] : = 3 : 4 Diameter of the cicle with centre = 5 cm Its radius (r 1 ) = 1.5 cm Diameter of the circle with centre = 15 cm Its radius (r ) = 7.5 cm oth the circles touch each other externally at point Distance between and = r 1 + r = Distance between and = 0 cm If the circumcircle lies in the interior of the triangle, then it is 1 acute angled triangle... Solve the following : = = 1 R = 10 5 R = R
4 R = 8 4 R = 1 In and R,... = R = R [From, and ] R [y SSS test of similarity] (a) m = 1 m(arc ) [Inscribed angle theorem] m = 1 60 m = 30º (b) m = 1 80º m(arc ) [Inscribed angle theorem] 80 = 1 m(arc ) m(arc ) = 80 m(arc ) = 160º (Rough Figure) 40º 40º 6. cm M 80º 50º 50º 6. cm M mark for rough figure mark for drawing base angles mark for drawing an arc mark for drawing M
5 Solve the following : (Rough Figure) S S 4.9 cm 4.9 cm R 95º 5.9 cm N 95º R 5.9 cm N mark for rough figure mark for drawing SRN 1 mark for drawing the angle bisectors 1 mark for drawing the incircle D is a trapezium side side D [Given] n transversal, D [onverse of alternate angles test] D D... [   ] In and D, D [From ] D [Vertically opposite angles] ~ D [y test of similarity] 1 = D [c.s.s.t.] = D [y lternendo] D  M  [If two circles are touching circles then the common point lies on the line joining their centres] M
6 M = D = 6 cm... M = = 9 cm... [Radii of the same circle] = M + M [  M  ] = [From and ] = 15 cm... In, m = 90º [Radius is perpendicular to the tangent] ² = ² + ² [y ythagoras theorem] 15² = ² + 9² [From and ] 5 = ² + 81 ² = 5 81 ² = 144 = 1 cm [Taking square roots] In D, m D = 90º [Radius is perpendicular to the tangent] = D + D [y ythagoras theorem] 15 = 6 + D [From and ] 5 = 36 + D D = 5 36 D = 189 D = 9 1 D = 3 1 cm. [Taking square roots] The lengths of seg and seg D are 1 cm and 3 1 cm respectively..4. Solve the following : seg Y divides in two parts of equal areas (Y) = 1 () ( Y) ( ) = 1... Y seg Y side [Given] n transversal, Y... [onverse of corresponding angles test] In Y and, Y [From ] Y [ommon angle] Y ~ [y test of similarity]
7 ( Y) ( ) = [reas of similar triangles] 1 1 = = = 1 = 1 = 1 1 = 1 [From ] [Taking square roots] [   ] 1 1 = 1 = seg is a diameter [Given] arc is a semi circle. m = 90º [ngle subtended by D a semicircle] In and D D [Each is 90º] D [ngles inscribed in the same arc and  D  ] ~ D [y test of similarity] 1 = a, = c, = b D = c D = r b D = bc r [Given] [c.s.s.t.]...
8 () = 1 base height = 1 D = 1 a bc r () = abc 4r [From ] (Rough Figure) Z 115º 3.3 cm Z 115º Z 9.5 cm Y 3.3 cm 5º 9.5 cm 5º Y 130º mark for rough figure mark for drawing centre 1 mark for drawing the perpendicular bisector 1 mark for locating point Z 1 mark for drawing ZY
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