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10 Solved Paper SSC Maharashtra Exam March 207 Class - X Geometry Time : 2 Hours Max. Marks : 40 Note : (i) Solve all questions. Draw diagrams wherever necessary. (ii) Use of calculator is not allowed. (iii) Diagram is essential for writing the proof of the theorem. (iv) Marks of constructions should be distinct. They should not be rubbed off.. Solve any five sub-questions : [5] (i) In the following figure, seg E seg and seg seg D. If E = 6 and D = 9, find ( E ). ( D) E (ii) If two circles with radii 8 cm and 3 cm respectively touch internally, then find the distance between their centres. (iii) Find the height of an equilateral triangle whose side is 6 units. (iv) If the angle θ = 45, find the value of tan θ. (v) Find the slope and y-intercept of the line y = 3x 5. (vi) Find the circumference of a circle whose radius is 7 cm. 2. Solve any four sub-questions : [8] (i) In PQR, seg RS is the bisector of PRQ, PS = 6, SQ = 8, PR = 8, Find QR. P D S (ii) In the given figure P = 9, P = 6 and PC = 2. Find PD. C Q R P (iii) Draw C of measure 5 and bisect it. (iv) Find the sine ratio of θ in standard position whose terminal arm passes through (5, 2). D

11 0 OSWL SSC Maharashtra Question ank, Geometry Class - X (v) Find the slope of the line passing through the points P (, ) and Q ( 2, 5). (vi) The dimensions of a cuboid in cm are Find its volume. 3. Solve any three sub-questions : [9] (i) Prove that, "If the angles of a triangle are , then each of the perpendicular sides is 2 times the hypotenuse." (ii) Find the angle between two radii at the centre of the circle as shown in the figure. Lines P and P are tangents to the circle at other ends of the radii and PR = 20. O 20 P R S (iii) Construct tangents to the circle from the point, having radius 3.2 cm and centre 'C'. Point is at a distance 7 cm from the centre. (iv) From the top of a lighthouse, an observer looks at a ship and finds the angle of depression to be 60. If the height of the lighthouse is 87 metres, then find how far is that ship from the lighthouse? ( 3 =.73) (v) The volume of a cube is 729 cm 3. Find its total surface area. 4. Solve any two sub-questions : [8] (i) Prove that, "The opposite angles of a cyclic quadrilateral are supplementary". (ii) Eliminate θ, if x = 3 cosec θ + 4 cot θ, y = 4 cosec θ 3 cot θ. (iii) toy is a combination of a cylinder, hemisphere and a cone, each with radius 0 cm as shown in the figure. Height of the conical part is 0 cm and total height is 60 cm. Find the total surface area of the toy. (π = 3.4, 2 =.4) 5. Solve any two sub-questions : [0] (i) In the given figure, D is the bisector of the exterior of DC. Seg D intersects the side C produced in D. Prove that : D CD = C

12 Solved Paper K C D (ii) Construct the circumcircle and incircle of an equilateral XYZ with side 6.5 cm and centre O. Find the ratio of the radii of incircle and circumcircle. (iii) (5, 4), ( 3, 2) and C (, 8) are the vertices of a triangle C. Find the equation of median D and line parallel to passing through point C. Solutions. (i) ( E) = 2 E and ( D) = 2 D ( E) ( D) E = 2 D 2 = E D ½ = 6 9 (given D = 9, E = 6) = 2 3 ½ (ii) Since both the circles are touching internally, Therefore, the centres of touching circles and their point of contact are always collinear 8 C 2 3 C T let R = 8 cm and R 2 = 3 cm Distance between their centres = R R 2 = 8 3 = 5 cm. (iii) height of equilateral triangle = 3 2 side Given side = 6 units h = = 3 3 units.

13 2 OSWL SSC Maharashtra Question ank, Geometry Class - X (iv) tan ( 45 ) = tan 45 = (v) Given line : y = 3x 5 Comparing the given line with y = mx + c, we get m = 2 and c = 5 Slope = 2 ½ and y-intercept = 5 ½ (vi) circumference of circle = 2πr = 2 π 7 (given r = 7) = = 44 cm. 2. (i) y angle bisector property, we get PR QR = PS QS Given, PS = 6, SQ = 8 and PR = 8 8 QR = 6 8 or QR = or QR = 24. (ii) We have P P = CP PD Given P = 9, P = 6 and PC = = 2 PD or PD = or PD = 4.5 (iii) C O 5 2 (iv) Here, terminal arm is O Using pythagoras theorem in O, we get Y (5, 2) O X O 2 = O O 2 = or O 2 = or O 2 = 69 O = 3

14 Solved Paper Now, sin θ = O = 2 3 (v) Slope of a line passing through two points is given by m = y y x x 2 2 Given, x =, x 2 = 2, y = and y 2 = 5 m = 5 ( ) 2 or m = 6 3 = 2 (vi) Volume of cuboid = l b h Given, l = 20 cm, b = 8 cm and h = 0 cm volume = = 3600 cm 3 3. (i) Given : In C, = 45, C = 45, = 90 ½ To prove that : = C = 2 C C Proof : In C, = C (each is 45 ) = C (sides opposite to equal angle)...(i) ½ In C, = 90 (given) C 2 = 2 + C 2 (y pythagoras theorem)...(ii) Form (i) and (ii), ½ C 2 = ½ C 2 = = 2 C2 = 2 C (Taking square root on both sides)...(iii) From (i) and (iii) ½ = C = C 2 Hence Proved. ½ (ii) Lines P and P are the tangents to the circle and seg O and O are the radii of the circle at the point of contact OP = OP = 90...(i) (radius is to the tangent at point of contact) lso, P = 80 PR = [ PR = 20 (given)] = 60...(ii) ½ In the quadrilateral OP, PO + O + OP + P = O = 360 (From (i) and (ii)) O = 360 ( ) or O = or O = 20 ½

15 4 OSWL SSC Maharashtra Question ank, Geometry Class - X (iii) P 3.2 M 7 Q nalytical figure Steps of Construction : () Draw a circle with centre and radius 3.2 cm. (2) Take point at a distance of 7 cm from the centre. (3) Obtain mid-point M of seg by drawing perpendicular bisector. Draw a circle with centre M and radius M. (4) Let P and Q be points of intersection of these two circles. (5) Draw line P and Q which are the required tangents. P Q (iv) The height of the lighthouse is 87 meters. The observer is at and the ship is at C. The angle of depression = 60 Let, the distance of ship from the lighthouse be x. P C, PC = C = 60 In right angled triangle C. 60 P 2 87 x 60 C

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