# SSC EXAMINATION GEOMETRY (SET-A)

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1 GRND TEST SS EXMINTION GEOMETRY (SET-) SOLUTION Q. Solve any five sub-questions: [5M] ns. ns. 60 & D have equal height ( ) ( D) D D ( ) ( D) Slope of the line ns. 60 cos D [/M] [/M] tan tan 60 cos cos 60 cos 60 cos60 ns.4 Diagonal of a square side 0 0 cm

2 ns.5 volume of a cube is =000 cm Guruanklan/SS Examination/Grand Test/Geometry/Set-/Sol. l 000 l 000 Taking cube root ns.6 RS is a tangent and Q in a chord, l 0cm QS M arcmq Tangent-Secant theorem... [/M] 0 given M arc MQ QS 0 QS 65 M S Q R Q. Solve any four sub-questions: [8M] 7 given 5 ns. sin sin cos Trigonometric Identity cos cos Taking square root 65 4 cos 5 ns. (i) Draw and angle of 5 [M] (ii) isect it [M] 5 5

3 ns. Guruanklan/SS Examination/Grand Test/Geometry/Set-/Sol. Y... [M] X' O 5 X The terminal arm lies in I-quadrant.... [M] ns.4 Length of an arc l 0cm ns.5 In QR, Rdius r 5 cm r rea of the sec tor lengthof an arc... [M] seg RS bisector RQ [given] 5 0 5cm R S [ngle bisector property]... [M] QR SQ QR S [/M] QR Q R 6 QR 0 ns.6 is a tangent and is a secant y tangent secant property... [M]

4 Q. Solve any three sub-questions. [9M] ns. D M arc Inscribed angle theorem D [/M] 0 D 05 or D 05 E [/M] D 80 [Opposite angle of a cyclic quadrilateral are supplementary]... [/M] D E [Exterior angle of cyclic quadrilateral is congruent to the angle opposite to its D 05 adjacent interior angle] ns. E 05 T T O cm 70 M cm 70 M (i) Draw a triangle (ii) isect any two sides (iii) ircumcircle [M] [M] [M]. In QR 0 Q 60 R cm 60 [/M] QR is triangle R Q [/M] R Qside opposite to 60 4

5 8 R 4 cm QR Qside opposite to 0 8 QR 4cm ns.4 sec x tan x L. H. S. a ba b a b sin cos cos x cos x cos x cos x sin sec x tan x sec, tan cos cos R.H.S L.H.S. R.H.S ns.5 Radius of sphere = r = 4. cm. 4 Volume of a sphere r

6 cm Surface area of a sphere 4 r cm Q.4 Solve any two sub-questions: [8M] ns. Given : circle with centre O Fig - [/M] and external point are given and are the two tangents drawn from an external point To prove : =... [/M] onstruction : Draw seg O, seg O and seg O... [/M] roof : In O and O O O O [Radii of the same circle] O [common side] O O 90 Tangent erpendicularity theorem ns. O O [Hypotenuse side test] = [orresponding sides of congruent triangle] Hence proved. Let represents the height of the first building = 0 cm D represents the height of the second building D represent the width of the road = 0 m 45 E 0m fig.[/m] 0m D 6

7 E 45 ngle of elevation Expln. [/M] In DE D DE 90 objects are perpendicular to the ground ED 90 E 90 Remaining angle DE is a rectangle [Definition of rectangle] E D 0m () ED 0m () opposite sides of a rectangle.... [/M] In E E tan 45 E E [From()] 0 E 0m () D E ED E D ns. 0 0 from & D 40m The height of the second building is 40 m Length of ther semicircular tunnel = km h = 000 m Diameter of the semicircular tunnel = 7 m Radius of the tunnel Volume of semicirluar tunnel r 7 r. 5m volumeof circular tunnel = r h

8 8500 m Expenditure for digging (semicircular) the tunnel at the rate Rs. 600 m urved surface area of semicirlular tunnel Rs rh rh m Expenditure for plastering inner side of the tunnel at the rate of Rs.50 per sq.m Rs Q. 5 Solve any two sub-questions. [0M] ns. Given :- In, ray E bisects To prove :- E E onstruction :- E 8 D fig. [/M] [/M] Draw a line parallel to ray E,. passing through the point. Extend so as to meet it at D. [/M] roof : - Line E line D and D is its transversal. E D () [onverse of corresponding angle test]...[/m] Line E Line D and as transversal E D () [onverse of alternate angles test]...[/m] ut E E () [given] ut D D [from (), () & ()] In D Side In D side D (4) [side opposite to congruent angles] seg E side D onstruction

9 E E (5) [.. T] D E E [from (4) & (5)] ns., 6 x, y 4, x, y divides seg internally in the ratio : m : n : Let x, y mx nx 0 x 0 m n 5... [M] Slope my 4 6 ny 0 y m n 5... [M] 0, Equation of the line by slope point form y y m x x... [ M] y x 0 y x ns. y 4 x x y 4 0 The equation of the required line is x y 4 0 9

10 (i) nalytical figure nalytical figure (ii) onstruct Q (iii) utting 7 arts & jointing Q 7 to (iv) omtruct RQ 5 Q 7 (v) onstruct R [M] [M] [M] [M] [M] 0

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