MT EDURE LTD. EXTR HOTS SUMS HTER : 1 - SIMILRITY GEOMETRY 1. isectors of and in meet each other at. Line cuts the side at Q. Then prove that : + Q roof : In Q, ray bisects Q [Given] Q y property of an angle...(i) Q bisector of a triangle In Q, ray bisects Q [Given] Q Q y property of an angle...(ii) Q bisector of a triangle Q Q Q [From (i) and (ii)] + Q Q + Q [y Theorem on equal ratios] + Q [ - Q - ]. In QR, QR 90 0, s shown S in figure, seg QS side R. M seg QM is angle bisector of QR. rove that : M² MR² S SR roof : In QR, seg QM bisects QR [Given] Q M MR Q [r operty of an angle QR bisec tor of a triangle] M MR ² QR²...(i) [Squaring both sides] In QR, m QR 90º [Given] seg QS hypotenuse R [Given] QR ~ SQ ~ QSR...(ii) [Theorem on similarity of right angled triangles] SQ ~ QR [From (ii)] Q QR S [orresponding sides Q of similar triangles] Q² R S...(iii) lso, QSR ~ QR [From (ii)] QR R SR [orresponding sides QR of similar triangles] QR² R SR...(iv) M R S R SR MR M MR S SR [From (i), (iii) and (iv)] SHOOL SETION 41 x x R
GEOMETRY MT EDURE LTD.. Find the radius of a circle drawn by a compass when angle between two arms of compass is 10 0 and length of each arm is 4cm. Sol. In the adjoining figure, seg and seg represents the arms of compass. In, side side [Given]...(i) [Isosceles triangle theorem] In, m + m + m 180º [ Sum of the measures of the angles of a triangles is 180º] 10º + m + m 180º [From (i)] m 180º 10º m 60º m 0º...(ii) Draw seg D side, - D -. In D, m D 0º [From (ii) and - D - ] m D 90º [Given] m D 60º [Remaining angle] D is a 0º - 60º - 90º triangle. y 0º - 60º - 90º triangle theorem. D [Side opposite to 60º] D 4 D 1 cm Similarly, we can get D 1 cm D + D [ - D - ] 1 + 1 4 cm. The radius of the circle is 4 cm. 4. In, 90º, segments D, seg E and seg F are medians. rove : (D² + E² + F² ) ². roof : In, seg D is median on side. [Given] E ² + ² D² + D² [y ppollonius theorem] ² + ² D² D² ² + ² 1 D² [ D is the midpoint of seg ] 1 ² + ² D² 4 F 4 cm 10 0 D D 4 cm ² + ² 1 ² D²...(i) 4 SHOOL SETION
MT EDURE LTD. GEOMETRY Multiplying throughout by, we get + 4D Similarly, we can prove that ² + ² ² 4E²...(ii) ² + ² ² 4F²...(iii) dding (i), (ii) and (iii) we get + + ² + ² ² + ² + ² ² 4D + 4E + 4F + + 4D + 4E + 4F ( + + ) 4 (D + E + F )...(iv) In, m 90º [Given] +...(v) [y ythagoras theorem] ( + ) 4 (D + E + F ) [From (iv) and (v)] 4 (D + E + F ) 6 4 (D + E + F ) (D + E + F ) [Dividing throughout by ] (D² + E² + F²) ² 5. In D points, Q, R and S lies on sides,, D and D respectively such that seg S seg D seg QR and seg Q seg SR. Then prove that seg Q seg. S D roof : seg S seg D [Given] On transversal, S D...(i) [onverse of corresponding angles test] R In S and D, Q S D [ommon angles] S D [From (i)] S D [y - test of similarity] S D S...(ii) [c.s.s.t.] D seg QR seg D [Given] On transversal, QR D...(iii) [onverse of corresponding angles test] In QR and D, QR D [ommon angles] QR D [From (iii)] QR D [y test of similarity] Q QR D R D In QRS,...(iv) [c.s.s.t.] seg Q seg RS [Given] seg S seg QR [Given] QRS is a parallelogram [y definition] S QR...(v) Q S D R D Q Q...(vi) [From (iv) and (v)] [From (ii) and (vi) [y Invertendo] SHOOL SETION 4
GEOMETRY MT EDURE LTD. Q Q In, Q Q Q Q [y Dividendo]...(vii) [ - - and - Q - ] [From (vii)] seg Q seg [y converse of..t.] 6. In, m 90º. seg DE side, seg DF side, F prove (EDF) E E F F E roof : In D, m D 90º [Given] seg D side [Given] D DE E E...(i) [y property of geometric mean] In D, m D 90º [Given] seg DF side [Given] DF F F...(ii) [y property of geometric mean] Multiplying (i) and (ii), DE DF E E F F DE DF E E F F...(iii) In EDF, m EF m ED m FD 90º [Given] m EDF 90º [Remaining angle] EDF is a rectangle [y definition] (EDF) DE DF...(iv) From (iii) and (iv), (EDF) E E F F [From (iii) and (iv)] HTER : - IRLE 1. From the end points of a diameter of circle perpendiculars are drawn to a tangent of the same circle. Show that their feet on the tangent are equidistant from the centre of the circle. Given : (i) circle with centre O. (ii) seg is the diameter of the circle. D (iii) Line l is tangent to the circle at point. (iv) seg D line l. (v) seg E line l. To rove : OD OE. onstruction : Draw seg O. O roof : seg D line l [Given] seg O line l [Radius is E perpendicular to the tangent] seg E line l [Given] l seg D seg O seg E [erpendiculars drawn to the same line are parallel to each other] On transversal and DE, O O D...(i) [y property of intercepts made by E three parallel lines] 44 SHOOL SETION
MT EDURE LTD. GEOMETRY ut, O O [Radii of the same circle] O O 1...(ii) D 1 E D E...(iii) [From (i) and (ii)] In OD and OE, seg O seg O [ommon side] OD OE [Each is a right angle] seg D seg E [From (iii)] OD OE [y SS test of congruence] seg OD seg OE [c.s.c.t.] OD OE. The bisectors of the angles, of intersect in I, the bisectors of the corresponding exterior angles intersect in E. rove that IE is cyclic. I roof : Take points and Q as shown in the figure. m + m 180º [Linear pair axiom] 1 m + 1 m 1 180º[Multiplying throughout by 1 ] m I + m E 90º [ Ray I and ray E bisects and respectively] m IE 90º...(i) [ngle addition property] Similarly, m IE 90º...(ii) m IE + m IE 90º + 90º [dding (i) and (ii)] m IE + m IE 180º IE is cyclic [If opposite angles of a quadrilateral are supplementary then quadrilateral is cyclic]. In the adjoining figure, 0 0 line is a tangent to a circle with centre O at point. O seg F is angle bisector of. rove that : seg seg E. E roof : F E E [Ray E bisects ] Let, m E m E x...(i) [ngles in alternate segment] Let, m m y...(ii) m E m + m E [ngle addition property] m E (y + x)...(iii) [From (i) and (ii)] E is a exterior angle of E, m E m E + m E [Remote interior angle theorem] m E m + m E [ - E - ] m E (y + x)...(iv) [From (i) and (ii)] In E, E E [From (iii) and (iv)] seg seg E [onverse of isosceles triangle theorem] SHOOL SETION 45 ) ) E Q
GEOMETRY MT EDURE LTD. 4. In the adjoining figure, is diameter of a circle with centre O, seg is tangent to the circle at point. line JD touches circle at point D, and intersects segment in point J. rove that : seg J seg J. roof : Take a point M on line DJ such that M - D - J. J seg J seg DJ...(i) [The lengths of two tangent segments from an external point to a circle are equal] m ODM 90 0...(ii) [Radius is perpendicular to the tangent] In OD, seg O seg OD [Radii of the same circle] OD OD [Isosceles triangle theorem] Let, m OD m OD xº...(iii) m ODM m OD + m DM [ngle addition property] 90 x + m DM [From (ii) and (iii)] mdm (90 x)º...(iv) ut, DM JD...(v) [Vertically opposite angles] m JD (90 x)º...(vi) [From (iv) and (v)] In, m 90º [Radius is perpendicular to tangent] m xº [From (iii) and - O -, - D - ] m (90 x)º [Remaining angle] m JD (90 x)º...(vii) [ - D - and - J - ] In JD, JD JD [From (vi) and (vii)] seg DJ seg J...(viii) [onverse of Isosceles triangle theorem] seg J seg J [From (i) and (viii)] 5. If four tangents of a circle determine a rectangle then show that it must be a square. Given : (i) Lines,, D and D are the tangents to the circle at points, Q, R and S respectively (ii) D is a rectangle. S Q To rove : D is a square roof : S...(i) [The lengths of the two D R Q...(ii) tangent segments from an R Q...(iii) external point to a circle DR DS...(iv) are equal] dding (i), (ii), (iii) and (iv), we get + + R + DR S + Q + Q + DS ( + ) + (R + DR) (S + DS) + (Q + Q) + D D +...(v) [ - -, - Q -, - R - D, - S - D] D is a rectangle [Given] D...(vi) D...(vii) O M [Opposite sides of a rectangle are congruent] D 46 SHOOL SETION
MT EDURE LTD. GEOMETRY + + [From (v), (vi) and (vii)] D is a square [ rectangle in which adjacent sides are congruent, is a square] 6. Two concentric circles with centre O. Seg, seg and seg are the tangents to the smaller circle at points, Q and R respectively and also they are chords of the bigger circle. rove that seg Q seg, Q 1. roof : R Q With respect to smaller circle, seg O seg...(i) [Radius is perpendicular to the seg OQ seg...(ii) tangent] With respect to bigger circle, seg O chord [From (i)]...(iii) [erpendicular drawn from centre of circle to chord bisects the chord] seg OQ chord [From (ii) Q Q...(iv) erpendicular drawn from centre of circle to chord bisects the chord] In, and Q are midpoints of sides [From (iii) and (iv)] and respectively. seg Q seg [y Midpoint theorem] Q 1 HTER : - GEOMETRI ONSTRUTION 1. oint I is the incentre of, I 10º, 4 cm, median cm. Draw. nalysis : Let m I m I x and m I m I y m x and m y [ngle addition property] In I, m I + m I + m I 180º 10 + x + y 180 x + y 180 10 x + y 60...(i) In, m + m + m 180 m + x + y 180 m + (x + y) 180 m + (60) 180 [From (i)] m + 10 180 m 180 10 m 60º Now, can be constructed with base, vertical angle and median. SHOOL SETION 47 O
GEOMETRY MT EDURE LTD. (Rough Figure) 60º cm O 10º 0º 0º 4 cm I 10º. oint O is the orthocentre of, m O 105º, seg D seg. 6.5 cm and D.5 cm. Draw. nalysis : O EOF [Vertically oppsoite angles] m O m EOF 105º In FOE, m FE + m FO + m EOF + m EO 60º [ngle sum property of a quadrilateral] m FE + 90 + 105 + 90 60 m FE + 85 60 (Rough Figure) m FE 60 85 m FE 75º m 75º [ - F -, - E - ] F E Now, O can be constructed with base, vertical angle and altitude D. 105º D 6.5 cm Y 75º.5 cm 150º 15º 15º 6.5 cm X.5 cm 48 SHOOL SETION
MT EDURE LTD. GEOMETRY. onstruct such that 8.8 cm, 50º, radius of incircle of is. cm. (Rough Figure) I 50º. cm M 8.8 cm Q 5 cm 4.8 cm O 5.8 cm Steps of construction : 1. Draw seg 8.8 cm.. t, draw m 50º.. Draw bisector of, as incentre lies on angle bisector. 4. Draw a line parallel to side at a distance of. cm from. 5. oint of intersection of line parallel to and angle bisector is incentre. Let incentre be I. 6. From I, draw seg IM side. seg IM is in-radius. 7. Draw incircle with IM as radius to touch sides and. 8. From draw the tangent to the circle to meet ray at. 4. Draw a sector O-X with radius 7 cm and m (arc X) 50º. Draw a circle touching the sides O and O and also the arc. (Rough Figure) 7 cm I X O 50º 7 cm SHOOL SETION 49
GEOMETRY MT EDURE LTD. X I 50º O Q 7 cm Steps of construction : 1. Draw m O 50º and arc of radius 7 cm.. Draw bisector of O. It intersects arc at X.. t X, draw the Q ray OX to cut ray O at and ray O at Q. 4. Draw bisector of Q. It intersects ray OX at I. 5. Draw incircle with I as centre and IX as radius. This circle touches the ray O, ray O and also arc X. 5. In, 5.8 cm, seg seg, seg Q seg, 5 cm, Q 4.8 cm. onstruct. (Rough Figure) Q 5 cm 5.8 cm 4.8 cm I Y. cm. cm 50º X 8.8 cm M 50 SHOOL SETION
MT EDURE LTD. GEOMETRY Steps of construction : 1. Draw seg of length 5.8 cm.. Draw a semicircle with seg as the diameter.. Taking as the centre and radius 5 cm cut an arc on the semicircle to get point and draw seg. 4. Taking as the centre and radius 4.8 cm cut an arc on the semicircle to get point Q and draw seg Q. 5. Extend seg Q and seg to intersect at point. is the required triangle. 6. Draw a line l. Take a point at a distance 5cm from line l. Draw a circle with radius cm such that the circle touches the line l and passes through point. (Rough Figure) cm cm 5 cm N O cm 5 cm N O M T cm l M T Steps of construction : 1. Draw line l.. Take a point M on line l and draw a perpendicular to line l at point M.. With point M as the centre, cut an arc of radius 5 cm on the perpendicular to get point. 4. With point M as the centre and radius cm cut an arc on seg M to get point N. 5. Draw a line m perpendicular to line M at point N. 6. With point as the centre cut an arc of radius cm on line m to get point O. 7. With point O as the centre and seg O as the radius, draw the required circle. 8. Draw a perpendicular from point O to line l to get point T. HTER : 4 - TRIGONOMETRY 1 1. If 1 + x sin x, prove that tan + cot x +. x roof : 1 + x sin x sin sin x 1 + x x [Squaring both sides] 1 + x SHOOL SETION 51
GEOMETRY MT EDURE LTD. sin + cos 1 cos 1 sin cos 1 cos cos x 1 + x 1 x x 1 + x 1 1 + x tan sin cos x 1 1 x 1 x x 1 x 1 x 1 x 1 cot tan 1 x L.H.S. tan + cot x + 1 x R.H.S. tan + cot 1 x + x tan cot. rove : + 1 + tan + cot 1 cot 1 tan tan cot roof : L.H.S. + 1 cot 1 tan sin cos cos sin 1 1 cos sin sin cos sin sin cos cos cos sin cos sin sin cos sin sin cos cos cos sin cos sin cos sin sin cos cos (sin cos ) sin (cos sin ) sin cos cos (sin cos ) sin (sin cos ) 1 sin cos (sin cos ) cos sin 1 sin cos (sin cos ) cos sin 1 (sin cos ) (sin sin. cos cos ) (sin cos ) cos sin 5 SHOOL SETION
MT EDURE LTD. GEOMETRY sin + sin. cos + cos cos. sin sin sin. cos cos cos. sin cos. sin cos. sin sin cos 1 cos sin tan + 1 + cot 1 + tan + cot R.H.S. tan cot + 1 + tan + cot 1 cot 1 tan. rove : sin 8 cos 8 (sin cos ) (1 sin cos ). roof : L.H.S. sin 8 cos 8 (sin 4 ) (cos 4 ) (sin 4 cos 4 ) (sin 4 + cos 4 ) (sin cos ) (sin + cos ) (sin 4 + cos 4 ) (sin cos ) (sin 4 + cos 4 ) [ sin + cos 1] (sin cos ) (sin 4 + cos 4 + sin cos sin cos ) (sin cos ) [(sin + cos ) sin cos ] (sin cos ) (1 sin cos ) [ sin + cos 1] R.H.S. sin 8 cos 8 (sin cos ) (1 sin cos ). 4. 1.5 m tall boy is standing at some distance from a 0 m tall building. The angle of elevation from his eyes to the top of building increases from 0º to 60º as he walks towards the building. Find the distance he walked towards the building. Sol. Let the distance he walked towards the building x m Height of tower () 0 m Height of boy (D) 1.5 m ut D EF G G 1.5 m G G G 0 1.5 G 8.5 m In right angled GF, tan 60º G GF 8.5 GF GF 8.5 GF 8.5 GF 8.5 GF 9.5 m SHOOL SETION 5 8.5 m G 1.5 m 60º E [y definition] 0º F 1.5 m x D 1.5 m
GEOMETRY MT EDURE LTD. In right angled GD, tan 0º G GD 1 8.5 9.5 x 9.5 x 8.5 x 8.5 9.5 x 19 [y definition] Distance he walked towards the building is 19 m. 5. rove : (sin + cosec ) + (cos + sec ) 7 + tan + cot. roof : L.H.S (sin + cosec ) + (cos + sec ) sin + sin. cosec + cosec + cos + cos. sec + sec (sin + cos ) + (cosec ) + (sec ) + sin. cosec + cos. sec 1 + (1 + cot ) + (1 + tan ) + sin 1 sin + cos 1 cos 1 + 1 + cot + 1 + tan + + 7 + tan + cot R.H.S. (sin + cosec ) + (cos + sec ) 7 + tan + cot. 6. rove that : roof : L.H.S. 7. 1 sin. cos sin cos sin cos (sec cosec ) sin + cos 1 sin. cos sin cos cos (sec cosec ) sin + cos 1 sin. cos (sin cos ) (sin cos ) 1 1 (sin cos ) (sin sin. cos cos ) cos cos sin (1 sin. cos ) (sin cos ) sin cos (1 sin. cos ) cos [ sin + cos 1] sin. cos 1 sin cos sin cos sin sin sin cos sin cos sin R.H.S. 1 sin. cos sin cos sin cos (sec cosec ) sin + cos tan sec - 1 + roof : L.H.S. tan sec + 1 tan sec - 1 + cosec tan sec +1 54 SHOOL SETION
MT EDURE LTD. GEOMETRY sin cos 1 + -1 cos sin cos (1 cos ) + cos sin cos 1 +1 cos sin cos (1 + cos ) cos sin 1 cos + sin 1 + cos 1 1 sin + 1 cos 1 + cos 1 + cos + 1 cos sin (1 cos ) (1+cos ) sin 1 cos² sin sin² sin cosec R.H.S. tan tan sec - 1 + sec + 1 cosec sin² + cos² 1 1 - cos sin ² ² 6. From the top of a light house, 80 metres high, two ships on same side of light house are observed. The angles of depression of the ships as seen from the lighthouse are found to be of 45 0 and 0 0. Find the distance between the two ships (ssume that the two ships and the bottom of the lighthouse are in a line). Sol. In the adjoining figure, seg represents the lighthouse. is the position of the observer D and are the position of the ships. Draw ray E seg D. ED and E are the angles of depression. m ED 0º and m E 45º On transversal D D m ED m D 0º On transversal m E m 45º In right angled D, tan 0 D 1 80 D D 80 m...(i) 0 0 45 0 SHOOL SETION 55 E 0 0 45 0 80 m [onverse of alternate angles test] [onverse of alternate angles test] [y definition]
GEOMETRY MT EDURE LTD. In right angled, tan 45 [y definition] 80 1...(ii) 80 m D + D [ D - - ] 80 80 + D [From (i) and (ii)] D 80 80 D 80 1 m The distance between the two ships is 80 1 m. HTER : 5 - O-ORDINTE GEOMETRY 1. (8, 5), (9, 7), ( 4, ) and D (, 6) are the vertices of a D. If, Q, R and S are the midpoints of sides,, D and D respectively. Show that QRS is a parallelogram, using the slopes. Sol., Q, R and S are the midpoints of side,, D and D of D. (8, 5), (9, 7), ( 4, ) and D (, 6) y midpoint formula, x1 x y1 y, 8 9, 5 7 17, 1 Q 9 4, 7 5, 5 R 4 6, ( 1, 4) S 8, 5 6 11 5, 5 ( 1) Slope of Q y x Slope of QR Slope of RS Slope of S y x 1 1 5 17 5 4 1 5 7 1 1 7 11 4 5 ( 1) 11 ( 1) 17 5 6 1 7 1 4 1 7 1 1 4 Slope of Q Slope of RS seg Q seg RS...(i) Slope of QR slope of S seg QR seg S...(ii) QRS is a parallelogram. [From (i), (ii) and by definition] 56 SHOOL SETION
MT EDURE LTD. GEOMETRY. Find the equations of the lines which through the point (, 4) and the sum of whose intercepts on the axes is 14. Sol. Let the intercepts made by the lines on the co-ordinate axes be a and b respectively. a + b 14...(i) The equation of the line is x y 1 a b Since the line passes through the point (, 4) 4 1 a b b + 4a ab...(ii) From (i), a 14 b Substituting a 14 b in (ii) we get, b + 4 (14 b) (14 b)b b + 56 4b 14b b b 15b + 56 0 b 8b 7b + 56 0 b (b 8) 7 (b 8) 0 (b 8) (b 7) 0 b 8 OR b 7 y (i) when b 8, c 14 8 6 and when b 7, c 14 7 7 Equations of the required lines are x y 1 and x y 1 6 8 7 7 4x + y 4 and x + y 7 4x + y 4 0 and x + y 7 0. Find the equation of a line which passes through the point (, 7) and makes intercepts on the co-ordinate axes which are equal in magnitude but opposite in sign. Sol. Let the intercepts made by the line on the co-ordinate axes be a and b. a b...(i) The equation of the line is x y 1 a b x y 1 b b x + y b This line passes through the point (, 7) ( ) + 7 b b 10 The equation of the line is x + y 10 x y + 10 0 4. Find the equation of a line which contains the point (4, 1) and whose x-intercept is twice its y-intercept. Sol. Let the intercepts made by the line on the co-ordinate axes be a and b respectively. a b The equation of the line is x y 1 a b x y 1 b b SHOOL SETION 57
GEOMETRY MT EDURE LTD. x + y b Since this line contains the point (4, 1) 4 + (1) b 6 b b Equation of the required line is x + y 6 x + y 6 0 5. Find the equation of side of an isosceles, if the equation of side is x y 4 0 and (4, 0) and (6, 4) are the extremities of the base. Sol. Let (h, k) Since lies on side i.e. on x y 4 0 h k 4 0 h k + 4 (k + 4, k) Since is an isosceles triangle with as base, l () l () (k 4 4) (k 0) (k 4 6) (k 4) On squaring both sides, k + k (k ) + (k 4) k + k k 4k + 4 + k 8k + 16 0 1k + 0 1k 0 k 0 1 k 5 k + 4 5 4 17 17, 5 Equation of side by two point form is x x1 x x y y 1 y y 1 x 6 17 6 1 y 4 5 4 x 6 y 4 17 6 5 4 x 6 y 4 1 7 x 6 y 4 1 7 7 (x 6) y 4 7x 4 y 4 7x y 8 0 58 SHOOL SETION
MT EDURE LTD. GEOMETRY 6. Find the equations of the line which cut off intercepts on the axes whose sum is 1 and product is 6. Sol. Let the intercepts made by the line on the co-ordinates axes be a and b respectively. a + b 1...(i) and ab 6...(ii) From (ii) b 6 a Substituting this in (i) we get, a 6 a 1 a 6 a a a 6 0 a a + a 6 0 a (a ) + (a ) 0 (a ) (a + ) 0 a or a y (i) when a, b 1 and when a, b 1 ( ) Now, equation of the line making intercepts a and b is x y 1 a b Equations of the required lines are x y 1 and x y 1 x y 6 and x + y 6 x y 6 0 and x y + 6 0 HTER : 6 - MENSURTION 1. tinmaker converts a cubical metallic box into 10 cylindrical tins. Side Sol. of the cube is 50 cm and radius of the cylinder is 7 cm. Find the height of each cylinder so made if the wastage of 1% is incurred in the process. (Given 7 ). Side of the cubical metallic box (l) 50 cm Total surface area of cubical box 6l 6 (50) 6 500 15000 cm Wastage incurred in the process of making 10 cylindrical tins 1% of 15000 1 15000 100 1800cm rea of metal sheet used to make 10 cylindrical tins Total surface area of cubical box Wastage incurred in the process 15000-1800 100 cm rea of metal sheet used to make each cylindrical tin 100 10 10 cm SHOOL SETION 59
GEOMETRY MT EDURE LTD. Radius (r) 7 cm rea of metal sheet used to make each cylindrical tin Total surface area of cylinder Total surface area of cylinder r (r + h) 10 7 (7 + h) 7 10 (7 + h) 10 7 + h 0 7 + h h 0 7 h cm Height of each cylinder is cm.. The three faces,, of a cuboid have surface area 450 cm, 600 cm and 00 cm respectively. Find the volume of the cuboid. Sol. Surface area of face 450 cm Surface area of face l h l h 450...(i) Surface area of face 600cm Surface area of face l b l b 600...(ii) Surface area of face 00 cm Surface area of face b h b h 00...(iii) Multiplying (i), (ii) and (iii), l h l b b h 450 600 00 l b h 450 00 00 l b h 900 00 00 [Taking square roots] l b h 0 00 l b h 9000cm ut, Volume of the cuboid l b h Volume of the cuboid 9000cm Volume of the cuboid is 9000 cm.. Oil tins of cuboidal shape are made from a metallic sheet with length 8 m and breadth 4 m. Each tin has dimensions 60 40 0 in cm and is open from the top. Find the number of such tins that can be made. Sol. Length of the metallic sheet (l) 8 m 8 100 800 cm its breadth (b) 4 m 4 100 h 400 cm b rea of metallic sheet l b 800 400 0000 cm l Length of the oil tin (l 1 ) 60 cm its breadth (b 1 ) 40 cm its height (h 1 ) 0 cm rea of metallic sheet required for each tin surface area of vertical faces + surface area of the base [ (l 1 + b 1 ) h 1 ] + [l 1 b 1 ] 60 SHOOL SETION
MT EDURE LTD. GEOMETRY [ (60 + 40) 0] + [60 40] ( 100 0) + (60 40) 4000 + 400 6400 cm Number of tins that can be made rea of metallic sheet rea of metal required for each tin 0000 6400 50 50 Oil tins can be made. 4. lastic drum of cylindrical shape is made by melting spherical solid plastic balls of radius 1 cm. Find the number of balls required to make a drum of thickness cm, height 90 cm and outer radius 0 cm. Sol. Outer radius of the drum (r 1 ) 0cm Its thickness cm inner radius of the drum (r ) 0 8 cm Outer height of cylindrical plastic drum (h 1 ) 90cm Inner height of cylindrical plastic drum (h ) Outer height thickness of base 90 88cm Volume of plastic required for the cylindrical drum Volume of outer cylinder Volume of inner cylinder r 1 h 1 r h [(0) 90 (8) 88] (900 90 784 88] (81000 6899) 1008 cm Radius of spherical solid plastic ball (r) 1cm Volume of each plastic ball 4 r 4 r 1 1 1 4 cm Number of balls required to make the drum Volume of plastic required for the drum 1008 4 1008 4 Volume of each plastic ball 9006 Number of plastic balls required to make the cylindrical drum is 9006. SHOOL SETION 61
GEOMETRY MT EDURE LTD. 5. Water drips from a tap at the rate of 4 drops in every seconds. Volume of one drop is 0.4 cm. If dripped water is collected in a cylinder vessel of height 7 cm and diameter is 8 cm. In what time will the vessel be completely filled? What is the volume of water collected? How many such vessels will be completely filled in hours and 40 minutes? Sol. Diameter of the cylindrical vessel 8cm Its radius (r) 4 cm its height (h) 7 cm Volume of the cylindrical vessel r h 4 4 7 7 16 5 cm Volume of water collected 5 cm Volume of one drop of water 0.4 cm Volume of 4 drops of water 4 0.4 1.6 cm Water drips from the tap at the rate of 4 drops in every seconds Volume of water collected in seconds 1.6 cm Volume of water collected in 1 seconds 1.6 cm Time required to fill the cylindrical vessel Volume of cylindrical vessel Volume of water collected in each sec ond 5 1.6 5 1.6 5 10 1.6 10 5 10 16 660 seconds 11 minutes [ 1 minutes 60 seconds] hours and 40 minutes 60 min + 40 min 180 + 40 0 minutes Number of vessels that can be completely filled in 0 minutes 0 11 0 0 vessels can be filled in hours and 40 minutes. 6. In the adjoining figure, DEF is a regular hexagon with each side 14 cm. From each vertex, arcs with radius 7 cm are drawn. Find the area of the shaded portion. Sol. Draw seg N chord Q Radii of each arc 7 cm [Given] i.e. Q 7 cm DEF is a regular hexagon... F E D X N Q 6 SHOOL SETION
MT EDURE LTD. GEOMETRY In Q, Q mq m Q...(i) [Isosceles triangle theorem] m Q + m Q + m Q 180º [Sum of the measures of angles of a triangle is 180º] m Q + m Q + 10 180 [From (i) and angle of regular hexagon] mq 180 10 mq 60 m Q 0º...(ii) In N, N 90º [onstruction] N 0º [From (ii) and - N - Q] N 60º [Remaining angle] N is 0º - 60º - 90º triangle y 0º - 60º - 90º triangle theorem, N 1 [Side opposite to 0º] N 7 N.5 cm N N [Side opposite to 60º] 7 N.5 cm Q N [erpendicular drawn from the centre of the circle to the chord bisects the chord] Q.5 Q 7 cm (Q) 1 Q N 1 7.5.5.5.5.5 1.7 0.5 1.7 1.195 1.19 cm (-XQ) r 60 10 7 7 60 7 154 51. cm (segment XQ) ( XQ) (Q) 51. 1.19 0.14 cm SHOOL SETION 6
GEOMETRY MT EDURE LTD. (shaded portion) 6 (segment XQ) 6 0.14 180.84 cm The area of shaded portion is 180.84 cm. 7. metallic right circular cylindrical disc is of height 0 cm and the diameter of the base is one half time the height. This metallic disc is melted and moulded into the sphere. ssuming that no metal is wasted during moulding, find the radius and total surface area of the sphere. Sol. Height of cylindrical disc (h) 0 cm 1 its diameter 1 times the height 0 45 cm Radius of cylindrical disc (r) 45 cm Volume of cylindrical disc r h 45 45 0 60750 cm 4 Let the radius of the sphere be r 1 cm The metallic disc is melted and moulded into the sphere [Given] Volume metallic sphere Volume cylindrical disc 4 r 1 60750 cm 4 4 r 45 45 1 0 r 1 45 45 0 4 r 1 45 45 45 45 r 1 [Taking cube roots] r 1.5 cm Total surface area of sphere 4r 4.5 7 4 506.5 7 88 506.5 7 4455000 7 664.8 cm (pproximately) Radius of the sphere is.5 cm and total surface area of the sphere is 664.8 cm. 64 SHOOL SETION