MATHS LECTURE # 09. Plane Geometry. Angles

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1 Mthemtics is not specttor sport! Strt prcticing. MTHS LTUR # 09 lne eometry oint, line nd plne There re three sic concepts in geometry. These concepts re the point, line nd plne. oint fine dot, mde y shrp pencil on sheet of pper, resemles geometricl point very closely. The shrper the pencil, the closer is the dot to the concept of geometricl point. lne The surfce of smooth wll or the surfce of sheet of pper or the surfce of smooth lckord re close exmples of plne. The surfce of lckord is, however, limited in extent nd so re the surfces of wll nd sheet of pper ut the geometricl plne extends endlessly in ll directions. Lines line in geometry is lwys ssumed to e stright line. It extends infinitely fr in oth directions. It is determined if two points re known. It cn e expressed in terms of the two points, wh ich r e written in cpitl letter s. Th e followin g lin e is c ll ed. line segment is prt of line etween two endpoints. It is nmed y its endpoints, for exmple, nd. is line segment. It hs definite length. If point is on the line nd t the sme distnce from s well s from, then is the midpoint of segment. When we sy =, we men tht the two line segments hve the sme length. prt of line with one endpoint is clled ry. is ry of which is n endpoint. The ry extends infinitely fr in the direction wy from the endpoint. ngles n ngle is formed y two rys with the sme initil point. So n ngle looks like the one in the figure elow. T duction, ll rights reserved. r, line my e given one nme with smll letter. The following line is nmed line k. k efinition: n ngle is the union of two non colliner rys with common initil point. Regd. ffice: Indore T centres spred cross Indi ~ stlished 1993 ur motto Kr Ke ikhyenge is delivered through our unique Technology riven rocess ngine (Tpro engine). mil: pthq@teduction.com We: I : Ttkmml09 (1) of (0)

2 The two rys forming n ngle re clled the rms of the ngle nd the common initil point is clled the vertex of the ngle. Nottion: The ngle formed y the two rys nd, is denoted y the symol or. Supplementry ngles If the sum of two ngles is stright ngle (180 ), the ngles re supplementry nd ech ngle is the supplement of the other. It is t times convenient to refer the ngle, simply s. However this cnnot e done if there re more thn one ngle, with the sme vertex. lso, it is convenient to denote ngles y symols such s 1,, 3, etc. is stright ngle = = 180. nd re supplementry ngles. Types of ngles Wht is the supplement of n ngle whose mesure is 57? Sol. The supplement = = 13. Verticl ngles When two lines intersect, four ngles re formed. The ngles opposite to ech other re clled verticl ngles nd re equl to ech other. d c nd c re verticl ngles, = c. nd d re verticl ngles, = d.. The supplement of n ngle is one third of the ngle. ind the supplement of the ngle. Sol. Let the ngle e x. So, the supplement is x/3. Now, x + x/3 = 180 or 4x/3 = 180 or x = 135. Hence, the supplement = 135/3 = 45. Right ngles If two supplementry ngles re equl, they re oth right ngles. right ngle is one hlf of 180. Its mesure is 90. right ngle is symolised y. Stright ngle stright ngle hs its sides lying long stright line. It is lwys equl to 180. is stright ngle. + = nd =. nd re right ngles. omplementry ngles K = = 180. i.e., is stright ngle. If the sum of two ngles is 90, the ngles re complementry nd ech ngle is the complement of the other. djcent ngles Two ngles re djcent if they shre the sme vertex nd common side ut no ngle is inside the other ngle. nd re djcent ngles. ven though they shre common vertex nd common side, nd re not djcent ngles ecuse one ngle is inside the other. () of (0) Y or exmple, if Y is right ngle nd + = Y = 90, then nd re complementry ngles. 3. Wht is the complement of n ngle whose mesure is 47? Sol. The complement = = 43. I : Ttkmml09

3 4. The complement of n ngle is two third of the ngle. ind the complement of the ngle. Sol. Let the ngle e x. So, the complement is x/3. Now, x + x/3 = 90 or 5x/3 = 90 or x = 54. Hence, the complement = 54 = Some properties of ngles Liner pir of ngles rom the figure nd re djcent ngles. Look t the non common rms, nd. These re two opposite rys or these re colliner. We hve nme for them lso. cute ngles cute ngles re the ngles whose mesure is less thn 90. No two cute ngles cn e supplementry. Two cute ngles cn e complementry ngles. or efinition: Two djcent ngles re sid to form liner pir of ngles if their non common rms re two opposite rys. Now, look t the figure elow is n cute ngle. tuse ngles tuse ngles re the ngles whose mesure is greter thn 90 nd less thn 180. (i) (ii) or In ech of the figures ove, nd form pir of djcent ngles. In figure (ii) we hve liner pir of ngles. If djcent ngles nd form liner pir, then is n otuse ngle. Reflex ngle n ngle with mesure more thn 180 nd less thn 360 is clled reflex ngle. 180 < < tegorise the following ngles s cute, right, otuse, reflex nd stright ngle : 37, 49, 138, 90, 180, 89, 91, 11, 45, 35. Sol. cute ngles : 37, 49, 45, 89. Right ngle : 90. tuse ngles : 138, 91, 11. Reflex ngle : 35. Stright ngle : = 180. Imp. Two djcent ngles re liner pir of ngles if nd only if they re supplementry. rllel lines Two lines meet or intersect if there is one point tht is on oth lines. Two different lines my either intersect t single point or never intersect, ut they cn never meet in more thn one point. Two lines in the sme plne tht never meet no mtter how fr they re extended re sid to e prllel, for which the symol is. In the following digrm. I : Ttkmml09 (3) of (0)

4 Imp. Two lines intersected y trnsversl re prllel, if If two lines in the sme plne re prllel to third line, then they re prllel to ech other. Since nd c, we know tht c. c Two lines tht meet ech other t right ngle re sid to e per pendiculr, for which th e symol is. Line is perpendiculr to line Two lines in the sme plne tht re perpendiculr to the sme line re prllel to ech other. 1. the corresponding ngles re equl, or. the lternte interior ngles re equl, or 3. the lternte exterior ngles re equl, or 4. interior ngles on the sme side of the trnsversl re supplementry. If, then Line Line q p r s y x z w 1. y = q. z = r, x = p nd w = s.. z = p nd w = q. 3. y = s nd x = r. 4. z + q = 180 nd p + w = 180. Since verticl ngles re equl, p = r, q = s, y = w nd x = z. If ny one of the four conditions for equlity of ngles holds true, then the lines re prllel c 6. In the given figure, two prllel lines re intersected y trnsversl. ind the mesure of ngle y. Line line c nd line line c.. line intersecting two other lines is clled trnsversl. Line c is trnsversl intersecting lines nd. x c 3x+50 y The four ngles etween the given lines re clled interior ngles nd the four ngles outside the given lines re clled exterior ngles. If two ngles re on opposite sides of the trnsversl, they re clled lternte ngles. Sol. The two lelled ngles re supplementry. x + (3x + 50 ) = x = 130 x = 6. Since y is verticl to the ngle whose mesurement is 3x + 50, it hs the sme mesurement. y = 3x + 50 = 3(6 ) + 50 = 18. y x z w q p r s z, w, q nd p re interior ngles. y, x, r nd s re exterior ngles. z nd p re lternte interior ngles, so re w nd q. y nd s re lternte exterior ngles, so re x nd r. irs of corresponding ngles re y nd q; z nd r; x nd p nd w nd s. (4) of (0) I : Ttkmml09

5 Mini Revision Test # 01 IRTINS: ill in the lnks. 1. prt of line with one end point is clled.... Two djcent ngles re liner pir of ngles if nd only if they re line intersecting two other lines is clled Two ngles re sid to e supplementry if their sum is Two ngles re sid to e complementry if their sum is The ngle formed y the two rys nd, is denoted y the symol... or When two lines intersect, four ngles re formed. The ngles opposite ech other re clled Two ngles re... if they shre the sme vertex nd common side ut no ngle is inside nother ngle. 9. If ry stnds on line, the sum of the two djcent ngles so formed is Two lines in the sme plne tht re perpendiculr to the sme line re... to ech other. ircle The following figure shows the prts of circle. imeter Rdius entre hord of circle is stright line tht divides the circle into two prts. In cse of the dimeter, it is the lrgest possile chord tht cn e drwn in circle. Imp. The prolem of finding lengths nd res relted to figures with curved oundries is of prcticl importnce in our dily life. If n d den ote the r e n d per imeter (ls o clled circumference) respectively, of circle of rdius r, then = πr, = πr Here π ( letter of reek lphet) is the rtio of circumference to the dimeter of circle. It stnds for prticulr irrtionl numer whose vlue is given, pproximtely to two deciml plces, y 3.14 or y the frction ind the circumference of circle with re 5π sq ft. Sol. re of the circle = πr = 5π. r = 5 ft. circumference = πr = 10π ft. 8. ind the re of circle with circumference 30π cm. Sol. ircumference = πr = 30π r = 15 cm. re = πr = π 15 = 5π sq. cm. Sector nd segment of circle: See the figures given elow. S ircle entre of circle is the point within tht circle from which ll stright lines drwn to the circumference re equl in length to one nother. θ R Q ircumference of circle is the outer periphery of the circle. It is lso known s the perimeter of the circle. Rdius of circle is the distnce from the centre of the circle to the circumference. imeter of circle is line tht runs from the edge of the circle, through the centre, to the edge on the other side. The dimeter is twice the length of the rdius. (i) (ii) In figure (i) note tht the re enclosed y circle is divided into two regions, viz., (shded) nd (non shded). These regions re clled sectors. minor sector hs n ngle θ, sutended t the centre of the circle. The oundry of sector consists of rc of the circle nd the two lines nd. In fig (ii), the re enclosed y circle is divided into two segments. Here segment RQ (shded) is the minor one nd segment QS (non shded) is the mjor one. The oundry of segment consists of n rc of the circle nd the chord dividing the circle into segments. I : Ttkmml09 (5) of (0)

6 re of sector nd length of rc The rc length l nd the re of sector of ngle θ in circle of rdius r, re given y π r θ π r θ l =, =. We lso note tht = 1 lr In circle with centre, nd re the ngles t the circumference, y the sme rc, then =. ( s in sme rc or s in sme seg.) In circle with centre nd dimeter, if is ny point t the circumference, then = 90. Theorems on ircles If N is from the centre of circle to chord, then N = N. ( in semi circle) N If = Q nd if, Q re on the sme side of, then,, Q, re concyclic. ( from centre isects chord) If N is the midpoint of chord of circle with centre, then N = 90. (onverse, from centre isects chord) Q qul chords in circle re equidistnt from the centre. If the chords nd of circle re equl nd if X, Y, then X = Y. X Y ( sutends equl s t nd Q on the sme side) In equl circles (or in the sme circle) if two rcs sutend equl ngles t the centre (or t the circumference), then the rcs re equl. Q X Y (q. chords re equidistnt from centre) If X chords, Y chord nd X = Y, then chords = (Refer the figure given ove). (hords equidistnt from centre re eq.) circle with centre, with t the centre, t the circumference, stnding on the sme rc, then =. If = XY, then = XY or if = XQY, then = XY. (q. s t centre or t circumference stnd on equl rcs) In the sme or equl circles if chords =, then rc = rc. ( t centre = t ircumference on the sme rc ) (q. chords cut off equl rcs) (6) of (0) I : Ttkmml09

7 ew importnt results If two chords & intersect externlly t then = = 1 [m (rc ) m (rc )] If is secnt nd T is tngent then, 10. The length of the minute hnd of clock is 7 cm. ind the re swept y the minute hnd in 10 minutes. (Tke π = /7). Sol. The minute hnd descries circle of rdius 7 cm. It descries n ngle of 360 = 6 per minute. 60 Thus, we get sector of ngle 6 nd rdius 7 cm per minute. Its re π r θ = = =.567 sq. cm. 30 re of the sector descried in one minute =.567 sq. cm. re descried in 10 minutes = sq. cm = 5.67 sq. cm. 11. sector is cut from circle of rdius 1 cm. The ngle of the sector is 150. ind the length of its rc nd re. X Y Sol. Length of the rc of the sector, l 150 = 180 π r cm. 150 = 1 cm = 55 cm = T = 1/[m(rc XT) m(rc YT)] T 150 re of the sector = 360 πr sq. cm. 9. omplete the following tle. Sol. Rdius r Rdius r r re ircumference r re = πr = πr 1 1 π π 4 4π 4π 3 9 9π 6π π 8π 150 = 1 1 cm = sq. cm Mini Revision Test # 0 IRTINS: etermine whether the given sttements re TRU or LS. 1. The sum of the interior ngles of pentgon is The perimeter of regulr hexgon with side is The circumference of circle whose rdius is 14 cm is 176 cm. 4. The perimeter of regulr hexgon with side 8 cm is 48 cm. 5. The re of qudrnt of circle with rdius 14 cm is 148 sq. cm. IRTINS: ill in the lnks. 6. ll circles mesure... degrees. 7. The re of regulr hexgon with side 8 cm is If the rdius of circle is decresed y 50%, its re will decrese y If the dimeter of circle is incresed y 100%, its re is incresed y The length of minute hnd on wll clock is 14 cm. The re swept y it in 30 minutes is... I : Ttkmml09 (7) of (0)

8 olygon efinition: polygon is closed figure ounded y stright lines. ch of the line segments forming the polygon is clled its side. The ngle determined y two sides meeting t vertex is clled n ngle of the polygon. When two sides of the polygons re equl, we indicte the fct y drwing the sme numer of short lines through the equl sides. H This indictes tht = nd = H. Two polygons with equl corresponding ngles nd corresponding sides in proportion re sid to e similr. The symol for similr is ~. H ~ Similr figures hve the sme shpe ut not necessrily the sme size. Remrk: or n = 3, the polygon hs the specil nme tringle nd similrly for n = 4, the polygon hs the specil nmequdrilterl. Some other specil nmes re entgon (n = 5). Hexgon (n = 6). Tringles nd their types efinition: polygon 1... n is clled convex if for ech side of the polygon, the line contining tht side hs ll the other vertices on the sme side of it. ll the three figures given (tringle, pentgon nd hexgon) re convex polygons. Regulr polygon: polygon is clled regulr polygon if ll its sides nd ngles re equl. In the given figure, the hexgon is regulr polygon. ongruent nd similr polygons If two polygons hve equl corresponding ngles nd equl corresponding sides, they re sid to e congruent. ongruent polygons hve the sme size nd shpe. The symol for congruence is. H tringle is polygon with three sides. Tringles re clssified y mesuring their sides nd ngles. The sum of the ngles of plne tringle is lwys 180. The symol for tringle is. The sum of ny two sides of tringle is lwys greter thn the third side. quilterl quilterl tringles hve equl sides nd equl ngles. ch ngle mesures 60. In the, = =. = = = 60. Isosceles Isosceles tringles hve two sides equl. The ngles opposite to the equl sides re equl. The two equl ngles re sometimes clled the se ngles nd the third ngle is clled the vertex ngle. Note tht n equilterl tringle is isosceles. In the H, = H. H. = H. is vertex ngle. nd H re se ngles. H (8) of (0) I : Ttkmml09

9 Sclene roperties of tringles Sclene tringles hve ll three sides of different length nd ll ngles of different mesure. In sclene tringle, the shortest side is opposite the ngle of smllest mesure nd the longest side is opposite the ngle of gretest mesure. In the, > >. > >. Sum of the three interior ngles of tringle is 180. When one side is extended in ny direction, n ngle is formed with nother side. This is clled the exterior ngle. Interior ngle + corresponding exterior ngle = 180. n exterior ngle = Sum of the other two interior ngles not djcent to it. Sum of ny two sides is lwys greter thn the third side. tringle must hve t lest two cute ngles. Right ngled tringle Right ngled tringle contins one right ngle. Since the right ngle is 90, the other two ngles re complementry. They my or my not e equl to ech other. The side of right tringle opposite the right ngle is clled the hypotenuse. The other two sides re clled legs. The ythgoren theorem sttes tht the squre of the length of the hypotenuse is equl to the sum of the squres of the lengths of the legs. In the, is the hypotenuse. nd re legs. = = c =. 1. is right tringle with = 90. If = 1 nd = 5, then wht is the length of? Sol. + =. = = = 169. = 13. c roperties of similr tringles If two tringles re similr, then rtio of sides = rtio of heights = rtio of perimeters = rtio of medins = rtio of ngle isectors = rtio of inrdii = rtio of circumrdii. In the ove result, rtio of sides refers to the rtio of corresponding sides etc. lso, rtio of res = rtio of squres of corresponding sides. Right tringle is right tringle with = 90. If is perpendiculr to, then () Tringle ~ Tringle nd =. () Tringle ~ Tringle nd =. (c) Tringle ~ Tringle nd =. Two tringles re similr 13.Show tht tringle with sides 15, 8 nd 17 is right tringle. Sol. The tringle will e right tringle if + = c.(s shown ove) = = 89. the tringle is right tringle nd 17 is the length of the hypotenuse. Two tringles re similr if the ngles of one of the tringles re r es pectively equ l to the n gles of the other nd the corresponding sides re proportionl. I : Ttkmml09 (9) of (0)

10 Theorems (quingulr tringles)... If two tringles re equingulr, their corresponding sides re proportionl. In tringles nd XYZ, if = X, = Y, = Z, then /XY = /XZ = /YZ. X (ngle. side. ngle).s.. If = Y, = YZ, = Z, then tringle tringle XYZ. (Side. side. side) S.S.S. If = XY, = XZ, = YZ, tringle tringle XYZ. (Right ngle hypotenuse side) (R.H.S) X (3 sides proportionl) If tringles hve their corresponding sides proportionl, then they re equingulr. In tringles nd XYZ, if /XY = /XZ = /YZ, then = X, = Y, = Z. (Rtio of two sides nd included ngle) If one ngle of the tringle is equl to one of the ngles of the other tringle nd the sides contining these ngles re proportionl, then the tringles re similr. If = X nd /XY = /XZ, then = Y, = Z. If tringles nd XYZ re similr, then it is denoted y ~ XYZ. ongruency of tringles Two tringles re sid to e NRUNT if they re equl in ll respects (sides nd ngles). Y Z If = Y = 90, Hypotenuses = XZ, = XY, then tringle tringle XYZ. Note: ongruent tringles re definitely similr ut similr tringles my not e congruent. Mini Revision Test # 03 IRTINS: nswer the questions. 1. Wht is the mesure of ech cute ngle of n isosceles right ngled tringle?. ne of the cute ngles of right tringle is 4. ind the mesure of the other cute ngle. 3. Two ngles of tringle mesure 9 nd 58 respectively. ind the mesure of the third ngle. 4. ne of the cute ngles of right tringle is doule the other. ind the mesure of its smllest ngle. Y Z X 5. ch of the equl ngles of n isosceles tringle is doule the third ngle. Wht is the mesure of the third ngle? Y Z 6. The mesures of the ngles of tringle re in the rtio : 3 : 4. ind the mesure of its gretest ngle. The three sides of one of the tringle must e equl to the three sides of the other, respectively. The three ngles of the first tringle must e equl to the three ngles of the other respectively. Thus, if nd XYZ re congruent, (represented s XYZ), then = XY, = XZ, = YZ nd = X, = Y, = Z. Theorems (Side. ngle. side) S..S. If = XY, = X, = XZ, then tringle tringle XYZ. (ngle. ngle. side)..s. If = Y, = Z, = XZ, then tringle tringle XYZ. 7. ne of the ngles of tringle is equl to the sum of the other two. ind the mesure of the gretest ngle. 8. Wht is the mesure of ech ngle of n equilterl tringle? 9. The mesures of the cute ngles of right tringle re in the rtio 1 : 5. ind the mesure of its ngles. 10. Mtch the following i. Right tringle i. 10, 8, 3. ii. Isosceles right tringle ii. 60, 60, 60. iii. tuse sclene tringle iii. 90, 4, 48. iv. quilterl tringle iv. 108, 36, 36. v. cute ngled isosceles v. 45, 45, 90. tringle vi. tuse isosceles tringle vi. 65, 50, 65. (10) of (0) I : Ttkmml09

11 Qudrilterls Squre qudrilterl is polygon of four sides. The sum of the interior ngles of qudrilterl is 360. rllelogrm Some importnt properties = = = = 90. = = =. Trpezium (Trpezoid) + = 180 = = = = = =. nd re ses. nd re legs. h = ltitude. 1 h Rhomus Imp. If in prllelogrm, ll the sides re equl, then the prllelogrm so formed is rhomus. If qudrilterl is not prllelogrm or trpezoid ut it is irregulrly shped, its re cn e found y dividing it into tringles, ttempting to find the re of ech nd dding the results. Some importnt properties The chrt elow shows t glnce the properties of ech type of qudrilterl. = & =. =, = & = 90. Qudrilterl =, =. + = 180 & + = 180. Rectngle rllelogrm (pposite sides nd opposite ngles re equl. igonls isect ech other) Trpezium (ne pir of opposite sides re prllel) l Some importnt properties Rectngle pposite sides re equl. ch ngle is 90. igonls isect ech other. igonls re of equl length. Squre ll sides re equl. ch ngle is 90. igonls isect ech other t right ngles. igonls re equl. Rhomus ll sides re equl. pposite ngles re equl. igonls isect ech other t right ngles. =, =, = = = = 90. = = =. I : Ttkmml09 (11) of (0)

12 (1) of (0) Mini Revision Test # 04 IRTINS: ill in the lnks. 1. qudrilterl hs... vertices, no three of which re.... qudrilterl is..., if for ny side of the qudrilterl, the line contining it hs the remining The digonls of rectngle re The digonls of prllelogrm... ech other. 5. The line segment joining ny two points lying on two... sides of qudrilterl, except the two... lies in the interior of the qudrilterl hs one pir of opposite sides prllel to ech other. 7. In prllelogrm, pir of opposite sides re... nd In rhomus, the digonls... ech other ch digonl of rhomus... the ngles through which it psses. 10. digonl of prllelogrm divides it into two... tringles. IRTINS: or ech sttement given elow, indicte whether it is true (T) or flse (). 11. digonl of qudrilterl is line segment joining the two opposite vertices. 1. squre is rhomus. 13. squre is prllelogrm. 14. rectngle is squre. 15. The digonls of rectngle re perpendiculr. + Importnt results nd formule nly one circle cn pss through three given points. There is one nd only one tngent to the circle pssing through ny point on the circle. rom ny exterior point of the circle, two tngents cn e drwn on to the circle. The lengths of two tngents segment from the exterior point to the circle, re equl. The tngent t ny point of circle nd the rdius through the point re perpendiculr to ech other. When two circles touch ech other, their centres & the point of contct re colliner. If two circles touch externlly, distnce etween centres = sum of rdii. If two circles touch internlly, distnce etween centres = difference of rdii ircles with sme centre nd different rdii re concentric circles. oints lying on the sme circle re clled concyclic points. Mesure of n rc mens mesure of centrl ngle. m(minor rc) + m(mjor rc) = 360. ngle in semicircle is right ngle. The sum of ll the ngles of tringle is 180. The sum of ny two sides of tringle is greter thn the third side. xterior ngle of tringle is equl to the sum of opposite interior ngles. onditions of congruence of tringles re SS, SSS, S nd RHS. ngles opposite to equl sides in tringle re equl nd vice vers. If two sides of tringle re unequl, the greter side hs greter ngle opposite to it nd vice vers. If the squre of the longest side of tringle is equl to the sum of squres of the other two sides, the tringle is rightngled. ngle opposite to the longest side is right ngle. If the squre of the longest side of tringle is greter thn the sum of squres of the other two sides, the tringle is otuse ngled. ngle opposite to the longest side is otuse. If the squre of the longest side of tringle is less thn the sum of squres of the other two sides, the tringle is cute ngled. erimeter of tringle is greter thn the sum of its medins. Tringles on the sme or congruent ses nd of the sme ltitude re equl in re. Tringles equl in re nd on the sme or congruent ses re of the sme ltitude. rllelogrms hve opposite sides nd opposite ngles equl. igonl of prllelogrm divides it into two congruent tringles. igonls of prllelogrm isect ech other. If pir of opposite sides of qudrilterl re prllel nd equl, the qudrilterl is prllelogrm. The line segment joining midpoints of two sides of tringle is prllel to the third side nd hlf of it. onversely, the line drwn through the midpoint of one side of tringle prllel to the other, psses through the midpoint of the third. onditions of similrity of two tringles re SS, SSS, S nd. res of similr tringles re in the rtio of squres of corresponding sides, ltitudes or medins. The digonls of rhomus isect ech other t right ngles. The figure formed y joining the midpoints of the sides of rhomus is rectngle. The digonls of rectngle re equl nd isect ech other. The digonls of squre re equl nd isect ech other t right ngles. I : Ttkmml09

13 SLV XMLS LTUR # 09 The following questions will help you get thorough hold on the sic fundmentls of the chpter nd will lso help you develop your concepts. The following questions shll e covered in the clss y the techer. It is lso dvised tht sincere student should solve ech of the given questions t lest twice to develop the required expertise. IRTINS: or ech of the following questions, plese give the complete solution. 5. lculte the vlue of x. 1. iven tht, nd, then will e x 6. In the given figure, find x nd y.. In the pentgon given elow, sum of the ngles mrked is 40 y x In the given figure, is prllel to, = 65, = 50, = 30. ind. 3. lculte the vlue of x x 8x x 4. lculte the vlue of x. 8. In the given figure,, nd re stright lines, find the mesure of () the ngle x, if x = c + d. () the ngle e, if = c = d = e = f = x. (c) the vlues of + + c + d + e + f. 37 x x 68 x I : Ttkmml09 (13) of (0)

14 9. In the figure,, nd re the ltitudes of intersecting t. ind the sum of nd. r 10. ind the numer of sides of polygon, if the sum of its ngles is three times tht of n octgon. 11. ch interior ngle of regulr polygon exceeds the exterior ngle y 150. ind the numer of sides of the polygon. 1. In the figure,,, re three sides of regulr polygon nd is 0. lculte () () (c) the exterior ngle of the polygon, the numer of sides of the polygon, s 15. In the following figure, equl sides nd of tringle re produced to Q nd respectively, so tht = Q. rove tht = Q. 16. is n isosceles tringle such tht =. Side is produced to point such tht =. Wht is the mesure of? Q 17. In the figure, is n equilterl tringle of side 3., Q nd R re points on, nd respectively such tht R = = Q =. x 0 c R Q 13. In the figure nd re prllelogrms. rove tht nd re congruent. rove tht QR is n equilterl tringle. 18. In trpezium,, =. If the digonls intersect t, show tht re of = 4 re of. 19. In the given figure, Q nd = 90. If Q = 15 cm, = 5 cm nd = 10.5 cm, then clculte the lengths of () () nd. 14. In the figure, if : = 1 : 3, = 4.5 nd Q = 7.5, R Q, then find R Q R Q is point on the se of the equilterl such tht = 1 3. rove tht 9 = 7. (14) of (0) I : Ttkmml09

15 nswers Mini Revision Tests Success is where preprtion nd opportunity meet... nswers to Mini Revision Test # ry. supplementry. 3. trnsversl or 7. verticl ngles 8. djcent prllel nswers to Mini Revision Test # 0 1. True. True 3. lse 4. True 5. lse sq.cm 8. 75% % sq.cm nswers to Mini Revision Test # , 75, i iii; ii v; iii i, iv ii; v vi; vi iv nswers to Mini Revision Test # four, colliner. convex, vertices 3. equl 4. isect 5. different, end points 6. trpezium 7. equl, prllel 8. intersect, t isects 10. congruent 11.True 1. True 13. True 14. lse 15. lse Solutions Solved xmples Lecture # =, + = 180, = 30, hence = Required sum = Sum of ngles of pentgon = = We hve 8x + 3x + x = x = 180 x = We hve 37 + x + x x = x = x = 55 x = We hve x = x = 360 x = x = 40 (verticlly opposite s) y = x (djcent s on stright line) = = rw. rw H, H. s = 30 (lternte s, H ). r = = 0. x = r = 0. (lternte s, H ) y = 65 (lternte s, ). = x + y = = ( ) x = (verticlly opposite s). c + d + = 180 ( sum of ). c + d + x = 180. x + x = 180 (given x = c + d). x = 180 x = y x 50 r s H 30 I : Ttkmml09 (15) of (0)

16 ( ) x = (verticlly opposite s). x + c + d = 180 ( sum of ). 3x = 180 ( Q c = d = x). x = 60. e = f = 60 (given e = f = x). (c) + + c + d + e + f = onsider H 360 o I K J = = r + s = 180. = 180 r s...(1) onsider = 180. = 90...() dding eqution (1) nd (), + = 70 ( + r + s). = = Let n e the numer of sides of the polygon. (n ) 180 = [(8 ) 180 ] 3. (n ) 180 = n = 18 n = Let n e the numer of sides of the polygon. ch exterior ngle = 360 o. n ch interior ngle = c h s per the given condition c h. 360 n = n n n 180 n n = 180n n = 70 n = ( ) c = 0 (se s, isosceles ). = ( sum of ) = 140. the exterior ngle = (djcent ngles on line). i.e. x = 40. ( ) Let n e the numer of sides. c h or 180n 360 = 140n. n 180 = 140 n 40n = 360 n = 9. (c) roduce nd to meet t. = = = 40. = (sides opposite equl ngles). = = = + =. = (se ngles, isosceles tringle). 180 o 100 = o o. = 40 i.e. = (opposite sides of grm). = = (opposite sides of grm). is grm (opposite sides re ). = (opposite sides of grm). In nd, ( ) = (opposite sides of grm). ( ) = (opposite sides of grm). (c) = (proved). (S.S.S.). (16) of (0) 14. Join Q cutting R t K. onsider s K nd Q, K 1 K 7.5 = = 4. onsider KRQ nd Q. KR 3 KR 4.5 = = 4 R = K + KR = = 4 = K 15. In nd Q since, = Q, = nd Q = (verticlly opposite ngles) s nd Q re congruent. 16. Hence = Q (corresponding sides of congruent tringles). If =, then = = (sy). lso if =, then = = (sy). Now in, = = = 180/ = In s RQ, R nd Q, 18. R = = Q = (given). lso RQ = R = Q = 60 (given) Q = R = = 3 =. RQ R Q (S..S.). i.e. QR is equilterl. x x s nd re similr. The rtio of the proportionl sides is 1 :. Hence the rtio of the res of the tringles nd is 1 : ( ) In tringles Q nd (i) = (common) (i i) Q = (corresponding s, Q ) (iii) Q = (corresponding s, Q ) Q ~ (...) Q = = = 17.5 cm. 5 ( ) = = = 7 cm. 0. rw. In, = (ythgors Theorem) = 5 7 = 576 = 4 cm. = = (given). = 1 (given). 3 = = 1 ( isector theorem). lso, =. 1 1 = = = lso, = + (ythgors Theorem). = 7 + = = 7. R Q I : Ttkmml09