GEOMETRY Properties of lines
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1 GEOMETRY Popeties of lines Intesecting Lines nd ngles If two lines intesect t point, ten opposite ngles e clled veticl ngles nd tey ve te sme mesue. Pependicul Lines n ngle tt mesues 90 o is igt ngle. If two lines intesect t igt ngels, te lines e pependicul to ec ote. Pllel Lines If two lines in te sme plne do not intesect, tey e pllel to ec ote. Lines nd D e pllel nd denoted y D. Pllel lines nd tnsvese: Tnsvese + 4 = (Sum of inteio ngles) 4 1 In te ove given figue, te two lines L 1 & L e pllel to ec ote nd T is te tnsvese to ot te lines. Ten we will ve, 1 = (Pi of coesponding ngles) = (Pi of ltente ngles) & L 1 L Polygons: closed plne figue mde up of sevel line segments tt e joined togete is clled polygon. Types of Polygons Equingul (ll ngles equl) Equiltel (ll sides equl) Regul (ll sides & ngles equl)
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4 Mid Point Teoem: line joining te mid points of ny two sides of tingle is pllel nd equl to lf of te tid side. If in, D & E e te mid points of & espectively, ten we ve DE nd DE = 1 D E Polem In EF, D is pllel to EF. D = DF, D = 4 nd DF =. Wt is EF? (1) () 4 () 8 (4) 6 Simil tingles: Two figues e sid to e simil, if tey ve te sme spe ut not te sme size. NOTE: onguent tingles e simil ut simil tingles need not e conguent. Popeties of simil tingles: If two tingles e simil, te following popeties old tue. () Te tio of te medins is equl to te tio of te coesponding sides. () Te tio of te ltitudes is equl to te tio of te coesponding sides. (c) Te tio of te intenl isectos is equl to te tio of coesponding sides. (d) Te tio of indii is equl to te tio of te coesponding sides. (e) Te tio of te cicumdii is equl to te tio of te coesponding sides. (f) Rtio of e is equl to te tio of sques of te coesponding sides. (g) Rtio of e is equl to te tio of sques of te coesponding medins. () Rtio of e is equl to te tio of te sques of te coesponding ltitudes. (i) Rtio of e is equl to te tio of te sques of te coesponding ngle isectos. sic Popotionlity Teoem: In tingle, if line dwn pllel to one side of tingle divides te ote two sides in te sme tio. So if DE is dwn pllel to, it would divide sides nd popotionlly i.e. D E = D E D E
5 Pytgos Teoem: Te sque of te ypotenuse of igt-ngled tingle is equl to te sum of te sques of te ote two sides. i.e. in igt ngled tingle, igt ngled t, = + ngle isecto Teoem: If in, D is te ngle isecto of, te tio of te lines D & D is equl to te tio of sides contining te ngle. D = D Qudiltel ( fou side closed figue): Popeties & Fcts: 1. In qudiltel, sum of ll fou ngles is equl to 60.. Te e of te qudiltel = ½ one digonl x sum of te pependicul to it fom vetices. Impotnt Results If we join te mid-points of te sides of qudiltel, we get pllelogm nd te mid-points of te sides of pllelogm will give ectngle. If we gin join te mid-points of te sides of ectngle, we get omus nd te mid points of te sides of omus will give us sque. Qudiltel Pllelogm Rectngle Romus icles Sque If O is fixed point in given plne, te set of points in te plne wic e t equl distnces fom O will fom cicle. D
6 Popeties of icle 1. If two cods of cicle e equl, tei coesponding cs ve equl mesue.. Mesuement of n c is te ngle sutended t te cente. Equl cs sutend equl ngles t te cente.. line fom cente nd pependicul to cod isects te cod. 4. Equl cods of cicle e equidistnt fom te cente. 5. Wen two cicles touc, tei centes nd tei point of contct e colline. 6. If te two cicles touc extenlly, te distnce etween tei centes is equl to sum of tei dii. 7. If te two cicles touc intenlly, te distnce etween te centes is equl to diffeence of tei dii. 8. ngle t te cente mde y n c is equl to twice te ngle mde y te c t ny point on te emining pt of te cicumfeence. Let O e te cente of te cicle. O = P, wen = P 9. If two cods e equl, te c contining te cods will lso e equl. 10. Te locus of te line joining te mid-points of ll te equl cods of cicle is lso cicle of dius, 1 4 d wee is te dius of te given cicle nd d is te lengt of equl cods. 11. Tee cn e one nd only one cicle tt touces tee non-colline points. 1. Te ngle inscied in semicicle is 90 o. 1. If two cods nd D intesect extenlly t P, P P = P PD P OR 14. If two cods nd D intesect intenlly t P, P P = P PD D P D P P O P T 15. If P is secnt nd PT is tngent, PT = P P
7 16. Te lengt of te diect common tngent (PQ) = ( Te dis tnce etween tei centes) ( 1 ) P O 1 Q O 17. Te lengt of te tnsvese common tngent (RS) = ( Te dis tn ce etween tei centes ) ( + O 1 R S O i yclic Qudiltel If qudiltel is inscied in cicle i.e. ll te vetex lies on te cicumfeence of te cicle, it is sid to e cyclic qudiltel. 1. In cyclic qudiltel, opposite ngles e supplementy.. In cyclic qudiltel, if ny one side is extended, te exteio ngle so fomed is equl to te inteio opposite ngle. ltente ngle teoem ngles in te ltente segments e equl. In te given figue, PT is tngent to te cicle nd mkes ngles P & T espectively wit te cods &. Ten, T = & = P 1 P T )
8 Fomule to clculte e of some geometicl figues: S.No Nme Figue Peimete in units of lengt e in sque units 1. Rectngle = lengt = edt ( + ). Sque. 4. Pllelog m Romus 5 Qudiltel = side = side = side djcent to = distnce etween te opp. pllel sides = side of omus; d 1,d e te two digonls D 1 is one of its digonls nd 1, e te ltitudes on fom D, espectively. d 1 d 4 1 (digonl) ( + ) 1 d1 d 4 Sum of its fou sides 1 () (1 + )
9 6. Tpezium,, e pllel sides nd is te distnce etween pllel sides Sum of its fou sides 1 ( + ) 7. Tingle is te se nd is Rigt tingle Equiltel tingle Isosceles tingle te ltitude.,, c e tee sides of. d(ypotenuse) = + = side = ltitude = c d c = unequl side = equl side c + + c = s wee s is te semi peimete. + + d + c 1 o s(s )(s )(s c) 1 (i) 1 (ii) 4 c 4 c 4 d 11. Isosceles igt tingle d(ypotenuse) = = Ec of equl sides. Te ngles e 90 o, 45 o, 45 o. + d 1
10 1. icle = dius of te cicle π = 7 o.1416 π π 1. Semicicle Ring (sded egion) = dius of te cicle R = oute dius = inne dius Secto of l cicle θ o = centl ngle of te secto θ o = dius of te secto l = lengt of te c R π + 1 π. π (R ) l + wee l = θ π θ 60 π 60
11 Volume of some solid figues S. No Ntue of te solid Spe of te solid Ltel/ cuved sufce e Totl sufce e Volume evitions Used 1. uoid (l + ) (l + + l) l l = lengt = et = eigt l. ue Rigt pism Rigt cicul cylinde Rigt pymid 6. Rigt cicul cone l 4 6 edge = lengt of (peimete of se) Heigt (e of one end) + ltel sufce e e of se eigt π π( + ) π 1 (Peimete of te se) (slnt eigt) π l e of te se + ltel sufce e π(l + ) 1 (e of se) eigt 1 π = dius of se = eigt of te cylinde = eigt = dius l = slnt eigt
12 S. No Ntue of te solid Spe of te solid Ltel/ cuved sufce e Totl sufce e Volume evitions Used = dius 7. Spee 4π 4 π Hemispee Speicl sell Volume of ucket π π π R R 4π (R ) 4 π(r ) π (R + + R) = dius R = oute dius = inne dius R = lge dius = smlle dius = eigt
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