S T P E C N & S O T C S TE

G E O M E T R Y GEOMETRY TRIANGLE CENTRES

Rajasthan AIR-24 SSC SSC (CGL)-2011 CAT Raja Sir (A K Arya) Income Tax Inspector CDS : 9587067007 (WhatsApp) Chapter 4 Triangle Centres fdlh Hkh triangle ds fy, yxHkx 6100 centres Intensive gSA defined Q. Alice the princess is standing on a side AB of buesa ls 5 Classical centres important gSa ftUgs ge ABC with sides 4, 5 and 6 and jumps on side bl chapter esa detail ls discuss djsaxsaA BC and again jumps on side CA and finally 1. Orthocentre (yEcdsUnz] H) comes back to his original position. The 2. Incentre (vUr% dsUnz] I) smallest distance Alice could have jumped is? jktdqekjh ,fyl] ,d f=Hkqt ftldh Hkqtk,sa vkSj 3. Centroid (dsUnzd] G) ABC 4, 5 lseh- gS fd ,d Hkqtk ij [kM+h gS] ;gk¡ ls og Hkqtk 4. Circumcentre (ifjdsUnz] O) 6 AB BC ij rFkk fQj Hkqtk CA ij lh/kh Nykax yxkrs gq, okfil vius 5. Excentre (ckº; dsUnz] J) izkjafHkd fcUnq ij vk tkrh gSA ,fyl }kjk r; dh xbZ U;wure nwjh Kkr djsaA 1. Orthocentre ¼yEcdsUnz] H½ : Sol. A fdlh triangle ds rhuksa altitudes (ÅapkbZ;ksa) dk Alice intersection point orthocentre dgykrk gSA Stands fdlh vertex ('kh"kZ fcUnw) ls lkeus okyh Hkqtk ij [khapk x;k F E perpendicular (yEc) altitude dgykrk gSA A

B D C F E Alice smallest distance cover djrs gq, okfil viuh H original position ij vkrh gS vFkkZr~ og orthic triangle dh perimeter (ifjeki) ds cjkcj distance cover djrh gSA Orthic triangle dh perimeter B D C acosA + bcosB + c cosC BAC + BHC = 1800 a = BC = 4 laiwjd dks.k (Supplementary angles- ) b = AC = 5 BAC = side BC ds opposite vertex dk angle c = AB = 6 BHC = side BC }kjk orthocenter (H) ij cuk;k 2 2 2 b +c -a 3 x;k angle. cosA = = 2bc 4 Proof : a2+c2-b2 9 cosB = = AEHF ,d cyclic quadrilateral ¼ pØh; prqHkqZt ½ gSA 2ac 16 0 0 F = 90 , E = 90 a2+b2-c2 cosC = = 1 FAE + EHF = 1800 2ab 8

FAE = BAC 3 9 105 Perimeter = 4 + 5 + 6 1 = cm. EHF = BHC 4 16 8 16 (vertically opposite angles - f'k"kkZfHkeq[k dks.k)  BAC + BHC = 1800

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 Orthic triangle 4. Orthic triangle dsoy acute angled triangles esa gh gS rFkk o Triangle ds rhuks altitudes ds feet dks feykus ls cuus possible right angled obtuse angled okyk dgykrk gSA ;g (vf/kd dks.k f=Hkqt) ds fy, orthic triangle ugh cuk;k triangle orthic triangle tk ldrk gSA triangle orthocentre ds fy, pedal triangle Hkh gksrk gSA A  Orthic triangle esa cuus okys cyclic quadrilaterals (pdzh; prqHkqZt)  AFHE, BFHD, CEHD  BCEF, ACDF, ABDE F E  Orthic triangle DEF dh inradius and circumradius H Inradius = 2RcosAcosBcosC Circumradius = R/2  Triangle ABC ds vertices ('kh"kZ fcUnw) pedal B C D triangle DEF ds excentres gksrs gSaA vFkkZr~ ABC dh Triangle DEF orthic triangle gSA Hkqtk;sa orthic triangle DEF ds fy, exterior angle cká dks.k lef}Hkktd gksrh gSA 1. Triangle ABC dk orthocentre H, orthic triangle bisectors ( ) DEF dk incentre (vUr% dsUnz) gksrk gSA vFkkZr~ original  Vertex A, B rFkk C ls orthic triangle DEF dh Hkqtkvksa triangle ABC ds altitudes orthic triangle DEF ds ij Mkys x, yEc (perpendiculars) ABC ds fy, interior angle bisectors gksrs gSa] vFkkZr~ circumcentre (ifjdsUnz) O ij feyrs gSa vFkkZr~ ABC dh ifjo`r dh f=T;k f=Hkqt dh Hkqtkvksa ∠BEF = ∠BED, ∠ADE = ∠ADF, ∠CFE = ∠CFD circumradius ( ) DEF ij yEcor~ gksrh gSA 2. Triangle ABC esa inscribed fd, tk ldus okys lHkh T A triangles esaa lcls de perimeter (ifjeki) orthic triangle dh gksrh gSA

Orthic triangle dh sides = F E (i) R Sin2A = a Cos A O Circumcentre (ii) R Sin2B = b Cos B ¼ifjdsUnz½

(iii) R Sin 2C = c Cos C B D C vr% orthic triangle dh perimeter =  Orthocentre and circumcircle ds e/; laca/k & acosA + bcosB + ccosC. A abc cosAcosBcosC Area of orthic triangle DEF = 2R K R = ABC dh circumradius F E

H 3. Orthic triangle DEF dk circumcircle original triangle ABC dh Hkqtkvksa ds mid points ls xqtjrk gS rFkk ;g circumcircle, orthocentre ls vertices dh B C nwjh ds e/; fcanqvksa ls Hkh xqtjrk gSA ;g bl triangle dk ij dk gSA nine point circle dgykrk gS] tks fd orthic triangle FK, line AB orthocentre H reflection vr% ,d gS rFkk dk DEF ds fy, circumcircle ¼ifjo`r½ gksrk gSA HBKA kite line segment HK midpoint F gSA Nine point circle dh radius, ABC dh circum radius dh vk/kh gksrh gSA ∠ABH = ∠ABK

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rFkk nksuks gS rFkk FE 1 A Orthic DEF JKL similar KJ = 2 A F E K J F E

H B D C bl case esa D, E rFkk F altitudes gSaA B D C BFC ~ BDA L (B common gS rFkk ∠F = ∠D = 900, vr% AA property ls)  AEB ~ AFC ⇨  CDA ~ CEB  Concurrent Lines BF BC AC AB CD CA vr% = ; = ; = 3 ;k 3 ls vf/kd lines tks fdlh ,d point ij intersect BD BA AF AC CE BC djrh gS vFkkZr~ fdlh ,d gh fcanw ij vkdj feyrh gS] bu rhuksa equations dks multiply djus ij] concurrent lines dgykrh gSA rFkk og fcanw ftl ij ;s BF . AE . CD = BC . AB . CA = 1 vkil esa dkVrh gSa BD AF CE BA AC BC lines point of concurrency dks djus ij dgykrk gSA Left Hand Side rearrange Line a AF . BD . CE = 1 Eg.: FB DC EA vr% ABC esa rhuksa altitudes ,d single point ij feysaxsa Line b m ftls orthocentre (yEcdsUnz) dgrs gSaA  Incentre ds fy,& Line c fdlh ABC esa incentre exist djrk gS bldk proof dk gS ;gka fcanw m, point of concurrency dgykrk gSA Ceva's theorem trigonometric version ftlds vuqlkj AD, BE rFkk CF rHkh concur djsaxh tc& fdlh triangle ds fy, fofHkUu lines (vyx&vyx ds vuqlkj fdlh ,d fcanw ij vkil esa dkVrh gSA Sin ∠ BAD . Sin ∠CBE . Sin ∠A CF = 1 concepts ) Sin ∠ABE Sin ∠BCF Sin ∠CAD bl fcanw dks mu lines ds vuqlkj vyx&vyx centres A (Eg. orthocentre, incentre, centroid, circumcentre, θ θ excentre etc.) ds :Ik esa define fd;k x;k gSA 2 2 F E Ceva'a theorem fdUgh Hkh rhu lines ds fy, concurrence crkrh gS vr% ;g orthocentre, incentre rFkk circumcentre ds fy, Hkh true gS rFkk Ceva's theorem dh lgk;rk ls ge bu centres dk B D C existence prove dj ldrs gSaA Incentre ds case esa D, E rFkk F angle bisectors ds feet gSa vr%  Proof of existence ∠BAD = ∠CAD, ∠ABE = ∠CBE rFkk ∠ACF = ∠BCF fdlh Hkh esa yEcdsUnz triangle orthocentre ( ) exist vr% Sin ∠ BAD . Sin ∠ABE . Sin ∠A CF = 1 : 1.1 = 1 djrk gS bldk lcls simple proof Ceva's theorem gSA Sin ∠CAD Sin ∠CBE Sin ∠BCF Left Hand Side dks rearrange djus ij] Ceva's theorem ds vuqlkj] Sin ∠ BAD . Sin ∠CBE . Sin ∠A CF = 1 Rkhu lines AD, BE rFkk CF rHkh concur gksaxh vFkkZr~ fdlh Sin ∠ABE Sin ∠BCF Sin ∠CAD ,d fcanw ij feysaxh tc & vr% rhuksa angle bisectors ,d point ij feysaxsa ftls AF. BD . CE = 1, vUr%dsUnz dgrs gSaA FB DC EA incentre ( )

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 Orthocentric System ADB esa Triangle ABC rFkk bldk orthocentre H feydj ,d AB > AD ...... (i) orthocentric system cukrs gSa vFkkZr~ fdUgh Hkh rhu  BEC esa points dks feykus ls cuus okys triangle ds fy, pkSFkk BC > BE ...... (ii) gksrk gSA point orthocentre CFA esa A AC > CF ...... (iii) adding (i), (ii), (iii) we get

F E AB+BC+AC > AD+BE+CF H F fdlh triangle ABC esa altitudes dk vuqikr %& 1 1 1 Area of ABC = 2 ABCF = 2 BC  AD = 2 AC  BE  AB : BC : AC = 1 : 1 : 1 CF AD BE B D C Equilateral triangle ¼leckgq f=Hkqt½ esa rhuksa Hkqtk,sa leku gksrh gS] vr% altitudes ¼yEc½ Hkh leku gksrs gSaA F ABC dk orthocentre = H ABH dk orthocentre = C  Position of Orthocentre- BCH dk orthocentre = A (i) Acute angled triangle (U;wu dks.k f=Hkqt) :- ACH dk orthocentre = B blesa orthocentre f=Hkqt ds vUnj fLFkr gksrk gSA F AH.HD = BH.HE = CH.HF A Orthocentre, altitudes dks ftu nks lengths esa divide djrk gS mudk product ges'kk ,d constant gksrk gSA H Similarly, (i) AD.DH = BD.DC (ii) BE.EH = AE.EC B C 0 (iii) CF.FH = AF.FB A, B, C rhuksa 90 ls de gSA ledks.k f=Hkqt F AF  BD  CE = 1 (ii) Right angled triangle ( ) FB DC EA F Pair of similar triangles A DBH ~ EHA DHC ~ FHA FHB ~ EHC F Sum of three altitudes of a triangle is less than sum of three sides of a triangle. BH C fdlh f=Hkqt ds rhuksa altitudes ¼yEcksa½ dk ;ksx rhuksa Hkqtkvksa Orthocentre ds ;ksx ls de gksrk gSA fdlh ledks.k f=Hkqt esa] Right angled triangle esaa orthocentre ledks.k okys ij fLFkr gksrk gSA Hypotenuse ¼d.kZ½ > altitude ¼yEc½ vertex

Like Our FB Page : @neon.classes 4 Web.: www.neonclasses.com (iii) Obtuse angled triangle (vf/kd dks.k f=Hkqt) (2) Incentre ¼vUr% dsUnz] I ½ Obtuse angled triangle esa orthocenter vf/kd dks.k okys 'kh"kZ dh rjQ f=Hkqt ls ckgj fudy tkrk gSA f=Hkqt ds rhuksa angles ds angle bisectors (dks.k

0 lef}Hkktd) dk intersection point (dVku fcUnw) B > 90 , B obtuse angle gSA Incentre dgykrk gSA A A 2 / 2  / E  I r F C B r

D  /2 r /2 H /2 /2 B C

0 ABC + AHC = 180 Incentre ls rhuksa sides dh nwjh leku gksrh gS bl nwjh dks A vUr% o`r dh f=T;k dgrs gSa rFkk blls cuus inradius (r ) E okyk o`r incircle (vUr%o`r) dgykrk gSA F Incentre ges'kk f=Hkqt ds vUnj gh fLFkr gksrk gSA rFkk F C B okLro esa ml f=Hkqt ds medial triangle ds vUnj fLFkr gksrk gSA A D

H

K I BAC + KHC = 1800

Proof /2 /2 EBA ~ DBH B C 0 E = D = 90 fdlh Hkqtk }kjk I ij cuus okyk angle rFkk Hkqtk ds EBA = DBH (vertically opposite angles) opposite vertex ij cuus okys angle dk laca/k&  EAB =DHB (similar triangles)  0  = 90 + KHC + DHB = 180 (Straight line ij cus angles) 2 EAB = BAC Proof : 0 KHC + BAC = 180 BIC esa ⇒  +  +  = 1800 ...... (i) 2 2 A ABC esa ⇒ + + = 1800 ...... (ii)

 +  +  = 900 C 2 2 2   B  + = 900 - ...... (iii) 2 2 2 (iii) dh value (i) esa put djus ij  + 900 -  = 1800   = 180 - (900-  ) H K 2 2   = 900 +  ACB + KHA = 1800 2 Download our app : NEON CLASSES 5 geometry NEON CLASSES, JAIPUR • 9828728833/34

Incentre (I) ds co-ordinates:  Incentre ls cus gq, rhuksa triangles ds area dk vuqikr mudh Hkqtkvksa ds vuqikr ds cjkcj gksrk gS& A (x1,y1) Area AIB Area BIC Area AIC AB = BC = AC D;ksafd rhuksa f=Hkqtksa esaa ÅapkbZ;ka leku gS rFkk ÅapkbZ r ds equal gh gSA c b  Lkedks.k f=Hkqt dh var%o`r dh f=T;k I r = Base + Perpendicular - Hypotenuse (x,y) 2 A r a l u

a c B C i (x , y ) (x , y ) d E Hypotenuse 2 2 3 3 n r e

ax1 + bx2 + cx3 ay1 + by2 + cy3 p I r r

I (x, y) = [ , ] e

a+b+c a+b+c P F  Incentre, Angle bisector line dks fuEu vuqikr esa r divide djrk gSA A B D Base C

θ θ B+P-H 2 2 r = AI b+c 2 c b IK = a

I Proof : AF = AE (fdlh ckº; fcUnw A ls circle ij [khaps x;s tangents, Li'kZ js[kk,a) blh argument ls] B K C rFkk a BF = BD = r CD = CE Area of ABC = r. s AB + BC - CA = (AF + r) + (r + DC) - (CE + EA) = 2r

Proof : A (D;ksafd AF = EA rFkk DC = CE fdlh ckº; point ls circle ds fy, tangents gSaA) r = AB+BC-CA 2 AB = perpendicular ¼yEc½ (P) r r ¼ vk/kkj½ I BC = Base (B) AC = Hypotenuse ¼ d.kZ½ (H) r  r = B+P-H 2

B C Area of ABC Note : = Ar AIB + Ar BIC + Ar AIC tc angle bisector line lkeus okyh Hkqtk ij yEcor~ gksrh gS rks og = 1 AB. r+ 1 BC. r + 1 AC. r perpendicular ( ) triangle 2 2 2 1 isosceles (lef)ckgq ftlesa AB = AC gksa rFkk vertex A = r [ (AB + BC + AC)] 2 ls Hkqtk BC ij [khapk x;k angle bisector yEcor gksxkA) = r . s ( AB+BC+A C semiperimeter) 2 = ;k equilateral (leckgq) gksrk gSA s = semiperimeter (v)Zifjeki)

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 Interior Angle bisector theorem (vUr%dks.k bls ge ,sls Hkh ns[k ldrs gSa fd ;fn ,d line tks BD ds lef}Hkktd izes;) : parallel gks rFkk point C ls xqtjs rks]

A line C || line BD || line KA rhu parallel lines ds e/; ds line segments (transverses fr;Zd js[kk;sa) dk ratio leku gksrk gSA

A K D e A in L D B /2 e  in L C /2 e D in B C L

AB = AD BC DC K B C

2 BD = (AB x BC) - (AD x DC) vr%

Proof KB = AD ysfdu gSA vr% AB = AD A BC DC KB = AB BC DC

2 /

 A D D

2 / /2 /2 2 θ/2 / 2 K B C / θ/2 B C ABC esa] BD, dks.k B dk bisector (lef}Hkktd) gSA ls ds ,d djrs Vertex A BD parallel line AK draw = 2 gSa tks extended CB dks K fcanw ij dkVrh gSA  BD = interior angle bisector ABD = CBD = 2 (bisected angles)  CD = exterior angle bisector KAB = ABD = 2 (alternate angles) AKB = DBC =  (corresponding angles) Proof- 2 ,d lef}ckgq Triangle ABK (isosceles triangle) ABC esa] f=Hkqt gS rFkk AB = KB (D;ksafd A = K) α + β + γ = 1800 ...... (i) KCA ~ BCD (D;ksafd K = B, C nksuks triangles esa common gS BDC esa] rFkk KA || BD)  θ + γ + + ω ...... (ii) vr% 2 2 θ α  θ = α + β , + + ...... (iii) BK = AD 2 2 2 BC DC θ dh value (ii) esa j[kus ij But BK = AB 2  α  0 vr% 2+ γ + 2 + 2 + ω = 180 ...... (iv)

AB AD α 0 α =  BC DC 2 + ω + + γ = 180 ∴ω = 2

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 Exterior Angle bisector theorem (ckádks.k AE = AF = s - a, BD = BF = s - b, CD = CE = s - c lef}Hkktd izes;) : A (AE = AF nksuks equal lengths dh gSa D;ksafd fdlh point ls circle ds fy, tangents gSaA) K AE + AF + CD = s /2 r = AE.BF.CD /2 AE+BF+CD B C /2 AD, BE rFkk CF ,d single point ij intersect djrh /2 gS ftls Gergonne Point dgrs gSaA ;fn fdlh ds gSa rks& E triangle altitudes h1, h2 and h3 D 1 + 1 + 1 = 1 h1 h2 h3 r

AB = AD BC DC  C Proof ABC esa] BD, exterior angle (cká dks.k) B dk bisector (lef}Hkktd) gS rFkk ;g extended AC dks fcanw D ij dkVrk gSA DB ds parallel ,d line CK draw djrs gSaA A F T Z B KCB = CBD =  (alternate angles) 2 fdlh right angled triangle ABC esaa] C = 900, C dk  CKB = DBE = 2 (Corresponding angles) angle bisector, C ls altitude rFkk median ls cus    angle dks bisect (lef}Hkkftr) djrk gSA CBD = EBD = 2 (Bisected angles) vr%  CF = altitude, CT = angle bisector KCB = CKB = 2 vr% triangle KBC ,d isosceles triangle (lef}ckgq CZ = median then ∠FCT = ∠ZCT f=Hkqt) gS rFkk BC = KB. Basic proportionality Theorem ls] Proof -

AB AD AB AD CT angle bisector gS vr% ∠TCA = ∠TCB = But KB = BC, vr% = KB DC BC DC Let's say ∠CAB = a 0 /;ku ls ns[ksa bldk Hkh And ∠CFA = 90 NOTE - result interior Angle 0 Bisector Theorem ds leku gh gS ;gka point D, AC ∴∠FCA = 90 - a Hkqtk ij ckgj vk x;k gSA ABC esa] ∠ABC = 900 - a = ∠ZBC circle dk diameter circumference ij 900 dk A angle cukrk gS vr% ABC ds circumcircle dk Z  centre gS rFkk AB O;kl gS vkSj Z, AB dk midpoint gS D;ksafd CZ ,d median gSA vr% ZA = ZB = ZC = R c F E b r ZBC ,d isosceles (lef}ckgq) triangle gS vr% ∠ZBC = ∠ZCB = 900 - a ∴∠FCA = 900 - a = ∠ZCB B D C ∴∠FCT = ∠ZCT a

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 Incircle : f=Hkqt ds area dk inradius ls relation-  fdlh triangle ds fy, inradius (r) rFkk abc A circumradius (R) dk relation  R.r = 2(a+b+c)  fdlh f=Hkqt dh sides - a, b, c rFkk r o R esa relations ab + bc + ca = s2 + (4R + r) r F E 2 2 2 2 I a + b + c = 2s -2 (4R + r) r  ,slh line tks triangle ds area rFkk perimeter dks /2 B r vk/kk&vk/kk djrh gS og ges'kk ds B/2 divide triangle B D C incentre ls xqtjrh gSA  BID esa] ∠D = 900, side ID = r A B BD Cot ( 2 ) = ID , Cot ( B ) = BD 2 r B I incentre  BD = r . Cot ( 2 ) vr% Area of BID = 1 base  height 2 1 B = 2 . r. r . Cot ( 2 ) B C BIF congruent gS BID dsA IA .IB IA .IC IB .IC + + = 1 vr% nksuks dk area same gSA vr% quadrilateral CA .C B BA .BC AB .AC (prqHkqZt) BDIF dk area = r2 Cot ( B ) 2 similarly, 2 A IA . IB. IC = 4Rr Quadrilateral AFIE dk area = r2 . Cot ( ) 2 A QuadrilateralCDIE dk area = r2. Cot ( C )  2 vr% triangle ABC dk area = x x  = r2 ( Cot A + Cot B + Cot C ) 2 2 2   Incircle ds area dk triangle ds area ls ratio ls 3 3 de ;k equal gksrk gSA z y area of incircle 

area of triangle  3 3  Equilateral triangle (leckgq f=Hkqt) esa ;g ratio ds 3 3 B z y C equal gksrk gSA x = AB+AC-BC y = CA+CB-AB  Incircle dk tangency point rhuksa sides dks (x, y), 2 2 z = BA+BC-AC (y, z) and (z, x) lengths esa divide djrk gS rks& 2 A ;fn ∠B = 900 rks z = r x x A xyz r = x+y+z Hypotenuse B+P-H r r = a l 2 u

z c i

y d

r n r e z p r e P B z y C B z Base C Area of triangle ABC = x.y.z (x+y+z)

Download our app : NEON CLASSES 9 geometry NEON CLASSES, JAIPUR • 9828728833/34  A 3. Centroid (dsUnzd] G) Centroid fdlh triangle dh rhuksa medians 2 2 (ekf/;dkvksa) dk intersection point gksrk gSA bls I centre of mass Hkh dgrs gSaA

/2 /  2 B /2 /2 C B C G

M ABC ,d triangle gS ftlesa I incentre (vUr%dsUnz) gSA dks vkxs djrs gSa fd ;g ds AI extend ABC A circumcircle dks point M ij feyrk gSA A M minor arc BC dk mid point gS rFkk BIC dk circumcentre gSA vr% MB = MI = MC = Radius of circumcircle of BIC ;g relation fact 5 ds uke ls tkuk tkrk gSA

B D C Proof Ekkf/;dk lkeus okyh ij ;fn M BIC dk circumcentre gS rks MB = MI (median) side perpendicular rHkh gksrh gS tc ABC ,d equilateral vFkkZr~ ∠MBI = ∠MIB triangle (leckgq f=Hkqt) gks ;k ABC ,d isosceles ∠MBI = ∠IBC + ∠MBC triangle gksa rFkk AB rFkk AC Hkqtk;sa leku gksaA bl case esa (∠MBC=∠MAC ∴chord MC }kjk circle dh median, ∠A ds fy, angle bisector ¼dks.k lef}Hkktd½ circumference ij cus angles Hockey Theorem) Hkh gksrh gSA α Properties : ∠MBC = ∠MAC = 2  α  izR;sd ds dks esa ∠IBC = ∴∠MBI = + (1) median triangle area 2 equal parts 2 2 2 divide djrh gSA ∠BIA + ∠BIM = 1800 (straight line) A ∠ABI esa]

Exterior angle MIB = ∠IAB+∠IBA (sum of h remote interior angles) α  vr% ∠MIB = 2 + 2 , ∠MBI = ∠MIB vr% MB = MI B M C blh argument ls] MI = MC vr% MB = MI = MC = Radius of circumcircle of BIC. ABM rFkk AMC dk base equal gS rFkk nksuks  ds ij bl izdkj fLFkr gS fd triangles dh height Hkh leku gS vr% area Hkh leku BCI circumcircle E BC = EC gksxkA then ∠BCE = ∠BAC ABM rFkk AMC nksuks leku parallel lines ds e/; ij Hkh gS tgka dk EC, CO perpendicular O, ABC fLFkr gSA vr% nksuks dh height leku gSA circumcentre gSA (2) rhuksa medians fdlh triangle dks 6 leku area okys triangles esa divide djrh gSA Like Our FB Page : @neon.classes 10 Web.: www.neonclasses.com

A (4) A

P N G P N

B M C B M C

 Area AGB = Area BGC = Area AGC AM+BN+CP 1 3 = Area ABC = AB+BC+CA (Perimeter) < 3 2 Area AGP = Area PGB = Area BGM AM2+BN2+CP2 3 = 2 2 2 = ¼bl relation dks Apollonius AB +BC +CA 4 = Area GMC = Area GCN = Area AGN theorem ls prove fd;k tk ldrk gSA = 1 Area ABC 6 Medians ls cuk;s x;s triangle dk area original (3) Centroid izR;sd ekf/;dk dks 2 : 1 esa divide djrk gS triangle ds area dk 3 gksrk gSA ftlesa vertex okyk Hkkx 2 rFkk side okyk Hkkx 1 gksrk gSA 4 ges'kk f=Hkqt ds vUnj fLFkr gksrk gSA A (5) Centroid (6) Centroid, actual esa rhuksa vertices dk average dgykrk gSA

A(x1, y1)

AG = 2 GM 1 G

G B M C (x, y)

Proof : A B C

(x2, y2) (x3, y3) x + x + x y + y + y G (x, y) = [ 1 2 3 , 1 2 3 ] 3 3 P N (7) Apollonius Theorem : A

G

c b B C m PNG ~ CBG GN = GP GB GC B M a/2 C PN = 1 (mid point theorem ls) a BC 2 vr% b2 + c2 = 2 [m2 + ( a )2] BG = CG = 2 2 GN GP 1 ekf/;dk AM = m = 1 2b2 + 2c2 - a2 2

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(8) ;fn rhuksa medians nh xbZ gksa rks sides dh length find Proof : dh tk ldrh gS& AM2 = AB2 + BM2 ………….(1) A CP2 = PB2 + BC2 ……………(2) PM2 = PB2 + BM2 ………….(3)

2 2 2 P N AC = AB + BC …………..(4) (1) + (2) AM2 + CP2 = AB2 + BM2 + PB2 + BC2

2 2 2 2 2 2 B M C AM + CP = AB + BC + BM + PB BC = a, AC = b, AB = c AM2 + CP2 = AC2 + PM2 ...... (i)

Medians, AM = m1 BN = m2 CP = m3 Mid point theorem ls] 2 Apollonious Theorem ls] AC 2 AC PM = 2 , PM = 4 2 2 2 2 AC2 a = 3 . 2m2 + 2m3 - m1 2 2 2 AM + CP = AC + 4 2 2 2 2 b = . 2m + 2m - m 2 2 2 3 1 3 2 4 (AM + CP )= 5AC ...... (ii) 2 2 2 2 c = . 2m + 2m - m 2 2 3 1 2 3 AC = 2PM, AC = 4PM 4 (AM2 + CP2) = 5 (4PM)2 (9) Medians ds fy, Heron's formula : Rkhu AM2 + CP2 = 5PM2 ...... (iii) medians = m1, m2, m3 ds fy, Area of ABC Right angled triangle circumcentre (ifjdsUnz) ds according: = 4 m .(m -m ).(m -m ).(m -m ). 3 s s 1 s 2 s 3 A

m1+m2+m3 R ms = 2 N R = circumradius (10.) A R R

b c P B C AC AC BN = 2 , PM = 2 vr% BN = PM

2 2 2 B M C AM + CP = 5BN ...... (iv) a

3 2 2 2 2 (11) rhuksa medians dk sum triangle dh perimeter ds ls (i) AM + CP = AC + PM T;knk gksrk gSA 4 (ii) 4 (AM2 + CP2 )= 5AC2 3 m1 + m2 + m3 > 4 (a + b + c) (iii) AM2 + CP2 = 5PM2 (iv) AM2 + CP2 = 5BN2

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(12) fdlh triangle dh fdUgh nks Hkqtkvksa dk ;ksx rhljh side dh (14) ;fn nks medians perpendicular gksa rks] ds nqxqus ls T;knk gksrk gSA median A A

b c P N

P N

B C a

B M C b2 + c2 = 5a2 AB + AC > 2AM, AB + BC > 2BN, BC + AC > 2CP ;fn nks medians perpendicular (yEcor~) gksa rks rhuksa Rkhuksa dks add djus ij] medians pythagorean triplets cukrh gSA 2(AB + BC + AC) > 2 (AM + BN + CP) (AB + BC + AC) > (AM + BN + CP) Q. ABC is an equilateral triangle. A line DE passes through its centroid, segmenting it into two (13) median ds e/; fcanw dks vertex ls feykus okyh js[kk regions. Find the minimum possible ratio of opposite side dks 1 : 2 esa divide djrh gSA the area of the smaller region to the area of A larger region. ,d leckgq f=Hkqt gSA ,d js[kk blds dsUnzd ls D ABC DE xqtjrh gS rFkk mls nks Hkkxksa esa ckaVrh gSA rks NksVs Hkkx ds {ks=Qy dk cM+s Hkkx ds {ks=Qy ls lcls de laHkkfor vuqikr K Kkr djks\ A

B M C AD 1 AK = KM DC = 2 A E Proof : G

D D

B C K E ;fn line DE vertex B ls xqtjrh gS rks ABC ds area dks 2 equal parts esa divide djrh gS vr% nksuks parts ds dk gksxk rFkk ;g ds B C area ratio maximum 1 equal M gksxkA ME || BD, BDC esa] M, side BC dk mid point gS rks ;fn ds gksxh rks rFkk E, side DC dk mid point gksxkA (Mid point theorem) line DE side BC parallel smaller ds dk gksxkA vr% DE = EC larger region area ratio minimum AME esa] DK || EM ADE ~ ABC 2 nksuks triangles dh medians dk ratio = K Hkqtk AM dk mid point gS rks D Hkh Hkqtk AE dk mid 3 gksxkA vr% area dk ratio = 4 point (Mid point theorem) 9 dk vr% vFkkZr~ AD 1 Smaller region area 4 DE = AD, DE = EC = AD =  dk = = 0.8 DC 2 Larger region area 5 tks fd minimum ratio gksxkA Download our app : NEON CLASSES 13 geometry NEON CLASSES, JAIPUR • 9828728833/34

16. Proof of existence (iv) side lengths ls medians dh yEckbZ find dh tk ldrh gS& Ceva's Theorem - bl theorem ds according, AD, BE rFkk CF rHkh concur (intersect) djsaxh tc AD = 2b2+2c2-a2 4 AE CD BF . . = 1 2 2 2 EC DB FA BE = 2a +2c -b 4 A CF = 2a2+2b2-c2 4 Centroid, median dks 2 : 1 esa divide djrk gS vr% F E 2 2 2 AG = 2b +2c -a 3

2 2 2 BG = 2a +2c -b 3

B C 2 2 2 D CG = 2a +2b -c 3 ;gka D, E rFkk F respective sides ds mid points gSa 18. Centroid (G) dh circumcentre (O) ls nwjh vr% AE = EC, CD = DB rFkk BF = FA. 2 2 1 2 2 2 AE CD BF OG = R - (a + b + c ) vr% . . = 1 9 EC DB FA tks ;g prove djrk gS fd fdlh triangle esaa centroid Proof :

exist djrk gSA 2 2 2 2 2 2 2 PA + PB + PC = GA + GB + GC + 3PG Substitute P = O 17. Centroid : OA2 + OB2 + OC2 = GA2 + GB2 + GC2 + 3OG2 A 9R2 = 3 (GA2 + GB2 + GC2) + 9OG2

(∵OA = OB = OC = R)

2 2 2 2 2 L 9R = a + b + c + 9OG (∵ AB2 + BC2 + CA2 = 3 (GA2 + GB2 + GC2)) G K 2 2 1 2 2 2 OG = R - 9 (a + b + c )

B C 19. ABC esa G centroid rFkk O circumcentre gSA rFkk X, Y, Z, BCG, ACG rFkk ABG ds circumcentres gksa fdlh ds fy,] (i) point P rks XYZ dk centroid O rFkk G bl triangle dk 2 2 2 2 2 2 2 PA + PB + PC = GA + GB + GC + 3PG symmedian point gksxkA 2 2 2 2 2 2 (ii) AB + BC + CA = 3 (GA + GB + GC ) XYZ rFkk original ABC ds medians original (iii) ;fn dksbZ line centroid ls xqtjs rFkk AB ds k fcUnw ij ABC dh Hkqtkvksa ds midpoints ij intersect djrs gSaA rFkk AC ds L fcUnw ij intersect djs rks 20. Larger polygons esa centroid, vertices dk centre BK CL KA + LA = 1 of mass rks gksrk gS but t:jh ugh gS fd ;g medians dk intersection point gh gksA

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Q. A line passes through the centroid G of ABC. ∵ABC esa CD median gS rFkk G centroid gS If AL = 2, BM = 6 then CN = ? centroid ekf/;dk dks 2 : 1 esa divide djrk gS vr% DG = 1 CG 2 ,d js[kk ABC ds dsUnzd G ls xqtjrh gSA ;fn AL = 2, rFkk DE = 4 gS vr% CN = 8. BM = 6 gS rks CN = ? B ;fn ABC dks co-ordinate A 6cm plane ij bl izdkj j[kk tk;s 2cm N G fd line LM dh equation y=o. Neon L M Centroid G line l ij fLFkr gS vr% bldk y-coordinate 0 Approach gksxkA

C "007" fdlh Hkh centroid dk y B Sol. cordinate mlds rhuksa D vertices ds y cordinates dk 6cm A gksrk gSA K F average 2cm G A dk y coordinate = 2 N line l E L M B dk y coordinate = 6.  2+6+y = 0  y = -8 3 vr% ls dh nwjh C C centroid G AF || line l, D is the midpoint of AB. = 8.

B In ABF Friends, vc vki le> x;s gksxsa fd eS Geometry dh bl esa ds D;ksa 4 booklet coordinate concepts D explain dj jgk gwaA Co-ordinate ds concepts ls ge geometry ds questions dks cgqr gh fast speed ls solve dj ldrs gSa rFkk /khjs&/khjs esaa dk A K F SSC co-ordinate waitage Hkh c<+rk gh tk jgk gSA ADK ~ ABF Geometry ds questions dks tYnh solve djus ds DK = AD = 1 fy, fuEu ij vkidh idM+ cgqr BF AB 2 7 concepts strong gksuh pkfg,& DK = 2. B rFkk DE = DK + 2 = 4. D (i) Co-ordinate Geometry A (ii) Cyclic Quadrilaterals (iii) Similar and Congruent Triangles G N line l (iv) Isosceles Triangles E M (v) Neon Triangle (vi) Equal Angles and Lengths (vii) Ice Cream Cones C tYn gh esa eSa bu ij DE DG 1 next booklet concepts DEG ~ CNG ⇨ = = CN CG 2 discuss d:axkA

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F  Medial triangle - fdlh triangle ds mid A points dks feykus ls cuus okyk triangle medial triangle dgykrk gSA A K P N

G

P N G B M C M, N rFkk P are mid points.

AK = 3 B M C KG 1 MNP, medial triangle dgykrk gSA

• Area PGM = Area MGN = Area NGP Proof : = 1 Area ABC Medial MNP esa G centroid gSA 12 1 P K N PN = 2 BC (midpoint theorem) GPN ~ GCB x 1 2 2 G Area GPN (PN) (2 BC) 1 Ar ea  GCB = (BC) 2 = BC 2 = 4 Area GPN = 1 Area GCB 4 2x 1 Area GCB = 3 Area ABC M Area GPN = 1 [ 1 Area ABC] = 1 Area ABC 4 3 12 • ABC rFkk medial triangle MNP nksuks dk centroid ABC es G centroid gSA ,d gh point G gSA A A F 2y

P N G y

B M C

B C M y = 2x; 2y = 4x Area APN = area BPM = area MNC = area AG = 4x MNP = 1 area ABC 4 AK + KG = 4x rFkk  Quadrilaterals APMN, PMCN PBMN But KG = x parallelogram (lekUrj prqHkqZt) gSA AK = 3x  Triangle ABC ds altitudes ds feet medial vr% ds ij fLFkr gksrs gSaA AK = 3 triangle MNP circumcircle KG 1

Like Our FB Page : @neon.classes 16 Web.: www.neonclasses.com 4. Circumcentre (ifjdsUnz] O) (ii) Right angled triangle (ledks.k f=Hkqt) Triangle dh rhuksa sides ds perpendicular bisectors A dk intersection point circumcentre dgykrk gSA R A

O O

B C B C Circumcentre ls f=Hkqt ds rhuksa vertices dh nwfj;ka leku ledks.k f=Hkqt esa circumcentre hypotenuse (d.kZ) ds gksrh gS rFkk ;g length circumradius (R) dgykrh gSA mid point ij fLFkr gksrk gSA A (iii) Obtuse angled triangle (vf/kd dks.k f=Hkqt)

A R O

B C

fdlh Hkqtk }kjk circumcentre ij cuk;k x;k angle bl Hkqtk }kjk opposite vertex ij cuk;s x;s angle dk B C double gksrk gSA vf/kd dks.k f=Hkqt esa circumcentre f=Hkqt dh lcls cM+h A Hkqtk dh rjQ ckgj fudy tkrk gSA  F Alternate segment theorem- c O b fdlh dh Li'kZjs[kk rFkk ds vUnj abc circle tangent ( ) circle Area of ABC = 4R 2 triangle dh fdlh ,d side ls cuk angle ml triangle dh 'ks"k nks sides ls cus angle ds equal gksrk gSA B a C

Position of circumcentre- (i) Acute angled triangle (U;wu dks.k f=Hkqt) A

O

B C U;wu dks.k f=Hkqt esa circumcentre f=Hkqt ds vanj fLFkr gksrk gSA Download our app : NEON CLASSES 17 geometry NEON CLASSES, JAIPUR • 9828728833/34

F The Cosine formula : a = b = c = 2R SinA SinC In a ABC, where a, b and c are the lengths of SinB the opposite sides of BAC, ABC, ACB. Q. 1. In ABC, let D be the midpoint of BC. If ADB 0 0 b2+c2-a2 = 45 and ACD = 30 , what is BAD in cosA = 2bc degrees? a2+c2-b2 cosB = 2ac A

2 2 2 cosC = a +b -c 2ab  bls fuEu forms esaa Hkh rearrange fd;k tk ldrk gS& a2 = b2 + c2 - 2bc cosA b2 = c2 + a2 - 2ca cosB c2 = a2 + b2 - 2ab cosC 0 450 30 Q. In a ABC, b = 10, c = 5 and A = 1200. What is B D C the length of a = ? Sol. Let BC = a, AC = b, AB = c Sol. Cosine formula ls] 0 DAC = 15 (exterior D dh value remote a2 = b2 + c2 - 2bc cosA interior angles ds sum ds cjkcj gksrh gSA) 0 0 2 2 0 vFkkZr~ 45 = 30 + DAC = 10 + 5 - 2  10  5 cos120 0 1 ∴DAC = 15 = 125 - 100 (- ) 0 2 ∴BAC =  + 15 = 175 ∴ a = 175 = 13.2 Using Sine Rule, 0 Sin  = Sin45 a/2 c Q. In a ABC, a = 5, b = 7 and c = 10. Find the a ⇨ Sin = 2. 2.c value of A = ? 0 Sin (+150) Sin30 2 2 2 ⇨ = Sol. CosA = b +c -a a c 2bc 0 a 2 2 2 ⇨ Sin (+15 ) = 7 +10 -5 2c 1 1 = 0 2710 Sin (+150) 2  2 2 Sin45 ⇨ = 2 = = = 0 124 Sin 1 1 Sin30 = 2 2 140 ⇨  = 300 -1 124 0 A = Cos ( 140 ) = 27.7 The Sine formula : Method II A A

c b F

R

0 45 0 B a C 30 B D C

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B ls side AC ij ,d perpendicular BF draw djrs gSaA vr% angle B dh 2 values possible gSa& 48.60 and 0 BFC esa] 131.4 0 BC vr% bl ls cuk;s tk ldrs gSaA BF = BCsin30 = = BD = DC information 2 triangles 2 0 0 0 0 0 BFD esa] (i) C = 30 , B = 48.6 , A = 180 - 78.6 = 101.4 0 0 0 0 0 BF = BD rFkk FBD = 600 vr% BFD ,d (ii) C = 30 , B = 131.4 ,A = 180 - 161.4 = 18.6 1 equilateral (leckgq) triangle gSA B vr% FDB = 600 FDA = 600 - 450 = 150 FAD esa] FAD = FDA = 150 B vr% ,d lef)ckgq gSA FAD isosceles triangle ( ) 8 ftlesa AF = FD

;k FD = BF (equilateral triangle BFD) 300 FD = AF = BF C 12 A vr% AFB ,d isosceles triangle gS rFkk Case 1. 0 AFB = 90 a = b SinA SinB FAB = FBA = 450 12.Sin101.40 0 0 0 a = 0 = 15.7 vr% BAD = FAB - FAD = 45 - 15 = 30 Sin48.6 ;g triangle ACB1 gSA Q.2. In a ABC, AB = 8, AC = 12 and ACB = 300. Case 2. Find the side BC. a b = Sol. C SinA SinB 12.Sin18.60 a = 0 = 5.1 Sin131.4 300 2 ;g gSA 1 triangle ACB = a b vr% side BC dh nks values possible gS& 5.1 rFkk 15.7

A c = 8 B Application of properties of circumcentre Tkc fdlh dh nks rFkk ,d triangle sides non- circumcentre cgqr important point gS D;ksafd ;g fn;k x;k gks rks ls vU; included angle Sine rule angles rFkk lengths dks interrelate djrk gSA cgqr ls fd;s tk ldrs gSa rFkk bl esa nks angles find condition questions tgka circumcentre directly involve gSaA distinct triangles possible ugh gksrk ogka Hkh bldh application ls questions dh Sine formula : cgqr fast solve fd;k tk ldrk gSA SinA SinB SinC ABC dk circumcentre O gksxk a = b = c SinB Sin300 If and only if :  = 12 8 0 3 1 (1) AO = BO = CO ∴ SinB = 12  Sin30 = = .75 (Sin300 = ) 8 4 2 (2) BO = CO and ∠BOC = ∠2A, (tgka ∠A = acute (U;wu B = sin-1(0.75) = 48.6 dks.k) rFkk A o O, BC dh same side esa gh gksA) But sinθ = sin (180 - ) (3) BO = CO and ∠BOC = 2 (180-∠A), (tgka ∠A= obtuse Sin48.60 = sin (180 - 48.6) = sin131.40 (vf/kd dks.k) rFkk A o O, BC dh opposite side esa gksA)

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2 2 2 Q.1. Quadrilateral ABCD satisfies ∠A = 760, ∠B = cosA = A C +AB -BC 2AC.AB 720, ∠C = 1420 and AD = AB = 10cm. Find BC = A C = AB = 2R ∠ACB-∠ACD. SinA SinB SinC BC Sol. C sinA = 2R 2 2 2 0 AC +AB -BC AC AB 142 2 2 2 AB +BC +AC = 2A C.AB . 2R . 2R

0 0 D 70 72 B 2 2 2 2 2 2 + BC . AB +BC -A C . AB + BC . A C . BC +A C -AB 2R 2AC.AB 2R 2R 2R 2AB.BC 760 2 2 2 2 2 2 AB +BC -AC A AB +BC +AC = 8R2 1 0 1 = 2 ∠A + ∠B + ∠C + ∠D = 360 8R ∠D = 3600 - (760 + 720 + 1420) R = 1 8 0 ∠D = 70 a+b = 1+8 = 9. rFkk 1 dk ckgjh dks.k AB = AD ∠C = 2 ∠A vr% A circumcentre gSA vFkkZr~ AB = AD = AC Method - II bl esa fuEu C question identity sin (A + B) = sinAcosB + cosAsinB 1420 Hkh apply dj ldrs gSa blds fy, equation dks 2 ls 0 D 70 720 B multiply djus ij] 0 76 2 2 2 0 2(AB + BC + AC ) A 284 = 2 CosA.sinB.sinC + 2 sinA.cosB.sinC +2 sinA.sinB.cosC = sinA (sinB cosC + sinC cosB) Triangle ABC esa] AB = AC vr% buds opposite +sinB (sinAcosC + sinC cosA) angles Hkh equal gksaxsa ∠ABC = ∠ACB = 720 + sinC (sinA cosB + sinB cosA) blh izdkj triangle ACD esa] AC = AD vr% buds = sinA. sin (B + C) + sinB. Sin (A + C) + sinC. Sin opposite angles Hkh equal gksaxsaA ∠ADC = ∠ACD = (A + B) 700 = sin2A+ sin2B + sin2C vr% ∠ACB -∠ACD = 720 - 700 = 20 (∴A+B+C = 1800 rFkk sin (x) = sin(180-x) Sine Rule ls] 2 2 2 2 2 2 Q.2. AB + BC + AC = CosA.sinB.sinC + AB AC BC 2 2 + 2 + 2 = 4R sinA.cosB.sinC + sinA.sinB.cosC Sin C Sin B Sin C 2 2 2 1 AB +BC +AC 2 If the area of the circumcircle of ABC can be 2 2 2 = 4R = Sin C+Sin B+Sin C 2 represented as a  , what is the value of a+b? b 2 1 a ∴ R = ;fn ABC ds ifjo`r ds {ks=Qy dks ls n'kkZ;k tkrk gS rks 8 b a+b dk eku D;k gksxk\ Area of circumcircle = R 2 =  8 Sol. cosA, cosB rFkk cosC dks law of cosines ls rFkk  = a ∴ a = 1, b = 8 sinA, sinB rFkk sinC dks extended law of sines ls 8 b solve djds value put djus ij] ∴ a+b = (1+8) = 9.

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(a) 3 (b) 1 ;g ,d general statement (c) 2 (d) 4 g S Sol. According to question. (SSC Pre - 2013) tks fdlh Hkh triangle ds fy, Neon A true gS vr% ABC dks 2 / Approach specific triangle assume z djds answer fd;k tk ldrk I gSA O "007" y0 /2 Ekkuk ABC ,d isosceles x z0 right angle triangle gSA B C A 2 2 2 1 1 x + x + ( 2x ) = 2 + 0 + 2 0 D 45  x = 1 2 Given ABC = x°, BID = y°, BOD = z° 2x rFkk r = 1 x 2 2  I is the incentre  area = ABI = 1 ABC 8 2 0 0 45  a+ b = 9 ABI = 1 x° = x  BAD = 1 BOD B x C 2 2 2  Angle subtended on the circumcircle is half dks leckgq ekudj Hkh ABC equilateral triangle ( ) the angle subtended on the centre of circle. solve fd;k tk ldrk gSA BAD = 1 BOD Ekkuk ABC dh izR;sd Hkqtk = a 2 0 2 2 BAD = z 3a = 3cos60 (sin60) 2 x0 z0 a = 3  R = a = 1  y° = + (Exterior angle) 8 3 8 2 2 0 0 2 1 Area = R =   a+ b = 9  y° = x +z  2y° = x° + z° 8 2 Now, NOTE : ges'kk /;ku j[ksa ;fn triangle esa dksbZ general 0 0 0 z +x 2y = 0 = 2 relation fn;k x;k gks rks ml question dks ges'kk y y equilateral triangle ;k isosceles right triangle ekudj solve djuk pkfg,A D;ksafd ;fn statement general triangle ds fy, true gS rks og specific triangle ds fy, Hkh true gksxkA ;g question SSC Pre- 2013 esa vk;k gqvk gSA Neon ;fn ABC dks equilateral Q.3. I and O are respectively the incentre and Approach triangle ¼leckgq½ eku ysa rks I o circucentre of a triangle ABC. The line AI O ,d gh fcUnq ij fLFkr gksaxsA produced intersects the circumcircle of ABC "007" 0 0 at the point D. If ABC = x°, BID = y° and ABC = x = 60 x+z BOD = z°, then = ? 0 0 y BID = Y = 60 ABC esa I rFkk O Øe'k% vUr% dsUnz o ifjdsUnz gSa] js[kk AI dks BOD = Z0 = 600 vkxs c

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Relative Position of different (i) Acute angled triangle (U;wu dks.k f=Hkqt) centres : C  Euler Line : Orthocentre (yEcdsUnz] H), centroid (dsUnzd] G) rFkk ifjdsUnz] ges'kk ,d O circumcentre ( O) line segment I G ij fLFkr gksrs gSa] bls Euler line dgrs gSaA Orthocenter rFkk circumcentre dh nwjh dks H centroid 2 : 1 esa divide djrk gSA A B A (ii) Right angled triangle (ledks.k f=Hkqt)

C (H yp ot I en H us e d G s m O id O I p G oi nt ij )

B C O G H A B H ij x 2x H (Right angle vertex ) vf/kd dks.k f=Hkqt F Isosceles triangle (lef}ckgq) esa ;s pkjksa centres (iii) Obtuse angled triangle ( ) Euler line ij gh fLFkr gksrs gSaA C A O (la rg es t H sid G e d s c kgj I d h r I jQ G ) θ > 900 O A B

B C ds ihNs dh rjQ F Equilateral triangle (leckgq) esa ;s pkjksa centres ,d H (Largest angle ) gh ij gh fLFkr gksrs gSaA point F Isosceles triangle (lef)ckgq) A (i) Acute angle Isosceles C

H I O G O G B C I F Euler line ij OH dk midpoint Nine point circle H B dk centre gksrk gSA A Like Our FB Page : @neon.classes 22 Web.: www.neonclasses.com

(ii) Right angled Isosceles A Centroid C (H yp ot en us e d P N s md G O i p G oi nt ij I )

450 450 B M C A B H H (Right angle vertex ij) ABC dk dsUnzd = G MNP dk dsUnzd = G (iii) Obtuse angled Isosceles F Medial triangle MNP dk orthocenter (H), dk gksrk gSA C original triangle ABC circumcentre O

O (la A rg es t sid G e d s c kgj I d h r jQ H ) P N 0 Circumcentre (ABC) A θ > 90 B G Orthocentre (MNP) O

H (Largest angle ds ihNs dh rjQ) B M C F Medial Triangle : F Medial triangle MNP rFkk Original triangle ABC A ds fy, incentre rFkk circumcentre vyx&vyx fcUnqvksa ij fLFkr gksrs gSA

H A P N G

O P N I O B M C o I Original triangle rFkk medial triangle dk centroid (G) ,d gh point gksrk gSA vr% fdlh Hkh B M C triangle ds fy, centroid ges'kk ml triangle ds medial triangle ds vanj gh fLFkr gksrk gSA

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 Euler's theorem :  Orthocentre (H) o circumcentre (O) ds e/; nwjh & Circumcentre (O) rFkk incentre (I) ds e/; dh nwjh dks Euler's theorem )kjk Kkr fd;k tkrk gS& A (i) d2 = R (R - 2r) tgka d = circumcentre o incentre ds e/; dh nwjh H R R = circumradius r = inradius O 1 1 1 (ii) + = R R+d R-d r R (iii) (R - r)2 = d2 + r2 B C Excircles ds fy, Hkh similar equation gS& (iv) (R + r )2 = d2 + r 2 ex ex OH2 = 9R2 - (a2 + b2 + c2) esa ls fdlh ,d dh rex = excircles excircle radius OH = circumcentre o orthocentre ds e/; nwjh d = ml excircle ds centre ls circumcentre dh nwjh Euler inequality- Q. Let a triangle ABC with sides a, b, c. It is given that a2 + b2 + c2 = 29 and the circumradius is 9. fdlh Hkh ds fy,] triangle Find the distance between circumcentre and R  2r orthocenter? f=Hkqt dh Hkqtk,sa rFkk gS] ;fn 2 2 2 Circumradius, ml triangle dh inradius ds nqxqus ls ABC a, b c a + b + c = 29 cM+h ;k nqxqus ds cjkcj gksrh gSA rFkk ifjo`r dh f=T;k 9 gS rks ifjdsUnz rFkk yEc dsUnz ds chp dh nwjh Kkr djks\ Equilateral triangle (leckgq) esa] R = 2r Sol. OH2 = 9R2 - (a2 + b2 + c2)  rFkk ds Euler Line (centroid circumcentre OH = circumcentre o orthocentre ds e/; nwjh e/; distance) OH2 = 9 (9)2 - (29) 2 2 1 2 2 2 2 (GO) = R - 9 (a + b + c ) OH = 729 - 29 (GO) = centroid o circumcentre ds e/; nwjh OH = 700 R = circumradius a, b, c = sides of a triangle Q. Let a triangle ABC with sides a, b, c. It is given that a2 + b2 + c2 = 29 and the circumradius is 9. A Find the distance between circumcentre and centroid? f=Hkqt ABC dh Hkqtk,sa a, b rFkk c gS] ;fn a2 + b2 + c2 = 29 rFkk ifjo`r dh f=T;k 9 gS rks ifjdsUnz rFkk dsUnzd ds chp dh G R nwjh Kkr djks\ Sol. (GO)2 = R2 - 1 (a2 + b2 + c2) 9 O GO = centroid o circumcentre ds e/; nwjh R 2 2 1 R (GO) = (9) - 9 (29) (GO)2 = 81 - 29 = 729-29 B C 9 9 700 GO = 3 Like Our FB Page : @neon.classes 24 Web.: www.neonclasses.com

(5) Excentre fdlh dks.k dk internal bisector mlds external ds gksrk gS vr% rhuksa fdlh triangle ds nks external angle bisectors rFkk bisector perpendicular ls cuk rFkk ,d internal angle bisector dk dVku fcanw Excentre excentres triangle (JA JB JC) incentre ,d cukrs gSaA dgykrk gSA ftl vertex ('kh"kZ fcanw) ls internal angle (I) orthocentric system bisector line (vUr% dks.k lef)Hkktd js[kk) vkrh gS mlh A ds respect esaa excentre dks define fd;k tk ldrk gSA α Excentre ls cuus okyk circle f=Hkqt dh ,d Hkqtk rFkk 'ks"k I nks Hkqtkvksa ds extensions dk tangent (Li'kZjs[kk) gksrk ω gSA A B /2 /2 C λ/2 δ/2 λ/2 2 2 δ/2 / /

 

θ

JA B C K /2 /2 ds fy,] /2 Vertex B /2  λ 0 0  + λ = 180  2 + 2 = 90 vr% internal bisector rFkk external bisector ,d

JA nwljs ds perpendicular (yEcor~) gSaA prqHkqZt IBKC esaa ∠B o ∠C dk ;ksx 1800 gS vr% ;g ,d cyclic quadrilateral (pØh; prqHkqZt) gSA vr% ω +  = 1800 α ω = 90 + 2 α 0 90 + 2 +  = 180

α bl ds fy, ls excentre vertex A internal angle  = 90 - 2 gS vr% ;g dgykrk bisector line excentre of A (JA) gSA Excircles dh f=T;k;sa exradii dgykrh gSA

A JC A JB

c b

I B C B C

E D r JA ra a J

rhuksa dks feykus ls cuus okyk f=Hkqt ,d excentres JA JB JC excentral triangle dgykrk gSA

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Excircle tks side BC ds tangent gS] extended side mijksDr formulas ls Li"V gS fd excircles ges'kk AC dks point D ij Li'kZ djrk gSA vr% JD side AC ij incircle ls cM+s gksrs gSaA largest excircle, lcls cM+h Hkqtk yEcor~ gS vr% dh ds gksrk gS rFkk lcls NksVh perpendicular ( ) ACJ height = ra tangent smallest excircle 1 Hkqtk ds tangent gksrk gSA Area of ACJ = . b . ra 2 bu lHkh formulas dks combine djus ij] By similar argument,  = r. ra. rb. rc Area of ABJ = 1 . c . r 2 a fdlh triangle ds incentre ds fy, trilinear 1 coordinates gksrs gSa& 1 : 1 : 1 Area of BCJ = 2 . a . ra rFkk ds fy, Area of ABC, excentres trilinears = ACJ + ABJ - BCJ (-1 : 1 : 1), (1 : -1 : 1), (1 : 1 : -1) 1 1 1 r + r + r = 4R + r = 2 b ra + 2 c ra - 2 a. ra a b c 1 2 2 2 2 2 ra + rb + rc = (4R + r) - 2s = 2 (b + c - a) . ra a+b+c  rhuksa excircles ds centres ls xqtjus okys circle dh = (s - a) r (s = ) a 2 radius = 2R Symmetry ls]  ;fn H fdlh triangle ABC dk orthocentre gS rks& Area of ABC = s . r = (s - a) ra = (s - b) rb = (s - c) rc r + r + r + r = AH + BH + CH + 2R Cosine ds rule ls] a b c b2+c2-a2 NOTE : CosA = 2bc (i) dqN quadrilaterals esa Hkh incircle cuk;k tk ldrk gSA Identity Sin2A + Cos2A = 1 ls combine djus ij] ;s tagential quadrilaterals dgykrs gSaA 4 4 4 2 2 2 2 2 2 SinA = -a -b -c +2a b +2b c +2a c budh opposite sides ds nks pairs dk sum equal gksrk 2abc gSA ;g pitot theorem dgykrh gSA 1 Area of triangle ABC = 2 bc sin A d D 1  = -a4-b4-c4+2a2b2+2b2c2+2a2c2 4 A a 1 d  = (a+b+c). (-a+b+c) (a-b+c) (a+b-c) 4 a  = s(s-a) (s-b) (s-c) O ;g Heron's formula gSA bl dks] Heron's formula b c  = s . r ls combine djus ij] B b 2 c 2  (s-a) (s-b) (s-c) C r = 2 = s s both pairs of opposite sides sum to a+b+c+d. similarly,  = (s - a). ra

2  2  ;g t:jh ugha gS fd lHkh ¼cgqHkqt½ esa r = r = 2 (ii) polygons incircle a (s-a) a (s-a) exist djrk gks] ;fn fdlh polygon esa incircle gksrk gS rks s(s-b) (s-c) 2 = og tangential polygondgykrk gS] rFkk polygon ds ra s-a angle bisectors dk intersection point incentre s(s-b) (s-c) ra = dgykrk gSA s-a

Like Our FB Page : @neon.classes 26 Web.: www.neonclasses.com 1  ΔOAB ds fy;s 5 classical centres ds co - r = Ar ea Δ ABC = 2×12×5 s 15 ordinates  I dscoordinates , I = (r, r) = (2, 2) ,d ledks.k f=Hkqt gS 2 2 2 ΔOAB right angled triangle ( )  Incircle ⇒ (x - 2) + (y - 2) = 2 blds fy;s 5 classical centres ds co - ordinates  fdlh Δ ABC ds fy;s ftldh sides a, b, c gS rFkk 6 B inradius r gks rks (s - a) (s -b)(s-c) 4 r = s b = 5 B 3. Centroid (dsUnzd , G) 2 6 B 0 A 0 0 2 4 6 8 10 12 4 Origin a = 12 Q P G O (0, 0), A (12, 0), B (0, 5) 2 1. Orthocentre (yEc dsUnz H) 0 A -2 O 0 2 4 6 8 10 12 14 6 B -2

D 0+12+0 0+0+5 4 G (x, y) = , 3 3 = (4, 5 ) 2 3 Centroid ds coordinates fUkEu rjhds ls Hkh find fd;s 0 A tk ldrs gS - O 0 2 4 6 8 10 12 H OR : RA = 1: 1 origin (0) tks fd right - angle okyk vertex gS blh vr% R = (6, 0) vertex ij ortho centre fLFkr gksxkA BG : GR = 2 : 1 ortho centre H = 0 (0, 0) 2×6+1×0 2×0+1×2.5 Centroid , G = , 2+1 2+1 = (4 , 5 ) 2. Incentre (avar% dsUnz , I) 3 4. Circumcentre ( ifjdsUnz, 0) 5 B 8 4 Z 3 r 6 B X 2 r I 4 1 O' r 2 0 A A -1 O 0 1 2 3 4 5 6 7 8 9 10 11 12 1 0 Y O 0 2 4 6 8 10 12 -2 OA = a = 12 -4 OB = b = 5 AB = c = 122 + 52 = 13 right angled triangle ds fy;s circumcentre mlds a + b + c Semi perimeter , S = = 15 hypotenuse (d.kZ) ds e/; fcUnw ij fLFkr gksrk gS A 2 Incircle dh radius , R = IX =IY = IZ O'(x , y) = 12+0 , 0+5 = (6, 2.5) Area of Δ ABC = r.s 2 2

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 a, b, c Hkqtkvksa okys triangle dk area  fdlh Hkh triangle (not only for right angled

2 triangle) ds fy, excircles dh f=T;k;sa & Area = abc = 2R sin A sinB sinC 4R r = s (s -b)(s-c) R = ifjdsUnz dh f=T;k 1 s-a A , B , C = ΔABC ds angles s (s-a)(s-c) r2 = s-b s (s -a)(s-b) ckg~; dsUnz r3 = 5. Ex - centre ( , J) s-c 30 25 20  Important Points r3 J3 15 10 1 5 B A r2 J2 A 0 r1 O 0 5 10 15 20 25 30 35 40 -5 -10 r 1 J1 -15 -20 r r2 r3 Excircle f=Hkqt ds ckgj fLFkr gksrk gS rFkk f=Hkqt dh fdlh ,d Hkqtk rFkk 'ks"k nks Hkqtkvksa ds extensions ds tangent B C (Li'khZ;) gksrk gS A fdlh triangle esa izR;sd f=Hkqt ds 3 ex-circles gksrs gS rFkk izR;sd Hkqtk ds ,d r = r 1 . r 2 + r 2 . r 3 + r3. r1 distinct excircle tangent gksrk gSA excentre f=Hkqt ds fdlh ,d angle ds internal bisector (var% dks.k ;fn ;g triangle equilateral gks rks lef}Hkktd js[kk) rFkk 'ks"k nks Hkqtkvksa ds external r1 = r2 = r3 bisectors (ckg~; dks.k lef}Hkktd js[kk) ds r = 3r intersection point (dVku fcUnw) ij fLFkr gksrk gSA 1 Area of ΔOAB 2 ,d incredibly useful property ;g gS fd

= Area of ΔJ1 AB + ΔJ1OB - ΔJ1OA orthocentre dk reflection fdlh Hkh side ds fy, ges'kk triangle ds circumcircle ij fLFkr gksrk gSA ab cr1 br1 ar1 b+c-a a+b+c-2a = + - = [ ] r1 = [ ] 2 2 2 2 2 2 bl property dks vU; visual way esa Hkh ns[kk tk ldrk a+b+c gS& fdlh circular piece of paper esa ;fn ,d = -a = (s - a )r 2 1 triangle inscribed fd;k tk;s rFkk bl triangle dh ab 12x5 rhuksa sides ds according paper dh arcs dks eksM+k tk;s ∴r = = = 10 1 2 (s-a) 2 (15-12) rks ;s rhuksa arcs ges'kk orthocentre ij gh feyrh gSA Similarly, r = ab = 12x5 esa] fdUgh Hkh nks rFkk 2 2 (s-a) 2 (15-5) 3 Triangle ABC vertex orthocentre H ls cuus okys triangles ds r = ab = 12 x5 = 15 3 2 (s-a) 2 (15-13) circumcircles rFkk original triangle ABC dk

Conclusion : Ex-centres : J1 = (r1,-r1) = (10, - 10) circumcircle congruent (equal radius) gksrs gSa D;ksafd BHC dk circumcircle, orthocentre H dk J2 = (-r2, r2) = (-3, 3) Locus gksrk gS rFkk vertex A original circumcircle

J3 = (r3, r3) = (15, 15) ds around ?kqerk gSA

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Q.1. How many triangles are there with integer Øe'k% 19, 22 rFkk 23 gSA rks AF dh yEckbZ D;k gksxh\ side lengths such that the area of the triangle (a) 14 (b) 11 (c) 12 (d) 13 formed by joining the orthocentre, circumcentre and centroid of ABC = 44 Q.5. Find the area of circumcircle of an isosceles square units? 0 ,sls fdrus f=Hkqt gSa ftudh Hkqtk dh yEckbZ iw.kkZad gS] rFkk triangle ABC, where a = b and A = 15 , mudss yEcdsUnz] ifjdsUnz rFkk dsUnzd dks feykus ij cus f=Hkqt provided that the perimeter of the triangle is ABC dk {ks=Qy 44 oxZ ,dd gS \ 25cm. ,d lef}ckgq ds ifjo`r dk {ks=Qy D;k gksxk] tgka (a) 3 (b) 2 ABC a = b rFkk A = 150 gS rFkk f=Hkqt dk ifjeki 25 lseh gSA (c) 1 (d) 0 (a) .384 cm2 (b) .259 cm2 (c) .121 cm2 (d) .304 cm2 Q.2. In ABC with centroid G, if AG = BC, what is the angle BGC in degrees? dsUnzd G okys ABC esa] ;fn AG = BC rks ∠BGC dk eku Q.6. ABC is an acute angle triangle with points D fMxzh esaa gksxk\ and E on BC and AC respectively. Such that BE (a) 900 (b) 650 and AD are altitudes. AD and BE intersect at H. 0 0 0 if ∠BCA = 42 and ∠EBA = 2 ∠DAB. What is the (c) 48 (d) 70 measure of ∠ABC? ,d U;wu dks.k ABC esa] Hkqtk BC o AC ij fcUnw dze'k% D Q.3. Let A = (2, 12), B = (10, 0) and C = (0, 0) be the rFkk E gSA BE rFkk AD ÅapkbZ;ka gSa rFkk ,d nwljs dks H ij vertices of triangle ABC. If G is the centroid of 0 dkVrh gSaA ;fn ∠BCA = 42 rFkk ∠EBA = 2∠DAB gSA the triangle. What are the coordinate of G and rks ∠ABC = ? the area of  AGB. (a) 780 (b) 760 (c) 750 (d) 770 ;fn A= (2,12), B = (10,0) rFkk C = (0,0), ABC ds 'kh"kZ gSaA ;fn G dsUnzd gks rks] dsUnzd ds funsZ'kkad rFkk AGB dk {ks=Qy D;k gksxk\ Q.7 In a ABC, point O is the circumcentre. If ∠AOB (a) (2, 4), 20 (b) (4, 4), 26 : ∠BOC : ∠COA = 2 : 3 : 4. What is the measure of ∠BAC? (c) (4,4) , 20 (d) (2, 4), 26 B ABC esaa] O ifjdsUnz gSA ;fn ∠AOB : ∠BOC : ∠COA = 2 : 3 : 4 gS rks ∠BAC = ? Q.4. (a) 900 (b) 450 0 0 F D (c) 60 (d) 75 I

Q. 8. A triangle has vertices at A = (0, 0), B = (14, 0) and C = (5, 12). What are the coordinates of the A E C orthocentre? ,d f=Hkqt ds 'kh"kZ A = (0,0), B = (14,0) rFkk C = (5, I is incentre of ABC. The lengths of BC, CA 12) gSaA rks yEcdsUnz ds funsZ'kkad D;k gksaxsa\ and AB are 19, 22 and 23 respectively. What is 13 13 (a) (12, ) (b) ( , 15 ) the length of AF? 2 2 4 (c) (5, 13 ) (d) (5, 15 ) I ABC dk vUr%dsUnz gSA BC, CA rFkk AB dh yEckbZ;ka 2 4

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Q.9 The sides of triangle ABC are 5, 6 and 7. P is a Q.12. What is the circumcentre of ABC with point in the plance of the triangle such that PA2 vertices A = (1,4), B = (-2, 3), C = (5, 2) ? 2 2 + PB + PC = 70. Find the distance of P from the 'kh"kZ A = (1, 4), B = (-2, 3), C (5, 2) okys f=Hkqt dk centroid? ifjdsUnz gS\ ABC dh Hkqtk,a 5, 6 rFkk 7 gSA f=Hkqt ds vUnj fcUnw P bl (a) (1, -1) (b) (0, 1) 2 2 2 izdkj gS fd PA + PB + PC = 70 gSA rks dsUnzd ls fcUnw P (c) (2,0) (d) (0, 4) dh nwjh Kkr djksA (a) 9 (b) 10 2 3 Q.13. In an isosceles ABC, AB = AC and ∠A is two 13 9 (c) 3 (d) 4 times of ∠B. If AB = 3cm, then ratio of inradius to the circumradius is. 4 ft. Q.10. C B ,d lef}ckgq ABC esa] AB = AC rFkk ∠A = 2∠B gSA ;fn AB = 3 lseh gks rks vUr%f=T;k o ifjf=T;k dk vuqikr D;k gksxk\ (a) 1 : 2 (b) 2 -1 : 1 (c) 2 2-1 : 1 (d) 1 : 2 2-1 4 ft. 4 ft. O. Q14. Let a, b, c be the side lengths of ABC and let d, e, f be the distances from its centroid G to the a2+b2+c2 vertices. What is the ratio 2 2 2 ? A d +e +f ABC is an equilateral triangle with sides ;fn ABC dh Hkqtk,a a,b,c gSa rFkk blds dsUnzd G ls 'kh"kksZa 2 2 2 equal to 4ft. A soccer ball with radius one foot dh nwfj;ka d, e, f gSA rks a +b +c vuqikr gS\ d2+e2+f2 is adjusted according to the figure. Find the (a) 7 (b) 3 distance from the centre of ball O to the vertex 3 (c) 1 (d) 7 A. 3 3 ABC, 4ft Hkqtk okyk leckgq f=Hkqt gSA fp=kuqlkj] ,d 1 ft f=T;k okyh QqVckWy xsan dks lek;ksftr fd;k gSA xsan ds dsUnz Q.15. Consider a right angled triangle with inradius O ls 'kh"kZ A dh nwjh Kkr djks 2 cm and circumradius of 7 cm. What is the area of the triangle? (a) 5 ft (b) 1 ft ,d ledks.k f=Hkqt ftldh vUr%f=T;k 2 lseh rFkk ifjf=T;k (c)2 ft (d) 6 ft 7 lseh gSA f=Hkqt dk {ks=Qy Kkr djksA (a) 32 sq cms (b) 31.5 sq cms Q.11. Three circles of equal radii all intersect at a (c) 32.5 sq cms (d) 33 sq cms single point P. What is point P for ABC? (a) Incentre (b) Circumcentre Q.16. If O be the orthocenter of ABC, OF ⟘ r AB and (c) Centroid (d) Orthocentre OE ⟘ r AC. If OE = 2cm and BE = 5cm then find leku f=T;k okys rhu o`r ,d gh fcUnw P ij dkVrs gSaA the value of OF × OC. ABC ds fy, fcUnw P gS\ ;fn O, ABC dk yEcdsUnz gS rFkk OF ⟘r AB, OE ⟘r vUr%dsUnz ifjdsUnz A (a) (b) gSA ;fn lseh rFkk lseh gS rks z . AC OE = 2 BE = 5 OF × OC (c) dsUnzd (d) yEcdsUnz x . dk eku Kkr djksA (a) 10 cm (b) 3 cm C P B . y (c) 6 cm (d) 2 cm

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17. In a ABC, a random line FE passes through its (a) 5.24 cm2 (b) 3.25 cm2 centroid, segmenting it into two regions. Find (c) 2 cm2 (d) 0 cm2 the minimum possible ratio of the area of the

smaller region to the area of the larger region. 2 2 2 20. A triangle has sides a , b and c . Then the ,d esa] ,d js[kk blds dsUnzd ls xqtjrh gqbZ bls ABC FE triangle with sides a, b, c has to be:- nks Hkkxksa esa ckaVrh gSA rks NksVs Hkkx ds {ks=Qy dk cM+s Hkkx ds {ks=QYk ls lcls de laHkkfor vuqikr Kkr djksA (a) Right-angled (b) Acute-angled A (c) Obtuse-angled (d) Can be any of these three ,d f=Hkqt dh rhu Hkqtk,a a2, b2 rFkk c2 gSaA rks a, b, c Hkqtkvksa okyk f=Hkqt gksxk&

E (a) ledks.k (b) U;wu dks.k (c) vf/kd dks.k (d) rhuksa eas ls dksbZ Hkh G 21. A triangle has vertices at A = (0, 0), B = (14, 0) and C = (5, 12). What are the coordinates of the B F C incentre? (a) 0.70 (b) 0.80 ,d f=Hkqt ds 'kh"kZ A = (0,0), B = (14, 0) rFkk C = (5, (c) 0.25 (d) 0.75 12) gS rks vUr%dsUnz ds funsZ'kkad D;k gksaxsa\ (a) (4, 6) (b) (6, 4) 18. Two circles are placed in an equilateral (c) (5, 3) (d) (2, 5) triangle as shown in the figure. What is the ratio of the area of the smaller circle to that of 22. An interior designer is creating a custom the equilateral triangle? fp= esa fn[kk, vuqlkj] ,d leckgq f=Hkqt esa nks o`r lekfgr coffee table for a client. The top of the table is a gSaA rks NksVs o`r ds {ks=QYk dk leckgq f=Hkqt ds {ks=Qy ls glass triangle that needs to balance on a single vuqikr Kkr dhft,A support. If the coordinates of the vertices of A the triangle are at (3, 6), (5, 2), (7, 10), at what point should the support be placed? ,d baVhfj;j fMtkbuj fdlh xzkgd ds fy, dkWQh Vscy cukrk gSA Vscy dk VkWi dk dkap f=Hkqtkdkj gS ftls fdlh ,d gh lgkjs dh t:jr gSA ;fn f=Hkqt ds 'kh"kksZa ds funsZ'kkad (3, 6), I (5, 2), (7, 10) gS rks lgkjk fdl fcanw ij j[kk tk,\ (a) (4, 5) (b) (5, 6) B D C (c) (3, 2) (d) (2, 3) (a) π : 36√3 (b) π : 18√3 (c) π : 27√3 (d) π : 42√3 23. In ABC, G is the centriod, AB = 15 cm, BC = 18cm, and AC = 25 cm. Find GD, where D is the 19. Consider an isosceles triangle ABC with mid point of BC. AB = AC = 5cm, BC = 6cm, where I, O, H denote ABC esa] G dsUnzd gS rFkk AB = 15 lseh] BC = 18 lseh its incentre, circumcentre, orthocentre rFkk AC = 25 lseh gSA ;fn D, BC dk e/; fcUnw gS rks GD respectively. Find the area of IOH. Kkr djksA ,d lef}ckgq ABC esa] AB = AC = 5 lseh] rFkk BC = 6 (a) 1 cm (b) 1 cm 2 86 3 86 lseh gS rFkk I, O, H dze'k% vUr%dsUnz] ifjdsUnz rFkk yEcdsUnz 7 2 (c) cm (d) cm gSaA rks IOH dk {ks=Qy Kkr djksA 6 86 3 86

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24. In a parallelogram, the lengths of adjacent 28. Triangle ABC is an isosceles triangle with sides sides are 12cm and 14 cm respectively. If one AB = BC and ∠ABC = 1230. Point D is the of the diagonals is 22 cm, find the length of the midpoint of AC, point E is the foot of the other diagonal, the perimeter and area of the perpendicular from D to BC, and point F is the parallelogram. midpoint of DE. The intersection point of AE ,d lekUrj prqHkqZt esa] laxr Hkqtkvksa dh yEckbZ;ka dze'k% 12 and BF is G. find ∠BGA? lseh rFkk lseh gSA ;fn ,d fod.kZ lseh gS rks nwljs 14 22 ,d lef}ckgq esa] Hkqtk rFkk 0 fod.kZ dh yEckbZ rFkk lekUrj prHkqZt dk ifjeki o {ks=Qy ABC AB = BC ∠ABC = 123 Kkr djksA gSA D, AC dk e/;fcanw gS] D ls BC ij Mkys x, yEc dk dVku fcanw gS rFkk dk e/;fcanw gSA rFkk dk dVku (a) 14 cm, 52 cm, 48 10 cm2 E F, DE AE BF 2 fcanw G gS rks ∠BGA = ? (b) 14 cm, 60 cm, 24 15 cm 0 0 (c) 14 cm, 52 cm, 45 10cm2 (a) 90 (b) 45 (d) 12 cm, 52 cm, 154 cm2 (c) 600 (d) 750

25. A = (17, 4) is a vertex of ABC and O = (0,0) is 29. ABC has vertices A (-11,4), B (-3,8), and C (3, its circumcentre. P, Q and R the midpoints of -10). The coordinates of the center of the circle sides AB, BC and CA respectively. If the circumscribed about ABC are orthocentre of PQR is H, then what is the equation of line AH? ABC ds 'kh"kZ A (-11,4), B (-3,8), and C (3,-10) gSaA rks ds ifjo`r ds dsUnz ds funsZ'kkad Kkr djksA ABC dk ,d 'kh"kZ A = (17, 4) rFkk ifjdsUnz O = (0,0) ABC gSA P, Q rFkk R dze'k% Hkqtkvksa AB, BC, CA ds e/; fcUnw gSaA (a) (-2, -3) (b) (-3, -2) ;fn PQR dk yEcdsUnz H gks rks js[kk AH dk lehdj.k gksxk\ (c) (3,2) (d) (-1, -1) (a) y = 17 x (b) y = 17x 4 (c) y = 21x (d) y = 4x 30. Three equal circles are placed inside an equilateral triangle such that any circle is 26. Let ∆PQR be the pedal triangle of ∆ABC. tangential to two sides of the equilateral Excentre of ∆PQR are (20,8), (4, 12) and (13, triangle and to two other circles. What is the 1). If one of the vertices of ∆PQR is (14, 2) then ratio of the areas of one circle to that of the find the area of ∆ABC. triangle? fdlh ∆ABC dk pedal triangle ∆PQR gSA ∆PQR dk Rkhu cjkcj o`r ,d leckgq f=Hkqt esa bl izdkj j[ksa x, gSa fd dksbZ Hkh o`r leckgq f=Hkqt dh nks Hkqtkvksa dks Li'kZ djrk gS rFkk excentre (20,8), (4, 12) rFkk (13, 1) gSA ;fn ∆PQR nwljs nks o`rksa dks HkhA rks ,d o`r ds {ks=Qy dk f=Hkqt ds dk ,d 'kh"kZ gS rks dk {ks=Qy Kkr djksA (14, 2) ∆ABC {ks=Qy ls vuqikr Kkr djkssA (a) 55 sq. unit (b) 62 sq. unit (a) π : (6+4√3) (b) 3π : (6+4√3) (c) 70 sq. unit (d) None (c) 2π : (6+4√3) (d) π : (6+2√3)

27. If G is the centriod of ABC and AG = BC, then ∠BGC is: ;fn G, ABC dk dsUnzd gS rFkk AG = BC gS rks ∠BGC gS\ (a)750 (b) 450 (c) 900 (d) 600

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31. Triangle ABC has a right angle at C. point D lies 35. In a ∆ABC, three sides are 5 cm, 4cm and 2cm. on hypotenuse AB such that CD is Find the length of median from smallest angle perpendicular to AB. if AD = 4 cm and BD = vertex. 9cm. what is the area of triangle? ∆ABC esa] rhu Hkqtk,a 5 lseh] 4 lseh rFkk 2 lseh gSA lcls ∆ABC, C ij ledks.k gSA fcanw D, d.kZ AB ij bl izdkj gS fd NksVs dks.k okys 'kh"kZ ls ekf/;dk dh yEckbZ Kkr djksA CD, AB ij yEcor gSA ;fn AD = 4 lseh rFkk BD = 9 lseh (a) 41 (b) 39 gS rks f=Hkqt dk {ks=Qy Kkr djksA 3 2 22 35 2 2 (c) (d) (a) 39 cm (b) 44 cm 3 3 (c) 26 cm2 (d) 20 cm2 36. In a ∆ABC, BD & CE are the two medians which intersect each other at right angle, if AB = 22, 32. In ∆ABC AD, BE and CF are the altitutdes in the AC = 19, find BC = ? ratio 1 : 2 : 3 respectively, then the ratio of AB : ∆ABC esa] nks ekf/;dk,a BD rFkk CE ,d nwljs dks ledks.k BC : CA is. ij dkVrh gSaA ;fn AB = 22, AC = 19 gS rks BC = ? ∆ABC esa] ÅapkbZ;ksa AD, BE, CF dk vuqikr 1 : 2 : 3 gSA rks (a) 9 cm (b) 11 cm AB : BC : CA dk vuqikr gS\ (c) 14 cm (d) 13 cm (a) 6 : 2 : 3 (b) 3 : 2 : 1 (c) 2 : 4 : 9 (d) 2 : 6 : 3 37. G is the centriod of ABC. The medians AD and BE intersect at right angles. If the lengths of AD 33. In triangle ABC, DE ∥ BC where D is a point on and BE are 9 cm and 12 cm respectively then the AB and E is point on AC. DE divides the area of length of AB (in cm) is? ∆ABC into two equal parts. Then DB : AB is ABC dk dsUnzd G gSA ekf/;dk,a AD rFkk BE ,d nwljs dks equal to ledks.k ij dkVrh gSA ;fn AD rFkk BE dh yEckbZ;ka dze'k% lseh rFkk lseh gks rks dh yEckbZ gksxh ¼lseh esa½ \ ∆ABC esa] DE || BC tgka D, AB ij rFkk E, AC ij dksbZ fcanw 9 12 AB gSA DE, ∆ABC ds {ks=Qy dks nks cjkcj Hkkxksa esa ckaVrh gSA rks (a) 11 (b) 10 DB : AB cjkcj gS\ (b) 10.5 (d) 85 (a) 2 : ( 2 + 1) (b) ( 2 - 1): 2 (c) 2 : ( 2 - 1) (d) ( 2 + 1): 2 38. In ∆ABC, BE, AD, CF is a median on AC, BC and AB respectively. AD = 10, BE = 6 and CF = 8 cm. Then find the area of ∆ABC ? 34. O is the circumcentre of a triangle ABC whose ∆ABC esaa] Hkqtkvksa AC, BC rFkk AB ij ekf/;dk,a dze'k% BE, ∠A = 50°. If bisector of ∠OBC and ∠OCB AD rFkk CF gSA AD = 10, BE = 6 rFkk CF = 8 cm gS rks intersect at P then what is the measure of ABC dk {ks=Qy gksxk\ ∠BPC? 2 2 0 (a) 24 cm (b) 48 cm ∆ABC dk ifjdsUnz O gS rFkk ∠A = 50 gSA ;fn ∠OBC 2 2 rFkk ∠OCB ds lef}Hkktd P ij dkVrs gSa rks ∠BPC dk eku (c) 52 cm (d) 32 cm gS\

0 0 39. The medians AD, BE CF of a triangle ABC (a) 125 (b) 105 0 0 intersect in G. Which of the following is true (c) 140 (d) 90 for any ∆ABC. ∆ABC dh ekf/;dk,a AD, BE rFkk CF fcanw G ij dkVrh gSaA rks ∆ABC ds fy, dkSulk lR; gS\ (a) GB + GC = 3GA (b) GB + GC < GA (c) GB + GC > GA (d) GB + GC = GA

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40. ∆ABC is an equilateral triangle (leckgq f=Hkqt) 44. The two sides of a triangle are 8 cm and 9 cm of Side 6 cm. Find the area of shaded region I and one angle is 600. Which of the following and II ? can be the length of its third side? 0 ∆ABC, 6 lseh Hkqtk okyk leckgq f=Hkqt gSA rks Nk;kafdr Hkkx ,d f=Hkqt dh nks Hkqtk,a 8 lseh o 9 lseh gS rFkk ,d dks.k 60 gSA rks rhljh Hkqtk dh yEckbZ D;k gks ldrh gSA I rFkk II dk {ks=Qy Kkr djksA A I. √23 cm I II. √73 cm III. (4.5-√3.25) cm IV. (4 + √33) cm II V. (9 + √13) cm (a) Only II and IV B C (b) Only I and III  11 (a) 2 , 4 3 18 3 18 (c) Only I, II and V (b) 1  , 4 11 3 9 3 18 (d) Only II, III and IV (c) 1  , 2  3 9 3 9   45. Alice made a collage with pictures of her (d) 2 , 2 3 9 3 18 friends. She wants to hang the collage from the ceiling in her room so that it is parallel to the 41. Find the area of rhombus ABCD given that the ceiling. A diagram of the collage is shown in radii of the circumcircles of triangles ABD and the graph at the right. At what point should ACD are 12.5 and 25 respectively. she place the string? leprqHkqZt ABCD dk {ks=Qy Kkr djks] fn;k gS fd ∆ABD ,fyl vius friends dh QksVks ls ,d dksykt cukrh gSA og rFkk ∆ACD ds ifjo`r dh f=T;k,a Øe'k% 12.5 rFkk 25 gSA bl dksykt dks flfyax ds lekUrj yVdkuk pkgrh gSA dksykt (a) 400 (b) 360 fp= esa fn[kk;k x;k gSA rks og jLlh dks fdl fcanw ij j[ksxh\ (c) 290 (d) 384 (0, 8)

42. The area of QED = 750. QE = 48 and QD = 52. To the nearest degree, what is the measure of the largest possible angle of QED? (6, 4) QED dk {ks=Qy 750 gSA QE = 48 rFkk QD = 52 gSA rks QED ds lcls cM+s laHkkfor dks.k dk eku Kkr djksA (3, 0) (a) 76 (b) 77 (c) 78 (d) 143 (a) (3, 4) (b) (4, 3) (c) (8, 3) (d) (3, 5) 43. If O is the circumcentre of a triangle ABC lying inside the triangle then ∠OBC + ∠BAC is equal to ;fn O, ∆ABC ifjdsUnz gS tks f=Hkqt ds vUnj fLFkr gSA rks ∠OBC + ∠BAC cjkcj gS\ (a) 120° (b) 110° (c) 90° (d) 60°

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46. An isosceles triangle ABC is right angled at B. cm gSA rFkk AD = DE = EF = FC rks BD2 + BE2 + BF2 D, is a point inside the triangle ABC. P and Q are dk eku Kkr djksA ¼lseh- esa-½ the feet of the perpendiculars drawn from D (a) 10,000 (b) 5,000 on the side AB and AC respectively of ∆ABC. If (c) 8,750 (d) 12,500 AP = acm, AQ = b cm and ∠BAD = 150. Find the value of sin 750 ? 49. The equation of the perpendicular bisector of ,d lef}ckgq f=Hkqt fcanw ij ledks.k gSA ds ABC B D, ABC the segment joining A (-9,2) to B (3,-4) is vUnj dksbZ fcanw gSA D ls Hkqtkvksa AB rFkk AC ij Mkys x, yEc A (-9,2) ls B (3,-4) dks feykus okys [kaM ds yEc Hkqtkvksa dks fcanw P rFkk Q ij dkVrs gSaA ;fn AP = a lseh rFkk lef}Hkktd dk lehdj.k Kkr djksA AQ = b lseh rFkk ∠BAD = 150 gS] rks sin750 dk eku gS\ -1 -1 (a) y − 1 = ​ 2 (x - 3) (b) y + 1 = 2 (x + 3) (a) AD (b) AP AP AD (c) y + 1 = 2(x + 3) (d) y + 3 = 2(x + 1) (c) AQ (d) AP T AD AQ 6 47. If in the given figure, ∠PQR = 900, O is the centriod of PQR. if PQ = 5cm and QR = 12cm, C then OQ is equal to. 10 ;fn fn, x, fp= esaa] ∠PQR = 900, PQR dk dsUnzd O gSA ;fn PQ = 5 lseh rFkk QR = 12 lseh gS rks OQ dk eku gSA A O B P

M 50. ABC is an equilateral triangle of Side 6 cm. O Find the area of shaded region I and II. ABC 6 lseh Hkqtk okyk ,d leckgq f=Hkqt gSA Nka;kfdr Hkkx I rFkk II dk {ks=Qy Kkr djksA Q R 1 1 A (a) 3 2 (b) 4 3 I 1 (c) 4 1 (d) 5 2 3

A r 48. D II B C

E 2  2  (a) r ( 3 - ) ; r (2 - ) 3 2 2  2  F (b) r ( 3 - ) ; r (4 - ) 2 3 2  2  (c) r (2 - ) ; r ( 3 - ) B C 2 3 2  2  In the above figure, ΔABC is right angled and (d) r (4 - ) ; r ( 3 - ) 3 2 AC = 100 cm. Also, AD = DE = EF = FC. Find the value of: BD2 + BE2 + BF2 (in cm2) fn, x, fp= esa] ΔABC ledks.k f=Hkqt gS rFkk AC = 100

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51. Find the coordinates of the orthocentre of cjkcj gSA rc Q T >2 dk eku gksxkA 1 TR 1 triangle ABC with vertices A(-3, 3), B (-1, 7) QT QT (a) TR < 3 (b) 3 < TR < 1 and C (3, 3). (c) Q T > 1 (d) Can't be determined 'kh"kksZa A (-3, 3), B (-1, 7) rFkk C (3, 3) okys f=Hkqt ds TR yEcdsUnz ds funsZ'kkad Kkr djksA

(a) (5, -1) (b) (-1, 5) 56. In the figure, segments AD and CE are medians (c) (3, -5) (d) (-5, 3) of ∆ACB, AD ⟘ CE. AB = 10 and CE = 9. Find CA. fp= esa] [kaM AD o CE, ∆ACB dh ekf/;dk,a gSa rFkk AD⟘ rFkk gS rks Kkr djksA 52. Isosceles ∆QRS has dimensions QR = QS = 60 CE. AB = 10 CE = 9 CA 5 E 5 and RS = 30. The centroid of QRS is located at A B 3 point T. What is the distance from T to QR? P Lkef)ckgq ∆QRS esa QR = QS = 60 rFkk RS = 30 gSaA 6 ∆QRS dk dsUnzd fcanw T ij gSA rks QR ls fcanw T dh nwjh gS\ D 5 (a) 2√15 (b) 2 √15 7 (c) 3√15 (d) 2 √15 C (a) 2 (b) 4 (c) 5 (d) 3.5 53. BL and CM are medians of ∆ABC right angled 3 5 at A and BC = 5 cm. If BL = 2 cm, then the length of CM is. 57. Find the altitude to side AC of triangle with side AB = 20 cm, AC = 20 cm, BC = 30 cm. ABC tks A ij ledks.k gS] dh ekf/;dk,a BL rFkk CM gS AB = 20 lseh, AC = 20 lseh, BC = 30 lseh Hkqtkvksa okys rFkk BC = 5 lseh gSA ;fn BL = 3 5 rks CM dh yEckbZ gS \ 2 f=Hkqt esa Hkqtk AC ij Mkys x;s yEc dh yEckbZ Kkr djksA (a) 4√2 (b) 3√5 (a) 10√7 (b) 8√7 (c) 2√5 (d) 3√2 (c) 7.5√7 (d) 15√7

54. Triangle ABC has angles A = 60° and B = 70°. 58. In a ∆AED, find the value of ∠ECD, If AC = 4x - 3, The incenter of this triangle is at I. Find angle BIC DC = 2x + 9, m ∠ECA = 15x + 2 and EC is a f=Hkqt ABC esa] dks.k A = 60° rFkk B = 70° gSA bl f=Hkqt median of ∆AED. dk vUr%dsUnz I gSA rks dks.k BIC Kkr djksA ∆AED esa]∠ECD dk eku Kkr djks ;fn AC = 4x - 3, DC = rFkk dh ekf/;dk gSA (a) 90° (b) 130° 2x + 9, m ∠ECA = 15x + 2 EC, ∆AED 0 0 (c) 80° (d) 120° (a) 92 (b) 88 (c) 780 (d) 900 55. A right angled triangle PQR is such that ∠PRQ 59. In ∆ABC, the internal bisector of the ∠A, ∠B = 90° and QR = 4 cm. T is a point on QR such and ∠C intersect the circumcircle at x, y and z that PT = 3 cm, and perimeter of triangle PQT = respectively. If ∠A = 500, ∠CZY = 300 then ∠BYZ Perimeter of triangle PTR Then, Q T >2 takes TR esa] ds vkUrfjd lef}Hkktd ifjo`r dks the value. ∆ABC ∠A, ∠B, ∠C 0 0 0 Øe'k% x, y, z ij dkVrs gSaA ;fn ∠A = 50 , ∠CZY = 30 gks rks ,d ledks.k f=Hkqt PQR esa] ∠PRQ = 90 rFkk QR = 4 ∠BYZ = ? lseh gS rFkk T, QR ij dksbZ fcanw bl izdkj gS fd PT = 3 0 0 lsehA f=Hkqt PQT dk ifjeki] f=Hkqt PTR ds ifjeki ds (a) 10 (b) 35 (c) 1000 (d) 1150 Like Our FB Page : @neon.classes 36 Web.: www.neonclasses.com

60. G is the Orthocentre of the scalene triangle 64. The circumcentre of triangle ABC is O. If ∠BAC ABC. What is the ratio of the area of =850 and ∠BCA=75° then the value of ∠OAC is. quadrilateral DECB to the area of triangle ∆ABC dk ifjdsUnz O gSA ;fn ∠BAC = 850 rFkk ∠BCA = ABC? 750 gS rks ∠OAC dk eku D;k gksxk\ O, fo"keckgq f=Hkqt ABC dk yEcdsUnz gSA prqHkqZt DECB ds (a) 200 (b) 550 (c) 700 (d) 900 {ks=Qy dk f=Hkqt ABC ds {ks=Qy ls D;k vuqikr gS\ A 4 65. O and C are respectively the orthocenter and u n it the circumcentre of an acute angled triangle s s it n PQR. The points P and O are joined and u D E 8 G produced to meet the side QR at S. If ∠PQS = 600 and ∠QCR = 1300, then ∠RPS = ? . O rFkk C ,d U;wudks.k f=Hkqt PQR ds yEcdsUnz rFkk ifjdsUnz B C 1 1 gSaA P rFkk O dks feykdj mls vkxs c<+kus ij ;g QR dks S ij (a) (b) 4 2 feyrh gSA ;fn ∠PQS = 600 rFkk ∠QCR = 1300 gS rks 3 (c) 4 (d) Data Insufficient ∠RPS gksxk\ (a) 500 (b) 450 (c) 100 (d) 350 61. In ∆ABC, ∠B = 60⁰ and ∠C = 40⁰. If AD and AE be respectively the internal bisector of ∠A and 66. Triangle ABC has area 15 and perimeter 20. perpendicular on BC, then the measure of Furthermore the product of the 3 side lengths ∠DAE is. is 225. If the three altitudes of the triangle ∆ABC esa] ∠B = 600, ∠C = 400 gSA AD, A dk vkUrfjd have lengths d, e and f, then the value of de + ef lef}Hkktd gS rFkk AE, BC ij yEc gSA ∠DAE dk eku gS\ + fd can be written as m . what is m + n? n (a) 300 (b) 150 ∆ABC dk {ks=Qy 15 o ifjeki 20 gSA rhuksa Hkqtkvksa dh (c) 100 (d) 250 yEckbZ;ksa dk xq.kuQy 225 gSA ;fn rhuksa Å¡pkb;ksa dh yEckbZ;ka d, e rFkk f gS rks de + ef + fd dks m fy[kk tk n ldrk gSA dk eku D;k gS\ 62. What is the distance between the orthocentre m + n and the circumcenter of a triangle whose sides (a) 1320 (b) 1217 measure 24 cm, 26 cm and 10 cm? (c) 1528 (d) 1023 ,d f=Hkqt ftldh Hkqtk,a 24 lseh, 26 lseh rFkk 10 lseh gS rks mlds yEcdsUnz o ifjdsUnz ds e/; nwjh Kkr djksA 67. Consider the ΔXYZ given below. (a) 13 cm (b) 12 cm X (c) 7.5 cm (d) √30 cm 63 16 63. The lengths of the sides of a triangle are 25, 29, and 34. To the nearest tenth of a degree, the Y Z measure of the largest angle is 65 P ,d f=Hkqt dh Hkqtkvksa dh yEckbZ;ka 25, 29 rFkk 34 gSaA rks Find the length of the perpendicular XP drawn lcls cM+s dks.k dk eku gS\ on the side YZ. (a) 102.3° (b) 77.7° fn;s x, ΔXYZ esa YZ Hkqtk ij Mkys x, yEc XP dh yEckbZ Kkr djksA (c) 87.6° (d) 87.7° (a) 15.5 units (b) 12.5 units (c) 13.5 units (d) 16.5 units

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68. In the following figure, the equilateral triangle A ABC has an area of 900 3 m2. Points P and Q are the mid-points of AB and AC respectively.

Find the area of the shaded region. F fn;s x, fp= esa leckgq f=Hkqt ABC dk {ks=Qy 900 3eh2 gSA P rFkk Q Hkqtk AB o AC ds e/;fcUnw gSaA Nk;kfdar Hkkx dk {ks=Qy Kkr djksA G A 1 B 2 D 3 E 3 C (a) 1 : 8 (b) 1 : 5

P Q (c) 1 : 3 (d) 1 : 4

71. It is given that AB and AC are the equal sides of an isosceles triangle ABC, in which an B C equilateral triangle DEF is inscribed. As (a) 64 3 m2 (b) 80 3m2 shown in the figure, ∠BFD = a and ∠ADE = b, (c) 75 3 m2 (d) 72 3 m2 and ∠FEC = c , then fn;k x;k gS fd lef}ckgw f=Hkqt ABC dh Hkqtk,a AB rFkk AC cjkcj gSa rFkk ftllsa leckgw f=Hkqt gSaA 69. In the figure given below , ABCD is a concave DEF inscribed 0 fn;s x, fp= ds vuqlkj quadilateral and the measure of ∠BAD = 90 , ∠BFD = a, ∠ADE = b, ∠FEC = rc \ BA = AD = 6 cm and BC = CD = 5 cm. What is the c A length (in cm) of the segment AC ? fn;s x, fp= esa ,d prqHkqZt gS rFkk 900 ABCD ∠BAD = , b BA = AD = 6 lseh rFkk BC = CD = 5 lseh gSA [k.M AC dh D E yEckbZ lseh esa Kkr djksA c A a B F C (a) a = b+c (b) b = a+c 2 2 (c) c = 2a + ab (d) None of these

C 72. In Δ ABC, G is the orthocentre and D is the

B D midpoint of BC, Area of ΔABC is five times the (a) 3 2 (b) 3 2 - 5 area of ΔGDC and ∠ABC = 600. If the minimum (c) 3 2 - 7 (d) 2 2 - 3 distance between any point on BC and point A is 10 cm, then find the length of GC. f=Hkqt esa] yEcdsUnz gS rFkk dk e/;fcanw gSA 70. In the following figue , CF || GE and DF || EA, ABC G D, BC Δ 0 BG = 1 unit, BD = 2 units and DE = EC = 3 units. ABC dk {ks=Qy] ΔGCD dk ikap xq.kk gS rFkk ∠ABC = 60 Find the ratio of areas of ΔBGE and ΔABE. gSA ;fn fcanw A ls BC ds fdlh fcanw dh lcls de nwjh 10 lseh gS rks dh yEckbZ Kkr djksA fn;s x, fp= esa CF || GE rFkk DF || EA, BG = 1 ,dd , GC 4 3 BD = 2 ,dd, DE = EC = 3 ,dd , gSaA rks ΔBGE rFkk Δ (a) 3 (b) 4 cm ABE ds {ks=Qyksa dk vuqikr Kkr djksA 8 3 (c) 8 cm (d) 3 Like Our FB Page : @neon.classes 38 Web.: www.neonclasses.com

73. In a triangle ABC, the incircle touches the three Answer Key sides AB, BC and CA at the points D, E and F 1 d 21 b 41 a 61 c respectively. If the length(in cm) of the sides 2 a 22 b 42 c 62 a AB, BC and CA are three consecutive even 3 c 23 d 43 c 63 b numbers, then which of the following cannot 4 d 24 a 44 d 64 c

be the radius (in cm) of the incircle ? 5 b 25 a 45 a 65 d

f=Hkqt esa] f=Hkqt dk vUr%o`r Hkqtkvksa rFkk ABC AB, BC CA 6 b 26 c 46 b 66 b dks dze'k% D, E rFkk F fcanw ij Li'kZ djrk gSA ;fn Hkqtkvksa 7 c 27 c 47 b 67 a AB, BC rFkk CA dh yEckbZ yxkrkj le la[;k,a (lseh- esa) gSa 8 d 28 a 48 c 68 c rks fuEu esa ls dkSulh vUr%o`r dh f=T;k ugh gks ldrhA (lseh- 9 b 29 b 49 c 69 c esa)? 10 c 30 a 50 a 70 d 11 d 31 a 51 b 71 a (a) 2 (b) 7 12 a 32 d 52 b 72 c (c) 15 (d) 3 13 b 33 b 53 c 73 d

14 b 34 c 54 d 74 b

74. Consider Triangle ABC with AB = 6 cms, BC = 7 15 a 35 b 55 a 75 b cms and CA = 8 cms. AD and AE are internal 16 c 36 d 56 b 2 and extenal bisectors of Angle A, what is AE + 17 b 37 b 57 c AD2 ? 18 c 38 d 58 b ,d ΔABC ftlesa AB = 6 lseh BC = 7 lseh rFkk CA = 8 19 d 39 c 59 b lseh gSA AD o AE, ∠A ds vUr% lef}Hkktd o cká 20 b 40 b 60 c lef}Hkktd gS] rc AE2 + AD2 = ? (a) 144 (b) 576 (c) 1296 (d) 2304 bu lHkh questions dk Detailed Explanation 75. In a triangle ABC, the internal bisector of angle tkuus ds fy, A meets BC at D. If AB = 4, AC = 3 and ∠A = 600. Then the length of AD is : Download djsa f=Hkqt ABC esa] dks.k A dk lef}Hkktd BC dks D ij feyrk gSA ;fn AB = 4, AC = 3 and ∠A = 600 gS rks AD dh Our App : yEckbZ Kkr djksA NEON CLASSES 12√3 (a) 2√3 (b) 7 (c) 15√3 (d) 6√3 8 7 Study Material

Geometry

Triangle Centres

Download our app : NEON CLASSES 39 SOLUTIONS x = 19 - y y = 22 - z 1 fdlh Hkh triangle esa orthocentre, circumcentre ;g esa j[kus ij rFkk centroid ,d gh line (Euler line) ij fLFkr gksrs gSa  x = 19 - (22 - z) value (i) vr% buls dksbZ triangle ugh cusxkA 19 - (22 - z) + z = 23 vr% lgh answer option (d) 0  Z = 13.

2. Let the midpoint of BC = M 5. Extended sine rule ls] a D;ksafd gSA = b = c = 2R AG = 2GM ( G centroid ) SinA SinB SinC ∵ BM = CM = 1 AG 2 (R = circumradius of  ABC)  BM = CM = GM R = a 2SinA vr% BGC esa] a = b rFkk a + b + c = 25 A c = 25 - 2a a = 25-2a SinA SinC ∠A = 15°, ∠B = 15°  ∠C = 150°

G a 25-2a 0 0 1 0 = 0 (Sin 150 = Sin 30 = ) Sin15 Sin150 2 a 25-2a 0 = Sin15 1/2 B M C 50 Sin150 25  a = 0 rFkk R = 0 1+4 Sin15 1+4 Sin15 vr% B, C rFkk G bl triangle BGC ds circumcircle ij  Area of circumcircle =  R2 fLFkr points gSa rFkk BC bl circle dk diameter (O;kl) gS vr% 0  ( 25 )2 = 474 cm2 ∠BGC = 90 1+4 Sin150 Sin150 = sin (45 - 300) 3. G = (2+10+0 , 12+0+0 ) = (4, 4) 0 0 0 0 3 3 = sin45 . Cos30 -cos45 . Sin30 1 Area of  AGB = 1 .  ABC = 1 . 3 - 1  3 2 2 2 2 1 AGB = 1  ( 10  12) = 20. 3 - 1 3 2 = = .259 2 2

4. B 6. A x x 

E

1380 z y 0 138

2  420 A z y C B D C 0 0 0 rFkk 0 x + z = 23 ----(i) 3 + 138 = 180 2 = 28  = 48  0 0 0 x + y = 19----(ii) ∠ABC = 2 + = 28 + 48 = 76 z + y = 22 ----(iii)

1 M 2 7. A 10. C B

4 ft. O. 1 O D x θ

B C A  AMB ~  ABD ∠AOB : ∠BOC : ∠COA = 2 : 3 : 4 vr% x = 4  x = 2 feet. 1 2 2x + 3x + 4x = 3600  ∠BOC = 3x = 1200 11. Symmetry ls] Circumcentre ij cuk angle opposite vertex ij XY = AC rFkk XY || AC cus angle dk double gksrk gSA XY  BP vr% BP  AC ------(i) vr% ∠BAC = 1  120 = 600 2 AB = ZY rFkk AB  ZY ZY  CP vr% CP AB------(ii) fdlh dk djus ds fy, gesa 8. triangle orthocentre find 2 XZ = BC, XZ  AP  AP  BC-----(iii) altitude lines gh required gksrh gSaA (i), (ii) rFkk (iii) ls] lcls igys ge vertex C ls side AB ij altitude find A djrs gSa ;g altitude line x = 5 gSA B ls AC ij altitude = line AC dh equation = 12 P y = 5 x AC ds perpendicular line dh form y = - 5 x + b 12 rFkk ;g line (14, 0) ls gksdj xqtjrh gS vr% altitude dh equation, y = - 5 x + 35 C B 12 6 vr% P, ABC dk orthocentre gSA vr% bl altitude rFkk line x (x = 5) dk intersection point = (5, 15 ) 4 12. Circumcentre O (a, b) ls ABC dh rhuksa vertices vr% ;g 15 dh point (5, ) orthocentre location dh distance equal gksrh gSA gksxhA 4  AO = BO = CO 2 2 2 2 2 2 2 2 2 AO = BO 9. PA + PB + PC = GA + GB + GC + 3PG (a - 1)2 + (b - 4)2 = (a + 2)2 + (b - 3)2 AB = 5, BC = 6, CA = 7 2 2 2 2 2 2 -2a + 1 8b + 16 = 4a + 4 6b + 9 AB + BC + CA = 3 (GA + GB + GC ) 2 2 2 3a + b = 2 ------(1) 25 + 36 + 49 = 3 (GA + GB + GC ) 2 2 2 2 2 BO = CO 110 = GA + GB + GC 3 (a + 2)2 + (b - 3)2 = (a - 5)2 + (b - 2)2 70 = 110 + 3PG2 3 4a + 4 6b + 9 = -10a + 25 4b + 4 100 = PG2 9 7a b = 8 ------(2) PG = 10 ans. (1) + (2) a = 1, b = -1 3  circumcentre of ABC is O(1, -1)

2 13. (b) (1) + (2) + (3) A 4a2 + 4b2 + 4c2 = a2 + b2 + c2 + 9d2 + 9e2 + 9f2 2 2 2 2 2 2 3 2x 3 3 (a + b + c ) = 9 (d + e + f ) a2+b2+c2  2 2 2 = 3 d +e +f

x x 15. r = 2, B C 1 ⇨ R = 7 = ( 2 of hypotenuse) x + x + 2x = 1800 Hypotenuse = 14 0 a+b-h x = 45 r = 2 a+b-14 ∠A = 2 × 45 = 90 ⇨ 2 = 2 BC = 32 + 32 = 3 2 cm a + b - 14 = 4 3+3-3 2 Inradius (r) = 2 a + b = 18 3(2- 3) 2 2 2 = 2 a + b = 14 3 2 2 2 Circumradius (R) = 2 a + (18 - a) = 196 R = 2- 2 = 2 - 1 : 1 a2 + 324 + a2 - 36a = 196 r 2 2a2 - 36a + 128 = 0 2 14. A a - 18a + 64 = 0 Now the 2 roots to this equation will effectively be a, 18- a. Product of the roots = 64. d Area = 1 × product of roots (How? Come on, you c F E b 2 G can figure this out.) e f = 32 sq. cms Answer choice (a)

B D C 16. (c) In OBF and OCE A BC = a, AC = b, AB = c AG = d, BG = e, CG = f d e f F GD = , GE = , GF = 2 E 2 2 2 esa] ls] O ABC Appolonious theorem 3

AD as a median. B C D AB2 + AC2 = 2AD2 + 2CD2 ∠OFB = ∠OEC = 90º 2 2 2 a 2 c + b = 2 ( 3 d) + 2 ( ) 2 2 And ∠BOF = ∠BOC 2c2 + 2b2 = 9d2 + a2 ------(1) OBF = OCE BE as a median, Hence, OB = OF or, AB2 + BC2 = 2BE2 + 2AE2 OC OE OB = OC c2 + a2 = 2 ( 3 e)2 + 2 ( b )2 OF OE 2 2 Or, OB × OE = OF × OC 2c2 + 2a2 = 9e2 + b2 ------(2) OF × OC = OB × OE CF as a median, = (BE - OE) (OE) 2a2 + 2b2 = 9f2 + c2 ------(3) = (5 - 2) × 2 = 6 cm²

3 3 a 17. A AR = 2 a - 3 3a-2a a = 2 3 = 2 3 1 AR = 3 AD. D E 1 Radius of smaller circle = 3 radius of larger circle 1 a a G Radius of smaller circle = 3  2 3 = 6 3 Area of smaller circle = πr2

a 2 πa2 B F H C π ( ) = 6 3 108 Let AD = DG = GH Area of Δ = 3 a2 1 4  πa2 2 AD = 3 AD Ratio = : 3 a 1 108 4 DE = 3 HC π : 27 3 . Answer choice (c) 1 Area of ADE = 18 ABC Figure ls easily understood gS fd F dks left or 19. ABC ,d isosceles (lef)ckgq) triangle gS vr% blesa right move djus ls right side esa T;knk area gks tk;sxk I, O rFkk H co-linear gksaxsa vr% triangle ugh cusxk vr% vr% figure esaa nh xbZ F dh position minimum area = 0.00 condition dks represent djrh gSA Right side dk area half ls 1 de gS rFkk left side dk 18 ls T;knk gSA 2 2 2 area half 1 1 20. Assuming a ≤ b ≤ c, we have a + b > c . vr% 2 18 = 9-1 = 8 = 0.8 required ratio = 1 1 9+1 10 This implies the triangle with sides a, b, c has 2 18 to be acute-angled.

18. A Choice (B)

O 21. The lengths of sides are (distance formula ls) P R Q fdUgh nks rFkk ds e/; dh nwjh points p (x1, y1) Q (x2, y2) = I 2 2 PQ = (x1 - x2) + (y1 - y2)

2 2 B D C a = (14-5) + (12-0) = 15 a In-radius of equilateral triangle of side a = 2 2 3 b = (5-0) + (12-0) = 13 Diameter of larger circle = a 2 3 2 Let us say common tangent PQ touches the c = (14-0) + (0-0) = 14 ax +bx +cx ay +by +cy two circle at R, center of smaller circle is O. I = ( 1 2 3 , 1 2 3 ) Now, PQ is parallel to BC. AR is perpendicular a+b+c a+b+c 150+1314+145 150+130+1412 to PQ. Triangle PQA is also an equilateral I = ( , ) 15+13+14 15+13+14 triangle and AORID is a straight line. (Try to I = (6, 4). establish each of these observations. Just to

maintain the rigour.) 3

AD = 2 a RD = a 3

4 22. the triangle has a balance point at centroid. 25. A A .(17,4) Find the centroid of the triangular coffee table. Use the centroid formula x +x +x y +y +y P R ( 1 2 3 , 1 2 3 ) 0(0,0) O 3 3 The centroid of the triangle is ( 3+5+7 , 6+2+10 ) or (5, 6) 3 3 B Q C F Let the point be A (3, 6), B (5, 2), C (7, 10). O is orthocenter of ΔPQR. The midpoint D of BC is O and H are a single point.. ( 5+7 , 2+10 ) or (6, 6). Note that AD is a line that Equation of line AH. 2 2 connects the vertex A and D, the midpoint of y-0 17 17 = x-0 = 4 ⇨ 4y = 17x ⇨ y = x BC. The distance from D (6, 6) to A (3, 6) is 6-3 4 or 3 units. 26. fdlh triangle ABC ds pedal triangle PQR ds If P is the centroid of the triangle ABC, then AC excentres original triangle ABC ds vertices ('kh"kZ = 2 AD. So, the centroid is 2 (3) or 2 units to the 3 3 fcanw) gh gksrs gSa vr% area of ABC = right of A. the coordinates of the centroid (P) x1 x2 x3 are (3+2, 6) or (5, 6). = 1 2 y1 y2 y3 Answer (5, 6) [ 1 1 1 ]

= [x1 (y2 -y3) - x2 (y1 - y3) + x3 (y1 - y2)]

A 20 4 13 23. 1 8 12 1 = 2 [ 1 1 1 ] 1 = [20 (12 - 1) - 4 (8 - 1) + 13 (8 - 12)] 15 cm 25 cm 2 1 = [220 - 28 + (-52)] ⇨ = 1 [140] = 70 units2. G 2 2

B C A 18 cm 27. (c) AB² + AC² = 2 (AD² + BD²)

⇨ 225 + 625 = 2(AD² + 81) ⇨ AD² = 344 G 1 AD = 2 86 and ⇨ GD = AD 3 2 ⇨GD = 86 cm B C 3 D AG = BC = 2x (let) 24. d 2 = d 2 = 2(a2 + b2) 1 1 GD = x (centriod divides median in 2 : 1) 484 + d 2 = 2 (144 + 196) 2 Now in BDG, BD = GD = x 484 + d 2 = 2 (340) = 680 2 ∠DBG = ∠BGD = θ, (let) 2 d2 = 196, d2 = 14 cm. Similarly in DGC, CD = GD = x Perimeter = 2 (14 + 12) = 52 cm. ∠DCG = ∠DGC = θ, (let) s = 22+14+12 = 24 2 A D ∠BGC = θ1 + θ2

area = 2 24×2×10×12 Now in DBGC = θ1+θ2+Now, (θ1+θ2+θ1+θ2) = 180°

2 12 ⇨θ1 + θ2 = 90° = 48 10 cm ⇨ ∠BGC + 90° B 14 C 5 28. ABC ,d isosceles triangle gS vr% BD AC 30. Let radii of each circle be 'a'. BED ~ DEC (Neon triangle) Distance PQ = 2a. BD ED  DC = EC Let radii of each circle be 'a'. DFB rFkk CEA esa] Distance PQ = 2a. ∠FDB = 900 - ∠DBC = ∠ECA Draw PM perpendicular to AB. rFkk DB DB ED 2ED DF ∠MAP = 30° CA = 2DC = 2EC = 2EC = CE A By SAS DFB ~ CEA vr% ∠DAG = ∠DBG M vr% BGDA ,d cyclic quadrilateral (pdzh; prqHkqZt) gS P vr% ∠BGA = ∠BDA = 900 Q R FF bl question dks co-ordinate geometry ls Hkh solve dj ldrs gSa& B C 3 0 MP = a. AM = a ;gka ∠B = 123 dh condition irrelevant gSA AB = 2a + 2a√3 A = (-b, 0) After this we are done. B = (0, a) Area of circle = πa2 C = (b, 0) Area of equilateral triangle = 3  (2a + 2a√3)2 D = (0, 0) 4 r Area of circle = πa2 vr% E = (r, s) rFkk F = ( , s ) 2 2 Area of equilateral triangle = 3  (2a + 2a√3)2 ar + bs = ab (E, BC ij fLFkr gSA) 4 = a2(6 + 4√3) as = br (DE  BC) 2 Ratio = π : (6 + 4√3)  r = a b a2+b2 ab2 s = 2 2 31. A a +b s-2a r 4 vRk% s = -1 r+b D vr% BF rFkk AE nksuks perpendicular gS vr% 0 ∠BGA = 90 . 9

29. (b) The circumcenter of a triangle is the C intersection of the perpendicular bisectors of B 1 CD2 = AD  BD the sides of the triangle. The slope of AB is 2 , and its midpoint is (-7, 6). The equation of the  CD2 = 4  9 perpendicular bisector to AB is y = -2x -8. The  CD = 6cm. slope of AC is -1, and its midpoint is (-4, -3). Area of ABC= AB  CD The equation of the perpendicular bisector of = 1  13  6 = 39 cm2 AC is y = x + 1. The intersection of y = -2x -8 and 2 y = x + 1 is (-3, -2).

6 1 32. Area of ABC = 2 BC × AD 35. Smallest angle is opposite to smallest side. By 1 1 Apollonius theorm. = AC × BE = AB × CF 2 2 2 2 2 AD = 2AB +2AC -BC  AB : BC : AC = 1 : 1 : 1 CF AD BE 4 1 1 1 78 = 3 : 1 : 2 = 2 : 6 : 3 = 32+50-4 + 4 4 AD = 39 33. A 2 36. According to the question. 1 In right angle ∆COD, D E OC² + OD² = CD² 1 361 4x² + y² = 4 B C In right angle ∆BOE, area of ADE = 1, OB² + OE² = BE² area of ABC = 2 X² + 4y² = 121 area of ADE 2 Add Both Eq (i) & (ii) = area of ADE 1 5x² + 5y² = + 121 2 side of ADE (AB) 169 = 2 side of ADE (AD) x² + y² = 4 2 AB By square root 1 = AD In right angle ∆BOC, DB = AB - AD OB² + OC² = BC² = 2 -1 4y² + 4x² = BC² 4(y² + x²) = BC² DB : AB = ( 2 - 1) : 2 From Eq. (iii) BC² = 169 34. Since angle subtened at the centre of the circle is double the angle subtened at BC = 13 circumference. Alternate: ∠BOC = 500 × 2 = 1000 In this type of question we use direct formula.  OB = OC 5BC² = AB² + AC² 2 2 BC = AB +AC ∠BOC = ∠OCB 5 0 0 180 -100 2 2 = = 40° = (22) +(19) 2 5 In ∆BPC, = 484+261 ∠BPC + ∠PBC + ∠OCB = 180° 5 845 Or, ∠BPC + × 40° + × 40° = 180° = 5 = 169 = 13 cm ∠PC = 180° - 20° ∆ 20° = 140° Alternate: ∠BOC 100° ∠BPC = 90° = 2 = 90° +A = 90° 25 = 140°

7 37. (b) Medians AD and BE intersect at G on 900 39. (c) A i.e. ∠AGB = 900 and ∆AGB will be a right angled 2 triangle We know, In a triangle centroid F 1 E divides the medians in 2 : 1 Ratio. G 1 A 2 1 2 B D C E

12 40. A 9 900 D E B D C G 2 ⇨ Then BG = 3 × BE 2 BG = 3 × 12 B C BG = 8 aF 2 9 ⇨ AG = 3 × AD Inradius of ΔABC = 2 3 ⇨ AG = 6 cm. diameter of bigger circle = 9 In right angled triangle AB will be a 3 hypotenuse Using pythgoras theorem Median of ΔABC = 3 a 2 A ⇨ (AB)² = (AG)² + (BG)² 3 a a AG = AF - GF = a - = 300 ⇨ (AB)² = 6² + 8² 2 3 2 3 AO = 2r ⇨ AB = 10cm 2r Therefore, length of AB = 10 cm. AG = AO + OG r AG = 2r + r = 3r O 38. Solution: According to the question. r = A G = a G 3 6 3 By using this formula. 2 Area of shaded Region = r ×√3r = - πr We can calculate the area of 3 = r2 [√3 - π ] ∆ABC 3 a 3 π 4 = ( ) [√3 - ] 6 3 3 Area of ∆ABC = 3 S (S-a) (S-b) (S-c) Here, a, b & c are the length of the median. = 36 [√3 - π ] 36×3 3 S = a+b+c = 8+6+10 = 12 2 2 = 1 [√3 - π ] 4 3 3 Area = 3 12 (12-6) (12-8) (12-10) 4 Area of shaded Region - II = 3 12×6×4×2 = 4 × 24 = 32 cm2 Q 3 R Alternate: S 2 2 P When m 1 + m 3 then T O Area of triangle = 2 × m m 3 1 2 B 2 = 3 × 6 × 8 = 32 cm² 8 Area of shaded Region = Ar of ΔBPQ- Ar of Δ 42. (d) The area of a triangle is given by the formula A= 1 absin (θ). BOS - Ar of sector QTP - Ar of sector SOT 2 In ΔBPQ In ΔBOS Substituting the given information, 750= 1 (48) (52)sin (θ). Solving this PQ = R OS = r 2 BQ = √3 R Br = √3 r equation for Q yields that Q = 37° or 143°. If m∠Q = 37°, then ∠E would be the largest angle and R = 3 r 1 of the triangle, and that measure, found using 2 2 the Law of Cosines, would be 78°, less than Region - II = 1 ×R×√3R - 1 × R × R - π R - πR 2 2 3 √3 6 27 143°. 1 = R2 [ 3 - - π - π ] 2 6 3 6 27 43. (c) 2 = R2 [ 2 7√ 3 -3√3-9 π - π ] A 54 900 -x = R2 [ 2 4√ 3 -11 π ] 54 O = R2 [ 4 - 11 π ] 3 3 54 1800 -2c x x R = √3 B C

= 3 [ 4 - 11 π ] 3 3 54 (In ∆ BOC) = 4 - 11π OB = OC (circum radius) 18 √3 ∠OBC = ∠OCB = x (Let) 41. Radius of circumcircle of  ABD = Then, d 0 1 = 12.5 -----(1) (d = diagonal DB) ⇨∠BOC = 180 - 2x 2SinA 1  ∠BOC 180-2X Radius of circumcirlce of ACD = ⇨∠BAC = 2 = 2 d2 0 = 25------(2) (d = diagonal AC) 2Sin (180 -A) 2 ⇨∠BAC = 90 - x (2) ÷ (1) ⇨∠OBC + ∠BAC = 900 x + x = 900

d2 = 2  d2 = 2d1 ------(3) d1 44. Case I: Area of rhombus = 1 d .d 2 1 2 Let x be the side opposite the angle. ACDdk area = abc 4R A (a = b = s leprqHkqZt dh Hkqtk) leprqHkqZt dk fod.kZ (c = d2 ) 8 cm rhombus dh Hkqtk S = 1 d 2 + d 2 x 2 1 2 2 1 2 2  s = (d1 + d2 ) 1 2 2 4 2 s .d (d1 + d2 ).d2   ACD dk area = 2 = 4 4R 4 (25) Rhombus dk area = 2  ACD dk area 600 1 (d 2 + d 2) .d = d + d = 2 [ 1 2 2 ] B 9 cm C 2 1 2 1625 By cosine rule, x2 = 82 + 92 - 289cos 60 20d = d 2 (d = 2d )  d =20 rFkk d = 40 1 2 2 1 1 2 x2 = 145 - 72 = 73  area of rhombus = 400. or x = √73 9 AQ Case II: From ∆AQD, ⇨ Sin 60° = AD b Let 9 cm be the side opposite the angle. 3 = ⇨ AD = 2b 2 AD 3 A AP From ∆APD ⇨ Sin 75° = AD 600 8 cm x 47. (B) By pyahagoras theorem PF = PQ2+QR2 = 52+122 = 13 cm. O is the centriod ⇨ QM is median and M is the B 9 cm C midpoint of PR. 2 2 2 By cosine rule, 9 = 8 + x - 28xcos60  QM = PM = 13 2 2 2 17 = x - 8x ⇨ x - 8x - 17 = 0 Centriod divides median in ratio 2 : 1 or x = (8 +- √(64+68)) / 2  OQ = 2 QM 3 = 4 +- √33 ⇨ Only 4 + √33 = 1 × 13 = 13 = 4 1 cm. 2 2 3 3 Case III: Let 8 cm be the side opposite the angle. 48. E is the midpoint of hypotenuse AC so, BE = A 1  AC=50 cm 3 Now, applying Apollonius's theorem on ΔABE 8 cm 2 2 2 2 x ⇒ AB + BE = 2 x (BD + AD ) ⇒ AB2 + 502 = 2 x (BD2 + 252)……….(1) (As, AD = DE = EF = FC = 25 cm)

600 Similarly, in ΔBEC, B 9 cm C ⇒ BC2 + BE2 = 2 x (BF2 + EF2) By cosine rule, 82 = 92 + x2 - 29xcos 60 ⇒ BC2 + 502 = 2 x (BF2 + 252)………..(2) 2 2 -17 = x - 9x ⇨ x - 9x + 17 = 0 Adding (1) and (2), or x = (9 +- √(81−68)) / 2 ⇒ AB2 + BC2 + 2  502 = 2  (BD2 + BF2 + 2  252) = 4.5 +- (√13)/2 ⇨ =4.5 +- √3.25 ⇒ AC2 + 2  502 = 2  (BD2 + BF2) + 4  252 (As, AB2 + BC2 = AC2) 45. The triangle has a balance point at centroid. ⇒ 1002 + 2  502 - 502 = 2  (BD2 + BF2) Find the centroid of the picture. ⇒ 1002+ 502 = 2  (BD2 + BF2) Use the centroid formula 10,000+2,500 2 2 x +x +x y +y +y ⇒ = BD +BF =6250 ( 1 2 3 , 1 2 3 ) 2 3 3 2 2 2 2 The centroid of the triangle is ⇒ BD + BF + BE = 6250 + 50 = 6250 + 2500 = 2 (0+3+6 , 8+0+4 ) = (3,4) Ans. 8750 cm 3 3

49. (c) The slope of AB is -4-2 = -6 = -1 . 46. According to the question 3-(-9) 12 2 ∆ABC is a isosceles triangle The slope of the perpendicular line is 2. The midpoint of AB is (-3,-1), so the equation of the So, ∠B = 90°, ∠C = ∠A = 45° perpendicular bisector is y + 1 = 2(x + 3). In ∆AQD ∠DAC = ∠BAC - ∠BAD = 45 - 15 = 30° And ∠QDA = 180° - (90 + 30) = 60° 10 7 - 3 4 2 50. A 51. The slope of AB is = = . -1 + 3 2 1 So, the slope of the altitude, which is 1 perpendicular to AB is - 2 . Now, the equation of the altitude from C to AB is y - 3 = - 1 (x - 3) 2 y - 3 = - 1 x + 3 2 2 y -3 + 3 = - 1 x + 3 +3 2 2 B C y = - 1 x + 9 A 2 2 use the same method to find the equation of the altitude from A to BC. That is r√3 y - 3 = 1(x + 3) or y = x + 6. E D Solve the equations to find the intersection r O point of the altitudes. x + 6 = - 1 x + 9 2 2 2(x + 6) = 2 (- 1 x + 9 ) 1 2 2 Area of ∆AOD = 2 × OD ×AD 2x + 12 = -x + 9 1 2x - x + 12 = -x - x + 9 = 2 × r ×r √3 3 2 x + 12 = 9 = 2 r Area of shaded Region = Ar of ⧠AEOD - Ar of x + 12 - 12 = 9 - 12 sector EOD 3x = -3 2 x = -1 = 2 [ 3 r2] - πr 2 3 2 y = (-1) + 6 = 5. = √3r2 - πr 3 So, the coordinates of the orthocenter of ∆ABC = r2[√3 - π 3 is (-1, 5). Area shaded Region - II B (-1, 7) y = x+6

P 2r T Q x r r (-1, 5) R S (3, 3) (-3, 3) Area of shaded Region A C 1 1 = Area of PQRS = area of sector 2 2 Answer : (-1, 5) PRT - Area of sector QST 2 = 2r×r - π r2 - πr 4 4 52. (c) The centroid is the intersection of the = 2r2 - π r2 medians of the triangle. The median from Q to 4 2 π RS is also the altitude to RS. Use the r [2 - ] 2 Pythagorean Theorem to show that this altitude has length 602-152 = 15√15. The 2 centroid lies 3 of the way along the median, so 11 T is 5√15 units from RS. ⇒ PR + RT + PT = PT + TQ + PQ ∆RST, ∆QST, and ∆QRT are all equal in area. ⇒ PR + RT = QT + PQ 1 The area of ∆RST is 2 (5√15) (30) = 75√15. Since, PQ > PR (PQ is hypotenuse) 1 The area of ∆QRT = 2 (h)(60) = 30h = 75√15 ⇒ RT > OT QT 75 5 ⇒ > 1 So, h = 30 √15 = 2 √15 TR Also, given PT = 3 so RT <3 (PT is hypotenuse) Since, RT+ TQ = 4, when RT = 3, TQ = 1 53. According to question. But Since RT <3 According to figure, when two medians ∴ QT > 1 intersect each other in a right angled triangle TR 3 then we use, this equation. ⇨ 4 (BL² + CM²) = 5BC² 56. Since CE and AD are medians, the intersect at ⇨ 4 × 3 5 + 4CM² = 5BC² the centroid of the triangle. Let P be the 2 centroid, the intersection point of the 45 + 4CM² = 125 medians. By the Centroid Theorem, CP = 2 CE. CM² = 125-45 =20 3 4 That is, CP = 6 and PE = 9 - 6 =3. Because AD⟘ CM = 2 5 cm. CE, then two right triangles are formed, ∆APE and ∆APC. Using the converse of the 54. A Pythagorean Theorem, we can find the lengths

0 of AP and then CA, as shown below 60 In the right triangle APE, AP = 52 - 32 I = 25 - 9

0 0 50 70 = 4 350 250 B C A + B + C = 180 in a triangle 57. This is an isosceles triangle. So, let us find the Hence C = 180 - 130 = 50° altitude to BC first. Incenter is the meeting point of angle A bisectors. BIC will form a triangle. In this triangle B = 35, C = 25 Hence BIC or I = 180 - (35 + 25) = 120° 20 20

55. n the given figure P

B D 30 C Altitude to BC bisects BC as ADB and ADC 3 are congruent. {RHS congruence; AD is common, AB = AC} DC = 15 2 2 2 R T Q AD + DC = AC 4 2 2 2 Given P (∆PRT) = P (∆PTQ) AD + 15 = 20

12 2 A AD = 400 - 225 60. 4 u n 2 it AD = 175 s s it n u D E 8 AD = 5√7 G 1 Area of the triangle = 2  base  height 1 1 = 2  BC  AD or 2  AC  (altitude to AC, say, h) B C 1 1 ⧠DECB is a cyclic quadrilateral. = 2  30  5√7 = 2  (20 x h) 3 ∠ADE = ∠ACB h = 2  5√7 = 7.5√7 Answer choice (c). ∠AED = ∠ABC Therefore, ∆ADE is similar to ∆ACB. 58. Given: AC = DC  Ar ea of ∆ ADE = ( 4 )2 = 1 Area of ∆ACB 8 4 4x - 3 = 2x + 9 3 Ratio of area of ⧠DECB to area of ∆ABC = 4 4x - 2x - 3 = 2x - 2x + 9 2x - 3 = 9 61. According to the question. 2x - 3 + 3 = 9 + 3 2x = 12 X = 6 Substitute 6 for x in m ∠ECA. m ∠ECA = 15x + 2 = 15(6) + 2 = 92. Given: ∠B = 60° EC is not an altitude of ∆AED because m ∠ECA ∠C = 40° = 92. As we know that Ans. m ∠ECD = 880. ∠A +∠B + 60° + ∠C = 180° ∠A = 180° - 60° - ∠40° 59. ∠ZOY = ∠BOC = 90 + 50 = 1150 2 ∠A = 80° 0 (Property 2) 80 0 ∠BAD = = 40 in ∆ZOY, 2 In ∆AEB 0 ∠BYZ = 180 - (30+115) = 35 0 ∠A + ∠B + ∠E = 180 0 0 0 A ∠A = 180 - 60 - ∠180 ∠A = 300 y Then, ∠DAE = ∠DAB - ∠EAB z = 40 - 30 Alternate: 0 0 ∠B - ∠C 60 -40 0 ∠DAE = = 2 = 10 B C 2

x

13 62. The sides measure 24 cm, 26 cm and 10 cm. In ∆AOC 10, 24, 26 is a Pythagorean triple! We known OC = OA (Circum Radius) So it is a right angled triangle we are talking ∠OAC = ∠OCA about. ∠OAC + ∠OCA + ∠COA = 180° Draw the perpendicular bisectors to get the 2 ∠OAC = 180° - 40° circumcenter. Orthocenter is the point where 2∠OAC = 140° all altitudes meet. In a right angled triangle it 0 ∠OAC = 70° is the vertex that makes 90 angle.

65. According to question Given: ∠PQS= 600 ∠QCR= 1300

m C 1 c ∠QPR = ∠QCR 4 2 2 2 6 1 0 0 c m ∠QCR × 2 ×130 = 65 Now, ∠PQS + ∠PQS + ∠QPS = 1800 0 0 O 10 cm 60 + 90 + ∠QPS = 180º 0 Now it is easy to find OC! It is the length of the ∠QPS = 30 diagonal of the rectangle formed. ∠RPS = ∠QPR - ∠QPS Length = 12 cm, breadth = 5 cm = 650 - 300 Diagonal = √(122+52) = 13 ∠RPS = 350

66. 1 ad = 1 be = 1 cf = 15 63. With SSS information, use the Law of Cosines 2 2 2 to determine the measure of the largest angle d = 30 , e = 30 , f = 30 (which is located opposite the longest side). a b c de + ef + fd = 900 ( 1 + 1 + 1 ) 342 = 252 + 292 − 2 (25) (29) cos (θ). 2 (25) (29) ab bc ca cos (θ) = 252 + 292 − 342, ∴m+n = 1200 + 17 = 1217. 2 2 2 so that cos θ = 25 +29 -34 2 (25) (29) or cos θ = 31 145 67. Observe that ΔXYZ is a right - angled triangle θ = 77.7°. because 63, 16 and 65 form a Pythagorean triplet. 1 1 64. According to the question Given. Then, ( 2 )  XP  65 = ( 2 )  63  16 ∠BAC = 85º ⇒ XP = 15.5 (approximately) ∠BCA = 75º A ∠OAC = ? 68. In ∆ABC

∠ABC + ∠BAC + ∠CAB = 180º P Q ∠ABC = ∠0º O ∠COA = 2 × 20 = 40° B C 14 Here side of the equilateral triangle is 60m. In 71. A the equilateral triangle, PQ || BC because P and Q are the mid points of sides AB and AC. b D E So, BC = 2PQ ⇒ PQ = 30 m c Now AE = 3  60 = 30√3m. 2 a B C 'O' is the centroid, So AO : OE = 2 ; 1. F Since b + ∠EDF = a + B OE = 10 √3 m and AO = 20 √3 1 {Sum of two internal angles of a triangle is AF = AE = 15 √3 and FO = AO - AF = 5 √3m 2 equal to third exterior angle} 1 Δ 0 therefore area of PQO = 2  PQ  FO So b + 60 = a + B ....(i) = 1  305√3 = 75√3m2 Similarly, 2 a + ∠DEF = c + C So a + 600 = c + C ...(ii) 69. Extend AC to meet BD at the point E. It can be clearly understood that triangle ABC is Subtracting equation (ii) from (i), congruent to the triangle ADC. (b + 60) - (a + 60) = (a + B) - (c + C) A (b - a) = (a - c) (∵B = C) (because is isosceles triangle) 0 0 45 45 ∴ 2a = b + c or b+c 6 6 a = C 2 5 5 72. A B E D Also, in the right angle triangle BAD, AE = BE = ED. Q Therefore, AE = Be = 6cos450 = 3 √2 units P G CE = BC2 - BE2 = 25 - 18 = √7 units Therefore, AC = AE - CE = 3 √2 - √7 units B E D C Hence, option (3) is the correct choice. Minimum distance between A and BC is the perpendicular distance. If ⟘ AE BC, it passes 70. ΔGBE ∽ ΔBFC through G and AE = 10 cm. Δ [∵ CF || GE ] Area of ( GDC) 1 Ar ea of ( Δ ABC) = 2 ∵ BF = 8  1 = 8 But Area of (ΔGDC) = Area of (ΔGDB) 5 5 8 5 Δ 2 and ΔBFD ∽ ΔBAE ⇨ AB = 5  = 4 Area of ( GDC) 2 ⇒ Ar ea of ( Δ ABC) = 5 5 Δ 1 Area of GBE = 2 sinθ ×BC×GE or 2 = 2 Area of ΔABE = 10 sinθ 1 5 2 ×BC×AE [∠GBE = ∠ABE = θ] ⇒ GE = 2 AE 5 DGBE 1  = ⇒ GE = 2 ×10 = 4 cm DABE 4 5 Now in ΔCPB

15 ∠CPB = 900 AB = BD = BE {Angle Bisector Theorem for both AC DC EC ∠ PBC = 600 ⇒ ∠BCP = 300 internal and external bisectors} So in right angled ΔGEC, AB = 6 = 3 AC 8 4 0 GE = 4 cm and ∠ECG = 30 ⇒ GC = 8 cm BD 3 = {BC=7; Since, BC=BD+DC ⇒BD=3, CD=4} DC 4 BD = 3, Dc = 4 73. A BE BE BE 3 x x EC = BE+BC = BE+7 = 4 D r F r 4 BE = 3 BE + 21 ⇒ BE = 21 x+2 O x-2 Coming to the angles,

r Now ∠BAE = ∠ BAF C 0 B x+2 E x-2 ∠DAB = ∠BAC ⇒ ∠BAC + ∠BAF = 180 ∠BAE + ∠DAB = 1 ∠BAF + ∠BAC = 1 ×180 = 900 Let, side AC = (2x - 2) cm 2 2 0 Side CB = (2x) cm BAE + ∠DAB = ∠DAE = 90 2 2 2 Side BA = (2x + 2) cm, where x is a natural AE + DA = 24 number greater than 1. { ED = EB + BD = 21+3= 24} We know that Area of ΔABC = Area (ΔOBC +ΔOAC+ΔOAB) 75. Let Bc = x and AD = y 1 = r {(2x - 2) + (2x) + 2x + 2} BD AB 4 2 = = As per bisector theorem , 3 = (3rx) DC AC Hence, BD = 4x ; Dc = 3x Also, semiperimeter 7 7

(2x-2)+(2x)+(2x+2) In triangle ABD, 2 2 2 16x S = 2 (4) +y - 0 49 cos30 = Area of ΔABC = 3x(x+2)(x)(x-2) = 3rx 2×4×y ⇒ 2×4 × y × 3 = 16 = y2 - ⇒ (x+2)(x)(x-2) 3x 2 2 2 16x 2 2 ⇒ 4 3y = 16 + y - ...... (i) ⇒(x - 4) = 3r 49 2 2 Similarly from triangle ADc, ⇒ x = 3r + 4 ....(i) 2 2 9x 2 9+y - 9x2 For x to be a natural number (3r + 4) has to be 0 49 3y 2 cos30 = ⇒ 3 = 9 + y - 49 .....(ii) a perfect square. 2×3×y Now, subtracting equation (i) × 9 from Only option (4) does not make x a natural equation (ii) × 16 number Hence, option (4) is the correct 3y 2 2 12 3 choice. 36 = 9y - 16y ⇒ y = 7 Alternately

74. 1 0 4 F ∆ABD = × 4 × y sin 30 = y 2 4 1 0 3 A ∆ADC= × 3 × y sin30 = y 2 4 ∴ 7 ∆ABC = 4 y 8 6 Again area of ABC

1 0 2 3 = 2 × 4 × 3 × sin60 = 2×2 E C B D 7 12 3 12 3 7 ∴ y = ⇒ y = 4 7 7 16 NEON CLASSES 168, Near Kanha Restaurant, Gopalpura Bridge, Tonk Road, Jaipur Contact No.: 9828728833/34 • Website : www.neonclasses.co.in Coming Soon ... vius Doubts dks WhatsApp (95299 49007) djsa vkSj ?kj cSBs ik;sa Detailed Explanation & Conceptual Understanding by - Manisha Bansal Ma'am Author

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