προαγωγικές γεωμετρία α λυκείου 2011 ευαγγελική σχολή

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Page 1: προαγωγικές γεωμετρία α λυκείου 2011 ευαγγελική σχολή

Graptè Proagwgikè Exet�sei Maòou-Ioun�ou 2011T�xh: B'Exetazìmeno M�jhma: Gewmetr�aJema 11. 'Estw orjog¸nio tr�gwno ABΓ me A = 1x kai ∆ h probol  th koruf  A sthn upote�nousa BΓ. Na apode�xeteìti(aþ) To tetr�gwno mia k�jeth pleur� tou ABΓ e�nai �so me to ginìmeno th upote�nousa ep� thn probol th pleur� aut  sthn upote�nousa. Dhlad  ìtiAB2 = BΓ · B∆ kai AΓ2 = BΓ · Γ∆(bþ) To �jroisma twn tetrag¸nwn twn k�jetwn pleur¸n tou e�nai �so me to tetr�gwno th upote�nousa dhlad 

AB2 +AΓ2 = BΓ2 (Pujagìreio Je¸rhma)2. Na apode�xete ìti an se tr�gwno ABΓ isqÔei AB2 +AΓ2 = BΓ2 tìte A = 1x.Mon�de : 1. 8 2. 8 3. 9Jema 2JewroÔme orjog¸nio parallhlìgrammo ABΓ∆ meAB = 6, A∆ = 8 kai thn probol  E tou A sthn B∆.1. Na upolog�sete thn diag¸nio B∆.2. Na upolog�sete to tm ma ∆E.3. Na upolog�sete to συνB∆Γ.4. Na upolog�sete to tm ma ΓE. Mon�de : 1. 6 2. 6 3. 6 4. 7

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Jema 3Se kÔklo (O,R) jewroÔme thn di�metro AB. Proekte�noume thn AB pro to mèro tou B kai pa�rnoume tm maBP = 2R. Fèrnoume thn efaptomènh tou kÔklou sto shme�o A kai h opo�a tèmnei thn efaptomènh PΓ sto shme�o ∆.

1. Na apode�xete ìti PΓ = 2√

2R.2. Na apode�xete ìti PO · PA = PΓ · P∆.3. Na apode�xete ìti P∆ = 3√

2R.4. Na apode�xete ìti O∆ =√

3R.5. Na upolog�sete w sun�rthsh tou R thn di�meso ∆Z tou trig¸nou ∆OP.Mon�de : 1. 5 2. 5 3. 5 4. 5 5. 5Jema 4Se kÔklo (O,R) pa�rnoume diadoqik� tìxa ⌢

AB = 60◦, ⌢

BΓ = 90◦ kai ⌢

Γ∆ = 120◦.1. Na apode�xete ìti to tetr�pleuro ABΓ∆ e�nai isoskelè trapèzio.2. Na upolog�sete w sun�rthsh tou R ti pleurè tou trapez�ou ABΓ∆ .3. Na apode�xete ìti to Ôyo tou trapez�ou e�nai �so me R(√3+1)2

.4. Na upolog�sete w sun�rthsh tou R to embadìn tou ABΓ∆.5. Na upolog�sete w sun�rthsh tou R to embadìn tou koinoÔ mèrou tou kÔklou (O,R) kai tou perigegrammènoukÔklou tou trig¸nou OΓ∆. Mon�de : 1. 5 2. 5 3. 5 4. 5 5. 5Na apant sete se ìla ta jèmata. Kal  Epituq�aN. SmÔrnh 30 Maòou 20112

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Γραπτές Προαγωγικές Εξετάσεις Μαΐου-Ιουνίου 2011, 30 Μαΐου 2011Τάξη: Β΄, Εξεταζόμενο Μάθημα: ΓεωμετρίαJema 11. 'Estw orjog¸nio tr�gwno ABΓ me A = 1x kai ∆ h probol th koruf  A sthn upote�nousa BΓ. Na apode�xete ìti(aþ) To tetr�gwno mia k�jeth pleur� tou ABΓ e�nai �some to ginìmeno th upote�nousa ep� thn probol  th pleur� aut  sthn upote�nousa. Dhlad  ìti

AB2 = BΓ · B∆ kai AΓ2 = BΓ · Γ∆(bþ) To �jroisma twn tetrag¸nwn twn k�jetwn pleur¸ntou e�nai �so me to tetr�gwno th upote�nousa dhlad AB2 +AΓ2 = BΓ2 (Pujagìreio Je¸rhma)2. Na apode�xete ìti an se tr�gwno ABΓ isqÔei AB2 + AΓ2 =

BΓ2, tìte A = 1x.Mon�de : 1. 6 2. 6 3. 13Apanthsei 1. (aþ) Sqolikì bibl�o sel�da 183(bþ) Sqolikì bibl�o sel�da 1842. Sqolikì bibl�o sel�da 184Jema 2JewroÔme orjog¸nio par-allhlìgrammo ABΓ∆ me AB =6, A∆ = 8 kai thn probol  Etou A sthn B∆.1. Na upolog�sete thn di-ag¸nio B∆.2. Na upolog�sete to tm ma

∆E.3. Na upolog�sete toσυνB∆Γ.4. Na upolog�sete to tm maΓE. Mon�de : 1. 6 2. 6 3. 6 4. 7Apanthsei 1. Efarmìzoume to je¸rhma tou Pujagìra sto orjog¸nio tr�g-wno AB∆:B∆2 = AB2 +A∆2 ⇒ B∆2 = 62 + 82 ⇒ B∆2 = 36 + 64 ⇒B∆2 = 100 ⇒ B∆ =

√100 ⇒ B∆ = 102. Sto orjog¸nio tr�gwno AB∆ to E e�nai h probol  tou Asthn upote�nousa B∆ epomènw

A∆2 = ∆E ·∆B ⇒ 82 = ∆E · 10 ⇒ ∆E = 82

10= 32

53. Apì to orjog¸nio tr�gwno B∆Γ e�nai συνB∆Γ = ∆Γ

∆BkaiafoÔ ∆Γ = 6 èqoume συνB∆Γ = 6

10= 3

5.

4. Efarmìzoume to nìmo twn sunhmitìnwn sto tr�gwno ∆ΓEkai èqoume: ΓE2 = ∆Γ2 +∆E2 − 2∆Γ ·∆E · συνB∆Γ opìteΓE2 = 62+

(32

5

)2−2·6· 325· 35�ra ΓE2 = 772

25kai ΓE =

√772

25=

2

5

√193Jema 3Se kÔklo (O,R) jewroÔme thn di�metro AB. Proekte�noume thn

AB pro to mèro tou B kai pa�rnoume tm ma BP = 2R. Fèrnoumethn efaptomènh tou kÔklou sto shme�o A kai h opo�a tèmnei thnefaptomènh PΓ sto shme�o ∆.1. Na apode�xete ìti PΓ2 = PB · PA.2. Na apode�xete ìti PO · PA = PΓ · P∆.3. Na apode�xete ìti P∆ = 3

√2R.4. Na apode�xete ìti O∆ =

√3R.5. Na upolog�sete w sun�rthsh tou R thn di�meso ∆Z toutrig¸nou ∆OP.Mon�de : 1. 5 2. 5 3. 5 4. 5 5. 5Apanthsei 1. E�nai PΓ2 = PB ·PA epomènw PΓ2 = 2R · 4R dhlad  PΓ2 =

8R2 kai epomènw PΓ =√8R2 = 2

√2R.

2. AfoÔ oi PΓ, A∆ e�nai efaptìmene tou kÔklou sta Γ, A jae�nai k�jete sti akt�ne OΓ, OA kai epomènw OΓ ⊥ PΓkai OA ⊥ A∆. 'Ara oi apènanti gwn�e Γ kai A toutetrapleÔrou OA∆Γ e�nai orjè epomènw kai paraplhrw-matikè . 'Ara to OA ⊥ A∆ e�nai eggr�yimmo kai giautìPO · PA = PΓ · P∆.

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3. Antikajist¸nta sthn sqèsh PO·PA = PΓ·P∆ ta PO = 3R,PA = 4R, PΓ = 2

√2R br�skoume ìti (3R) · (4R) =

(2√2R

P∆ kai epomènw 3 · 2R =√2 · P∆ epomènw P∆ = 3·2R√

2=

3√2R.4. Apì to orjog¸nio OP∆ èqoume ìti O∆2 = OΓ2+∆Γ2. Al-l� ∆Γ = P∆ − PΓ = 3

√2R − 2

√2R =

√2R kai epomènw e�nai O∆2 = R2 +

(√2R

)2= 3R2 apì thn opo�a prokÔpteiìti O∆ =

√3R.5. Efarmìzoume to (pr¸to) je¸rhma twn diamèswn sto tr�gwno

O∆P kai èqoume:∆O2 +∆P2 = 2∆Z2 +

OP2

2Epomènw (√

3R)2

+(3√2R

)2

= 2∆Z2 +(3R)2

2�ra:21R2 = 2∆Z2 +

9

2R2Apì thn teleuta�a sqèsh br�skoume ìti ∆Z2 = 33

4R2 kaiepomènw ∆Z =

√33R2

.Jema 4Se kÔklo (O,R) pa�rnoume diadoqik� tìxa ⌢

AB = 60◦, ⌢

BΓ = 90◦kai ⌢

Γ∆ = 120◦.1. Na apode�xete ìti to tetr�pleuro ABΓ∆ e�nai isoskelè trapèzio.2. Na upolog�sete w sun�rthsh tou R ti pleurè toutrapez�ou ABΓ∆.3. Na apode�xete ìti to Ôyo tou trapez�ou e�nai �so me R(√

3+1)2

.4. Na upolog�sete w sun�rthsh tou R to embadìn tou ABΓ∆.5. Na upolog�sete w sun�rthsh tou R embadìn tou koinoÔ mèr-ou tou kÔklou (O,R) kai tou perigegrammènou kÔklou toutrig¸nou OΓ∆.Mon�de : 1. 5 2. 5 3. 5 4. 5 5. 5Apanthsei

1. E�nai ⌢

∆A = 360◦ −⌢

AB−⌢

BΓ−⌢

Γ∆ = 360◦ − 60◦ − 90◦ −120◦ = 90◦. 'Ara ta tìxa ⌢

BΓ kai ⌢

∆A e�nai �sa. Autìsunep�getai ìti• Oi qordè A∆ kai BΓ e�nai �se kai• Oi qordè AB kai Γ∆ e�nai par�llhle .Epomènw to tetr�pleuro ABΓ∆ e�nai isoskelè trapèzio.2. Ston kÔklo (O,R) e�nai AOB = 60◦ = ω6 kai ΓO∆ =

120◦ = ω3. Epomènw oi qordè AB, ∆Γ dèqontai antis-to�qw ep�kentre gwn�e �se me ti kentrikè gwn�e toukanonikoÔ exag¸nou kai tou kanonikoÔ trig¸nou pou e�nai

eggegrammèna ston kÔklo (O,R). 'Ara e�nai �se me ti pleurè tou kanonikoÔ exag¸nou kai tou kanonikoÔ trig¸noudhlad  AB = λ6 = R kai ∆Γ = λ3 = R√3. 'Omoia oi qordè

BΓ, A∆ dèqontai orj  ep�kentrh gwn�a dhlad  �sh me thnkentrik  gwn�a tou ω4 tou kanonikoÔ tetrag¸nou pou e�naieggegrammèno ston (O,R). 'Ara ja e�nai BΓ = A∆ = λ4 =R√2.3. Fèrnoume OE⊥AB pou tèmnei thn∆Γ sto Z. AfoÔ AB//∆Γja e�nai kai OZ⊥∆Γ. To EZ e�nai Ôyo tou trapez�ou kaiapart�zetai apì ta tm mata OE kai OZ pou e�nai apost matatwn qord¸n AB, ∆Γ en¸ ta apost mata tou e�nai �sa me taant�stoiqa apost mata kanonikoÔ exag¸nou kai tou kanon-ikoÔ trig¸nou. Dhlad  OE = α6 = R

√3

2kai OZ = α3 = R

2.'Ara to Ôyo tou trapez�ou e�nai

EZ = OE+OZ =R√3

2+

R

2=

R(√

3 + 1)

24. QrhsimopoioÔme ton tÔpo tou embadoÔ trapez�ou kai br�sk-oume: (ABΓ∆) = AB+Γ∆

2· EZ = R+R

√3

2· R(

√3+1)2

=R2(2+

√3)

25. H akt�na r tou perigegrammènou kÔklou C tou trig¸nou OΓ∆upolog�zetai apì ton nìmo twn hmitìnwn: Γ∆

ηµΓO∆= 2r �ra

R√

3

ηµ120◦= 2r dhlad  R

√3

√3

2

= 2r kai epomènw r = R. 'Ara tokèntro K tou C an kei ston (O,R). To embadìn tou koinoÔmèrou twn dÔo kÔklwn e�nai dipl�sio tou embadoÔ tou kuk-likoÔ tm mato S dhlad  e�nai2 (tomèa ΓO∆− tr�gwnoΓO∆) =

2

(πR2

3− 1

2R · R · ηµ120◦

)=

1

6R2

(4π − 3

√3)

Allo tropo : To kèntro K tou perigegrammènou kÔklou Ctou trig¸nou OΓ∆ ja an kei sthn mesok�jeto tou A∆ poue�nai diqotìmo th ΓO∆ kai sthn mesok�jeto tou OΓ.

To tr�gwno ΓOK ja e�nai isoskelè me m�a apì ti par� thnb�sh gwn�e tou 60◦ kai epomènw ja e�nai isìpleuro �raKO = R kai suneq�zoume ìpw pio p�nw.

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