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Teng yonli trapetsiyaning asoslari va yon tomoniga ko’ra yuzini
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trapetsiyaning yuzi uchun turli formulalar
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- Teng yonli trapetsiyaning diagonali va shu diagonal bilan katta asosi orasidagi burchagiga kura yuzini formulasi
- Teng yonli trapetsiyaning asoslari va utkir burchagiga ko’ra yuzini hisoblash formulasi
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Teng yonli trapetsiyaning asoslari va yon tomoniga ko’ra yuzini formulasi.
2 2 ) ( 4 4 b a c b a S tr − − + =
25
ABCD teng yonli trapetsiyaning a va b, yon tomoni c bo’lsin, uning yuzini topamiz.
D b C
A x E a F x B
, 2
b x = +
2
− = ADE ∆ dan [ ] 2 2 2 2 2 ) ( 4 4 1 ) 2 ( b a c b a c h − − = − − =
) 2 )( 2 ( 2 1 ) ( 4 2 1 2 2
c b c b a b a c h − + + − = − − =
) 2 )( 2 ( 4 ) ( 4 4 2 2 c c b c b a b a b a c b a S ABCD + − + − + = − − + =
Misol. ABCD teng yonli trapetsiyada
, 24sm a =
, 10sm b =
sm c 25 =
? −
S
= ⋅ = + − + − + = 36 64 4 34 ) 50 24 10 )( 50 10 24 ( 4 10 24 tr S
408 12 34 6 8 4 34 = ⋅ = ⋅ ⋅ = 2 408sm S tr =
Eslatma. Agar trapetsiyaning ichki va tashqi aylanalar chizish mumkin bo’lsa, u holda b a с + = 2 va
, d c =
ab c S tr =
Ixtiyoriy to’rt burchaklar uchun ayrim ularning yuzlari bilan bog’liq bo’lgan masalalarni qaraymiz. Teorema: har qanday to’g’rituburchakning yuzi uning diagonallari bilan ular orasidagi sinuslar kupaytmasining yarmiga teng 26
Isbot. ABCD-ixtiyoriy to’trburchak, unda D C o
2
1
A B d BD = , , d AC =
α = ∠AOB ABCD to’rtburchak o’zining diagonallari bilan 4 ta uchburchak bo’lishga, ularning yuzalarini topamiz α sin 2 1
AO S AOB ⋅ = α α sin 2 1 ) 180
sin( 2 1 BO CO BO CO S AOB ⋅ = − ° ⋅ =
α sin 2 1 DO CO S AOB ⋅ = α α sin 2 1 ) 180
sin( 2 1 DO AO DO AO S AOB ⋅ = − ° ⋅ =
ABCD to’rtburchakning yuzi [ ] α sin 2 1 ) )( ( 2 1
AC DO BO CO AO S ABCD ⋅ = + + = Demak, α sin
2 1 2 1 d d S ABCD =
Agar to’rtburchakning diagonallari o’zaro perpendikulyar bo’lsa, , 90 ° = α 1 sin = α bo’ladi va 2 1 2 1
d S =
Teng yonli trapetsiyaning diagonali va shu diagonal bilan katta asosi orasidagi burchagiga kura yuzini formulasi
x d S tr 2 sin 2 1 2 =
ABCD teng yonli trapetsiyada ,
BD AC = = α = ∠AOB
27
В С
d o d
α
ABCD S ni toppish kerak . AOB ∆ uchburchakni qaraymiz, unda x AOB 2 180 − ° = ∠
Teoremaga ko’ra x d x d S ABCD 2 sin 2 1 ) 2 180
sin( 2 1 2 2 = − ° = Teng yonli trapetsiyaning asoslari va utkir burchagiga ko’ra yuzini hisoblash formulasi
β
4 2 2 + = Trapetsiyaning yuzi
⋅ + = 2
D b C
h h c
β
A x E a F x B
, 2
b x = + 2
a x − = 28
, 2 β β tg b a xtg h − = =
β cos
2 b a c − = β β tg b a tg b a b a S tr 2 2 2 2 2 − = − ⋅ + = Aylanaga tashqi chizilgan teng yonli trapetsiyaning yon tomoni l va
asoslaridan biri a ga teng. Trapetsiyaning yuzini toping
h b a S tr ⋅ + = 2
Masalaning ko’rinishiga ko’ra trapetsiyaning o’rta chizig’i uning yon tomoniga teng D C
E h F A x a x B , 2
b a EF = + =
b CD = belgilash olingan l b a 2 = +
a l b − = 2 Shakildan
2 2 2 ) 2 ( 2 2 − = + − = − − = − = = +
l a x − = ) 2 ( 2 2 ) ( 2 2 2 2 2 2 2 2 2
l a a al l al a l l a l x l h − = − = − + − = − − = − =
) 2 ( a l a h − = ) 2 ( a l a l S tr − = Misol. ABCD teng yonli trapetsiyada , 9sm l =
. 9sm a =
Yuzini topamiz 2 81 9 9 9 9 9 ) 9 9 2 ( 9 9 sm S ABCD = ⋅ = ⋅ = − ⋅ = To’g’riburchakli burchakli trapetsiya berilgan. Trapetsiyaning asoslari parallel biror to’g’ri chiziq uni ikkita trapetsiyaga bo’ladi , bu trapetsiyalarning xar biriga ichki aylana chizish mumkin. 29
Agar berilgan trapetsiyaning yon tomonlari a va
d
) ( c d > bo’lsa, uning yuzini xisoblang. Trapetsiyaning yuzi parallel bo’lmagan kesishguncha davom P
ettiramiz, natijada uchta o’xshash uchburchaklar hosil bo’ladi, bunda o’rta va katta D c uchburchaklarning o’xshashlik koeffsentini kichik va o’rta uchburchaklarning uxshashlik c d koeffisenti bilan birxil bo’ladi. Bu koeffsentni λ bilan, trapetsiyaning katta asosini x bilan belgilaymiz, katta aylananing A E D radiusi R bo’lsin. U holda x asosga parallel kesmalarning uzunliklari
λ va x 2 λ bo’ladi pastki trapetsiyaning katta yon tomoni , 2 c d R ikkinchi aylananing radiusi
λ bo’ladi shunday qilib 2
R R = + λ
Tashqi chizilgan to’rtburchakning xossasiga ko’ra . 2 2 c d R R x x + = + λ
Berilgan trapetsiyaning kichik asoslari C uchidan katta asosga CE perpendikulyar tushuramiz, natijada katetlari c, x x 2 λ − va gepotenuzasi d bo’lgan to’g’riburchakli uchburchak hosil qilamiz. B’lardan ushbu sestemaga ega bo’lamiz 30
= + − = − + − = + 2 ) 1 ( ) 1 ( 2 2 ) 1 ( 2 2 2
R c d x d c R x x λ λ Bundan
c c d d x 2 2 − − = Berilgan ABCD trapetsiyaning asoslari uchun quyilgan ifodalarga ega bo’lamiz.
, 2
2 c d d a x − + = =
2 2 2 c d d b − − =
) )( )( )( ( ) ( 4 d b a c d b a c d c b a d c b a b a b a S tr + + − − − + − + − + + − − + =
, 2 2
d b a = = +
2 2
d b a − = −
2 2 c d d c d c b a − + + = + + −
2 2
d c d c d b a − + − = − + −
2 2
d d c d b a c − + − = − − +
2 2
d d c d b a c − − + = + + −
= − − + − + − − + − − + + ) )( )( )( ( 2 2 2 2 2 2 2 2
d d c c d d c c d c d c d d c
[ ][ ] = − − − − − + = 2 2 2 2 2 2 2 2 ) ( ) ( ) ( ) (
d c d c d d c
[ ][ ] ) ( 4 ) 2 )( 2 ( ) ( ) ( ) ( ) ( 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
d c c cd d c d c d d cd c c d c d c d d c − = − + − − + − + + = = − − − − − +
2 4 2 ) ( 4 4 2 2 2 2 2 2 2 2 2 cd c d c d cd c d c c d d S = − − = − ⋅ − = ∆
Trapetsiyaning diagonallari va asoslari o’rtalarini birlashtiruvchi tog’ri chiziq kesmasining uzunligiga ko’ra yuza formulasi:
( )( )( )( )
d d l d d l d d l d d S TR 2 2 2 2 4 1 2 1 2 1 2 1 2 1 + − − + + + + − = 31
ABCD trapetsiyada va d 1
2 d diagonallari hamda asoslarining o’rtalari orasidagi masofa
berilgan. Trapetsiyaning yuzini topamiz. C nuqtadan BD ga parallel qilib to’g’ri chiziq o’tkazamiz va uning AD to’g’ri chiziq bilan kesishgan nuqtasini K bilan belgilaymiz. B C
1 d
2
A L D K
M ACK uchburchakning yuzi berilgan trapetsiyaning yuziga teng, uchburchakning AC va CK tomonlari trapetsiyaning diagonallariga teng, CL medianasi esa trapetsiyaning AD va BC asoslari o’rtalaridagi masofaga teng. CL ni davomida shunday M nuqta olamizki,
= bo’lsin. CKM uchburchakni hosil qildik. Bu uchburchak ABCD trapetsiyaga tengdoshdir. Uchburchakning tomonlari trapetsiyaning diagonallari va asoslari o’rtalari orasidagi masofaning ikkilanganiga tengdir. Bunday uchburchakning yuzini Geron formulasi bo’yicha topilishi mumkin:
(
)( )( ) l d d l d d l d d l d d S S ABCD CKM 2 2 2 2 4 1 2 1 1 2 2 1 2 1 − + + − − + + = = Misol. 3 1
d sm,
5 2 = d sm,
= l 2 4 sm. 2 6 2 12 4 1 4 6 2 12 4 1 sm S tr = ⋅ ⋅ = ⋅ ⋅ ⋅ = ; 2 6sm S S tr uch = = 32
Trapetsiyaning asoslari a va b (a>b). Bu trapetsiyaga ichki va tashqi aylanalar chizish mumkin bo’lsa, uning yuzi hamda ichki va tashqi chizilgan aylanalarning radiuslarini toping. Trapetsiyaga ichki va tashqi aylanalar chizish mumkinligidan uning o’rta chizig’i yon tomoniga teng bo’lgan teng yonli trapetsiya ekanligi kelib chiqadi. B b C
E h F
A x K a L x D c b a EF = + = 2
Chizmadan
a b x = + 2 ,
2 b a x − = ABK uchburchakdan:
( ) ab ab b ab a b ab a b a b a x c h = = = − + − + + = − − +
= − = 4 4 1 2 2 4 1 2 2 2 2 2 2 2 2 2 2 2
ab h = 2 , ab h =
Trapetsiyaning yuzi:
ab c ab b a h b a a S tr = ⋅ + = + = 2 2 Misol. Trapetsiyaning asoslari 4 sm va 16 sm, unga ichki va tashqi aylanalar chizish mumkinligi ma’lum. U vaqtda
2 80
8 10 4 16 2 4 16 sm S S tr tr = = ⋅ = ⋅ ⋅ + = 33
Ichki va tashqi chizilgan aylanalarning radiuslari
sm ab h r 4 2 8 2 1 2 = = = = , sm R 41 4 5 =
Teng yonli trapetsiyaning o’rta chizig’ining uzunligi m va diagonallari o’zaro perpindikulyar. Yuzini toppish kerak. Trapetsiyaning diagonallari o’zaro perpindikulyar, shunga ko’ra uning o’rta chizig’i balandligiga teng. B C
E m F
A D
m h EF = = Bunday trapetsiyaning yuzi:
2
m m h EF S ABCD = ⋅ = ⋅ = Misol: sm EF 5 = diagonallari BD AC ⊥ , yuzini toping. 2 25 5 5
h EF S ABCD = ⋅ = ⋅ = 2) Trapetsiyaning yuzi 2
, diagonallari o’zaro perpindikulyar bo’lsa, balandligini toping. Yuqoridagiga ko’ra:
2
h EF S tr = ⋅ = , ya’ni 2 2 a h = , a h =
3) Trapetsiyaning asoslari a va b. Trapetsiya yuzini teng ikkiga bo’luvchi va asoslariga parallel bo’lgan to’g’ri chiziq kesmasining uzunligi 2 2
b a +
bo’ladi. 34
4) Trapetsiyaning asoslariga parallel va uning diagonallarining kesishgan nuqtasidan o’tgan to’g’ri chiziq kesmasining uzunligi b a ab + 2 bo’ladi. 5) Trapetsiya yon tomonlarining o’rtalarini birlashtiruvchi to’g’ri chiziq kesmasi (trapetsiyaning o’rta chizig’i) uning asoslariga parallel va
2
+ = 6) Trapetsiyaninig asoslari orasidagi har qanday to’g’ri chiziq kesmasi uning uning o’rta chizig’I bilan teng ikkiga bo’linadi. D a C
E F A b B 7) Trapetsiya asoslarining o’rtalarini birlashtiruvchi to’g’ri chiziq kesmasi uni ikki tengdosh trapetsiyaga bo’ladi 8) Trapetsiya diagonallari o’rtalarini birlashtiruvchi to’g’ri chiziq kesmasi uning asoslariga parallel va uzunligi
2 1 1 b a C B − = ga teng bo’ladi.
B C
1 B
1 C
A D 35
9) Teng yonli trapetsiya tomonlarining o’rtalari rombning uchlari bo’lib xizmat qiladi. Trapetsiyaning asoslari a va b, diagonallari asosidagi o’tkir burchaklarinig bissektrisalari bo’lsa, trapetsiyaning yuzini toping. D C
β
α
α O β
α
β
A B 2 2 ) ( 4 4 b a b b a S ABCD − − + =
ABCD trapetsiyada a AB = , b CD = (a>b), AC va BD diagonallar A ∠ va B ∠ burchaklarning bissektrisalari bo’lib xizmat qiladi. α = ∠BAC bo’lsin, u holda α =
. CD AB bo’lgani uchun α =
bo’ladi, (ichki almashinuvchi burchaklar sifatida), shuning uchun ACD uchburchak – teng yonli, ya’ni
b CD AD = = . Huddi shuningdek, agar β =
bo’lsa, β = ∠ = ∠ = ∠
CBD ABD
bo’ladi, ya’ni BCD uchburchak teng yonli, shuning uchun b BC CD = = Shunday qilib, ABCD - teng yonli traprtsiya va uning yuzi:
2 2 ) ( 4 4 b a c b a S ABCD − − + =
yoki
2 2 2 2 2 3 4 ) ( 4 4 a ab b b a b a b b a S ABCD − + + = − − +
Misol. Teng yonli trapetsiyaning asoslari 6 sm va 15 sm, uning diagonali o’tkir burchagini teng ikkiga bo’ladi. Tapetsiyaning yuzini va perimetrini toping.
36
Yechish. Yuqorida isbot qilinganidek,
sm AB 15 = , sm CD 6 = D 6 sm C
6 = =
A 15 sm B
Perimetri: sm P 33 6 3 15 = ⋅ + = Trapetsiyning yuzi esa,
63 4 21 3 21 4 21 ) 9 12 )( 9 12 ( 4 21 9 6 4 4 21 ) 6 15 ( 6 4 4 6 15 ) ( 4 4 2 2 2 2 2 2 = ⋅ = − + = − ⋅ = = − − ⋅ + = − − +
a b b a S ABCD
2 7 4 63 sm S ABCD =
Misol. Teng yonli trapetsiya asoslari 12 sm va 16 sm, balandligi 14 sm. Trapetsiyaga tashqi chizilgan doira yuzi topilsin. Yechish. Trapetsiyaning yuzi
196 14
2 12 16 2 = = ⋅ + = tr S
2 sm
sm h r 7 2 14 2 = = =
Π = Π = 49 7 2 r S
sm R 10 = Π = Π = 100 10 2
S
O p
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