O’zbekiston respublikasi oliy va o’rta maxsus ta’lim vazirligi alisher navoiy nomidagi samarqand davlat universiteti
-§. Trapetsiyaning diagonali va balandligiga ko’ra
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trapetsiyaning yuzi uchun turli formulalar
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- 7-§. Teng yonli trapetsiyaning yuzasini hisoblash formulalari
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6-§. Trapetsiyaning diagonali va balandligiga ko’ra yuzini hisoblash formulasi Trapetsiyaning diagonali va balandligiga ko’ra yuzini formulasi. ABCD trapetsiyaning , 1
2
diagonallari va h balandligi berilgan. yuzini topamiz. h b a S tr ⋅ + = 2
a AB =
b CD =
BDE va ACF to’g’ri burchakli uchburchaklardan
2 2 1 h d BE − = ) ( ) ( ) ( EF FB BE AF BF EF EF AE BE AF + + + = + = +
2 2 1 h d AF − = , CD EF =
2 2 2 2 2 1 h d h d b a CD AB AF − + − = + = + =
) (
2 2 2 2 2 1 h d h d h S tr − + − =
Misol. Trapetsiyada , 20 1 sm d =
, 15 2 sm d =
sm h 12 = ? − tr S
20
150 25 6 ) 9 16 ( 6 ) 81 256
( 6 ) 12 15 12 20 ( 2 12 2 2 2 2 = ⋅ = + = + = − + − = tr S
2 150sm S tr =
Diagonallari o’zaro perpendikulyar trapetsiyaning asoslari va yon tomonlari orasidagi burchagiga ko’ra yuzini formulasi
α tg b a b a ab S tr − + = 2 ) (
Trapetsiyaning asoslari a va b diagonallari o’zaro pependikulyar va yon tomonlari orasidagi burchagi α bo’lsin ? − tr S
α
y
A a-b E a B ABCD trapetsiyada BC DE o’tkazamiz bunda α =
,
DE =
y BC DE = = Bo’lsin shartga ko’ra BD AC ⊥ , uholda AOD va BOC to’g’ri burchakli uchburchaklardan.
2 2
2 2 2 CO BO y DO AO x + = + =
Bulardan
2 2
2 2 2 2 2 2 2 2 2 ) ( ) ( ) ( ) (
a DO CO BO AO DO AO CO BO y x + = + + + = = + + + = +
dan konus tenglamasiga asosan. , cos
2 cos
2 ) ( 2 2 2 2 2 2 α α
b a xy y x b a AE − + = − + = − = α cos ab xy =
ADE ∠
21
α α α α
ab xy h b a S ADE 2 1 sin cos
2 1 sin 2 1 ) ( 2 1 = = = − = ∆ Bunda
, α
b a ab h − = demak α
b a b a ab h b a S ABCD − + = ⋅ + = 2 ) ( 2
7-§. Teng yonli trapetsiyaning yuzasini hisoblash formulalari
Teng yonli trapetsiyaning katta asosi yon tomoni va diagonaliga ko’ra yuzini formulasi. 2 2
2 2 2 2 2 2 2 ) ( 4 4
d a d a a c d a S tr − + − − + =
D b C d c h h c
A x E a F x B Trapetsiyaning yuzi
h b a S tr ⋅ + = 2
a AB =
c BC AD = = d AC BD = = Shakildan quyidagilarni topamiz x BF AE = = bo’lsin u holda a b 2x = + , 2
a x − = b CD =
, 2 2 b a b a b x b AE EF BF EF BE + = − + = + = + = + = 2
a AF BE + = =
∆ dan
2 2 2 2 ) 2 (
a x c h + − − =
Bulardan 2 2 2 2 2 ) 2 ( b a d BE d h + − = − = 22
2 2 2 2 ) 2 ( ) 2 (
a c b a d − − = + − 2 2 2 2 4 ) ( 4 ) (
d b a b a − = − − + , 4 4 2 2 c d ab − = 2 2 c d ab − = a c d b 2 2 − =
, 2 2 2 2 2 a c d a a c d a b a − + = − + = +
a c d a b a 2 2 2 2 2 − + = +
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 ) ( 4 4 ) ( ) 2 ( a c d a d a a c d a d b a d h − + − = − + − = + − = 2 2 2 2 2 2 2 2 2 2 ) ( 4 4 c d a d a a c d a S tr − + − − + =
Ildiz ostidagi ifodani bunday ko’rinishga keltirish mumkin [ ][ ] [ ][ ] ) )( )( )( ( ) ( ) ( ) ( 2 ) ( 2 ) ( ) 2 ( ) ( 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 a d c d c a c d a c d a d a c c d a c d a ad c d a ad c d a ad c d a d a − + − + − + + + = = − − − + = − + − − + + = = − + − = − + −
Demak trapetsiyaning yuzi ) )( )( )( ( 4 2 2 2 2 a d c d c a c d a c d a a c d a S tr − + − + − + + + − + = Misol. Teng yonli trapetsiyaning katta asosi 44sm yon tomoni 17sm va diagonali 39sm. Yuzini toping Yechish a=44sm, c=17sm, d=39sm. 100 17
44 = + + = + + c d a
66 17 39 44 = − + = − + c d a
22 39 17 44 = − + = − + d c a
12 44 17 39 = − + = − + a c d
22 9 44 18 11 44 198 44 44 792 44 4 3168 44 4 289 1521 1936
4 2 2 2 2 2 = = ⋅ = ⋅ = ⋅ = ⋅ − + = − +
c d a
540 60 9 1320 22 9 12 22 66 100 22 9 = ⋅ = ⋅ = ⋅ ⋅ ⋅ ⋅ = tr S
23
2 540sm S tr =
Boshqacha usul, ABCD- teng yonli trapetsiya unda katta asos 44sm yon tomon 17sm va diagonali 39sm. Yuzini topamiz. , 44sm AB =
sm AD BC 17 = =
sm BD 39 =
h b a S ABCD ⋅ + = 2
D C 39 17 h h 17
A x E 44 F x B , x BF AE = = ,
CD =
c BC =
, 2
b x = + 44 2 = + b x
2 44
x − = 2 44 2 44
b b x b BF EF BE + = − + = + = + =
∆ ga
2 2 2 2 2 2 2 ) 2 44 ( 17 17
x x c h − − = − = − =
BDE ∆ ga 2 2 2 2 2 ) 2 44 ( 39
BE d h + − = − =
2 2
) 2 44 ( 17 ) 2 44 ( 39 b b − − = + −
1232 44
b
cm b 28 = 72 28 44 = + = + b a
36 2
+ b a
, 225 3 75 ) 36 39 ( ) 36 39 ( ) 36 ( 39 2 2 2 = ⋅ = = − ⋅ + = − = h
sm h 15 = 2 540 15 36
S ABCD = ⋅ =
Teng yonli trapetsiyaning asoslari va diagonaliga ko’ra yuzini formulasi. 24
2 2 ) ( 4 4 b a d b a S tr + − + =
ABCD teng yonli trapetsiyada a va b asoslar hamda diagonali malum. ? − tr S
D b C h d h
A x E a F x B x BF AE = = 2 2 b a b a b x b AF y + = − + = + = = ACE ∆ to’g’ri burchakli uchburchakdan [ ] 2 2 2 2 2 2 2 ) ( 4 4 1 ) 2 (
a d b a d y d h + − = + − = − = 2 2 ) ( 4 2 1
a d h + − =
) 3 )( 2 ( 4 ) ( 4 4 2 2
a d d b a b a b a d b a S tr − − + + + = + − + =
Misol. ABCD teng yonli trapetsiyada
a 24 = 10 = b
865 =
408
24 17 48 2 17 2304 2 17 1156 3460 4 34 34 865
4 4 34 2 = ⋅ = ⋅ = = − = − ⋅ = tr S
2 408sm S tr =
Eslatma. Keyingi chiqarilgan ikkita formula ixtiyoriy trapetsiyalar uchun chiqarilgan formulalarning xususiy hollari sfatida hosil qilingan edi.
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