O’zbekiston respublikasi oliy va o’rta maxsus ta’lim vazirligi alisher navoiy nomidagi samarqand davlat universiteti
-§. Trapetsiyaning asoslari va dioganallariga ko’ra
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trapetsiyaning yuzi uchun turli formulalar
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- 5-§. Trapetsiyaning asoslari, yon tomoni va dioganallariga ko’ra yuzini hisoblash formulasi
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4-§. Trapetsiyaning asoslari va dioganallariga ko’ra yuzini hisoblash formulasi 3) Trapetsiyaning asoslari va diagonallariga ko’ra yuzining formulasi. ) )(
)( ( 4 1 2 1 2 1 2 1 2 1 d d b a d d b a d d b a d d b a S tr − − + + − + − + + + + + =
ABCD trapetsiyaning asoslari a va b, diagonallari esa 2 1 vad d
bo’lsin yuzi h b a S tr 2 + =
formula bo’yicha hisoblanishi mumkin bunda h balandlikni topish kerak x BF = desak x b EF BF BE + = + = bo’ladi x a AF − = BDE va ACF tug’ri bo’rchakli uchbo’rchakdan topamiz h 2
1 2 ) ( x b d + − =
2 2 2 2 ) (
b d h − − =
2 2 2 2 1 ) ( ) (
b d x b d − − = + − 2 2 2 2 2 2 2 1 2 2
ax a d x bx b d − + − = − − −
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2 2 2 2 2 1 ) ( 2 b a d d x b a − + − = + ) ( 2 ) ( ) ( 2 2 2 1 2 2
a d d b a x + − + − = [ ]
] [ ] [ ] [ ][ ] { } [ ] { } [ ][ ] ) )( )( )( ( ) ( ) ( ) ( 2 ) ( ) ( 2 ) ( ) ( ) ( ) ( 2 ) ( ) ( ) ( 2 ) ( ) ( ) ( 2 ) ( ) ( ) ( 4 ) ( ) ( ) ( 2 ) ( 4 ) ( 4 2 1 2 1 2 1 2 1 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 1 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 1 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2
d b a d d b a d d b a d d b a d b a d d d b a d d b a b a d d d d b a b a d d b a d b a d d b a d b a d d b a d b a d d b a d b a d d b a b b a d b a h b a − − + + − + − + + + + + = = − + − − + + = + − − + − − + + + − = = − − + − + − + − + + = = − + − − + = = − + − − + = − + − + + − + = +
2
ning ikkinchi ifodasi uchun ham xuddi shunday natija kelib chiqadi. Shunday qilib ) )(
)( ( 4 1 2 1 2 1 2 1 2 1 d d b a d d b a d d b a d d b a S tr − − + + − + − + + + + + =
Misollar. 1) Trapetsiyaning balandligini boshqacha topish mumkin
2
2 2 2 2 ) 8 ( ) 17 5 ( ) 24 ( 13 x h x h − − = − − =
2 2 ) 24 ( 13 x − − = 2 2 ) 8 ( ) 17 5 ( x − − 169-576+48x- 25 144
169 2 = − =
64x=768 x=12 sm h 5 = 2 80 5 16 5 2 32 5 2 24 8 2 sm b a S tr = ⋅ = ⋅ = ⋅ + = + =
Trapetsiyaning balandligini boshqacha topish mumkin = + − + − = + − + − = ) 8 24 ( 2 ) 169 425
( ) 64 576 ( ) ( 2 ) ( ) ( 2 2 2 1 2 2 b a d d b a x
17
12 64 768 = =
sm x 12 = , 25 ) 12 13 )( 12 13 ( ) ( 2 2 2 2 = − + = − − =
a d h h=5
2 80 5 16 5 2 8 24
S tr = ⋅ = ⋅ + =
Endi masalani formuladan foydalanib echamiz ) 8 24 13 17 5 )( 13 17 5 8 24 )( 13 17 5 8 24 )( 17 5 13 8 24 ( 4 1 + + + + − + − + + + + + ⋅ ⋅ = tr S
= − − + + = ) 19 17 5 )( 17 5 45 )( 17 5 19 )( 17 5 45 ( 4 1 = = − + − + ) 19 17 )( 19 17 )( 17 9 )( 17 9 ( 5 4 1 2 = 5 64 4 1 64 64 5 4 1 ) 361 425
)( 17 81 ( 5 4 1 ⋅ ⋅ = ⋅ ⋅ = − − ⋅ = 2 80 5 16 sm = ⋅ Eslatma . trapetsiya uchun (2) va (3) formulalar quydagicha yoziladi. 2 2
( 4 4 b a c b a S tr − − + = (c=d) 2 2 ) ( 4 4 b a c b a S tr + − + =
) ( 2 1 d d d = = Misol , 69sm a =
, 51sm b =
sm c 41 = = ⋅ − ⋅ = − ⋅ = − − ⋅ + = 2 2 2 2 2 2 ) 9 2 ( ) 41 2 ( 30 18 41 4 4 120 ) 51 69 ( 41 4 4 5 69 tr S
= − + ⋅ = − ⋅ = − = ) 9 41 )( 9 41 ( 60 9 41 2 30 ) 9 41 ( 2 30 2 2 2 2 2
2400 40 60 1600 60 = ⋅ =
2 2400sm S tr =
5-§. Trapetsiyaning asoslari, yon tomoni va dioganallariga ko’ra yuzini hisoblash formulasi Trapetsiyaning asoslari, yon tomoni va diagonaliga ko’ra yuzini formulasi 18
) )( )( )( ( 4 d a c d c a d c a d c a a b a S tr + − − + + − − + + =
ABCD trapetsiyaning asoslari a,b (a>b), yon tomoni c va diagonal d berilgan. yuzni topamiz x BF = bo’lsin, u holda x b BE + = AE =a-x
x b a b x a EF AF AE − − = − − = − = ) (
BDE ∆ dan 2 2 2 ) ( x b d h + − =
∆ dan
2 2 2 ) (
b d h + − =
Bulardan 2 2 2 ) ( x b d h + − = = 2 2 2 ) ( x b d h + − =
2 2
2 2 2 ) ( 2 ) ( 2 x x b a b a c x bx b d − − + − − = − − −
[ ] 2 2 2 2 ) ( 2 ) ( 2 b a c b d x b b a − + − − = + −
a b a c b d x 2 ) ( 2 2 2 2 − + − − =
= − + − − + − = 2 2 2 2 2 2 2 ) 2 ) ( ( a b a c b d b d h
= + − + − − + − = 2 2 2 2 2 2 2 2 4 ) 2 2 ( a b ab a c b d ab d
[ ][ ] = − − − − = − + − = 2 2 2 2 2 2 2 2 2 2 2 2 ) ( ) ( 4 1 4 ) ( 4 d a c c d a a a c d a d a
[ ] ) )( )( )( ( 4 1 2 d a c d c a d c a d c a a + − − + + − + + =
Demak trapetsiyaning yuzi ) )( )( )( ( 4 d a c d c a d c a d c a a b a S tr + − − + + − + + + =
Agar trapetsiyaning boshqa diagonali berilgan bo’lsa, formula bunday ko’rinishni oladi ) )(
)( ( 4 d b c d c b d c b d c b a b a S tr + − − + + − + + + =
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Misol. Trapetsiyaning asoslari a=24sm b=8sm, yon tomoni c=13sm va digonali d=13sm bo’lsa yuzini toping = +
− + + − + + ⋅ + = ) 13 24 13 )( 13 13 24 )( 13 13 24 )( 13 13 24 ( 24 4 8 24 tr S
80 10 24 24 4 32 2 24 24 50 24 4 32 = ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ =
2 80sm S tr =
Bu misolda trapetsiyaning yon tomoni uning diagonaliga teng c=d=13sm. Bu holda chiqarilgan formula birmuncha sodda ko’rinish oladi. 2 2
2 4 4 4 4
d b a a c b a S tr − + = − + =
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