A
B
C
B
A
C
A
C
B
a
b
d
149
23- mavzu. ROMBNING YUZI
A
B
C
D
AC
=
d
1
BD
=
d
2
150
O
)
30
°
A
P D
B
C
151
a
=
AB
=
BC
=
CD
=
AD
= 20 sm,
BP
⊥
AD
,
.
A
=
30°.
h
83
(
)
+
=
+
=
⋅
⋅
=
⋅
+ ⋅
= ⋅ ⋅ ⋅
=
⋅
1
1
1
1
1
1
1
1
1
1
1
1
1
4
4
.
ABCADC
5 5
5
@
@
@
@
@ @
@ @
@ @
@ @
Demak,
5
@ @
=
⋅
. Teorema isbotlandi.
1- m a s a l a .
ABCD
rombning tomoni 20 sm ga, otkir burchagi esa 30° ga
teng. Shu rombning yuzini toping (151- rasm).
Y e c h i l i s h i . 1)
ABP
togri burchakli.
h
=
BP
=
0,5
a
=
0,5 · 20
=
10 (sm)
(30° li burchak qarshisidagi katet gipotenuzaning yarmiga teng).
2)
S
=
ah
=
20 · 10
=
200 (sm
2
).
J a v o b :
S
=
200 sm
2
.
2- m a s a l a .
Rombning diagonallaridan biri ikkinchisidan 1,5 marta katta,
rombning yuzi esa 27 sm
2
ga teng. Shu rombning diagonallarini toping.
B e r i l g a n :
ABCD
romb;
S
ABCD
=
27 sm
2
;
AC
=
1,5
BD
(150- rasmga q.)
T o p i s h k e r a k :
AC
,
BD.
Y e c h i l i s h i .
BD
=
x
sm bolsin, u holda
AC
=
1,5
x
sm boladi.
=
⋅
1
2
)*+,
5
)+ *,
,
bunga belgilashlarni qoyamiz:
⋅
=
⋅
1
2
27
1,5
N N
. U holda
x
2
=
36 boladi, bundan
x
=
6 (sm). Shunday qilib,
BD
=
6 sm va
AC
=
1,5 · 6
=
9 (sm) ga
teng.
J a v o b : 9 sm, 6 sm.
305.
1) Rombning yuzi tomoni va balandligi boyicha qanday topiladi?
2) Rombning yuzi diagonallari orqali qanday topiladi? Uni ifodalang.
306.
Rombning yuzi 40 sm
2
, balandligi esa 5 sm ga teng. Shu rombning
perimetrini toping.
307.
Rombning balandligi 16 sm, otkir burchagi esa 30° ga teng. Shu romb-
ning yuzini toping.
308.
Rombning tomoni 1,8 dm, otkir burchagi esa 30° ga teng. Shu romb-
ning yuzini toping.
309.
Diagonallari: 1) 1,5 dm va 1,8 dm; 2) 24 sm va 15 sm; 3) 2,5 dm va
4 sm; 4) 3,2 sm va 0,5 dm bolgan rombning yuzini toping.
310.
Rombning tomoni 6 sm ga, yuzi esa 18 sm
2
ga teng. Shu rombning
katta burchagini toping.
311.
Romb burchaklarining nisbati 1 : 5 ga, tomoni esa
a
ga teng. Shu romb-
ning yuzini toping.
312.
Rombning tomoni 8 sm ga, otkir burchagi 30° ga teng. Shu rombning
diagonallari kopaytmasini toping.
313.
Rombning yuzi 60 sm
2
, diagonallaridan biri 10 sm ga teng. Shu
rombning ikkinchi diagonalini toping.
314.
Romb diagonallarining nisbati 1 : 2 ga, uning yuzi esa 32 sm
2
ga teng.
Shu rombning tomonini toping.
315.
Rombning yuzi 30 sm
2
, perimetri esa 24 sm ga teng. Shu rombning
balandligini toping.
Savol, masala va topshiriqlar
84
T e o r e m a .
24- m a v z u . TRAPETSIYANING YUZI
Malumki, har qanday kopburchakni diagonallar otkazish yoli bilan uch-
burchaklarga ajratish mumkin. Ixtiyoriy kopburchakning yuzini hisoblash uchun
uni avval uchburchaklarga ajratib olamiz. Songra uchburchaklar yuzi hisob-
lanadi. Kopburchak yuzi esa uni tashkil qilgan bir-birini qoplamaydigan uchbur-
chaklar yuzlari yigindisiga teng boladi. Trapetsiya yuzlarini hisoblashda shu
usuldan foydalanamiz.
Trapetsiyaning yuzi uning asoslari yigindisining yarmi bilan balandligi
kopaytmasiga teng:
=
⋅
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