76
2- m a s a l a .
Togri burchakli uchburchakning katetlari 12 sm va 24 sm ga
teng. Gipotenuzaning ortasidan uchburchakning
katetlariga perpendikularlar
otkazilgan. Hosil bolgan togri tortburchakning yuzini toping.
B e r i l g a n : togri burchakli
ABC
da:
AO
=
OB
,
OE
⊥
AC
,
OF
⊥
CB
,
AC
=
24 sm,
BC
=
12 sm (138- rasm).
T o p i s h k e r a k :
S
CEOF
.
Y e c h i l i s h i . Bizga malumki, bir togri chiziqqa otkazilgan ikki perpen-
dikular ozaro parallel boladi. Fales teoremasiga kora:
AE
=
EC
=
0,5
AC
=
0,5
·
24
=
12 (sm) va
CF
=
FB
=
0,5
BC
=
0,5
·
12
=
6 (sm).
Demak,
S
CEOF
=
C E · C F
=
12
·
6
=
72 (sm
2
).
J a v o b : hosil bolgan
CEOF
togri tortburchakning yuzi 72 sm
2
ga teng.
272.
1) Togri tortburchakning yuzi nimaga teng?
2) Togri tortburchakning yuzi haqidagi teoremani isbotlashda qanday
xossalardan foydalanildi?
273.
Togri tortburchakning ikki tomoni: 1) 60 sm va 5,8 sm; 2) 3,4 dm va
6 sm; 3) 4 m va 1,4 m; 4) 2,5 dm va 1,2 dm. Uning yuzini toping.
274.
Agar togri tortburchakning yuzi va tomonlaridan biri mos ravishda:
1) 270 sm
2
va 15 sm; 2) 142 dm
2
va 35,5 dm; 3) 16 m
2
va 400 sm;
4) 0,0096 km
2
va 300 m. Uning ikkinchi tomonini toping.
275.
Togri tortburchakning perimetri 26 sm ga teng, tomonlaridan biri esa
9 sm. Shu tortburchakning yuziga teng yuzli kvadratning tomonini toping.
276.
Togri tortburchakning yuzi 40 sm
2
, tomonlarining nisbati 2 : 5 ga teng.
Shu togri tortburchakning perimetrini toping.
277.
Togri tortburchakning boyini
n
marta va enini
k
marta
uzaytirilsa,
uning yuzi qanday ozgaradi?
278.
ABCD
togri tortburchak
B
burchagining bissektrisasi
AD
tomonni
K
nuqtada kesadi,
AK
=
5 sm va
KD
=
7 sm. Shu togri tortburchak-
ning yuzini toping.
279.
Togri tortburchakning ikki tomoni: 1) 24 sm va 20 sm; 2) 3,5 dm va
8 sm; 3) 8 m va 4,5 m; 4) 3,2 dm va 1,5 dm. Uning yuzini toping.
280.
Togri tortburchakning bir tomoni 36 dm, ikkinchisi esa 16 dm. Unga
tengdosh kvadratning tomonini toping.
281.
139- rasmda tasvirlangan shakllarning perimetri va yuzini toping.
Kvadratchaning olchamini 1 kv. sm deb oling.
Savol, masala va topshiriqlar
kv. birlik
a
139
b
d
77
T e o r e m a .
2 1- m a v z u . PARALLELOGRAMMNING YUZI
Parallelogrammning
istalgan tomonini uning
asosi
deb olish mumkin, u
holda shu tomonning ixtiyoriy nuqtasidan qarama-qarshi tomongacha bolgan
masofa uning
balandligi
boladi. Balandlik tomonga
yoki tomonning davomiga
tushishi mumkin. 140- rasmda
BP
va
CF
ABCD
parallelogrammning baland-
liklaridir.
Parallelogrammning yuzi asosi bilan balandligining kopaytmasiga teng:
S
=
a
·
h
.
I s b o t .
ABCD
parallelogrammni korib chiqamiz (140-
a
rasm). Bu paralle-
logrammning asosi qilib
AD
=
a
tomonini olamiz, balandligi esa
h
ga teng bolsin.
S
=
a
·
h
ekanini i s b o t q i l i s h t a l a b e t i l a d i .
Asosi parallelogrammning
BC
tomoniga teng, balandligi esa
h
dan
iborat
bolgan
PBCF
togri tortburchak yasaymiz.
ABP
va
DCF
uchburchaklar teng
(gipotenuzasi va otkir burchagiga kora:
AB
=
DC
gipotenuzalar,
∠
1
= ∠
2
mos burchaklar).
ABCD
parallelogramm
PBCD
trapetsiya bilan
ABP
uchbur-
chakdan,
PBCF
togri tortburchak esa osha
PBCD
trapetsiya bilan
ABP
ga
teng bolgan
DCF
uchburchakdan tuzilgan. Demak,
ABCD
parallelogramm bilan
yasalgan
PBCF
togri tortburchak teng tuzilgandir (yani, tengdoshdir).
Bundan,
ABCD
parallelogrammning yuzi
PBCF
togri tortburchakning yuziga,
yani
ah
teng, degan natija chiqadi.
Shunday qilib,
asosi
a
va unga tushirilgan balandligi
h
bolgan parallelo-
grammning
S
yuzi quyidagi formula boyicha hisoblanadi:
S
=
a
·
h
.
Shuni isbotlash talab qilingan edi.
N a t i j a .
Agar ikki parallelogramm bitta asosga
ega va balandliklari teng
bolsa, ular teng tuzilgandir.
1- m a s a l a .
Parallelogrammning tomonlari 25 sm va 20 sm, birinchi tomo-
niga tushirilgan balandlik esa 8 sm. Shu parallelogrammning ikkinchi tomoniga
tushirilgan balandligini toping.
Y e c h i l i s h i .
ABCD
parallelogrammda:
AD
=
a
=
25 sm,
DC
=
b
=
20 sm,
h
a
=
8 sm (140-
b
rasm).
h
b
=
?
Birinchidan,
S
=
ah
a
=
25 · 8
=
200 (sm
2
).
Ikkinchidan,
S
=
bh
b
, yani 200
=
20 ·
h
b
. Bundan
h
b
=
200 : 20
=
10 (sm).
J a v o b : 10 sm.
A
P
D
F
B
C
25 sm
h
a
h
b
140
20
sm
A
P
D
F
B
C
h
h
)
)
1
2
a
b
78
2- m a s a l a .
B e r i l g a n :
ABCD
parallelogramm,
AD
=
20 sm,
BD
=
16 sm,
∠
BDA
=
30°.
T o p i s h k e r a k :
S
ABCD
.
Y e c h i l i s h i . 1) Berilgan parallelogrammning
BP
balandligini otkazamiz
va
BDP
uchburchakni korib chiqamiz (141- rasm). U togri burchakli, chunki
BP
⊥
AD
.
BP
balandlikni topamiz. 30° li burchak qarshisidagi katet gipotenuza-
ning yarmiga teng, shuning uchun
BP
=
0,5
BD
=
0,5 · 16
=
8 (sm).
2) Shunday qilib,
ABCD
parallelogrammning yuzi
S
=
AD
·
BP
=
20 · 8
=
160 (sm
2
)
ga teng boladi.
J a v o b :
S
=
160 sm
2
.
282.
1) Parallelogrammning asosi va balandligi deganda nimani tushunasiz?
2) Parallelogrammning yuzi haqidagi teoremani ifodalang.
283.
a
parallelogrammning asosi,
h
balandligi,
S
yuzi.
1) Agar
a
=
60 sm,
h
=
0,5 m bolsa,
S
ni; 2) agar
a
=
250 m,
S
=
200 m
2
bolsa,
h
ni; 3) agar
a
=
0,25 m,
h
=
100 sm bolsa,
S
ni; 4) agar
h
=
2 m,
S
=
2 000 sm
2
bolsa,
a
ni toping.
284.
Perimetri 80 sm ga teng bolgan parallelogrammning tomonlari nisbati
2 : 3 ga, otkir burchagi esa 30° ga teng. Parallelogrammning yuzini toping.
285.
1) Parallelogrammning yuzi 72 sm
2
, balandliklari 4 sm va 6 sm. Paral-
lelogrammning perimetrini toping.
2) Parallelogrammning tomonlari 12 sm va 16 sm, balandliklaridan biri
esa 15 sm. Parallelogrammning ikkinchi balandligini toping. Masalaning
nechta yechimi bor?
286.
BP
ABCD
parallelogrammning balandligi (142-rasm). Agar
AB
=
13 sm,
AD
=
16 sm va
∠
B
=
150° bolsa,
S
ABCD
ni toping.
287.
BP
ABCD
parallelogrammning balandligi (143- rasm). Agar
AP
=
PD
,
BP
=
6,4 sm va
∠
A
=
45° bolsa,
S
ABCD
ni toping.
288.
Yuzi 41 sm
2
bolgan parallelogrammning tomonlari 5 sm va 10 sm.
Uning ikkala balandligini toping.
289.
1) Parallelogrammning
a
va
b
tomonlari orasidagi burchak 30°. Shu pa-
rallelogrammning yuzini toping.
2) Parallelogrammning diagonallari kesishish nuqtasidan otgan ixtiyoriy
togri chiziq uni ikkita tengdosh qismga ajratadi. Shuni isbotlang.
Savol, masala va topshiriqlar
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