a
b
103
379.
Uchburchakning tomonlari: 1)
a
=
5 sm,
b
=
7 sm,
c
=
6 sm; 2)
a
=
13 dm,
b
=
14 dm,
c
=
15 dm; 3)
a
=
24 sm,
b
=
25 sm,
c
=
7 sm ga teng. Katta
tomonga tushirilgan balandlikni toping.
380.
1) Agar teng tomonli uchburchakning tomoni 12 sm ga teng bolsa,
uning balandligini; 2) agar teng tomonli uchburchakning balandligi
16 sm ga teng bolsa, uning tomonini toping.
381.
Balandligi
h
ga teng bolgan teng tomonli uchburchakning tomonini
toping.
382.
Uchburchakning tomonlari 16 sm, 12 sm va 8 sm ga teng. Shu uchbur-
chakning kichik balandligini toping.
383.
Uchburchakning tomonlari 8 sm, 10 sm va 12 sm ga teng. Shu uchbur-
chakning eng katta va eng kichik balandliklarini toping.
384.
Tomonlari: 1) 17, 65, 80; 2) 8, 6, 4; 3) 24, 25, 7; 4) 30, 34, 16;
5) 15, 17, 8 ga teng bolgan uchburchakning kichik balandligini toping.
Malumki, uchburchakning yuzi uning asosi bilan balandligi kopaytmasining
yarmiga teng:
1
1
1
2
2
2
a
>
?
5
a
D
> D
? D
=
⋅
=
⋅
=
⋅
.
Balandlik orniga uning uchburchak tomonlari orqali ifodasini qoyib, uni
soddalashtirib ushbu formulani hosil qilamiz:
=
−
−
−
5
F F
=
F
>
F
?
.
Bu formula milodning I asrida yashagan qadimgi yunon olimi iskandariyalik
Geron
tomonidan topilgan bolib, u
Geron formulasi
deb ataladi.
Geron formulasi uchburchakning uchala tomoni uzunligi malum bolganda
uning yuzini hisoblash uchun ishlatiladi.
385.
Uchburchakning yuzi uchun Geron formulasini keltirib chiqaring. Uch-
burchakning yuzini yana qanday formulalar yordamida hisoblash mum-
kin? Ularning ifodasini keltiring.
386.
Uchta tomoniga kora uchburchakning yuzini toping:
1) 17, 65, 80;
2) 15, 15, 18;
3) 4, 13, 15;
4) 29, 25, 6.
387.
Rombning tomoni 26 sm ga, diagonallaridan biri esa 48 sm ga teng. Shu
rombning yuzini toping.
388.
Teng tomonli uchburchakning yuzi
3
4
a
S
=
formula boyicha hisob-
lanadi, bunda
a
uchburchakning tomoni. Shuni isbotlang.
3 1- m a v z u .
UCHBURCHAK YUZI UCHUN GERON FORMULASI
Savol, masala va topshiriqlar
104
389.
Rombning diagonallari 18 dm va 24 dm. Shu rombning perimetri va
parallel tomonlar orasidagi masofani toping.
390.
Teng tomonli uchburchakning tomoni: 1) 15 sm; 2) 3,2 dm; 3) 20 sm;
4)
"
2
sm; 5) 6 sm. Uchburchakning yuzini toping.
391.
Tomonlari: 1) 39, 42, 45; 2) 35, 29, 8; 3) 8, 10, 14; 4) 45, 39, 12;
5) 20, 20, 32 ga teng bolgan uchburchakning yuzini toping.
Ushbu mavzuda Pifagor teoremasiga bogliq
amaliy masalalarni korib chiqamiz.
1- m a s a l a .
Ustunni tik ornatish.
Y e c h i l i s h i
.
Pifagor teoremasi amaliy ma-
salalarni hal qilishda juda kop ishlatiladi. Ushbu
masala ham shular jumlasidandir. Buning uchun
3, 4 va 5 sonlaridan iborat Pifagor uchligidan foy-
dalanamiz. Bu sonlar uchun 3
2
+ 4
2
= 5
2
tenglik
orinlidir. Bundan katetlari 3 va 4 uzunlik birligiga
teng bolgan togri burchakli uchburchakning gi-
potenuzasi 5 birlikka teng boladi.
Ustunni tik ornatish uchun ustun uzunligini ip bilan olchaymiz. Songra bu
ipni ikki marta teng ikkiga bolamiz. Bunda ustunga nisbatan bir uzunlik birligini
hosil qilamiz. Ustun esa tort birlikka teng boladi. Ustun asosidan boshlab uch
birlik olchaymiz va bu nuqtadan ustun uchigacha masofani olchaymiz. Agar bu
masofa besh birlikka teng bolsa, ustun tekislikka nisbatan tik turgan boladi.
Faqat bu ishni kamida ikki yonalishda bajarish lozim (182- rasm).
2- m a s a l a .
Tomonining har biri 10 birlikka teng bolgan teng tomonli
uchburchakning yuzi topilsin (183- rasm).
Y e c h i l i s h i
.
Al-Xorazmiy uchburchakning yuzini asosi bilan balandligi
kopaytmasining yarmiga teng ekanini, turli tomonli uchburchakda biror uchdan
tushirilgan balandlik ozi tushgan tomonni teng bolaklarga bolmasligini, teng
yonli va teng tomonli uchburchaklarda esa asos
teng ikkiga bolinishini aytadi, songra teng
tomonli uchburchakning yuzini quyidagi tartibda
hisoblashni tavsiya qiladi, yani masalani quyi-
dagicha yechadi:
uchburchakning balandligi:
100
5
75
N
h
=
−
=
,
uchburchakning yuzi:
# # # %# #
3 43,3
5
),
=
= ⋅
−
= ⋅
=
≈
3 2- m a v z u .
MASALALAR YECHISH
4
3
2
1
1
2
3
O
182
C
D
B
A
10
10
10
183
7
5
105
yoki
2
# %#
1
8
%#
43,3
5
=
⋅
=
≈
.
Bularning hammasi soz bilan tushuntirilgan.
392.
1) Ustunning tik ekani qanday tekshiriladi?
2) Tomonlari 5, 6 va 9 birlikka teng bolgan uchburchakning yuzini
toping.
393.
Teng yonli trapetsiyaning diagonali 25 sm ga, balandligi esa 15 sm ga
teng. Trapetsiyaning yuzini toping.
394.
ABCD
kvadratning tomoni 12 sm ga teng. Uning
AB
tomonida
P
nuqta
shunday belgilanganki, unda
PC
=
13 sm.
APCD
tortburchakning yuzini
toping.
395.
Togri tortburchakning perimetri 62 sm, diagonallarining kesishish
nuqtasidan tomonlardan birigacha bolgan masofa esa 12 sm ga teng.
Shu togri tortburchakning diagonalini toping.
396.
ABCD
togri tortburchakning
BC
tomonida
P
nuqta shunday belgi-
langanki, unda
AP
=
15 sm,
BA
=
12 sm,
PC
=
6 sm.
APCD
tortbur-
chakning yuzini toping.
397.
Uchburchakning balandligi 36 sm, yon tomoni 85 sm va 60 sm. Shu
uchburchakning yuzini toping (ikki holni korib chiqing).
398.
Togri tortburchakning tomonlari 8 sm va 15 sm. Uning diagonalini
toping.
399.
Rombning diagonallari 14 sm va 48 sm. Rombning perimetrini va paral-
lel tomonlar orasidagi masofani toping.
400.
ABC
uchburchakda
A
burchak otmas,
BP
uchburchakning balandligi.
BC
2
=
AB
2
+
AC
2
+ 2
AP
·
AC
ekanini isbotlang.
401.
Tomonlari: 1)
5
6
,
5
6
, 6; 2) 13,
!
!
7
,
!
4
%
ga teng bolgan uchbur-
chakning eng katta balandligini toping.
402.
Rombning tomoni 20 sm ga, diagonallaridan biri esa 24 sm ga teng. Shu
rombning yuzini toping.
403.
Biror trapetsiyaning otmas burchagi uchidan chiqqan diagonali va yon
tomoni, mos ravishda, 26 sm va
#
77
sm ga, uning balandligi 24 sm,
kichik asosi esa 7 sm ga teng. Trapetsiyaning yuzini toping.
404.
Teng yonli trapetsiyaning asoslari 7 sm va 13 sm ga, otmas burchagi esa
135° ga teng. Shu trapetsiyaning yuzini toping.
3- § ga doir qoshimcha mashqlar
Savol, masala va topshiriqlar
106
1.
Togri burchakli uchburchakning katetlaridan biri 12 sm, gipotenuzasi esa
ikkinchi katetdan 6 sm uzun. Gipotenuzaning uzunligini toping.
A) 15 sm;
B) 25 sm;
D) 26 sm;
E) 18 sm.
2.
Togri burchakli uchburchakning gipotenuzasi 25 sm, katetlari ozaro 3 : 4
nisbatda. Shu uchburchakning kichik katetini toping.
A) 10 sm;
B) 15 sm;
D) 9 sm;
E) 20 sm.
3.
Togri burchakli uchburchakning katetlaridan biri 12 sm, ikkinchisi esa
gipotenuzadan 8 sm qisqa. Shu uchburchakning gipotenuzasini toping.
A) 15 sm;
B) 16 sm;
D) 13 sm;
E) 25 sm.
4.
Tomonlari 13 sm, 14 sm va 15 sm bolgan uchburchakning eng kichik
balandligi necha santimetr?
A) 11,5 sm;
B) 11,1 sm;
D) 11 sm;
E) 11,2 sm.
5.
Rombning diagonallari 14 sm va 48 sm ga teng. Shu rombning perimetrini
toping.
A) 60 sm;
B) 100 sm;
D) 80 sm;
E) 120 sm.
6.
Togri burchakli
ABCD
(
∠
D
=
90°) trapetsiyaning asoslari 17 sm va 9 sm,
kichik yon tomoni esa 15 sm ga teng.
AB
tomonni toping.
A) 15 sm;
B) 17 sm;
D) 9 sm;
E) 8 sm.
«Bilki,
deb yozadi Xorazmiy,
har bir togri burchakli uchburchak
shundayki, agar kichik tomonlarining har biri oz-oziga kopaytirilsa va bu ko-
paytmalar qoshilsa, bu katta tomonining oz-oziga kopaytmasiga teng boladi»
.
Buni isbotlash uchun Xorazmiy
ABDC
kvadrat shakl yasaydi (184- rasm).
Uning
AC
tomonini
E
nuqtada teng ikkiga bolib, unga
EG
perpendikular
otkazadi.
AB
ni
F
nuqtada teng ikkiga bo-
lib, unga
FH
perpendikular otkazadi. U
holda
ABDC
shakl tortta ozaro teng shakl-
lardan iborat boladi. Songra
EF
,
FG
,
GH
,
HE
chiziqlarni otkazib, sakkizta ozaro teng
uchburchaklar hosil qiladi.
AF
chiziqning
oz-oziga kopaytmasi bilan
AE
chiziqning
oz-oziga kopaytmasi birgalikda tortta
ozaro teng uchburchaklar yuzlarini hosil
qiladi.
FE
chiziqning oz-oziga kopayt-
masi ham xuddi shunday ozaro teng uch-
burchaklar yuzlarini tashkil etadi. Isbot ana
shundan iboratdir.
T a r i x i y m a l u m o t l a r
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