Planimetriya: murakkab masalalar

Bu mavzuda DTM nimani so'raydi

DTM testlarida planimetriyadan 2–3 ta savol tushadi. Ular bir nechta xossani birga qo'llashni talab qiladi:

1. Uchburchaklar: Bissektrisa va mediana xossalari, yuzalar, ichki va tashqi aylanalar radiuslari ($r$ va $R$).
2. To'rtburchaklar: Trapetsiyada diagonallar hosil qilgan yuzalar, romb va parallelogrammda diagonallar va tomonlar bog'liqligi.
3. Aylana: Urinma va kesuvchi teoremalari, vatarlar proporsiyalari, to'rtburchakka aylana chizish shartlari.
4. O'xshashlik: Uchburchaklar o'xshashligidan tomonlar va yuzalar nisbatini topish.

Asosiy formulalar va usullar

Figura	Formulalar	Eslatma
Uchburchak yuzi	$S = \frac{1}{2}ah = \frac{1}{2}ab\sin\gamma = pr = \frac{abc}{4R}$	Geron: $S = \sqrt{p(p-a)(p-b)(p-c)}$
Bissektrisa va mediana	$\frac{BD}{DC} = \frac{AB}{AC}$; $l_a = \frac{2bc\cos(\alpha/2)}{b+c}$	Medianalar $2:1$ nisbatda bo'linadi
To'g'ri burchakli uchburchak	$c^2 = a^2 + b^2$; $h_c = \frac{ab}{c}$; $R = \frac{c}{2}$	$r = \frac{a+b-c}{2} \implies r + R = \frac{a+b}{2}$
Parallelogramm va romb	$d_1^2 + d_2^2 = 2(a^2 + b^2)$; rombda: $d_1^2 + d_2^2 = 4a^2$	Rombda: $S = \frac{1}{2}d_1 d_2 = ah$; $r = \frac{h}{2}$
Trapetsiya yuzalari	$S_{yon} = \sqrt{S_1 S_2}$; $S = (\sqrt{S_1} + \sqrt{S_2})^2$	$S_1, S_2$ — asoslardagi uchburchaklar
Aylana va to'rtburchak	$AK^2 = AC \cdot AD$; $AP \cdot PB = CP \cdot PD$	Tashqi: $a+c = b+d$; Ichki: $\angle A + \angle C = 180^\circ$

Yechilgan misollar

1-misol

$ABC$ uchburchakda $AB = 6\text{ sm}$, $BC = 8\text{ sm}$ va $\angle B = 60^\circ$. $BD$ bissektrisasining uzunligini toping.

A) $\frac{24\sqrt{3}}{7}\text{ sm}$
B) $\frac{12\sqrt{3}}{7}\text{ sm}$
C) $\frac{18\sqrt{3}}{7}\text{ sm}$
D) $4\sqrt{3}\text{ sm}$

Yechim:
Yuzalar orqali: $S_{ABC} = S_{ABD} + S_{CBD}$. $BD = l$ bo'lsin. $\angle ABD = \angle CBD = 30^\circ$.
$\frac{1}{2} \cdot 6 \cdot 8 \cdot \sin 60^\circ = \frac{1}{2} \cdot 6 \cdot l \cdot \sin 30^\circ + \frac{1}{2} \cdot 8 \cdot l \cdot \sin 30^\circ \implies 12\sqrt{3} = \frac{7}{2}l \implies l = \frac{24\sqrt{3}}{7}\text{ sm}$.

Tezkor yo'l:
Formulaga qo'yish: $l = \frac{2ac\cos(\beta/2)}{a+c} = \frac{2 \cdot 6 \cdot 8 \cdot \cos 30^\circ}{14} = \frac{24\sqrt{3}}{7}\text{ sm}$.
Javob: A

2-misol

$ABCD$ trapetsiyada asoslar $BC = 3\text{ sm}$ va $AD = 6\text{ sm}$, diagonallar $O$ nuqtada kesishadi. Agar $\triangle BOC$ yuzi $6\text{ sm}^2$ bo'lsa, trapetsiya to'la yuzini toping.

A) $36\text{ sm}^2$
B) $48\text{ sm}^2$
C) $54\text{ sm}^2$
D) $60\text{ sm}^2$

Yechim:

1. $\triangle BOC \sim \triangle DOA$, koeffitsiyent $k = \frac{AD}{BC} = 2$.
2. $S_{AOD} = k^2 \cdot S_{BOC} = 4 \cdot 6 = 24\text{ sm}^2$.
3. $S_{AOB} = S_{COD} = \sqrt{6 \cdot 24} = 12\text{ sm}^2$.
4. Umumiy yuza: $S = 6 + 24 + 12 + 12 = 54\text{ sm}^2$.

Tezkor yo'l:
$S = (\sqrt{S_1} + \sqrt{S_2})^2 = (\sqrt{6} + \sqrt{24})^2 = (3\sqrt{6})^2 = 54\text{ sm}^2$.
Javob: C

3-misol

Gipotenuzasi $25\text{ sm}$, bir kateti $20\text{ sm}$ bo'lgan to'g'ri burchakli uchburchakka ichki va tashqi chizilgan aylanalar radiuslari yig'indisini ($r + R$) toping.

A) $15\text{ sm}$
B) $17{,}5\text{ sm}$
C) $18\text{ sm}$
D) $20\text{ sm}$

Yechim:

1. Ikkinchi katet: $b = \sqrt{25^2 - 20^2} = 15\text{ sm}$.
2. $R = \frac{c}{2} = 12{,}5\text{ sm}$, $r = \frac{a+b-c}{2} = \frac{20+15-25}{2} = 5\text{ sm}$.
3. Yig'indi: $r + R = 5 + 12{,}5 = 17{,}5\text{ sm}$.

Tezkor yo'l:
To'g'ri burchakli uchburchakda har doim $r + R = \frac{a+b}{2} = \frac{20+15}{2} = 17{,}5\text{ sm}$.
Javob: B

4-misol

Aylana tashqarisidagi $A$ nuqtadan $AB$ urinma va $ACD$ kesuvchi o'tkazilgan. Agar $AB = 8\text{ sm}$ va $AC = 4\text{ sm}$ bo'lsa, $CD$ vatar uzunligini toping.

A) $8\text{ sm}$
B) $10\text{ sm}$
C) $12\text{ sm}$
D) $16\text{ sm}$

Yechim:
Teoremaga ko'ra: $AB^2 = AC \cdot AD \implies 8^2 = 4 \cdot AD \implies AD = 16\text{ sm}$.
$CD = AD - AC = 16 - 4 = 12\text{ sm}$.

Tezkor yo'l:
$CD = \frac{AB^2}{AC} - AC = \frac{64}{4} - 4 = 12\text{ sm}$.
Javob: C

5-misol

Parallelogrammning tomonlari $7\text{ sm}$ va $9\text{ sm}$, bir diagonali $8\text{ sm}$. Ikkinchi diagonal uzunligini toping.

A) $12\text{ sm}$
B) $13\text{ sm}$
C) $14\text{ sm}$
D) $16\text{ sm}$

Yechim:
$d_1^2 + d_2^2 = 2(a^2 + b^2) \implies 8^2 + d_2^2 = 2(49 + 81) = 260$.
$d_2^2 = 260 - 64 = 196 \implies d_2 = 14\text{ sm}$.

Tezkor yo'l:
$d_2 = \sqrt{260 - 64} = 14\text{ sm}$. Oxirgi raqamdan tekshirish: $260 - 64 = \dots 6$, demak $14^2 = 196$.
Javob: C

Mashq

1. Uchburchak tomonlari $13\text{ sm}$, $14\text{ sm}$, $15\text{ sm}$. Eng kichik balandligini toping.

Javob va yechim

Eng kichik balandlik $15\text{ sm}$ li tomonga tushadi. $p = 21\text{ sm}$. $S = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84\text{ sm}^2$. $h = \frac{2S}{a} = \frac{168}{15} = 11{,}2\text{ sm}$. *Javob:* $11{,}2\text{ sm}$.

2. To'g'ri burchakli trapetsiyaning asoslari $4\text{ sm}$ va $7\text{ sm}$, katta yon tomoni $5\text{ sm}$. Yuzini hisoblang.

Javob va yechim

Asosdagi proyeksiya: $7 - 4 = 3\text{ sm}$. Balandlik: $h = \sqrt{5^2 - 3^2} = 4\text{ sm}$. $S = \frac{4 + 7}{2} \cdot 4 = 22\text{ sm}^2$. *Javob:* $22\text{ sm}^2$.

3. Rombning diagonallari $12\text{ sm}$ va $16\text{ sm}$. Unga ichki chizilgan aylana radiusini toping.

Javob va yechim

Tomon: $a = \sqrt{6^2 + 8^2} = 10\text{ sm}$. $S = \frac{1}{2} \cdot 12 \cdot 16 = 96\text{ sm}^2$. $h = \frac{96}{10} = 9{,}6\text{ sm}$. Radius: $r = \frac{h}{2} = 4{,}8\text{ sm}$. *Javob:* $4{,}8\text{ sm}$.

4. Aylananing kesishuvchi ikki vatari $AB$ va $CD$ o'zaro $M$ nuqtada kesishadi. Agar $AM = 6\text{ sm}$, $MB = 8\text{ sm}$, $CM = 4\text{ sm}$ bo'lsa, $MD$ kesmani toping.

Javob va yechim

$AM \cdot MB = CM \cdot MD \implies 6 \cdot 8 = 4 \cdot MD \implies MD = 12\text{ sm}$. *Javob:* $12\text{ sm}$.

5. Teng yonli uchburchakning asosi $12\text{ sm}$, yon tomoni $10\text{ sm}$. Tashqi chizilgan aylana radiusini toping.

Javob va yechim

Balandlik: $h = \sqrt{10^2 - 6^2} = 8\text{ sm}$. $S = \frac{1}{2} \cdot 12 \cdot 8 = 48\text{ sm}^2$. $R = \frac{abc}{4S} = \frac{10 \cdot 10 \cdot 12}{4 \cdot 48} = 6{,}25\text{ sm}$. *Javob:* $6{,}25\text{ sm}$.

6. Teng yonli trapetsiyaning asoslari $8\text{ sm}$ va $18\text{ sm}$. Unga ichki aylana chizilgan bo'lsa, yuzini hisoblang.

Javob va yechim

$2c = a + b \implies c = 13\text{ sm}$. Balandlik: $h = \sqrt{13^2 - 5^2} = 12\text{ sm}$. $S = \frac{8+18}{2} \cdot 12 = 156\text{ sm}^2$. *Javob:* $156\text{ sm}^2$.

7. $ABC$ uchburchakda $AB = 5\text{ sm}$, $AC = 7\text{ sm}$, va $A$ burchak bissektrisasi $BC$ tomonni $D$ nuqtada kesadi. Agar $BC = 6\text{ sm}$ bo'lsa, $BD$ kesmani toping.

Javob va yechim

$\frac{BD}{DC} = \frac{AB}{AC} = \frac{5}{7} \implies BD = \frac{5}{12} \cdot 6 = 2{,}5\text{ sm}$. *Javob:* $2{,}5\text{ sm}$.

8. Parallelogramm tomonlari $6\text{ sm}$ va $10\text{ sm}$, o'tkir burchagi $30^\circ$. Yuzini toping.

Javob va yechim

$S = ab\sin\alpha = 6 \cdot 10 \cdot \sin 30^\circ = 30\text{ sm}^2$. *Javob:* $30\text{ sm}^2$.

9. To'g'ri burchakli uchburchak katetlari $9\text{ sm}$ va $12\text{ sm}$. Gipotenuzaga tushirilgan balandlikni toping.

Javob va yechim

Gipotenuza: $c = \sqrt{9^2 + 12^2} = 15\text{ sm}$. Balandlik: $h_c = \frac{ab}{c} = \frac{9 \cdot 12}{15} = 7{,}2\text{ sm}$. *Javob:* $7{,}2\text{ sm}$.

10. Radiusi $10\text{ sm}$ bo'lgan aylanada markazdan $6\text{ sm}$ masofada yotuvchi vatar uzunligini toping.

Javob va yechim

Vatarning yarmi: $\sqrt{10^2 - 6^2} = 8\text{ sm}$. Butun vatar: $2 \cdot 8 = 16\text{ sm}$. *Javob:* $16\text{ sm}$.

Keng tarqalgan tuzoqlar

1. Urinma va kesuvchi teoremalarida tashqi qism o'rniga ichki qismni ko'paytirish:

   • ✗ Xato: $AK^2 = AC \cdot CD$ deb yozish.
   • ✓ To'g'ri: Urinma kvadrati butun kesuvchi va tashqi qism ko'paytmasiga teng: $AK^2 = AC \cdot AD = AC \cdot (AC + CD)$.

2. To'g'ri burchakli uchburchak balandligini gipotenuzaning yarmi deb olish:

   • ✗ Xato: Balandlikni $h = \frac{c}{2}$ deb hisoblash.
   • ✓ To'g'ri: Gipotenuzaning yarmiga faqat mediana teng ($m_c = \frac{c}{2}$). Balandlik esa $h_c = \frac{ab}{c}$ orqali topiladi.

3. O'xshashlikda chiziqli nisbat va yuza nisbatini teng deb o'ylash:

   • ✗ Xato: Tomonlar $2$ marta ortganda yuza ham $2$ marta ortadi deb hisoblash.
   • ✓ To'g'ri: Yuzalar nisbati chiziqli o'lchamlar nisbatining kvadratiga teng: $\frac{S_1}{S_2} = k^2$. Tomonlar $2$ marta ortsa, yuza $4$ marta ortadi.

4. Ichki va tashqi to'rtburchak qoidalarini adashtirish:

   • ✗ Xato: Aylanaga ichki chizilgan to'rtburchakda tomonlar yig'indisini tenglash.
   • ✓ To'g'ri: $a + c = b + d$ tengligi aylanaga tashqi chizilgan to'rtburchak uchundir. Aylanaga ichki chizilgan to'rtburchakda qarama-qarshi burchaklar yig'indisi $180^\circ$ bo'ladi ($\angle A + \angle C = 180^\circ$).