26) ABCDEFGH nuntazam sakkizburchak berilgan. Agar ABCD to'rtburchak yuzi 12 ga teng bo'lsa, sakkizburchak yuzini toping. 7) ABCDE beshburchakda $$ \mathrm{AE}=\mathrm{AB} ; \mathrm{BC}=\mathrm{CD} $$ va $$ \angle A=\angle C=90^{\circ} $$ bo'lsin. Agar $$ \mathrm{AC}=6 $$ bo'lsa, bu beshburchak yuzini toping. 18) ABC uchburchakda $$ \angle A C B=100^{\circ}, \angle A B C= $$ $$ 50^{\circ} $$. Markazi $$ C $$ nuqtada bo'lgan, radiusi AC ga teng aylana AB va CB tomonlarni mos ravishda $$ D $$ va $$ E $$ nuqtalarda kesadi. AD va DE to'g'ri chiziqlar orasidagi o'tkir burchakni toping. 19) $$ A B C $$ uchburchak yuzasi 60 ga teng. $$ \overrightarrow{B E}=-\frac{1}{2} \overrightarrow{B A} \text { bo'lsa, } $$ $$ A C E $$ uchburchak yuzini toping.

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UpStudy AI · Accepted Answer

Albatta, keltirilgan geometriya masalalarini yechib chiqamiz.

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### **19) Masala:**

**Berilgan:**
- $$ \triangle ABC $$ uchburchagining yuzi 60 ga teng.
- $$ \overrightarrow{BE} = -\frac{1}{2} \overrightarrow{BA} $$.

**Topshiriq:**
- $$ \triangle ACE $$ uchburchagining yuzini toping.

**Yechim:**

1. **Vektorlar yordamida joylashuvni belgilash:**
   - B punktini boshlang'ich nuqta (tenglama uchun $$ \vec{B} = \vec{0} $$).
   - $$ \vec{A} = \vec{a} $$, $$ \vec{C} = \vec{b} $$.
   - $$ \vec{E} = \vec{B} + \overrightarrow{BE} = \vec{0} - \frac{1}{2}\vec{a} = -\frac{1}{2}\vec{a} $$.

2. **Uchburchak $$ ACE $$ yuzini hisoblash:**
   - Vertex koordinatalari:
     - $$ A = \vec{a} $$
     - $$ C = \vec{b} $$
     - $$ E = -\frac{1}{2}\vec{a} $$
   
   - Vektorlarni topish:
     - $$ \overrightarrow{AC} = \vec{b} - \vec{a} $$
     - $$ \overrightarrow{EC} = \vec{b} - (-\frac{1}{2}\vec{a}) = \vec{b} + \frac{1}{2}\vec{a} $$
   
   - Yuzaning formulasi:
     $$
     S_{\triangle ACE} = \frac{1}{2} |\overrightarrow{AC} \times \overrightarrow{EC}| 
     $$
   
   - Hisoblash:
     $$
     \overrightarrow{AC} \times \overrightarrow{EC} = (\vec{b} - \vec{a}) \times (\vec{b} + \frac{1}{2}\vec{a}) = -\frac{3}{2} (\vec{a} \times \vec{b})
     $$
     $$
     |-\frac{3}{2} (\vec{a} \times \vec{b})| = \frac{3}{2} |\vec{a} \times \vec{b}| 
     $$
   
   - Dastlabki uchburchak $$ ABC $$ yuzasidan:
     $$
     S_{\triangle ABC} = \frac{1}{2} |\vec{a} \times \vec{b}| = 60 \Rightarrow |\vec{a} \times \vec{b}| = 120
     $$
   
   - $$ \triangle ACE $$ yuzi:
     $$
     S_{\triangle ACE} = \frac{1}{2} \times \frac{3}{2} \times 120 = \frac{3}{4} \times 120 = 90
     $$

**Natija:**
$$ \triangle ACE $$ uchburchak yuzasi **90** ga teng.

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Agar boshqa masalalarni ham yechishni istasangiz, iltimos, alohida qilib so'rang.
$$ \triangle ACE $$ uchburchak yuzasi 90 ga teng.

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