16) ABCDEFGH nuntazam sakkizburchak berilgan. Agar ABCD to'rtburchak yuzi 12 ga teng bo'lsa, sakkizburchak yuzini toping. 17) 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.

Question

UpStudy AI · Accepted Answer

Solve the equation by following steps:
- step0: Solve for $$S$$:
  $$S=12$$
Solve the system of equations $$ AE=AB; BC=CD; angle A=90; angle C=90; AC=6 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}AE=AB\BC=CD\angleA=90\angleC=90\AC=6\end{array}\right.$$
- step1: Calculate:
  $$\left\{ \begin{array}{l}AE=AB\BC=CD\eanglA=90\eanglC=90\AC=6\end{array}\right.$$
- step2: Solve the equation:
  $$\left\{ \begin{array}{l}A=0\BC=CD\eanglA=90\eanglC=90\AC=6\end{array}\right.$$
- step3: Substitute the value of $$A:$$
  $$\left\{ \begin{array}{l}BC=CD\eangl\times 0=90\eanglC=90\0\times C=6\end{array}\right.$$
- step4: Multiply the numbers:
  $$\left\{ \begin{array}{l}BC=CD\0=90\eanglC=90\0=6\end{array}\right.$$
- step5: Simplify the expression:
  $$\left\{ \begin{array}{l}BC=CD\0=90\eanglC=90\end{array}\right.$$
- step6: Calculate:
  $$\left\{ \begin{array}{l}A \in \varnothing \B \in \varnothing \C \in \varnothing \D \in \varnothing \E \in \varnothing \a \in \varnothing \g \in \varnothing \l \in \varnothing \n \in \varnothing \end{array}\right.$$
- step7: Rewrite:
  $$(A, B, C, D, E, a, g, l, n) \in \varnothing$$
Solve the system of equations $$ angle ACB=100; angle ABC=50; D=AC; E=AC; angle ADE=angle ABC $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}angleACB=100\angleABC=50\D=AC\E=AC\angleADE=angleABC\end{array}\right.$$
- step1: Calculate:
  $$\left\{ \begin{array}{l}eanglACB=100\eanglABC=50\D=AC\E=AC\anglADE=anglABC\end{array}\right.$$
- step2: Substitute the value of $$D:$$
  $$\left\{ \begin{array}{l}eanglACB=100\eanglABC=50\E=AC\anglA\times ACE=anglABC\end{array}\right.$$
- step3: Simplify:
  $$\left\{ \begin{array}{l}eanglACB=100\eanglABC=50\E=AC\anglA^{2}CE=anglABC\end{array}\right.$$
- step4: Substitute the value of $$E:$$
  $$\left\{ \begin{array}{l}eanglACB=100\eanglABC=50\anglA^{2}CAC=anglABC\end{array}\right.$$
- step5: Simplify:
  $$\left\{ \begin{array}{l}eanglACB=100\eanglABC=50\anglA^{3}C^{2}=anglABC\end{array}\right.$$
- step6: Solve the equation:
  $$\left\{ \begin{array}{l}A=\frac{100}{eanglCB}\eanglABC=50\anglA^{3}C^{2}=anglABC\end{array}\right.$$
- step7: Substitute the value of $$A:$$
  $$\left\{ \begin{array}{l}eangl\times \frac{100}{eanglCB}\times BC=50\angl\left(\frac{100}{eanglCB}\right)^{3}C^{2}=angl\times \frac{100}{eanglCB}\times BC\end{array}\right.$$
- step8: Simplify:
  $$\left\{ \begin{array}{l}100=50\\frac{100^{3}}{e^{3}a^{2}n^{2}g^{2}l^{2}CB^{3}}=\frac{100}{e}\end{array}\right.$$
- step9: Calculate:
  $$\left\{ \begin{array}{l}A \in \varnothing \B \in \varnothing \C \in \varnothing \D \in \varnothing \E \in \varnothing \a \in \varnothing \g \in \varnothing \l \in \varnothing \n \in \varnothing \end{array}\right.$$
- step10: Rewrite:
  $$(A, B, C, D, E, a, g, l, n) \in \varnothing$$
Keling, har bir masalni alohida ko'rib chiqamiz.

### 16) Sakkizburchakning yuzi
Berilgan to'rtburchak ABCD ning yuzi $$ S_{ABCD} = 12 $$ ga teng. Sakkizburchak ABCDEFGH ning yuzi to'rtburchakning yuziga teng bo'lsa, sakkizburchakning yuzi ham $$ S_{ABCDEFGH} = 12 $$ bo'ladi.

### 17) Beshburchakning yuzi
Beshburchakda $$ AE = AB $$, $$ BC = CD $$ va $$ \angle A = \angle C = 90^{\circ} $$ bo'lsa, $$ AC = 6 $$ ga teng. Beshburchakning yuzi uchun formulani qo'llaymiz. Beshburchakni to'rtburchak va uchburchaklarga bo'lishimiz mumkin.

1. $$ AB = AE = x $$
2. $$ BC = CD = y $$

To'rtburchakning yuzi:
$$
S_{rectangle} = AB \cdot BC = x \cdot y
$$

Uchburchakning yuzi:
$$
S_{triangle} = \frac{1}{2} \cdot AC \cdot AB = \frac{1}{2} \cdot 6 \cdot x
$$

Beshburchakning umumiy yuzi:
$$
S_{ABCDE} = S_{rectangle} + S_{triangle} = x \cdot y + \frac{1}{2} \cdot 6 \cdot x
$$

Bu yerda $$ x $$ va $$ y $$ ni aniqlashimiz kerak. Biroq, berilgan ma'lumotlar yetarli emas, shuning uchun bu masalani hal qila olmaymiz.

### 18) Uchburchakdagi burchak
Uchburchakda $$ \angle ACB = 100^{\circ} $$ va $$ \angle ABC = 50^{\circ} $$ bo'lsa, $$ \angle BAC $$ ni topamiz:
$$
\angle BAC = 180^{\circ} - \angle ACB - \angle ABC = 180^{\circ} - 100^{\circ} - 50^{\circ} = 30^{\circ}
$$

Aylana AB va CB tomonlarni D va E nuqtalarda kesadi. AD va DE to'g'ri chiziqlar orasidagi o'tkir burchakni topish uchun:
$$
\angle ADE = 90^{\circ} - \angle ABC = 90^{\circ} - 50^{\circ} = 40^{\circ}
$$

### 19) Uchburchakning yuzi
Uchburchak $$ ABC $$ ning yuzi $$ S_{ABC} = 60 $$ ga teng. Agar $$ \overrightarrow{BE} = -\frac{1}{2} \overrightarrow{BA} $$ bo'lsa, $$ A $$ nuqtasi $$ B $$ nuqtasidan $$ E $$ nuqtasiga nisbatan 2 marta uzoqroq.

Uchburchak $$ ACE $$ ning yuzi:
$$
S_{ACE} = S_{ABC} - S_{ABE}
$$
Bu yerda $$ S_{ABE} $$ ni topishimiz kerak. Biroq, berilgan ma'lumotlar yetarli emas, shuning uchun bu masalani hal qila olmaymiz.

Umuman olganda, 16-masala bo'yicha sakkizburchakning yuzi 12 ga teng, 17 va 19-masala uchun esa qo'shimcha ma'lumotlar kerak. 18-masala bo'yicha o'tkir burchak 40° ga teng.
16) Sakkizburchakning yuzi 12 ga teng. 17) Beshburchakning yuzini topish uchun qo'shimcha ma'lumotlar kerak. 18) AD va DE to'g'ri chiziqlar orasidagi o'tkir burchak 40° ga teng. 19) Uchburchak $$ ACE $$ning yuzini topish uchun qo'shimcha ma'lumotlar kerak.

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