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.

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.