1. Muammo: To‘rtburchak A(2; 2\sqrt{3}), B(5; 5\sqrt{3}), C(9; 3\sqrt{3}), D(3; \sqrt{3}) nuqtalarida berilgan. \n\n a) \sin \beta ni toping, bu yerda \beta = \angle BAD. \n\n b) ABCD to‘rtburchak yuzini toping.\n\n2. \angle BAD burchagini topish uchun, avvalo vektorlarni aniqlaymiz:\n\n \quad \vec{AB} = (5-2, 5\sqrt{3} - 2\sqrt{3}) = (3, 3\sqrt{3})\n\quad \vec{AD} = (3-2, \sqrt{3} - 2\sqrt{3}) = (1, -\sqrt{3})\n\n3. \sin \beta formulasi: \n\n$$\sin \beta = \frac{|\vec{AB} \times \vec{AD}|}{|\vec{AB}| \cdot |\vec{AD}|}$$\n\nBu yerda \times vektorlar orasidagi 2D kross mahsulotni bildiradi: \n\n$$\vec{u} = (u_x, u_y), \vec{v} = (v_x, v_y) \Rightarrow \vec{u} \times \vec{v} = u_x v_y - u_y v_x$$\n\n4. Kross mahsulotni hisoblaymiz:\n\n$$\vec{AB} \times \vec{AD} = 3 \cdot (-\sqrt{3}) - (3\sqrt{3}) \cdot 1 = -3\sqrt{3} - 3\sqrt{3} = -6\sqrt{3}$$\n\nModulini olamiz:\n\n$$|\vec{AB} \times \vec{AD}| = 6\sqrt{3}$$\n\n5. Vektorlarning uzunliklarini topamiz:\n\n$$|\vec{AB}| = \sqrt{3^2 + (3\sqrt{3})^2} = \sqrt{9 + 27} = \sqrt{36} = 6$$\n\n$$|\vec{AD}| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$$\n\n6. Endi \sin \beta ni hisoblaymiz:\n\n$$\sin \beta = \frac{6\sqrt{3}}{6 \times 2} = \frac{6\sqrt{3}}{12} = \frac{\cancel{6}\sqrt{3}}{\cancel{12} \times 2} = \frac{\sqrt{3}}{2}$$\n\n7. To‘rtburchak yuzini topish uchun, parallelogramma yuzasi formulasi:\n\n$$S = |\vec{AB} \times \vec{AD}| = 6\sqrt{3}$$\n\nAmmo berilgan javob \sqrt{13} ga teng, shuning uchun ehtimol to‘rtburchak parallelogram emas, balki umumiy to‘rtburchak.\n\n8. To‘rtburchak yuzasini topish uchun, uni ikkita uchburchakka bo‘lamiz: ABC va ADC.\n\n9. Uchburchak yuzasi formulasi: \n\n$$S = \frac{1}{2} |\vec{u} \times \vec{v}|$$\n\n10. \n\n$$S_{ABC} = \frac{1}{2} |\vec{AB} \times \vec{AC}|$$\n\n$$\vec{AC} = (9-2, 3\sqrt{3} - 2\sqrt{3}) = (7, \sqrt{3})$$\n\n$$\vec{AB} \times \vec{AC} = 3 \times \sqrt{3} - 3\sqrt{3} \times 7 = 3\sqrt{3} - 21\sqrt{3} = -18\sqrt{3}$$\n\n$$S_{ABC} = \frac{1}{2} \times 18\sqrt{3} = 9\sqrt{3}$$\n\n11. \n\n$$S_{ADC} = \frac{1}{2} |\vec{AD} \times \vec{AC}|$$\n\n$$\vec{AD} \times \vec{AC} = 1 \times \sqrt{3} - (-\sqrt{3}) \times 7 = \sqrt{3} + 7\sqrt{3} = 8\sqrt{3}$$\n\n$$S_{ADC} = \frac{1}{2} \times 8\sqrt{3} = 4\sqrt{3}$$\n\n12. To‘rtburchak yuzasi:\n\n$$S = S_{ABC} + S_{ADC} = 9\sqrt{3} + 4\sqrt{3} = 13\sqrt{3}$$\n\n13. Berilgan javob \sqrt{13} ga teng, shuning uchun ehtimol masalada xatolik bor yoki boshqa usul kerak.\n\n14. Agar to‘rtburchak parallelogram bo‘lsa, uning yuzasi \n\n$$S = |\vec{AB} \times \vec{AD}| = 6\sqrt{3}$$\n\nBu javob berilgan javobdan farq qiladi.\n\nNatija: \n\n a) \sin \beta = \frac{\sqrt{3}}{2} \n b) To‘rtburchak yuzasi = 6\sqrt{3} (berilgan javob bilan farq bor)