{"common":{"save":0,"post_id":"6706","level":2,"total":10,"point":10,"point_extra":0},"segment":[{"id":"7283","mon_id":"1158767","chapter_id":"1158998","post_id":"6706","question":"<p>Tam giác ABC có AB = 9 cm, BC = 15 cm, AC = 12 cm. Khi \u0111ó \u0111\u01b0\u1eddng trung tuy\u1ebfn AM c\u1ee7a tam giác có \u0111\u1ed9 dài là<\/p>","options":["<strong>A.<\/strong> 10 cm","<strong>B.<\/strong> 9 cm","<strong>C.<\/strong> 7,5 cm","<strong>D.<\/strong> 8 cm"],"correct":"3","level":"2","hint":"","answer":"<p>AM² = <span class=\"math-tex\">$\\dfrac{AB^2+AC^2}{2}-\\dfrac{BC^2}{4}$<\/span> = <span class=\"math-tex\">$\\dfrac{9^2+12^2}{2}-\\dfrac{15^2}{4}$<\/span> = <span class=\"math-tex\">$\\dfrac{225}{4}$<\/span> ⇒ AM = <span class=\"math-tex\">$\\dfrac{15}{2}$<\/span> = 7,5 cm.<\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>C.<\/strong> 7,5 cm.<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-09-21 00:48:59","option_type":"txt","len":0},{"id":"7284","mon_id":"1158767","chapter_id":"1158998","post_id":"6706","question":"<p>Ch\u1ecdn công th\u1ee9c \u0111úng trong các \u0111áp án sau:<\/p>","options":["<strong>A.<\/strong> S = <span class=\"math-tex\">$\\dfrac{1}{2}.bc.sinA$<\/span>","<strong>B.<\/strong> S = <span class=\"math-tex\">$\\dfrac{1}{2}.ac.sinA$<\/span>","<strong>C.<\/strong> S = <span class=\"math-tex\">$\\dfrac{1}{2}.bc.sinB$<\/span>","<strong>D.<\/strong> S = <span class=\"math-tex\">$\\dfrac{1}{2}.bc.sinC$<\/span>"],"correct":"1","level":"2","hint":"","answer":"<p>S = <span class=\"math-tex\">$\\dfrac{1}{2}.bc.sinA$<\/span> = <span class=\"math-tex\">$\\dfrac{1}{2}.ac.sinB$<\/span> = <span class=\"math-tex\">$\\dfrac{1}{2}.ab.sinC$<\/span> .<\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"> <strong>A.<\/strong> S = <span class=\"math-tex\">$\\dfrac{1}{2}.bc.sinA$<\/span> .<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-09-21 00:52:43","option_type":"math","len":0},{"id":"7287","mon_id":"1158767","chapter_id":"1158998","post_id":"6706","question":"<p>Tam giác ABC có các trung tuy\u1ebfn <span class=\"math-tex\">$m_a=15$<\/span> , <span class=\"math-tex\">$m_b=12$<\/span> , <span class=\"math-tex\">$m_c=9$<\/span> . Di\u1ec7n tích S c\u1ee7a tam giác ABC b\u1eb1ng <\/p>","options":["<strong>A.<\/strong> 72","<strong>B.<\/strong> 144","<strong>C.<\/strong> 54","<strong>D.<\/strong> 108"],"correct":"1","level":"2","hint":"","answer":"<p>Theo bài toán ta có<\/p><p><span class=\"math-tex\">$\\begin{cases}m_a^2=\\dfrac{b^2+c^2}{2}-\\dfrac{a^2}{4}=15^2\\\\m_b^2=\\dfrac{a^2+c^2}{2}-\\dfrac{b^2}{4}=12^2\\\\m_c^2=\\dfrac{a^2+b^2}{2}-\\dfrac{c^2}{4}=9^2\\end{cases}$<\/span> ⇔ <span class=\"math-tex\">$\\begin{cases}2b^2+2c^2-a^2=900\\\\2a^2+2c^2-b^2=576\\\\2a^2+2b^2-c^2=324\\end{cases}$<\/span> ⇔ <span class=\"math-tex\">$\\begin{cases}a=10\\\\b=4\\sqrt{13}\\\\c=2\\sqrt{73}\\end{cases}$<\/span> .<\/p><p>p = <span class=\"math-tex\">$\\dfrac{a+b+c}{2}$<\/span> = <span class=\"math-tex\">$5+2\\sqrt{13}+\\sqrt{73}$<\/span> , áp d\u1ee5ng công th\u1ee9c He-rong, ta có<\/p><p><span class=\"math-tex\">$S_{ABC}=\\sqrt{p(p-a)(p-b)(p-c)}$<\/span> = 72 .<\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"> <strong>A.<\/strong> 72.<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-09-21 01:01:03","option_type":"txt","len":0}]}
