ÌâÄ¿ÄÚÈÝ

18£®ÎÒ¹úÄÏËÎÖøÃûÊýѧ¼ÒÇØ¾ÅÉØÔÚËûµÄÖø×÷¡¶ÊýÊé¾ÅÕ¡·Ò»ÊéÖУ¬¸ø³öÁËÖøÃûµÄÇØ¾ÅÉØ¹«Ê½£¬Ò²½ÐÈýбÇó»ý¹«Ê½£¬¼´Èç¹ûÒ»¸öÈý½ÇÐεÄÈý±ß³¤·Ö±ðΪa£¬b£¬c£¬Ôò¸ÃÈý½ÇÐεÄÃæ»ýΪS=$\sqrt{\frac{1}{4}[{a}^{2}{b}^{2}-£¨\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2}£©^{2}]}$£®ÏÖÒÑÖª¡÷ABCµÄÈý±ß³¤·Ö±ðΪ1£¬2£¬$\sqrt{5}$£¬Ôò¡÷ABCµÄÃæ»ýΪ1£®

·ÖÎö ¸ù¾ÝÌâÄ¿ÖеÄÃæ»ý¹«Ê½¿ÉÒÔÇóµÃ¡÷ABCµÄÈý±ß³¤·Ö±ðΪ1£¬2£¬$\sqrt{5}$µÄÃæ»ý£¬´Ó¶ø¿ÉÒÔ½â´ð±¾Ì⣮

½â´ð ½â£º¡ßS=$\sqrt{\frac{1}{4}[{a}^{2}{b}^{2}-£¨\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2}£©^{2}]}$£¬
¡à¡÷ABCµÄÈý±ß³¤·Ö±ðΪ1£¬2£¬$\sqrt{5}$£¬Ôò¡÷ABCµÄÃæ»ýΪ£º
S=$\sqrt{\frac{1}{4}[{1}^{2}¡Á{2}^{2}-£¨\frac{{1}^{2}+{2}^{2}-£¨\sqrt{5}£©^{2}}{2}£©^{2}]}$=1£¬
¹Ê´ð°¸Îª£º1£®

µãÆÀ ±¾Ì⿼²é¶þ´Î¸ùʽµÄÓ¦Ó㬽â´ð±¾ÌâµÄ¹Ø¼üÊÇÃ÷È·ÌâÒ⣬ÀûÓÃÌâÄ¿ÖеÄÃæ»ý¹«Ê½½â´ð£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø