Помогите пожалуйста решить задачку. Чему равна площадь восьмиугольника, вершины

Ответы 1

$\left \{ {{|x|+|y|=7,} \atop {x^2+y^2=25,}} ight. \\ (|x|+|y|)^2=x^2+2|x||y|+y^2, \\ x^2+y^2=(|x|+|y|)^2-2|x||y|=7^2-2|xy|, \\ \left \{ {{|x|+|y|=7,} \atop {49-2|x||y|=25,}} ight. \left \{ {{|x|+|y|=7,} \atop {|x||y|=12,}} ight. \left \{ {{|x|+\frac{12}{|x|}=7,} \atop {|y|=\frac{12}{|x|} ,}} ight. , \\ x^2-7|x|+12=0, \\ \left [ {{|x|=3,} \atop {|x|=4,}} ight. \left[\begin{array}{c}x=-3,\\x=3,\\x=-4,\\x=4,\end{array}ight.$ $\left[\begin{array}{c} \left \{ {x=-3,} \atop {|y|=4,}} ight. \\ \left \{ {x=3,} \atop {|y|=4,}} ight. \\ \left \{ {x=-4,} \atop {|y|=3,}} ight. \\ \left \{ {x=4,} \atop {|y|=3,}} ight. \end{array}ight.$ $\left[\begin{array}{c} \left \{ {x=-3,} \atop {y=-4,}} ight. \\ \left \{ {x=-3,} \atop {y=4,}} ight. \\ \left \{ {x=3,} \atop {y=-4,}} ight. \\ \left \{ {x=3,} \atop {y=4,}} ight.\\ \left \{ {x=-4,} \atop {y=-3,}} ight. \\ \left \{ {x=-4,} \atop {y=3,}} ight. \\ \left \{ {x=4,} \atop {y=-3,}} ight. \\ \left \{ {x=4,} \atop {y=3,}} ight. \end{array}ight.$ $S=S_\square-4S_\triangle=8\cdot8-4\cdot\frac{1}{2}\cdot1\cdot1=64-2=62$
- Автор:
  salomésingh
- 6 лет назад
- 0
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