[tex]3[/tex] и [tex]4[/tex]
Обозначим стороны прямоугольника за [tex]a[/tex] и [tex]b[/tex]
Тогда сумма [tex]a+b=7[/tex]
И по теореме Пифагора квадрат диагонали равен сумме квадратов сторон, т.е. [tex]5^2=a^2+b^2[/tex]
Составим и решим систему
[tex]\displaystyle \left \{ {{a^2+b^2=25} \atop {a+b=7}} \right. \Bigg.\\\\\left \{ {{a^2+b^2+2ab=25+2ab} \atop {a+b=7}} \right.\Bigg.\\\\\left \{ {{(a+b)^2=25+2ab} \atop {a+b=7}} \right. \Bigg.\\\\\left \{ {{2ab=49-25} \atop {a+b=7}} \right. \Bigg.\\\\\left \{ {{2ab=24} \atop {a+b=7}} \right. \Bigg.\\\\\left \{ {{ab=12} \atop {a+b=7}} \right. \Bigg.\\\\\left \{ {{a=3} \atop {b=4}} \right.[/tex]
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Answers & Comments
Ответ:
[tex]3[/tex] и [tex]4[/tex]
Объяснение:
Обозначим стороны прямоугольника за [tex]a[/tex] и [tex]b[/tex]
Тогда сумма [tex]a+b=7[/tex]
И по теореме Пифагора квадрат диагонали равен сумме квадратов сторон, т.е. [tex]5^2=a^2+b^2[/tex]
Составим и решим систему
[tex]\displaystyle \left \{ {{a^2+b^2=25} \atop {a+b=7}} \right. \Bigg.\\\\\left \{ {{a^2+b^2+2ab=25+2ab} \atop {a+b=7}} \right.\Bigg.\\\\\left \{ {{(a+b)^2=25+2ab} \atop {a+b=7}} \right. \Bigg.\\\\\left \{ {{2ab=49-25} \atop {a+b=7}} \right. \Bigg.\\\\\left \{ {{2ab=24} \atop {a+b=7}} \right. \Bigg.\\\\\left \{ {{ab=12} \atop {a+b=7}} \right. \Bigg.\\\\\left \{ {{a=3} \atop {b=4}} \right.[/tex]