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Show that the triangle of maximum area that can be inscribed in a given circle is an equilateral triangle. |
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Answer» OB=9 BD=`sqrt(a^2-x^2)` CD=`sqrt(a^2-x^2)` AD=a+x<*(a+x)br>Area=`1/2*(a+x)*2sqrt(a^2-x^2` A=`(a+x)sqrt(a^2-x^2` `(dA)/(dx)=sqrt(a^2-x^2)+(a+x)/(2sqrt(a^2-x^2)`-2x=0. `=2a^2-2x^2-2ax-2x^2=0` `a^2-ax-2x^2=0` `(a-2x)(a+x)=0` `a-2x=0 or a+x=0` `x=a/2,-x` BD=`sqrt(a^2-x^2)=sqrt3/2a` BC=`sqrt3a` `AB^2=3/4a^2+9/4a^2=3a^2` `AB=sqrt3a`. BC and AB are equal. |
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