Calculus 1

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find the area of the largest posible isosceles triangle with 2 sides equal to 6. thanks

  • Calculus 1 -

    h-hight , a-thirth side of triangle
    al-angle ALPHA , A-area
    sin(al)=(h/6) , h=6*sin(al)
    cos(al)=(a/2)/6 =( a/12) , a=12*cos(al)
    A=(1/2)a*h=(1/2)*6*sin(al)*12*cos(al)
    =(1/2)*72*sin(al)*cos(al) =36*sin(al)*cos(al) =18*(2*sin(al)*cos(al))=18*sin(2al)
    =dA/dal)=18*2*cos(2al=0 , cos(2al)=0
    2al=90° al=45°

  • Calculus 1 -

    h=6*sin(45‹)=6*(1/ã2)=6/ã2
    a=12*12*cos(45)=12*(1/ã2)=12/ã2
    A=(1/2)a*h=(1/2)6/ã2*12/ã2
    =(1/2)*(1/ã2)/*1/ã2)*72=(1/2)*(1/)*72
    =(1/4)*72= 18

    Largest posible area= 18

  • Calculus 1 -

    ã2 is square root
    Area=(1/2)*(1/2)*72=18

  • Calculus 1 -

    thanks

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