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

Answer 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°

Answer 2

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

Answer 3

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

Answer 4

thanks