Asked by Oswaldo
find the area of the largest posible isosceles triangle with 2 sides equal to 6. thanks
Answers
Answered by
Bosnian
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°
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°
Answered by
Bosnian
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
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
Answered by
Bosnian
ã2 is square root
Area=(1/2)*(1/2)*72=18
Area=(1/2)*(1/2)*72=18
Answered by
Oswaldo
thanks
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