Posted by **Oswaldo** on Tuesday, October 19, 2010 at 10:53am.

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

- Calculus 1 -
**Bosnian**, Tuesday, October 19, 2010 at 12:03pm
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 -
**Bosnian**, Tuesday, October 19, 2010 at 12:20pm
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 -
**Bosnian**, Tuesday, October 19, 2010 at 12:28pm
ã2 is square root

Area=(1/2)*(1/2)*72=18

- Calculus 1 -
**Oswaldo**, Tuesday, October 19, 2010 at 2:46pm
thanks

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