XxmathguyxX

hm..... not sure why but i think it is wrong....

i think there is something smaller..... maybe i am wrong?

thank you!

yup!!

\(\frac{1}{1+\frac{1}{\sqrt{3}+2}}\)

\(\frac{1}{\frac{\sqrt{3}+3}{\sqrt{3}+2}}\)     

\(\frac{1}{\frac{b}{c}}=\frac{c}{b}\)  \(\frac{\sqrt{3}+2}{\sqrt{3}+3}\)

\(\frac{3+\sqrt{3}}{6}\)

ok the area of a trapezoid = \(\frac{(base(1)+base(2))\times(height)}{2}\)

so \(\frac{(12+15)\times 6}{2}\)=\(27\times3=81\)

so 81

we were calculating different things i was calculating the arc ab and cphill was calculating the line ab

all the angles of a triangle add up to 180 degrees. 1+2+5=8 and 180/8=22.5 so angle a is 22.5, p is 45, and

i'll let you take it from here 

   hope this helps,

      XxmathguyxX

it's great you found it out yourself i also got 15 degrees which is a tiny bit of proof your answer is in fact correct

it would really help if you can provide a picture

ok..

i don't know if this is possible..

x can equal 1 to infinity and it works