In #DeltaABC, /_BAC=30^@, AB=AC=10cm#. How will you find the area of #DeltaABC# without using trigonometry?

Let #O# be the center of the circumcircle of #DeltaABC#,
let #r# be the circumradius, #=> OA=OB=OC=r#,
given #angleBAC=30^@, => angleBOC=2xx30=60^@#
#=> DeltaOBC# is an equilateral triangle
#=> OB=OC=BC=r#
#color(red)(i))# consider #DeltaOBM#,
#OM^2=r^2-(r/2)^2=3/4r^2, => OM=sqrt3/2r#
consider #DeltaABM#,
#10^2=(r+sqrt3/2r)^2+(r/2)^2#
#=> 100=(1+sqrt3+3/4+1/4)r^2=(2+sqrt3)r^2#
#=> r^2=(100)/(2+sqrt3)#
#=># area of #DeltaABC=1/2*BC*AM#
#= 1/2*BC*(AO+OM)#
#=1/2*r*(r+sqrt3/2r)#
#=1/2*r^2(1+sqrt3/2)#
#=1/2*(100)/(2+sqrt3)*(2+sqrt3)/2=25 " unit"^2#

#color(red)(ii))# we can also use the following formula to find the area of #DeltaABC# :
recall that area of a triangle with sides #a,b and c#, and its circumradius #=r# is given by : #A'=(abc)/(4r)#
Here, as #AB=AC=10, and BC=r#,
#=># area of #DeltaABC=(10xx10xxr)/(4r)=100/4=25 " unit"^2#

The above figure represents the situation as stated in the problem.

Obviously #/_ABC=/_ACB=1/2(180^@-30^@)=75^@#
A perpendicular #BD# is drawn on #AC# from #B#. If #AC=10cm# is taken as base then #BD=h# will represent its height w r to #AC#.

In #Delta BCD,/_DBC=15^@#

So #/_ABD=60^@#

Now if we take #DE=BE# then #DeltaBDE# will be equilateral and

hence #DE=BE=BD=h#

In #DeltaAED,/_EDA=/_EAD=30^@#

So #DeltaAED# is isosceles i.e = #AE=ED=h#

Now #AB=10cm#

#=>2h=10cm#

#=>h=5cm#

Now area of #Delta ABC=1/2xxACxxBD#
#=1/2xx10xxh=5xx5=25cm^2#