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

1 Answer
CW
Oct 25, 2017

see explanation

Explanation:

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Let #O# be the center of the circumcircle of #DeltaABC#,
given #angleBAC=30^@, => angleBOC=2xx30=60^@#
#=> DeltaOBC# is an equilateral triangle
#=> OB=OC=BC=10#
#=># radius #r=OA=OB=OC=10#
#=> OM=sqrt(10^2-5^2)=5sqrt3#
#=># area of #DeltaABC=1/2*BC*AM#
#= 1/2*BC*(AO+OM)#
#= 1/2*10*(10+5sqrt3)#
#=50+25sqrt3#
#=25(2+sqrt3) " unit"^2#
#~~93.3 " unit"^2#

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