In #DeltaABC, /_BAC=45^@, AB=AC and BC =10cm#. How will you find the area of #DeltaABC# without using trigonometry?

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Answer 1 · 2017-10-27T03:35:31

drawn

#AD# is the perpendicular dropped from #A# to #BC#.

As in #DeltaABC,AB=AC#,#AD# will bisect #/_BAC#.

So #/_BAD=/_CAD=22.5^@#

#/_ABE =/_BAD=22.5^@# is drawn.

So in #DeltaABE,AE=BE#

In #DeltaEAB,/_EAB=45^@ and /_EDB=90^@#

So #/_BED=45^@#
Hence #DE=BD=a=1/2xx10cm=5cm#

Now #BE^2=BD^2+ED^2#

So #AE=BE=sqrt(BD^2+ED^2)=sqrt2a=5sqrt2#

Finally area of #DeltaABC#

#=1/2xxBCxxAD#

#=1/2xx10xx(AE+ED)#

#=5xx(5sqrt2+5)#

#=25(sqrt2+1)cm^2#