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

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Answer 1 · 2017-10-26T15:18:36

drawn
#BD# is the perpendicular drawn from #B# on #AC#.

So in #Delta ABD,/_D=90^@,/_BAD=/_ABD=45^@#.

Hence #AD=BD= a (say)-># height of #DeltaABC# w r to base #AC#

Now by Pythagoras theorem

#AB^2=AD^2+BD^2=2a^2#

#=>100^2=2a^2#

#=>a=5sqrt2#

So area of #Delta ABC=1/2xxACxxBD#

#=1/2xx10xxa=5xx5sqrt2=25sqrt2cm^2#