Question

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

Answer 1

`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`