Question

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

Answer 1

area `=25 " unit"^2`

Explanation:

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`

Answer 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`