Question

please my dear sir ji question number 30.​

Answer 1

Step-by-step explanation:

let BD = x , then CD = 2x and angle B is alpha.

Now

use area of triangle =

$area = \frac{1}{2} ab \sin( \alpha )$

Area of triangle ∆ ABD =

$a _1 = \frac{1}{2} (x)(y) \sin( \alpha )$

And area of , ∆ABC =

$a_{2} = \frac{1}{2} (x + 2x)y \sin( \alpha ) = 3.( \frac{1}{2} xy \sin( \alpha ) ) \\ \: a_{2} =3 a_{1}$