In the figure shown, calculate the area of the triangular region MBN , if a=1u and b=2u.(S:$\frac{2}{5}u^2$)
(figure out of scale)
I didn't get much, does anyone have an idea for a solution?
Follow the relations I found
$B, M$ and $N$ are points of tangent
Let $O$ is the center of the minor circumference
$AB = BO + OA = AN = AM + MN$
$BO = a = u$
$OA + u = AM + 2u$
$MN = 2u$
$\triangle OAP:OP^2=R^2+r^2$
$\triangle OPH: HP^2=R^2+r^2+(R-r)^2 \implies HP=MN=\sqrt{2Rr}$
$MN = \sqrt{2.2u.u}=\sqrt{4u^2} = 2u$
$AB = AN$
$S\triangle BMN=\frac{MN.h}{2} = \frac{2u.h}{2} =\frac{u.h}{2}$
