Question

If lengths of tangents drawn from the vertices of triangles whose vertices are $(0,0),(5,0)$ and $(0,12)$ to it's incircle are $a,b,c(a<b<c)$ units. Then $(a+b)\times c$ is

Answer 1

Let the lengths of the tangents drawn from the given vertices to circle are $x,y,z$


From the diagram,
$x+y=5$
$x+z=12$
$y+z=\sqrt{{12}^2+5^2}=13$

Solving the above equation, we get
$x=2,y=3,z=10$
$\therefore a=2,b=3,c=10(\because a<b<c)$
Thus $(a+b)\times c=50$