Question

Example 4. In fig. 13.20, point M is middle point
of line segment AB, taking AM, MB and AB as diameter
of semi circles has been drawn in one side. Taking 'O' as
center and r radius draw a circle in such a way that it
touches the all three circles, then prove that r = - AB.​

Answer 1

Answer:

r = 1 / 6 AB

Step-by-step explanation:

Your question needs a correction.

Correction : r = 1 / 6 AB

In the given figure,

= > OM = RM - RO

= > OM = AB / 2 - RO { RM is radius of the circle whose diameter is AB, thus RM = AB / 2 }

= > OM = AB / 2 - r { OR = radius of the circle with center O }

= > OM = 2x - r ... ( 1 ) { Let AB / 4 = x }

Using Pythagoras Theorem :

= > OD^2 = MD^2 + OM^2

= > OD^2 - MD^2 = OM^2

= > ( r + QD )^2 - ( AB / 4 )^2 = OM^2 { from figure, OD = r + QD }

= > ( r + AB / 4 )^2 - ( AB / 4 )^2 = OM^2 { QD = radius of circle with center D, thus QD = DB = AB / 4 }

= > ( r + x )^2 - x^2 = OM^2 { Above we assumed AB / 4 = x }

= > r^2 + x^2 + 2xr - x^2 = OM^2

= > r^2 + 2xr = OM^2

= > r^2 + 2xr = ( 2x - r )^2 { from ( 1 ) , OM = 2x - r }

= > r^2 + 2xr = 4x^2 + r^2 - 4xr

= > 2xr = 4x^2 - 4xr

= > r = 2x - 2r

= > 3r = 2x

= > 3r = 2( AB / 4 )

= > 3r = AB / 2

= > r = AB / 6 = 1 / 6 AB

Hence proved.

*AB / 4 is assumed as x just to decrease the length.

Answer 2

Step-by-step explanation:

here you go thank you