Question

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Answer 1

Question :

In the given figure, M is the mid point of the line segment AB whose length is 2a. Semicircles having diameter AM, MB and AB drawn at the same side of the line. The radius of a circle touching all the three semicircle is a / 3 units

Answer :

Refer to attachment

Diameter of the bigger semicircle = AB = 2a units

Radius of the bigger semicircle = AB / 2 = 2a / 2 = a units

M is the mid point of AB

Diameter of smaller semi circles = AM = MB = AB / 2 = 2a / 2 = a units

Radius of the smaller semi circles = CM = MD = AM / 2 = a / 2

Now let's do some construction

Construction :

Join the center of the bigger semicircle and smaller semicircle to form a triangle. Draw a symmetry line to the semicircle GM at midpoint M

Consider ∆ OMD

Here ∠OMD = 90°

So, ∆ OMD is a Right angled triangle

Let the Radius of the smaller circle joining there semicircles be r units

By Pythagoras theorem

⇒ OD² = OM² + MD²

From figure

• OD = OF + FD = ( a / 2 + r ) units
• MD = a / 2 units
• OM = GM - OM = ( a - r ) units

⇒ ( a / 2 + r )² = ( a - r )² + ( a / 2 )²

Using ( a + b )² = a² + b² + 2ab and ( a - b )² = a² + b² - 2ab we get,

⇒ ( a / 2 )² + 2( a / 2 )r + r² = a² + r² - 2ar + a² / 4

⇒ a² / 4 + ar = a² - 2ar + a² / 4

⇒ ar = a² - 2ar

⇒ ar + 2ar = a²

⇒ 3ar = a²

⇒ 3r = a

⇒ r = a / 3

Therefore the radius of a circle touching all the three semicircle is a / 3 units.