Construction : - draw a perpendicular onchord EF from mid-point B.
Let chord EF of larger semi-circle and parallel to AC touches the smaller semi-circle at point D.
Also let radius of smaller semi-circle be r cm and radius of bigger semi-circle be R unit.
Since, B is mid-point of diameter AC.
\(\therefore\) AB = BC = R
Also r is radius of smaller semi-circle whose diameter is AB.
\(\therefore\) AB = 2r
\(\Rightarrow \frac{AB}{2} = \frac{R}{2}\) (\(\because AB = R\))
Also, BOPD is a square. Therefore BP = OD = OB = r cm.
Now, since, BP is perpendicular on chord EF.
\(\therefore\) \(\angle EPB = \angle FPB = 90^\circ.\)
Now, in triangle EBP & FBP,
\(\angle EPB = \angle FPB = 90^\circ.\)
BE = BF (Radius of larger semi-circle)
BP = BP (common side)
\(\therefore\) \(\Delta EBP \cong \Delta FBP\) (By RHS congruence rule)
\(\therefore\) EP = PF = \(\frac{EF}{2} = \frac{10}{2}\) = 5 cm (By property of congruence triangles)
Now, in right angled triangle \(\Delta EPB,\)
BE2 = EP2 + BP2 (By pythagoras theorem)
\(\Rightarrow R^2 = 5^2 + r^2\) (\(\because \) BE = R, BP = r and EP = 5 cm)
\(\Rightarrow R^2 - r^2 = 5^2 = 25\) ......(1)
Area of region between two semi-circles
= Area of larger semi-circle - Area of smaller semi-circle
\(=\frac{1}{2} \pi R^2 - \frac{1}{2} \pi r^2\)
\(=\frac{\pi}{2}(R^2 - r^2)\)
\(=\frac{25 \pi}{2}\) (from CB)
Hence, the value of k is k = 25.