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in Circles by (350 points)
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AC be a line segment with midpoint B. Two semicircles with diameter AB and AC are drawn such that one completely lies inside the other.

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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.

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