Semicircle inscribed in other Semicircle

i) Wish you have a neat sketch prepared for this question at your end.

ii) QR is the diameter of the semi circular arc through R-O-Q So if M is the midpoint of QR, OM = MQ = MR = r [Radii of the same circle] ---- (1)

But given PS is tangent to this circle at O; Hence OM perpendicular to PS. [Radius and tangent perpendicular at point of contact] ==> OM perpendicular to QR [Since QR is parallel to PS] ---- (2)

So from (1) & (2), OMQ is a right triangle and applying Pythagoras theorem, r² + r² = OQ² = 4 [Since OQ is the radius of the bigger circle of diameter 4 units] ==> r = √2 units. ---- (3)

Also <QOR = 90 deg [Angle in the semi circle of R-O-Q] ----- (4)

iii) Area of the semi circle ROQ = (1/2)(πr²) = (1/2)(2π) [Substituting r = √2 from (3)] ==> Area of the semi circle ROQ = π sq. units ----- (5)

iv) QXR is the minor segment of the circle whose radius is 2 units and the chord is QR. Angle subtended by the chord QR at the center O = <QOR = 90 deg [Angle in the semi circle of the arc QOR]

Area of minor segment = (1/2)(r²){Ө - sin(Ө)}, where Ө is the angle at the center in radians and r is the radius of the circle] Here Ө = 90 deg = π/2; so sin(Ө) = sin(π/2) = 1 r = radius of bigger circle = 2 units

Thus area of the segment QXR = (1/2)(4)(π/2 - 1) = π - 2 sq. units ----- (6)

Adding (5) & (6), Area of the region QXROQ = (2π - 2) sq. units = 2(π - 1) sq. units