given angle yxa =105 deg

join pts x and a with b

let angle xab =x , xyb =y

so angle angle yxb =y & axb =x (both are isoceles traingles)

so x+y =105

so in quadrilateral xyab we have

x+ y + x+ y + angle B =360

so angle B = 360 - 2(x+y)
=150 degs

now let diameter ABC = 8 so AB=4= diameter of smaller semicircle

so radius of smaller semicirle = 2

Now area of shaded region = 1/2 * pi * 16 - area of semicircle + area of semicircle

= 8pi

now area of shaded region above red line is

150/360 * pi*16 - area of semicirlce

= 20 pi /3 - 1/2*pi*4

=14pi/3

now area of shaded region below red line = 8pi - 14pi/3

=10 pi /3

so ratio is

14pi/3 /10pi/3

=7/5

so ans should be D

Suppose radius of the circle = x

angle YXA = 105. We know the property that angle formed by an arc at center is double of the angle formed ab the same arc on the opp side of the circle.

So, angle at center formed by ARC YCB = 105*2 = 210

angle YBC = 210-180 = 30
angle YBA = 180-30 = 150

area of the sector YBC = pi*x^2*30/360 = pi*x^2/12 -------(1)

area of the sector YBA = pi*x^2*150/360 = pi*x^2*5/12 -------(2)

area of the small semicircle (diameter x) = pi*(x/2)^2/2 = pi*x^2/8 -------(3)

area of the shaded region above red line = (2)-(3) = 7*pi*x^2/24 ------------(4)

area of the shaded regios below red line = (1)+(3) = 5*pi*x^2/24 --------(5)

ratio (4)/(5) = 7/5

... Yup... 7/5 is my answer too ...