a:b Ratio in given square and semicircle figure

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Take a look at figure given below. Give it a try and when you are ready then watch the solution.

As you can observe arc AE and arc ED are same That means E is at center of circle. i.e. Perpendicular to AD and align with center of circle.

Let assume side of square is x.

Diameter of semicircle is a and radius of semicircle is x/2.

Now construct a extension line to AB at top and horizontal line from E

Join A and F.

Angle AFD must be 90° as the angle inscribed in a semicircle.

FD is perpendicular to BE and AD is perpendicular to BG

⇒ Angle ADF = Angle EBG --------(1)

Now observe triangle BGE

\[\left(\frac{3a}{2}\right)^2+\left(\frac{a}{2}\right)^2=EB^2\]

\[\Rightarrow\frac{10a^2}{4}=EB^2\]

\[\Rightarrow EB=\frac{\sqrt{10}a}{2}\]

Now observe Triangles BGE and AFD:

Both triangle have one angle is θ (From (1)) and one angle is 90

°

⇒ Third angle is also same.

⇒ Triangles BGE and AFD are similar by A-A-A Test.

Now,

\[\frac{BE}{AD}=\frac{BG}{FD}\]

\[\frac{\sqrt{10}x/2}{x}=\frac{3x/2}{b}\]

\[b=\frac{3x}{\sqrt{10}}\]

Now, we can find a:b Ratio.

\[\frac{a}{b}=\frac{\sqrt{10}x/2}{3x/\sqrt{10}}\]

\[\frac{a}{b}=\frac{10}{6}\]

\[\frac{a}{b}=\frac{5}{3}\]

⇒ a:b = 5:3