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Take a look at figure given below. Give it a try and when you are ready then watch the solution.
°
to AD and AD is diameter That means E is at center of circle. i.e. Parallel with center of circle.Let assume side of square is a.
Diameter of semicircle is a and radius of semicircle is a/2.
AD is perpendicular to AB and FD is perpendicular to EB
⇒ Angle ABE = Angle ADF.
Now construct a extension line to AB at right side and Perpendicular to it up-to E
I am constructing diagonal just for reference. It will not play any role in this solution.
Let AH is y1 and HD is y2
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 BAH:
Both are similar.
\[\Rightarrow \frac{GE}{GB}=\frac{AH}{AB}\]
\[\Rightarrow \frac{a/2}{3a/2}=\frac{y_1}{a}\]
\[\Rightarrow y_1=\frac{a}{3}\]
AD = AH + HD
AD is 'a' and AH is 'a/3'
⇒ HD = 2a/3
Now observe Triangles BGE and DFH:
Both triangle have one angle is θ and one angle is 90
°
⇒ Third angle is also same.
⇒ Triangles BGE and DFH are similar by A-A-A Test.
Now,
\[\Rightarrow \frac{DH}{EB}=\frac{FD}{BG}\]
\[\Rightarrow \frac{2a/3}{\sqrt{10}a/2}=\frac{6}{3a/2}\]
\[\Rightarrow \frac{4}{3\sqrt{10}}=\frac{4}{a}\]
\[\Rightarrow a=3\sqrt{10}\]
Now, we can find area of square.
Area of Square = a^2
= (3√10)^2
⇒ Area of Square = 90 Sq. Units
