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Princ
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Refer to diagram 2 above of first post shared by wizakobaskar

We need to get sides of triangle ABF
let's calculate CF FIRST

∆EDA~∆ECF (by AA)
So AD/CF=ED/EC
27/CF=20.25/6.75
CF=9

Let's calculate AF=AB^2+ BF^2
=36^2+27^2
=45^2
AF=45

For right angle ∆, in which circle is inscribed radius r=
(Sum of other two sides -hypotenuse)/2

= 27+36-45/2
=18/2
=9

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OA: D

Point E is on side CD, Radius of circle is dependent on the position of E on CD.
It can be seen from below sketch( not drawn to scale).
Attachment:
Capture AC.PNG
Capture AC.PNG [ 23.63 KiB | Viewed 5621 times ]
When E is coincident with C , circle will have minimum radius.
As E starts moving towards D, radius of circle will keep on increasing until E coincides with D.

Case1 : Minimum radius,\(r_{min}\)
Using 1st figure of above sketch
\(AB=BC=a\)

\(AC= \sqrt{2}a\) , (using Pythagoras theorem in \(\triangle\)ABC)

\(BH=BI=r_{min}\)
\(AH=IC=a-r_{min}\)
\(AG=AH=a-r_{min}; GC=IC=a-r_{min}\) (Tangents from a point outside the circle are equal in length.)
\(AC = AG+GC\)
\(\sqrt{2}a= 2(a-r_{min})\)
\(r_{min}=\frac{{(2-\sqrt[]{2})a}}{{2}}\)

putting \(a = 27 cm\), we get \(r_{min}\approx{7.9}\)

Case 2: considering last figure of sketch
\(r_{max}= \frac{a}{2}=\frac{27}{2}=13.5\)

Only Option D lies between this range, hence it is the answer.
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Since the range of options is large we can use "easy peasy" approach :-)
I draw the figure and saw that are of the circle is about twice smaller than that of the trapezoid. See my attachment.

Answer (D)
Attachments

1.png
1.png [ 9.54 KiB | Viewed 5515 times ]

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OA : D

First we should simplify the dimensions specified in question stem.
Let take side dimension 27 cms = a, then DE(20.25cm) becomes =\(\frac{3}{4}a\)
Attachment:
Radius.PNG
Radius.PNG [ 14.76 KiB | Viewed 5442 times ]

\(AF=GC=a-r\)
\(AI=AF=a-r\) (Tangents from a point outside the circle are equal in length)

In right angle \(\triangle\)ADE
\(AE^2=AD^2+DE^2\)
\(AE^2 = a^2+\frac{{9a^2}}{16}\)= \(\frac{{25a^2}}{16}\)
\(AE = \frac{5a}{4}\)
\(IE = AE- AI = \frac{5a}{4}-a+r =\frac{a}{4}+r\)

As \(\triangle\)ADE and \(\triangle\)ECH are similar.
\(\frac{CH}{AD}=\frac{HE}{AE}=\frac{EC}{DE}\)

\(\frac{CH}{a}=\frac{HE}{\frac{5a}{4}}=\frac{\frac{a}{4}}{\frac{3a}{4}}\)
\(CH=\frac{a}{3} ; HE = \frac{5a}{12}\)

\(HI = GH\) (Tangents from a point outside the circle are equal in length)
\(IE+EH=GC+CH\)
\(\frac{a}{4}+r+\frac{5a}{12}=a-r+\frac{a}{3}\)
\(2r=a+\frac{a}{3}-\frac{a}{4}-\frac{5a}{12}\)
\(2r=\frac{{12a+4a-3a-5a}}{12}\)
\(2r=\frac{{8a}}{12}\)

\(r=\frac{{a}}{3}\)

putting \(a =27 cm\), we get \(r =9 cm\)
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what did they do in the last part.. hypotenuse - semi perimeter ? can anyone explain
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