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A circle is inscribed in the quadrilateral $\square ABCD$ .Given that $BC=38\quad cm,AB=27\quad cm\quad and \quad DC=25\quad cm$,and that $AD$ is perpendicular to $DC$ . Find the maximum limit of the radius and the area of the circle.

$a.) 10\quad cm,226\quad cm^2\\ b.) 14\quad cm,616\quad cm^2\\ c.) 14\quad cm,216\quad cm^2\\ \color{green}{d.) 28\quad cm,616\quad cm^2}\\$

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I have found that $PO=OS=DS=DP=r$

I am not getting any ideas.

I have studied maths upto $12th$ grade.

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    $\begingroup$ $SC= 25 - r$, $RC = 25 - r$, $BR=13 + r = QB$, $AQ = 14 - r$, $AP = AQ = 14 -r$, you can get length of $AC$, then find the area of $ABCD$ by adding the two triangles $ADC$ and $ABC$. the radius $r$ satisfies $r(AB + BC + CD + DA)/2 = AREA$ $\endgroup$ – Yimin May 18 '15 at 21:42
  • $\begingroup$ This method seems long , is their any short method, may be using any construction. $\endgroup$ – R K May 18 '15 at 22:30
  • $\begingroup$ as the following answer said, $r < AD = 14$, then there is only one option. $\endgroup$ – Yimin May 18 '15 at 22:57
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Being inscribed circumscribed to a circle is a special property for a quadrilateral, so only few quadrilaterals will be candidates here. You already have $DS=DP=r$. From that you get $SC=CR=25-r$ and from that $RB=BQ=38-(25-r)=13+r$ and finally $QA=AP=27-(13+r)=14-r$. So you know $AD=AP+PD=14$. There is nothing to maximize, this value is the only choice where all four edges touch the circle.

If you had to actually compute the radius of the circle, you might want to follow the approach Yimin suggested in a comment: compute the diagonal $AC$ using the right angle and Pythagoras, then compute the areas of the two triangles using Heron's formula and add them up. Then look for tangential quadrilaterals to find that $K=r\cdot s$ describes a relation between the area $K$ of the quadrilateral, the radius $r$ of the incircle, and the semipermieter $s$.

But in your question, you apparently have to choose between three given choices for the radius. So I'd pick a pair of compasses and a ruler and actually draw the situation. After all, given $A,C,D$ from the right angle and the corresponding lengths, constructing $B$ is simply intersecting two circles of given radius, and once you have that, you can try all three different circle sizes to see which one fits. There is some margin of error to this method, but the given numbers are so different that you should be able to use this approach nevertheless. Computing the intersection of two circles is hard, but performing the construction is easy.

The relation between radius and area of a circle is very easy to look up, and a pocket calculator or some pen-and-paper computation (if you know a few digits of $\pi$) will help you there. That should rule out at least one of the options, since you can't have two distinct areas for radius $14\,\text{cm}$.

After actually performing the computation, I find that none of the given answers exactly matches the computed result. Perhaps the question is meant in a different way, perhaps the circle doesn't have to touch all four sides, but only three of them? Toying around with the configuration in Cinderella, it seems as if the area should be maximal in a setup where the circle does not touch $CD$. But even the number I got for that isn't in your list. Looking at the pairs of numbers you are given, perhaps all of this is meant as some kind of trick question, since obviously radii and areas don't always match. But I'd consider this an ill-posed question as it stands.

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  • $\begingroup$ The correct option given by book is option $d.)$ $\endgroup$ – R K May 18 '15 at 23:12
  • $\begingroup$ @RK: What book? Yould there be some mistake in how you quoted the question? Otherwise, this looks simply wrong in the book. $\endgroup$ – MvG May 18 '15 at 23:40

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