Question 936388
<pre>

The other tutor doesn't understand the problem.  He doesn't
know that a cyclic quadrilateral is a quadrilateral with all
four vertices lying on a circle:

{{{drawing(200,200,-7,7,-7,7,

circle(0,0,5sqrt(65)/8+.09),

line(5sqrt(65)/8,0,3.45126649,3.671428144),
line(3.45126649,3.671428144, -1.410895105,4.837354649),
line(-1.410895105,4.837354649,-4.883867675,-1.240347345),
line(-4.883867675,-1.240347345,5sqrt(65)/8,0)

 )}}}



We use Brahmagupta's formula to find the area of
the cyclic quadrilateral.

{{{A=sqrt((s-a)(s-b)(s-c)(s-d))}}} where {{{semiperimeter=s=(a+b+c+d)/2}}},

and 

a=5, b=4, c=7, and d=10

{{{s=(a+b+c+d)/2=(5+4+7+10)/2=26/2=13}}}}}}

{{{A=sqrt((s-a)(s-b)(s-c)(s-d))=sqrt((13-5)(13-4)(13-7)(13-10))=sqrt(8*9*6*3)=sqrt(1296)=36}}}

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To find the radius.

Parameshvara's formula for the radius of the circumscribed circle is

{{{R=expr(1/4)*sqrt(

((ab+cd)(ac+bd)(ad+bc))/

( (s-a)(s-b)(s-c)(s-d) ) ) }}}

We have already calculated the denominator above as 1296, so we have

{{{R=expr(1/4)*sqrt(

(5*4+7*10)(5*7+4*10)(5*10+4*7)/1296)=expr(1/4)*sqrt(

(20+70)(35+40)(50+28)/1296)=""}}}
{{{expr(1/4)*sqrt((90)(75)(78)/1296) = expr(1/4)*sqrt(526500/1296)=expr(1/4)*sqrt(1625/4) = expr(1/4)*expr(sqrt(25*65)/sqrt(4))=expr(1/4)expr((5sqrt(65))/2)=(5sqrt(65))/8}}}

Edwin</pre>