Category : JEE Main & Advanced

A quadrilateral PQRS is said to be cyclic quadrilateral if there exists a circle passing through all its four vertices P, Q, R and S.

Let a cyclic quadrilateral be such that

$PQ=a,\,QR=b,\,RS=c$ and $SP=d$.

Then$\angle Q+\angle S={{180}^{o}}$,$\angle A+\angle C={{180}^{o}}$

Let  $2s=a+b+c+d$

Area of cyclic quadrilateral = $\frac{1}{2}(ab+cd)\sin Q$

Also, area of cyclic quadrilateral = $\sqrt{(s-a)(s-b)(s-c)(s-d)}$, where $2s=a+b+c+d$ and $\cos Q=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}-{{d}^{2}}}{2(ab+cd)}$.

(1) Circumradius of cyclic quadrilateral : Circum circle of quadrilateral PQRS is also the circumcircle of $\Delta PQR$. $R=\frac{1}{4\Delta }\sqrt{(ac+bd)(ad+bc)(ab+cd)}=\frac{1}{4}\sqrt{\frac{(ac+bd)(ad+bc)(ab+cd)}{(s-a)(s-b)(s-c)(s-d)}}$.

(2) Ptolemy's theorem : In a cyclic quadrilateral PQRS, the product of diagonals is equal to the sum of the products of the length of the opposite sides i.e., According to Ptolemy's theorem, for a cyclic quadrilateral PQRS $PR.QS=PQ.RS+RQ.PS.$

LIMITED OFFER HURRY UP! OFFER AVAILABLE ON ALL MATERIAL TILL 31st March 2018 ONLY!

You need to login to perform this action.
You will be redirected in 3 sec