In a triangle with one angle of $120^\circ$,the lengths of the sides form an $A.P.$ If the length of the greatest side is $7 \ cm$,the area of the triangle is:

In a triangle $ABC$,$\left(\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}\right)^2 \leq$

If $x, y$ and $z$ are the distances of the incentre from the vertices $A, B$ and $C$ of the triangle $ABC$ respectively,then $\frac{abc}{xyz}$ is equal to

Let $a, b, c$ be the lengths of sides of triangle $ABC$ such that $\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=k$. Then $\frac{(A(\triangle ABC))^2}{k^4}=$

In a triangle $ABC$,$AD$ is the altitude from $A$. Given $b > c$,$\angle C = 23^o$ and $AD = \frac{abc}{b^2 - c^2}$,then $\angle B = \dots^o$.

In a non-right-angled triangle $\triangle PQR$, let $p, q, r$ denote the lengths of the sides opposite to the angles at $P, Q, R$ respectively. The median from $R$ meets the side $PQ$ at $S$, the perpendicular from $P$ meets the side $QR$ at $E$, and $RS$ and $PE$ intersect at $O$. If $p=\sqrt{3}, q=1$, and the radius of the circumcircle of the $\triangle PQR$ equals $1$, then which of the following options is/are correct?
$(1)$ Area of $\triangle SOE = \frac{\sqrt{3}}{48}$
$(2)$ Radius of incircle of $\triangle PQR = \frac{\sqrt{3}}{2}(2-\sqrt{3})$
$(3)$ Length of $RS = \frac{\sqrt{7}}{2}$
$(4)$ Length of $OE = \frac{1}{6}$