If $R = \frac{65}{8}$,$r_1 = \frac{21}{2}$,and $r_2 = 12$ are the circumradius and the radii of the excircles opposite to the vertices $A$ and $B$ of a triangle $ABC$ respectively,then the area of the triangle (in square units) is

  • A
    $21$
  • B
    $28$
  • C
    $84$
  • D
    $168$

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Similar Questions

Let $PQRS$ be a quadrilateral in a plane,where $QR = 1$,$\angle PQR = \angle QRS = 70^{\circ}$,$\angle PQS = 15^{\circ}$ and $\angle PRS = 40^{\circ}$. If $\angle RPS = \theta^{\circ}$,$PQ = \alpha$ and $PS = \beta$,then the interval$(s)$ that contain$(s)$ the value of $4 \alpha \beta \sin \theta^{\circ}$ is/are
$(A)$ $(0, \sqrt{2})$
$(B)$ $(1, 2)$
$(C)$ $(\sqrt{2}, 3)$
$(D)$ $(2 \sqrt{2}, 3 \sqrt{2})$

If the incircle of the $\Delta ABC$ touches its sides at $L, M$ and $N$ respectively,and if $x, y, z$ are the circumradii of the triangles $\Delta MIN, \Delta NIL$ and $\Delta LIM$ respectively,where $I$ is the incentre,then the product $xyz$ is equal to:

In any triangle $ABC$,${\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} + {\sin ^2}\frac{C}{2}$ is equal to:

With usual notation in a $\Delta ABC$,$\left( \frac{1}{r_1} + \frac{1}{r_2} \right) \left( \frac{1}{r_2} + \frac{1}{r_3} \right) \left( \frac{1}{r_3} + \frac{1}{r_1} \right) = \frac{K R^3}{a^2 b^2 c^2}$ where $K$ has the value equal to

In a triangle,if $r_1 = 2r_2 = 3r_3$,then $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is equal to

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