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Q.

The vertex 'O' of an isosceles triangle OAB lies at the origin and the equation of the base AB is x - y +1 = 0. If OA = OB = 6, the area of the triangle OAB is $\frac{\sqrt{ab}}{c}$ square units. Then $\frac{a + b}{c} =$

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$Let M be the foot of ⊥ from O to AB$
$\begin{array}{l} OM = \frac{1}{\sqrt{2}}, (AM)^{2} = (OA)^{2} - (OM)^{2} \\ (AM)^{2} = 36 - \frac{1}{2} = \frac{71}{2} \end{array}$
$Area of △ OAB = \frac{1}{2} (AB) (OM) = \frac{\sqrt{71}}{2} sq.units a = 7, b = 1, c = 2$

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