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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 $\frac{\sqrt{ab}}{c} s . u$ Then $\frac{a + b}{c} =$

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Detailed Solution

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Let M be the foot of the $⊥$ from O on AB.
Question Image
$\begin{array}{l} ∴ OM = \frac{1}{\sqrt{2}} \\ ∴ (AM)^{2} = (OA)^{2} - (OM)^{2} = 36 - \frac{1}{2} = \frac{71}{2} \\ ∴ Area of △ OAB = \frac{1}{2} (AB) (OM) = (AM) (OM) = \frac{\sqrt{71}}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{71}}{2} sq. units. \end{array}$

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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 abcs.u Then a+bc=