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Q.

$In an isosceles triangle O A B, O is the origin and O A = O B = 6 . The equation of the side A B is x - y + 1 = 0 . Then the$
$area of the triangle is$

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a

$2 \sqrt{21}$

b

$\sqrt{142}$

c

$\frac{\sqrt{142}}{2}$

d

$\frac{\sqrt{71}}{2}$

answer is D.

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Detailed Solution

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$\begin{array}{l} Let D be the mid point of A B . \\ O D = \frac{|c|}{\sqrt{a^{2} + b^{2}}} = \frac{1}{\sqrt{2}} \\ A D = \sqrt{6^{2} - O D^{2}} = \sqrt{36 - \frac{1}{2}} = \sqrt{\frac{71}{2}} \\ ∴ Area of triangle = \frac{1}{2} A B \times O D = \sqrt{\frac{71}{2}} \cdot \sqrt{\frac{1}{2}} = \frac{\sqrt{71}}{2} \end{array}$

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In an isosceles triangle OAB,O is the origin and OA=OB=6 . The equation of the side AB is x−y+1=0 . Then the area of the triangle is