MHB What is the Area of Triangle ODG?

[Albert1]

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[Amer]

Spoiler

First I will find D coordinates D(x,y) .
the slope of OD: $\displaystyle S_{OD} = \frac{y}{x} $
Note that $\displaystyle y = \frac{-3x}{4} + 2 $ since D on the line
$\displaystyle AD \perp OD $ so we get $S_{AD} . S_{OD} = -1 $
$\displaystyle S_{AD} = \frac{-3}{4} $
$\displaystyle S_{OD} = \frac{4}{3} = \frac{-3x + 8}{4x} \Rightarrow 16x = -9x + 24 \Rightarrow x = \frac{24}{25} $
$D\left(\frac{24}{25} , \frac{32}{25} \right) $
$\displaystyle \bar{OD} = \sqrt{\left( \frac{24}{25}\right)^2 + \left( \frac{32}{25}\right)^2} = \frac{8}{5} $
The upper circle has radius 1 since the line AB cut the y-axis at (0,2), I think the lower circle has also radius 1, if so then N(1,1)
to find G coordinates, G on the line ON : y=x so $\displaystyle G\left(\frac{8}{7}, \frac{8}{7} \right) $
$\displaystyle\bar{DG} = \sqrt{ \left(\frac{24}{25} - \frac{8}{7}\right)^2 + \left(\frac{32}{25} - \frac{8}{7}\right)^2 }= 8 \sqrt{\left(\frac{3}{25} - \frac{1}{7}\right)^2 + \left(\frac{4}{25} - \frac{1}{7} \right)^2 } $
The area of the triangle = 1/2 OD (DG)

 

[Albert1]

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Spoiler

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