How to calculate solid angle of a rectangular detector of 20cm x 10cm? I have an detector of $20\mathrm{cm}\times 10\mathrm{cm}$, how can I calculate the solid angle subtended by the detector if the detector is placed at $30\ \mathrm{cm}$ apart? Because directly I can not use the formula $\frac{area}{distance^2}$. Results obtained from this method are approximation and not exact. 
 A: As you have not mentioned the location of the point with respect to the given point hence let's assume that the given point lies on the vertical axis passing through the center of rectangle.
The solid angle $(\omega)$ subtended by any rectangle of size $l\times b$ at any point lying on the perpendicular axis at a distance $d$ from the center is given by Standard formula [1]
$$\omega=4\sin^{-1}\left(\frac{lb}{\sqrt{(l^2+4d^2)(b^2+4d^2)}}\right)$$
Hence, by substituting $l=20cm$, $b=10cm$ & $d=30cm$ we can easily get the solid angle subtended by rectangle
$$\omega=4\sin^{-1}\left(\frac{20\cdot10}{\sqrt{(20^2+4\cdot30^2)(10^2+4\cdot30^2)}}\right)$$$$\omega=4\sin^{-1}\left(\frac{200}{\sqrt{14800000}}\right)$$ $$\omega=4\sin^{-1}\left(\frac{1}{\sqrt{370}}\right)\approx 0.208043883  \ \mathrm{sr}$$
[1]: https://www.academia.edu/8747694
A: Equations for the solid angle subtended by rectangles, including off-center rectangles, are derived and reported here: https://vixra.org/abs/2001.0603
If the detector is squarely facing the origin:
$$\alpha = \frac{20cm}{2*30cm} = 1/3$$
$$\beta = \frac{10cm}{2*30cm} = 1/6$$
$$\Omega = 4 \arccos \left(\sqrt{\frac{1+ \alpha^2 + \beta^2}{(1+\alpha^2)(1+\beta^2) }}\right) \approx 0.208\, steradian$$
(Identical to Mr. Rajpoot's answer)
Leading order approximation:
$$\Omega \approx 4\alpha\beta \approx 0.222\, steradian$$
Next to leading order approximation:
$$\Omega \approx 4\alpha\beta - 2\alpha\beta(\alpha^2+\beta^2) \approx 0.207\, steradian$$
