Surface of earth seen is 1/4 At what height from the ground can one see exactly 1/4 th of the Earth's surface ? 
Take earth to be sphere of radius r. 
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$$
{1 \over 4}\,\pars{4\pi r^{2}} = 2\pi r^{2}\int_{0}^{\alpha}\sin\pars{\theta}\,\dd\theta =
4\pi r^{2}\sin^{2}\pars{\alpha \over 2}\quad\imp\quad
\sin^{2}\pars{\alpha \over 2} = {1 \over 4}
$$

$$
\cos\pars{\alpha} = 1 - 2\sin^{2}\pars{\alpha \over 2} = \half ={r \over r + h}\quad\imp\quad \color{#00f}{\large h = r}
$$

A: Let Earth's radius be $R$, and the height above it be $r$. Moreover, denote the radius of the spherical cap visible as $a$. 
Form the right triangle with side $R$ and hypotenuse $R+r$, whose second side has length $\sqrt{(R+r)^2-R^2}$. By triangle similarity you have:
$$\frac{a}{\sqrt{(R+r)^2-R^2}}=\frac{R}{R+r}$$
Solve for $a$ and use the fact that the area of a spherical cap is:
$$A=\pi(a^2+(R-\sqrt{R^2-a^2})^2)=\pi (a^2+(R-\sqrt{R^2-a^2})^2)$$
And find for what $r$, $A= \pi R^2$ as requested.
