Question: I am stumped on the following question

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> > PQRS is a parallelogram and ST=TR. What is the ratio of area of triangle QST to the area of parallelogram (Ans 1:4)
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![enter image description here](https://i.stack.imgur.com/o9JYN.png)

I need the height of the triangle, how would I get that? Any suggestions would be appreciated.

Answer: 
The maps you gave are the coordinate charts on $\mathbb{C}\mathbb{P}^1$ that makes it into a manifold. In particular, they are bijective. 

If we take $\phi\_0^{-1}(D)$, we get all point $[1:z]$, with $z\in D$. Similarly, $\phi\_0^{-1}(D)$ is all points of the form $[z:1]$. Together, these sets cover $\mathbb{C}\mathbb{P}^1$. The inverse maps are continuous, because the maps $\phi\_i$ are bijective, so $\phi\_0^{-1}(D)$ and $\phi\_1^{-1}(D)$ are compact, as the image of a compact set under a continuous map is compact. It's not hard to show that if a space is a union of two compact sets, it is compact, so we are done.
The complex projective line is the union of the two open subsets
$$
D\_i=\{[z\_0:z\_1]\text{ such that }z\_i\neq0\},\quad i=0, 1.
$$
Each of them is homeomorphic to $\Bbb C$, hence to the open unit disk.

Compacteness follows from the fact that it is homeomorphic to the sphere $S^2$. You can see this as a simple elaboration of the observation that
$$
\Bbb P^1(\Bbb C)=D\_0\cup[0:1].
$$