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.
So you had two questions, one about the charts covering the space, and one about compactness.

The first has been answered, but i would like to add some details.
So you do see that the $U\_i = \phi\_i^{-1}(\mathbb{C})$ for $i=0,1$ cover $\mathbb{C} P^1$ right? My claim is now that we can restrict to the closed disks in $\mathbb{C}$. The cases $[0:1]$ and $[1:0]$ are simple and the only points outside $U\_{0,1} := U\_0 \cap U\_1$, so we restict to $[z:w] \in U\_{0,1}$. Now both $\phi\_i$ are defined at this point, and since $\phi\_0([z:w]) = \frac{1}{\phi\_1([z:w])}$ we can pick the one that has norm smaller or equal to one, so that the image lies in the unit disk. Since the point was arbitrary, every point has as $i$ such that its $\phi\_i$ image lies in the unit disk, q.e.d.

Regarding the second one, recall that the continuous image of a compact space is compact. Moreover it is well known that the sphere $S \subset \mathbb{C}^2$ is compact, and that $\mathbb{C}P^1$ is a quotient of this sphere. Summarizing, there is a surjective quotient map $\pi: S \rightarrow \mathbb{C}P^1 $, hence $\mathbb{C}P^1$ is the image of a compact space, hence compact.

The reason i add the second answer is that it is insightful and easily generalized to complex or real projective spaces of any dimension, even to Grassmanians (although that requires some adjustments) and i believe also you can replace $\mathbb{C}$ by any normed division algebra over $\mathbb{R}$.

Hope this helps.
Joachim