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 argument you have outlined appeals to the following lemma of general topology.

**Lemma.** Let $X$ be a topological space. If there are (quasi)compact subsets $Y\_1, \ldots, Y\_n$ such that $Y\_1 \cup \cdots \cup Y\_n = X$, then $X$ is itself (quasi)compact.

*Proof.* Let $\{ U\_\alpha : \alpha \in I \}$ be an open cover of $X$. For each $i$, $\{ U\_\alpha \cap Y\_i \}$ is an open cover of $Y\_i$, so has a finite subcover $\{ U\_{\alpha\_j} \cap Y\_i : \alpha\_j \in J\_i \}$. Thus we have a finite subcover $\{ U\_{\alpha\_j} : 1 \le j \le n, \alpha\_j \in J\_i \}$ for $X$.