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 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
$$
\Bbb P^1(\Bbb C)=D\_0\cup[0:1].
$$
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 $X$.