Least and greatest possible values of perimeter of a triangle. One side of a triangle is 5 cm and the other side is10 cm and its perimeter is P cm , where P is an integer. The least and the greatest possible values of P are respectively 


*

*21 and 29

*22 and 27

*19 and 29

*20 and 28
I have no idea how to approach this question. Please help me.
 A: Imagine two strips of paper, one $5$ cm long, the other $10$. Join them at one end. Now vary the angle between them to construct all the possible triangles with those two sides. When will the third side be shortest? When longest?
A: sum of 2 sides in a triangle is more than the third side.
If 10 the largest side then the third side should be at least 6. Therefore the minimum value of p is 5+6+10=21.
If the third side is the largest then it should be less than 15 i.e. greatest possible value of the third side is 14. Therefore the maximum possible value of p is 15+14=29.
A: Given the sides $5, 10, x$ and the perimeter $P=5+10+x=15+x$, use the triangle inequality:
$$\begin{cases}5+10>x \\ 5+x>10\\ 10+x>5\end{cases} \Rightarrow \\
5<x<15 \Rightarrow \\
5+15<x+15<15+15 \Rightarrow \\
20<P<30 $$ 
The sides are integer, so, can you find the smallest and largest values of $P$?
A: Consider the vertices of the triangle in the complex plane. Let the common point for the line segments with length $5$ and $10$ be $O$. Without loss of generality, assume $O$ to be the origin of the complex plane. So, you may assume that the other ends of the line segments with length $5$ and $10$ be $A$ and $B$ respectively. Suppose that $a$ and $b$ be complex numbers that corresponds to the vertices $A$ and $B$ respectively. Therefore, $|a|=5$ and $|b|=10$ and the length $AB=|a-b|$.Therefore the perimeter $P=|a|+|b|+|a-b|$. Clearly, the following inequality holds
$$|a|+|b|\geq |a-b|\geq ||a|-|b||$$and hence we get,$$2(|a|+|b|)\geq P\geq |a|+|b|+||a|-|b||$$and using the fact $2|a|=|b|=10$ we get $P\in(20,30)$. But since $P\in\mathbb{Z}$, we can say that $P\in[21,29]$.
