is it true?

No, they will not, because: 4+5 < 10.

First approuch (Theoritical approuch)

Well we have the law of cosine that states: a² = b²+c² - 2bc*cos(A) (where A is the angle across the side a) 

We can manipulate this formula to the form of $A = acos(\frac{(a²-b²-c²)}{-2bc}{})$

So if we fill in our formula we'll get: $A = acos(\frac{10²-4²-5²)}{-2*4*5}{})$

We'd get $A = acos(-1.475)$, which has no solution

Second approuch (Visual approuch)

You could look at the function $f(x) = \sqrt{4² + 5² - 2*4*5*cos(x)}$(where f(x) would be your third side and x would be the angle between the 4- and 5 side)

If you look at the plot of this function you'll see that it never reaches f(x) = 10

Third approuch (intuitional approuch)

A triangle has  three sides connected to eachother on their ends -> The sum of two sides must be bigger than or equal to the third side, else there would be no way to connect them to the ends of that side. 4+5 = 9 -> 9 < 10 thus the triangle would be impossible. 

Hope this helped!